To find the maximum and minimum velocities that the ball can have and still land in the trunk of the truck, we need to use the equations of motion for the horizontal and vertical positions of the ball as a function of time. Since the truck is moving away from the ball, the horizontal position of the ball will be given by:
x(t) = x0 + Vt - v0cos(θ)*t
Where x0 is the initial horizontal distance from the back of the truck to the ball, V is the velocity of the truck, t is the time, and v0 is the initial velocity of the ball.
Similarly, the vertical position of the ball will be given by:
y(t) = y0 + v0*sin(θ)*t - (1/2)gt^2
Where y0 is the initial height of the ball, v0 is the initial velocity of the ball, g is the acceleration due to gravity and t is the time.
We know that the ball will land in the trunk of the truck if the horizontal position of the ball is equal to the length of the trunk (L) at the same time that the vertical position of the ball is equal to the initial height of the ball (y0).
We can find the time of flight by setting y = 0,
t = (2v0sin(θ))/g
Now, we substitute this value of t and y0 = 0 in the equation of x(t) and equate it with L.
x0 + Vt - v0cos(θ)*t = L
We can solve this equation for v0 to get the initial velocity of the ball.
v0 = (L + x0cos(θ) - Vt)/(cos(θ) - t*sin(θ))
Finally, we know that the minimum and maximum initial velocities will correspond to the maximum and minimum values of sin(θ) and cos(θ). Since θ = 45°, we can find the maximum and minimum velocities as follows:
v0_max = (L + x0cos(45) - Vt)/(cos(45) - tsin(45))
v0_min = (L + x0cos(45) - Vt)/(cos(45) + tsin(45))
Note that in this case x0=5m, L=2.5m, θ=45°, V=9m/s, t = (2v0sin(45))/g and g is 9.8m/s^2
you can use these values to find out v0_max and v0_min.
Select the correct answer. Which graphs show the correct relationship between kinetic energy and mass? A. Graph representing a relationship between mass on the x-axis and kinetic energy on the y-axis. the line starts at the origin and increases upwards B. Graph representing the relationship between mass on the x-axis and kinetic energy on the y-axis. the curve starts at the origin and keeps on increasing C. Graph representing a linear relationship between mass on the x-axis and kinetic energy on the y-axis D. Graph representing a relationship between mass on the x-axis and kinetic energy on the y-axis. the line starts at the origin and increases as it goes with a small bend Reset Next
C. Graph representing a linear relationship between mass on the x-axis and kinetic energy on the y-axis.
What is kinetic energy?
Kinetic energy is the energy possessed by a body due to its motion.
K.E = ¹/₂mv²
where;
m is mass of the objectThe kinetic energy of a body is directly proportional to the mass of the object.
Thus, the correct relationship between kinetic energy and mass is Graph representing a linear relationship between mass on the x-axis and kinetic energy on the y-axis.
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Glucose solution is administered to a patient in a hospital. The density of the solution is 1.308 kg/l. If the blood pressure in the vein is 35.7 mmHg, then what is the minimum necessary height of the IV bag above the position of the needle?
The minimum necessary height of the IV bag above the position of the needle, if the density of the solution is 1.308kg/L, and the pressure in the vein is 35.7 mmHg is, 2.78m.
To find the answer, we need to know more about the pressure exerted by a liquid column.
How to find the height of IV bag above the position of needle?Consider a liquid of density ρ contained in a vessel of height h, the pressure exerted by the liquid column at the bottom of the vessel is given by ,[tex]P=[/tex] ρgh
In our question, it is given that,[tex]density=1.308 kg/L\\\\P=35.7 mmHg.\\[/tex]
Thus, the height of the bag h will be,[tex]h=\frac{P}{density*g} =\frac{35.7}{1.308*9.8}\\\\ h=2.78 m[/tex]
Thus, we can conclude that, the height of the Iv bag is 2.78 m above the position of needle.
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If the solution has a density of 1.308 kg/L and the vein pressure is 35.7 mmHg, the minimum height of the IV bag above the needle position is 2.78 m.
We need to learn more about the pressure that a liquid column exerts in order to determine the solution.
How can I determine the height of IV bag above where the needle is?The pressure exerted by the liquid column at the bottom of the vessel is given by for a liquid with density enclosed in a vessel of height h.[tex]P=[/tex] ρgh
In our inquiry, it is assumed that,[tex]Density=1.308kg/L\\P=35.7mmHg[/tex]
As a result, the bag's height will be,[tex]h=\frac{P}{density*g}\\\\ h=2.78m[/tex]
As a result, we may say that the I.V. bag is 2.78 m above the ground.
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Even if all stars were the same distance from Earth, their absolute magnitude and
apparent magnitude would be very different.
True
False
Answer: True
hope this helps!
A proton moving along the positive x-axis enters a uniform magnetic field which is directed along the
positive y-axis. If the magnitude of the field is 0.7 T, and the charge and speed of the proton are 1.5 x10-19 C
and 3.0 x 106 m/s respectively. Calculate the force acting on the proton
Answer: The force acting on the proton of charge 1.5 x10-19 C moving with velocity 1.5 x10-19 C under the influence of a magnetic field of 0.7 T will be 3.15×10^-13 N.
Explanation: To find the answer we need to know more about the Lorentz magnetic force.
What is the Lorentz magnetic force acting on the proton?Consider a proton of charge q moving with a velocity v in a magnetic field, then the Lorentz magnetic force exerted on the proton can be expressed as,F= q (v× B)
[tex]F= qvBsin\alpha[/tex] where, [tex]\alpha[/tex] is the angle between v and B.
In the question, it is given that,[tex]B=0.7 T\\q=1.5*10^{-19}C\\v= 3*10^{6}m/s.\\\alpha =90 degree.\\[/tex] because, from the question it is clear that the proton is moving along x axis and the magnetic field is along the y axis.
Thus, we can find the force acting on the proton as,[tex]F=qvBsin\alpha =1.5*10^{-19}C*3*10^6 m/s*0.7T*sin (90)\\F=3.15*10^{-13}N[/tex]
Thus, we can conclude that the Lorentz force acting on the proton will be 3.15×10^-13 N.
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In response to a magnetic field of 0.7 T, a proton with a charge of 1.5 x 10-19 C travelling at a speed of 1.5 x 10-19 C will experience a force of 3.15 x 10^-13 N.
We need to learn more about the Lorentz magnetic force in order to locate the solution.
What does the proton experience as the Lorentz magnetic force?If you imagine a proton with charge q travelling at speed v in a magnetic field, you can write down the Lorentz magnetic force acting on the proton as,F= Q (v× B)
[tex]F=QvBsin\alpha[/tex]
[tex]\alpha[/tex] the angle between v and B is, where.
It is stated in the query that, the magnetic field is along the y axis and the proton is travelling along the x axis. Thus, the angle will be 90 degrees.As a result, we can identify the proton's driving force as,[tex]F=3.15*10^{-13}N[/tex]
Thus, we can infer that the proton will be subject to a 3.15 10^-13 N Lorentz force.
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Consider the following isobaric process for air, modelled as a Calorically Perfect Ideal Gas
(CPIG), from state 1 to state 2. P1 = 200 kPa, T1 = 500 K, T2 = 600 K. Show that the
condition satisfies the second law of thermodynamics. (Show all relevant steps involved).
These given conditions satisfy the second law of thermodynamics.
As the process is isobaric
So there will be a straight line of P= 200kPa in P-v and P-T planes
P1 = P2 = 100kPa
For perfect ideal gas, v-T plane:
[tex]v = (\frac{R}{P}) T[/tex]
[tex]v_{1} = (\frac{R}{P_{1} }) T_{1}[/tex] = 287 × 500/200000 = 0.717 m³/kg
[tex]v_{2} = (\frac{R}{P_{2} }) T_{2}[/tex] = 287 × 600/200000 = 0.861m³/kg
As it is the calorically perfect gas
de = [tex]c_{v}[/tex]dT
Integration on both sides
e2 - e1 = [tex]c_{v}[/tex](T2 - T1)
= ( 716.5J/kg/K) (600-500)
= 71650 J/kg
also,
Tds = de + Pdv
Tds = [tex]c_{v}[/tex]dT +Pdv
For ideal gas
V = RT/P
dv = Rdt/P - RTdp/P²
Tds = [tex]c_{v}[/tex]dT + Rdt - RTdp/P
ds = ([tex]c_{v}[/tex] + R)dT/T - RdP/P
ds = ([tex]c_{v} + c_{p} -c_{v}[/tex])dT/T - RdP/P
ds = [tex]c_{p}[/tex]dT/T - RdP/P
Integration on both sides
s2 - s1 = [tex]c_{p}[/tex]ln (T2/T1) - R ln (P2/P1)
Since P is constant
s₂ - s₁ = [tex]c_{p}[/tex] ln (T2/T1)
= 1003.5 ln (600/500)
= 1003.5 × 0.182
= 182.95 J/kg/K
w = Pdv
[tex]w_{12}[/tex] = P(v₂ - v₁)
= 2,00,000 ( 0.861 - 0.717)
= 28,800 J/kg
de = δq -δw
δq = de + δw
q₁₂ = (e₂ - e₁) + w₁₂
= 71,650 + 28,800 = 1,00,450 J/kg
Now in this process, the gas is heated from 500 K to 600 K. We would expect at a minimum that the surroundings were at 600 K.
Let’s check for second law satisfaction.
s₂ - s₁ ≥ q₁₂ / Tₓ
182.95 ≥ 1,00,450 / 600 K
182.95 J/kg/K ≥ 167.41 J/kg/K
Hence this condition satisfies the second law of thermodynamics
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The decibel level of the sound from a certain hair dryer is measured at 60 dB. Find the intensity of the sound.
Based on the calculations, the sound intensity level is equal to 1.0 × 10⁻⁵ W/m².
How to determine intensity of the sound?Mathematically, sound intensity level can be calculated by using this formula:
[tex]\beta = 10 log(\frac{I}{I_{ref}} )[/tex]
Where:
I is the intensity of the sound.
Note: The reference value of sound intensity is equal to 1.0 × 10⁻¹² W/m².
Rewriting the formula, we have:
β/10 = logI - logIo
Substituting the parameters into the formula, we have;
60/10 = logI - log(1.0 × 10⁻¹²)
6 = logI + 12
logI = 6 - 12
logI = -6
I = 1.0 × 10⁻⁵ W/m².
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Two bullets are fired at the same time with the same kinetic energy.
If one bullet has twice the mass of the other, what is the ratio of the speed of the lighter bullet to the speed of the heavier?
Which can do the most work?
The ratio of the speed of the lighter bullet to the speed of the heavier will be four times the heavier bullet.
Both can do same amount of work.
What is kinetic energy?Kinetic energy of a body is the energy due to the motion of the body.
Kinetic energy = mv²/2
where m is mass and v is velocity of the object.
Since both objects have the same kinetic energy but one bullet has twice the mass of the other, the ratio of the speed of the lighter bullet to the speed of the heavier will be four times the heavier bullet.
The kinetic energy of both bullets is the same. Hence, they can do equal amount of work.
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A solid, homogeneous sphere with a mass of m0, a radius of r0 and a density of ρ0 is placed in a container of water. Initially the sphere floats and the water level is marked on the side of the container. What happens to the water level, when the original sphere is replaced with a new sphere which has different physical parameters? Notation: r means the water level rises in the container, f means falls, s means stays the same. Combination answers like 'f or s' are possible answers in some of the cases.
The new sphere has a density of ρ > ρ0 and a mass of m = m0.
The new sphere has a mass of m > m0 and a radius of r = r0.
The new sphere has a radius of r = r0 and a density of ρ > ρ0.
The correct response for each of the condition given in the questions are,
[tex]d > d_0 , m=m_0[/tex] ⇒ f or s [tex]m > m_0, r=r_0[/tex]⇒ r [tex]r=r_0,d > d_0[/tex] ⇒ rTo find the answer, we have to know about the Archimedes principle.
How to solve the problem for different conditions?The Archimedes principle states that the upthrust F on a body is equal to the weight W of the displaced liquid.The sum of forces must be zero for the sphere to be in equilibrium.[tex]F-W=0\\F=W\\W=mg, where.\\m=density*volume=d*V\\V=\frac{4}{3} \pi r^3[/tex]
Let's apply the idea of density to the body and water now. We are taking d instead of ρ.[tex]d_wVg = d_0(\frac{4}{3}\pi r^3 ) g \\d_wV = d_0(\frac{4}{3}\pi r^3 )[/tex] (1)
Let's examine each example for the initial condition with d₀, m₀, and r₀, where the height of the water is h.Case 1:
The new sphere's mass is m = m₀ .The new sphere's density, d > d₀.Here, the smaller, denser sphere with the same mass, If the sphere floats, the amount of water it displaces will be equal to its mass, which will be the same as the amount of water the original sphere displaces.Consequently, the water level is unchanged.But if the sphere descends, the water displaced is less than the sphere's mass, m = m0, and the level drops, f.Therefore, f or s is the appropriate response.Case 2:
The new sphere's radius, r = r₀, and mass, m > m₀.Consequently, the new sphere is denser than the old one.The right answer is r, because the mass of the water displaced where the sphere floats is m > m0, which is greater than the water displaced for the initial sphere.Case 3:
sphere with the same radius but a higher density.When the right side of equation (1) rises, the left side must also rise in order for the volume to rise and the height to rise as a result (r)Thus, we can conclude that,
[tex]d > d_0 , m=m_0[/tex] ⇒ f or s [tex]m > m_0, r=r_0[/tex]⇒ r [tex]r=r_0,d > d_0[/tex] ⇒ rLearn more about Archimedes principle here:
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Light is incident normally on the short face of a 30∘−60∘−90∘ prism (Figure 1). A drop of liquid is placed on the hypotenuse of the prism.
a) If the index of the prism is 1.50, find the maximum index that the liquid may have for the light to be totally reflected.
Express your answer using three significant figures.
1.06 is the maximum refractive index that the liquid may have for the light to be totally reflected.
Only when a light source passes from a denser to a rarer medium can it completely reflect.
When the angle of incidence surpasses a specific critical value, specular reflection occurs in the more highly refractive of the two mediums at their interface, and this reflection is known as total reflection.
sin [tex]i_{c}[/tex] = μ[tex]_{r}[/tex] / μ[tex]_{d[/tex]
From the diagram
Angle of incidence = 60°
sin60° ≥ sin[tex]i_{c}[/tex] = μ[tex]_{r}[/tex]/μ[tex]_{d}[/tex]
μ[tex]_{r}[/tex] ≤ μ[tex]_{d}[/tex] sin60°
μ[tex]_{r}[/tex] ≤ √1.5 × √3/2
= 1.06
Hence, the maximum index that the liquid may have for the light to be totally reflected is 1.06
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What is the Difference between accuracy and precision ?
Answer:
Accuracy measures how close results are to the true or known value. Precision, on the other hand, measures how close results are to one another.
Accuracy means the state of being accurate, without any mistakes and the results should be 100% true.
Whereas, precision means, approximately true or almost true.
Hope it help you
What force causes a bike to move forward?
A. Air resistance
B. Thrust
C. Friction
D. Gravity
The Answer is Option C. Friction.
Explanation:
The friction force acts in the forward direction on the rear wheel and it acts in the backward direction on the front wheel. The magnitude of friction force on the rear wheel can be more, equal or less than that on the front wheel.
Select the correct answer.
Using the statistical definition of entropy, what is the entropy of a system where W = 4?
Using the statistical definition of entropy, The entropy of a system W = 4 is 1.91×10²³.
What is entropy?Entropy is typically referred to as a measurement of a system's randomness or disorder. In 1850, a German physicist named Rudolf Clausius first proposed this idea. Entropy is a thermodynamic property used to characterize a system's behavior in terms of temperature, pressure, entropy, and heat capacity. This thermodynamic explanation took the systems' equilibrium condition into account.
Entropy can be calculated using a mathematical expression.
Entropy = Total change of heat / thermodynamic temperature
S= KblnW
S is the statistical entropy
The value of Boltzmann's constant is 1.38×10⁻²³.
S= 1.38×10⁻²³ ln(4)
S = 1.91×10⁻²³
Using the statistical definition of entropy, The entropy of a system W = 4 is 1.91×10²³.
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Answer: C.
1.91 × 10⁻²³ joules/kelvin
Explanation: edmentum
A 40.0 kg beam is attached to a wall with a hi.nge and its far end is supported by a cable. The angle between the beam and the cable is 90°. If the beam is inclined at an angle of θ = 31.0° with respect to horizontal.
The horizontal component of the force exerted by the hi.nge on the beam = 8.662×101 N
What is the magnitude of the force that the beam exerts on the hi.nge?
The magnitude of the force that the beam exerts on the hi.nge will be,261.12N.
To find the answer, we need to know about the tension.
How to find the magnitude of the force that the beam exerts on the hi.nge?Let's draw the free body diagram of the system using the given data.From the diagram, we have to find the magnitude of the force that the beam exerts on the hi.nge.For that, it is given that the horizontal component of force is equal to the 86.62N, which is same as that of the horizontal component of normal reaction that exerts by the beam on the hi.nge.[tex]N_x=86.62N[/tex]
We have to find the vertical component of normal reaction that exerts by the beam on the hi.nge. For this, we have to equate the total force in the vertical direction.[tex]N_y=F_V=mg-Tsin59\\[/tex]
To find Ny, we need to find the tension T.For this, we can equate the net horizontal force.[tex]F_H=N_x=Tcos59\\\\T=\frac{F_H}{cos59} =\frac{86.62}{0.51}= 169.84N[/tex]
Thus, the vertical component of normal reaction that exerts by the beam on the hi.nge become,[tex]N_y= (40*9.8)-(169.8*sin59)=246.4N[/tex]
Thus, the magnitude of the force that the beam exerts on the hi.nge will be,[tex]N=\sqrt{N_x^2+N_y^2} =\sqrt{(86.62)^2+(246.4)^2}=261.12N[/tex]
Thus, we can conclude that, the magnitude of the force that the beam exerts on the hi.nge is 261.12N.
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The hi.nge will be subjected to a force of 261.12N from the beam.
We must understand the tension in order to choose the solution.
How can the amount of force the beam applies on the height be determined?Let's use the provided information to create the system's free body diagram.We need to calculate the force the beam is exerting on the height using the diagram.For this, it is assumed that the horizontal component of force is 86.62N, the same as the horizontal component of the normal reaction that the beam exerts on the height.We need to identify the vertical component of the normal reaction the beam exerts on the height. We must equalize the total force acting in the vertical direction to achieve this.[tex]N_y=F_v=mg-Tsin59[/tex]
Finding the tension T is necessary to determine Ny. Thus, we can use the net horizontal force to equate this.[tex]F_H=N_x=Tcos59\\T=\frac{F_H}{cos59} =169.84N[/tex]
As a result, the normal reaction that the beam has on the height becomes, with a vertical component,[tex]N_y=(40*9.8)-(169.84*sin59)=246.4N[/tex]
As a result, the force the beam applies on the height will be of the order of,[tex]N=\sqrt{N_x^2+N_y^2} =261.12N[/tex]
Thus, we can infer that the force the beam applies to the height is 261.12N in size.
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A gas is confined to a vertical cylinder by a piston of mass 2 kg and radius 1 cm. When 5J of heat are added, the piston rises by 2.4 cm. Find: (a) the work done by the gas; (b) the change in its internal energy. Atmospheric pressure is 105Pa
The work done by gas is 0.753 J and change in internal energy is 4.247J
So we are given that mass is 2kg , radius 1 cm and the amount of heat is 5 cm
The piston raised by 2.4cm
As we know that Work done is PΔV
Where ΔV is change in volume
Therefore ΔV = πr^2 h = π x (.01)^2 x .024 =7.53×10^(-6)m^3
Here pressure is 10^5 pa
So W = [tex]10^5\times7.53\times10^-(6)[/tex]
Therefore W = 0.753 J
Now coming to change in internal energy
Change in Internal Energy = Heat Added - Energy lost in work
∴ 5J - 0.753 J = 4.247J
Hence the change in internal energy is 4.247 J
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Object A has the same mass as Object B but Object A is traveling faster. What can we say about the momentum of Object A compared to that of Object B?
Answer:
Object A's momentum is larger
Explanation:
as the formula for momentum goes:
P = M * V
where P is momentum, M is mass, V is velocity.
so where Va (Object A's velocity) is larger than Vb (Object B's velocity) we get:
( i ) Va > Vb
as the masses of both objects are equal, we mark:
( ii) Ma = Mb = M
we multiply both sides of ( i ) by M to get:
( iii ) Va × M > Vb × M
and we finally get:
( iv ) Pa > Pb
Answer: Object A's momentum of larger than compared to that of object B.
Explanation:
what can you say about the speed of a ball that travelled a distance of 20 m and displacement of 10 m in 5 seconds
The speed of the ball at the given distance and time of motion is 4 m/s.
What is speed?The speed of an object is the rate of change of distance traveled by the object with time.
Speed is a scalar quantity because it has only magnitude and no direction. It is measured in meters per second.
Speed of the ballFor a ball that travelled a distance of 20 m and displacement of 10 m in 5 seconds, the speed of the ball at the given distance and time of motion is calculated as follows;
Speed of the ball = distance traveled by the ball / time of motion
from the question, distance = 20 mtime of motion, t = 5 secondsspeed of the ball = 20 m / 5 s
speed of the ball = 4 m/s
Thus, the speed of the ball at the given distance and time of motion is 4 m/s.
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A baseball player holds a 1.42 N baseball in his hand, a distance of 34.0 cm from the elbow joint as shown in the figure below. The biceps attached at a distance of 2.75 cm from the elbow exerts an upward force of 12.6 N on the forearm. Consider the forearm and hand to be a uniform rod with a mass of 1.20 kg. Calculate the net torque acting on the forearm and hand.
Answer:
90.3N
Explanation:
⊥mg = (0.170 m)(1.20 kg) 9.81 m/s
τ ball = r⊥Wball = (0.340 m)(1.42 N) = − 0.483 N ⋅m
F − 2.001− 0.483 N ⋅m = 0
F = 2.484 N ⋅m
0.0275 m = 90.3 N
The net torque acting on the forearm and hand is 90.3N
What is torque?
Torque is a measure of the pressure that can motivate an object to rotate about an axis. simply as pressure is what causes an object to accelerate in linear kinematics, torque is what reasons an item to collect angular acceleration. Torque is a vector amount.
⇒mg = (0.170 m)(1.20 kg) 9.81 m/s
⇒torque = r⊥weight of the ball
⇒ (0.340 m)(1.42 N) = − 0.483 N ⋅m
⇒F = − 2.001− 0.483 N ⋅m = 0
⇒F = 2.484 N ⋅m
⇒0.0275 m = 90.3 N
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State the newton's law of motion and give
application of each law.
Answer:
In the first law, an object will not change its motion unless a force acts on it. In the second law, the force on an object is equal to its mass times its acceleration. In the third law, when two objects interact, they apply forces to each other of equal magnitude and opposite direction.
Explanation:
Glucose solution is administered to a patient in a hospital. The density of the solution is 1.308 kg/l. If the blood pressure in the vein is 35.7 mmHg, then what is the minimum necessary height of the IV bag above the position of the needle?
The minimum necessary height of the IV bag above the position of the needle is 0.37 m.
we know that,
P = ρgh
where,
P = 35.7mmHg
= 4759.609 Pa
g = 9.8[tex]\frac{m}{s^{2} }[/tex]
ρ = 1.308 kg / m^3
now, substituting all the values, we get,
4759.609 = 1.308 × 9.8 × h
h = 0.37 m
The minimum necessary height of the IV bag above the position of the needle = 0.37 m.
what is an IV bag ?
A reagent, also called as an analytical reagent, is a substance or compound that is added to a system in chemistry to bring about a chemical reaction or examine to see if one happens. Even though the terms "reagent" and "reactant" are frequently used synonymously, "reactant" refers to a substance that is consumed during a chemical reaction.
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What adaptation of a cactus protects it from predators? a round cactus with many spines Broad leaves Sharp spines Thick stems Yellow flowers
A round cactus with many spines is the adaptation of a cactus that protects it from predators.
A cactus, unlike other plants, has unique adaptations in its roots, leaves, and stems that allow it to flourish in hot and dry surroundings.
The one adaptation that protects the cactus from predators is spines.
A cactus does not have any parts that resemble leaves if you could look at one closely.
Instead, the leaves are transformed into spines, which protrude from the plant's tiny bumps known as areoles.
Herbivores that live in the desert may be enticed to eat the cactus. The spines prevent these predators by modifying leaves into spines.
Other than protection, Spines perform many functions like
1) Since evaporation is a problem in a desert since water is scarce, the spines prevent excessive evaporation.
2) The spines also impede airflow and prevent evaporation by trapping air.
3) Collecting dew from the early-morning fog is another crucial job that the spines do.
The gathered dew turned into liquid water and ran down to the earth below. The plant then absorbs this water.
Hence, the adaptation of a cactus that protects it from predators is round cactus with many spines.
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Tuning an Instrument. A musician tunes the C-string of her instrumeut to a fundamental frequency of 65.4 Hz. The vibrating portion of the string is 0.600 m long and has a mass of 14.4 g. (a) With what tension must the musician stretch it? (b) What percent increase in tension is needed to increase the frequency from 65.4 Hz to 73.4 Hz, corresponding to a rise in pitch from C to D?
(a) The tension the musician must stretch it is 147.82 N.
(b) The percent increase in tension is needed to increase the frequency is 26%.
Tension in the stringv = √T/μ
where;
v is speed of the waveT is tensionμ is mass per unit length = 0.0144 kg / 0.6 m = 0.024 kg/mv = Fλ
in fundamental mode, v = F(2L)
v = 2FL
v = 2 x 65.4 x 0.6 = 78.48 m/s
v = √T/μ
v² = T/μ
T = μv²
T = 0.024 x (78.48)²
T = 147.82 N
When the frequency is 73.4 Hz;v = 2FL = 2 x 73.4 x 0.6 = 88.08 m/s
T = μv²
T = (0.02)(88.08)²
T = 186.19 N
Increase in the tension= (186.19 - 147.82)/(147.82)
= 0.26
= 0.26 x 100%
= 26 %
Thus, the tension the musician must stretch it is 147.82 N.
The percent increase in tension is needed to increase the frequency is 26%.
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Planet-X has a mass of 4.74×1024 kg and a radius of 5870 km. The First Cosmic Speed i.e. the speed of a satellite on a low lying circular orbit around this planet is 7.34 km/s.
1. What is the Second Cosmic Speed i.e. the minimum speed required for a satellite in order to break free permanently from the planet?
2. If the period of rotation of the planet is 16.6 hours, then what is the radius of the synchronous orbit of a satellite?
10,378.82 m/s is the second cosmic speed.
69,801 km is the radius of the synchronous orbit of a satellite.
Given
Mass of planet = 4.74 × [tex]10^{24}[/tex] kg
Radius of planet = 5870 km = 5870000m
First Cosmic speed = 7.34 km/sec
1) Second cosmic speed i.e. the minimum speed required for a satellite to break free permanently from the planet is also known as the escape velocity of a satellite.
It can be calculated by
v = √2GM/r where,
v= Escape velocity of the satellite
G = Gravitational constant
M = Mass of planet
r = Radius of planet
v = √[( 2 x 6.67 x 10⁻¹¹ x 4.74 x 10²⁴) / (5870 x 10³)]
v = 10,378.82 m/s
2) Speed of the satellite at the given period
v = 2πr/T where,
T= Time period of rotation = 16.6 × 3600 seconds
r = v×T/2π
r = (7,338.93 x 16.6 x 3600 s) / (2π)
r = 69,801 km
Hence
The Second Cosmic Speed i.e. the minimum speed required for a satellite to break free permanently from the planet is 10,378.82 m/s.
And the radius of the synchronous orbit of a satellite is 69,801 km.
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A contestant in a winter games event pulls a 36.0 kg block of ice across a frozen lake with a rope over his shoulder as shown in Figure 4.29(b). The coefficient of static friction is 0.1 and the coefficient of kinetic friction is 0.03.
Figure 4.29
(a) Calculate the minimum force F he must exert to get the block moving.
40.873
(b) What is its acceleration once it starts to move, if that force is maintained?
m/s2
(a) The minimum force F he must exert to get the block moving is 38.9 N.
(b) The acceleration of the block is 0.79 m/s².
Minimum force to be applied
The minimum force F he must exert to get the block moving is calculated as follows;
Fcosθ = μ(s)Fₙ
Fcosθ = μ(s)mg
where;
μ(s) is coefficient of static frictionm is mass of the blockg is acceleration due to gravityF = [0.1(36)(9.8)] / [(cos(25)]
F = 38.9 N
Acceleration of the blockF(net) = 38.9 - (0.03 x 36 x 9.8) = 28.32
a = F(net)/m
a = 28.32/36
a = 0.79 m/s²
Thus, the minimum force F he must exert to get the block moving is 38.9 N.
The acceleration of the block is 0.79 m/s².
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What is the symbol for a variable resistor?
A. Horizontal wire
B. 2 parallel wires
C. Line with a zig zag pattern with a diagonal line drawn across it
D. Circle with an X through it
The symbol for a variable resistor is a line with a zig zag pattern with a diagonal line drawn across it.
What is a variable resistor ?
A resistor whose electric resistance value may be altered is known to as a variable resistor. A variable resistor, which works usually by sliding a contact (wiper) over a variable resistor, is basically an electro-mechanical transducer.
The flow of electrical current is restricted by a resistor. The resistivity of a fixed resistor is steady. By adjusting a slider's position, the resistance of this resistor can be altered. Some volume controls and dimmer switches employ variable resistors.
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A block of mass 2.60 kg is placed against a horizontal spring of constant k = 725 N/m and pushed so the spring compresses by 0.0500 m.
What is the elastic potential energy of the block-spring system (in J)?
_______J
If the block is now released and the surface is frictionless, calculate the block's speed (in m/s) after leaving the spring.
_____s/s
The elastic potential energy of the block-spring system is 0.906 J.
Velocity of the block , v = 0.83 m/s.
What is elastic potential energy?Elastic potential energy is the energy stored in a stretched or compressed elastic material.
Elastic potential energy = Ke²/2The elastic potential energy of the block-spring system = 725 * 0.05²/2 The elastic potential energy of the block-spring system = 0.906 J
The elastic potential energy of the spring is converted to kinetic energy of the block.
1/2 mv² = 0.906 J
where v is velocity
v = √(0.906 * 2)/2.6
v = 0.83 m/s.
In conclusion, elastic potential energy is present in compressed or stretched elastic materials.
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An action force is 50 N to the left. The reaction force must be:
A. 50 N right
B. 50 N down
C. 50 N left
D. 50 N up
All moving objects have ____________.
A. Force
B. Distance
C. Momentum
D. Time
Answer:
I think it's force
Explanation:
Mark as brainliest if it is right
Answer: A
Explanation:
Force can cause a stationary object to start moving or a moving object to change its speed or direction or both
Which element is a metalloid?
Answer:
The metalloids are located on the right side of the periodic table in a "step-like" arrangement.
All of the possible metalloids are:
boron (B), silicon (Si), germanium (Ge), arsenic (As), antimony (Sb), tellurium (Te), and polonium (Po)
A 1100-N crate rests on the floor.
1) How much work is required to move it at constant speed 5.0 m along the floor against a friction force of 310 N .
Express your answer to two significant figures and include the appropriate units.
2)
How much work is required to move it at constant speed 5.0 m vertically.
Express your answer to two significant figures and include the appropriate units.
(a) The work done in moving the crate along the floor is 7,050 J.
(b) The work done in moving the crate vertically is 0 J.
Work done in moving the crate horizontally
W = Fdcosθ
where;
F is total force to be appliedd is displacement of the crateθ is the horizontal angle = 0W = (1100 + 310) x 5
W = 7,050 J
Work done in moving the crate verticallyW = Fd cos(90)
W = 0
Thus, the work done in moving the crate along the floor is 7,050 J.
the work done in moving the crate vertically is 0 J.
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A 550-g squirrel with a surface area of 945 cm2 falls from a 4.0-m tree to the ground. Estimate its terminal velocity. (Use the drag coefficient for a horizontal skydiver. Assume that the squirrel can be approximated as a rectanglar prism with cross-sectional area of width 11.6 cm and length 23.2 cm. Note, the squirrel may not reach terminal velocity by the time it hits the ground. Give the squirrel's terminal velocity, not it's velocity as it hits the ground.)
m/s
What will be the velocity of a 55.0-kg person hitting the ground, assuming no drag contribution in such a short distance?
m/s
The velocity of a 55.0-kg person hitting the ground, is mathematically given as
vt=39.5983m/s
What will be the velocity of a 55.0-kg person hitting the ground, assuming no drag contribution in such a short distance?Generally, the equation for is mathematically given as
mass of squirrel,
[tex]m=550 \mathrm{~g}\\\\Surface area, $A=945 \mathrm{~cm}^{2}=88 \times 10^{-3}$\\\\Height, $h-4 \mathrm{~m}$\\[/tex]
Terminal velocity is given by:
[tex]$v_{i}=\sqrt{\frac{2 m g}{\rho A C}}$[/tex]
where \rho is the density of fluid that is falling and it is given by
[tex]$\rho=\frac{m}{V}$[/tex]
since, volume =area * height
[tex]^{\rho=} \frac{0.55 \mathrm{Kg}}{0.0945 \mathrm{~m}^{2} \times 4.0 \mathrm{~m}}\\\\$\rho=0.1455 \mathrm{Kg} / \mathrm{m}^{3}$[/tex]
A is the surface area of squirrels.
C is the drag coefficient.
The surface area facing the fluid is given by:
[tex]A_{f}=\frac{0.0945 \mathrm{~m}^{2}}{2} \\\\\\ A_{f}=0.04725 \mathrm{~m}^{2}[/tex]
so, terminal velocity is :
[tex]$v_{t}=\sqrt{\frac{2 \times 0.55 \mathrm{Kg} \times 9.8 \mathrm{~m} / \mathrm{s}^{2}}{0.1455 \mathrm{Kg} / \mathrm{m}^{3} \times 0.04725 \mathrm{~m}^{2} \times 1}}$[/tex]
Vt=39.5983
In conclusion, the terminal velocity of the squirrel is 39.5983m/s
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