Answer:
1) x = x₀ + vot - ½ c t² - 1/6 bt³, v = v₀ - ct - ½ b t²
2) v₁ = 5.25 m/s, v₂ = -8 m/s
Explanation:
1) For this exercise, the relationship of the body is not constant, so you must use the definition of speed and position to find them.
acceleration is
a = c + bt
a) the relationship between velocity and acceleration
a = [tex]\frac{dv}{dt}[/tex]
dv = -a dt
The negative sign is because the acceleration is contrary to the speed to stop the vehicle.
we integrate
∫ dv = - ∫ a dt
∫ dv = -∫ (c + bt) dt
v = -c t - ½ b t²
This must be valued from the lower limit, the velocity is vo, up to the upper limit, the velocity is v for time t
v - v₀ = -c (t-0) - ½ b (t²-0)
v = v₀ - ct - ½ b t²
b) the velocity of the body is
v = [tex]\frac{dx}{dt}[/tex]
dx = v dt
we replace and integrate
∫ dx = ∫ (v₀ - c t - ½ bt²) dt
x-x₀ = v₀ t - ½ c t² - ½ b ⅓ t³
Evaluations from the lower limit the body is at x₀ for t = 0 and the upper limit the body is x = x for t = t
x - x₀ = v₀ (t-0) - ½ c (t²-0) + [tex]\frac{1}{6}[/tex] (t³ -0)
x = x₀ + vot - ½ c t² - 1/6 bt³
2) The speed when you reach the traffic light
Let's write the data that indicates, the initial velocity is vo = 15 m / s, the initial position is xo = 315m, let's use the initial values to find the constants.
t = 25 s x = 20
we substitute
20 = 315 + 15 25 - ½ c 25² - 1/6 b 25³
0 = 295 + 375 - 312.5 c - 2604.16 b
670 = 312.5 c + 2604.16 b
we simplify
2.144 = c + 8.33 b
Now let's use the equation for velocity,
v = v₀ - ct - ½ b t²
v = 15 - c 25 - ½ b 25²
v = 15 - 25 c - 312.5 b
let's write our two equations
2.144 = c + 8.33 b
v = 15 - 25 c - 312.5 b
Let's examine our equations, we have two equations and three unknowns (b, c, v) for which the system cannot be solved without another equation, in the statement it is not clear, but the most common condition is that if the semaphore does not change, it follows with this acceleration (constant) to a stop
a = c + b 25
from the first equation
c = 8.33 / 2.144 b
C = 3.885 b
we substitute in the other two
v = 15 - 25 (3.885 b) - 312.5 b
v = 15 - 409.6 b
final acelearation
a = 28.885 b
let's use the cinematic equation
[tex]v_{f}^2[/tex]= v² - 2 a x
0 = v² - 2a 20
0 = v² - (28.885b) 40
v² = 1155.4 b
we write the system of equations
v = 15 - 409.6 b
v² = 1155.4 b
resolve
v²= 1155.4 ( [tex]\frac{15 -v }{409.6}[/tex] )
v² = 2.8 ( 15 -v)
v² + 2.8 v - 42.3 = 0
v= [ -2.8 ±[tex]\sqrt {2.8^2 + 4 \ 42.3) }[/tex] ]/2 = [-2.8 ± 13.3]/2
v₁ = 5.25 m/s
v₂ = -8 m/s
why do you have to soak the leaf. in boiling water after heating it in alcohol
Answer:
The hot water kills the leaf and the alcohol breaks down the chlorophyll, taking the green color out of the leaf.
Explanation:
Soaking the leaf in alcohol helps to remove the chlorophyll in the leaves and make it easy to identify the starch present in the them.
What is starch test in leaf ?The starch in plants formed in their leaves by photosynthesis can be identified using some analytical teste. For this the leaves have to placed inside a dark room for one day.
After that soak the leaf in hot water after heating it in alcohol. The alcohol degrades the chlorophyll, removing the leaf's green hue, while the hot water kills the leaf. One of the leaves will become blue-black and the other will turn reddish-brown when you apply iodine to them.
Iodine is an indicator that changes color from blue to black when starch is present. The leaf that was in the sun changes from green to blue-black, showing that it has been doing photosynthesis and making starch.
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A red pool ball is rolling directly east before it collides with the
white cue ball moving directly north. Due to conservation of
momentum the total momentum of both objects after the
collision would be in which direction?
Answer: North East
Explanation: Trust me, I was just doing this on the Ck-12 and this the answer I choose and It said I'm correct.
1. What types of natural phenomena could serve as time standards?
Answer:
The movement of Sun and moon
Explanation:
When the sun rise.it is am and when it sets .it is pm.
If you swing an object on the end of a string around a circle, the string pulls on the object to keep it moving in a circle. What is the name of this force?
A. inertial
B. centripetal
C. resistance
D. gravitational
Answer:
B
Explanation:
The centripetal force keeps an object moving in a circular path. Therefore option (B) is correct.
What is centripetal force?A centripetal force can be described as a force that makes a body follow a curved path and its direction is orthogonal to the motion of the body. Gravity offers the centripetal force causing astronomical orbits.
The centripetal force is directed perpendicular to the direction of the displacement of an object. It always acts towards the center of the circle on an object moving in a circular path. For example, When spinning a ball on a string, the tension on the rope pulls the object toward the center.
The Centripetal Force can be described as the product of mass and velocity squared, divided by the radius.
F = mv²/r
Where F is the Centripetal force, m is the mass, r is the radius of the circle and v is the velocity of the object.
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You exert a 138 N push the leftmost of two identical blocks of mass 244 g connected by a spring of stiffness 605 kg/s2. After pushing the block a distance 15 cm, you release it; by this time the rightmost block has moved a distance 5 cm. (a) What is the energy in the oscillations between the blocks
Answer:
the energy in the oscillations between the blocks is 3.025 J
Explanation:
Given the data in the question;
Force f = 138 N
stiffness of spring k = 605 kg/s²
mass of block = 202 g = 0.202 kg
pushing the block a distance 15 cm, the rightmost block has moved a distance 5 cm
i.e
x₁ = 15 cm
x₂ = 5cm
the energy in the oscillations between the blocks will be;
E[tex]_A[/tex] = E[tex]_B[/tex] = [tex]\frac{1}{2}[/tex]k( Δx )²
we substitute
= [tex]\frac{1}{2}[/tex] × k( 15 - 5 )² × 10⁻⁴
= [tex]\frac{1}{2}[/tex] × 605 × ( 10 )² × 10⁻⁴
= [tex]\frac{1}{2}[/tex] × 605 × 100 × 10⁻⁴
= 3.025 J
Therefore, the energy in the oscillations between the blocks is 3.025 J
why food cook faster with salt water than cook with pure water
Answer:
oil heats faster
Explanation:
The knee extensors insert on the tibia at an angle of 30 degrees (from the longitudinal axis of the tibia), at a distance of 3 cm from the axis of rotation at the knee. How much force must the knee extensors exert to produce an angular acceleration at the knee of 1 rad/s2 , given a mass of the lower leg and foot of 4.5 kg, and a radius of gyration of 23 cm
Answer:
the knee extensors must exert 15.87 N
Explanation:
Given the data in the question;
mass m = 4.5 kg
radius of gyration k = 23 cm = 0.23 m
angle ∅ = 30°
∝ = 1 rad/s²
distance of 3 cm from the axis of rotation at the knee r = 3 cm = 0.03 m
using the expression;
ζ = I∝
ζ = mk²∝
we substitute
ζ = 4.5 × (0.23)² × 1
ζ = 0.23805 N-m
so
from; ζ = rFsin∅
F = ζ / rsin∅
we substitute
F = 0.23805 / (0.03 × sin( 30 ° )
F = 0.23805 / (0.03 × 0.5)
F F = 0.23805 / 0.015
F = 15.87 N
Therefore, the knee extensors must exert 15.87 N
The gravitational potential energy of an object is defined as the energy it has due to its position in a gravitational field. A ball with a weight of 50 N is lifted to a height of 1 meter. Which graph correctly represents the change in gravitational potential energy (shaded in blue) as it is lifted to this height?
Answer:
athletic
Explanation:
because internet system has been down since we were in few days
(will give brainliest to whoever is first and gives reason) A mass is spun in a circle with a frequency of 40Hz. What is the period of its rotation?
Answer:
[tex]\huge\boxed{T = 0.025\ seconds}[/tex]
Explanation:
Given:
Frequency = f = 40 Hz
Required:
Time period = T = ?
Formula:
[tex]\sf T = 1 / f[/tex]
Solution:
T = 1 / 40
T = 0.025 seconds
[tex]\rule[225]{225}{2}[/tex]
Hope this helped!
~AH1807Why are protists difficult classify?
Answer:
Protists are difficult to characterize because of the great diversity of the kingdom. These organisms vary in body form, nutrition, and reproduction. They may be unicellular, colonial, or multicellular.
Answer:
There are many varying characteristics and exceptions to each type of protist.
They have been previously categorized based on what they are not
Recent studies show that protists have not descended from one common ancestor.
Explanation:
its right i took test :)
g Two long parallel wires are a center-to-center distance of 2.50 cm apart and carry equal anti-parallel currents of 2.70 A. Find the magnitude of the magnetic field at the point P which is equidistant from the wires. (R
Answer:
864 mT
Explanation:
The magnetic field due to a long straight wire B = μ₀i/2πR where μ₀ = permeability of free space = 4π × 10⁻⁷ H/m, i = current in wire, and R = distance from center of wire to point of magnetic field.
The magnitude of magnetic field due to the first wire carrying current i = 2.70 A at distance R which is mid-point between the wires is B = μ₀i/2πR.
Since the other wire also carries the same current at distance R, the magnitude of the magnetic field is B = μ₀i/2πR.
The resultant magnetic field at B is B' = B + B = 2B = 2(μ₀i/2πR) = μ₀i/πR
Now R = 2.50 cm/2 = 1.25 cm = 1.25 × 10⁻² m and i = 2.70 A.
Substituting these into B' = μ₀i/πR, we have
B' = 4π × 10⁻⁷ H/m × 2.70 A/π(1.25 × 10⁻² m)
B = 10.8/1.25 × 10⁻⁵ T
B = 8.64 × 10⁻⁵ T
B = 864 × 10⁻³ T
B = 864 mT
This question involves the concept of the magnetic field due to two current-carrying wires in the same direction, parallel to each other.
The magnitude of the magnetic field at the point P, which is equidistant from the wires is "8.64 x 10⁻⁵ T".
The following formula is used to find the magnetic field at the center distance between two parallel current-carrying wires in the same direction:
[tex]B = \frac{\mu_oI_1}{2\pi r}+\frac{\mu_oI_2}{2\pi r}\\\\But,\ I_1=I_2=I\\\\B = \frac{\mu_oI}{\pi r}[/tex]
where,
B = magnetic field at required point = ?
μ₀ = permeability of free space = 4π x 10⁻⁷ H/m
I = current = 2.7 A
r = distance from wires to the point = 2.5 cm/2 = 1.25 cm = 0.0125 m
Therefore,
[tex]B=\frac{(4\pi\ x\ 10^{-7}\ H/m)(2.7\ A)}{\pi (0.0125\ m)}[/tex]
B = 8.64 x 10⁻⁵ T
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(1) Which appliance is designed to transfer electrical energy to kinetic energy?
D)
A food mbuer
BB kettle
Clamp
D radio
Answer:
bb kettle
Explanation:
it transfres electricsl to kinetic
Explain how you think an asteroid impact could affect the tilt of Earth’s axis. Explain how this effect would change Earth’s seasons and why. Support your explanation with scientific reasoning and evidence.
Answer:
An asteroid impact could affect the tilt of the Earth due to the force it applies onto the planet. This would change Earth's seasons due to the fact that Earth's tilt causes seasons.
5. How much heat is needed to warm .052 kg of gold from 30°C to 120°C? Note: Gold has a specific heat of 136
J/kg °C
Answer:
Q = 636.48 J
Explanation:
Given that,
The mass of gold, m = 0.052 kg
The temperature increase from 30°C to 120°C.
The specific heat of gold is 136 J/kg °C.
We need to find the heat needed to warm the gold. The formula for heat needed is given by :
[tex]Q=mc\Delta T\\\\Q=0.052\times 136\times (120-30)\\\\Q=636.48\ J[/tex]
So, 636.48 J of heat is needed to warm gold.
A hair dryer draws a current of 12.8 A.
(a)How many minutes does it take for
6.8 x 10° C of charge to pass through the
hair dryer? The fundamental charge is
1.602 x 10-19 C.
Answer in units of min.
(b)How many electrons does this amount of
charge represent?
Answer in units of electrons.
Answer:
(a) 8.85×10⁻³ minutes
(b) 4.24×10¹⁹ electrons
Explanation:
(a) Using,
Q = it............................. Equation 1
Where Q = quantity of charge, i = current, t = time.
Make t the subject of the equation
t = Q/i............................. Equation 2
Given: Q = 6.8×10⁰ C, i = 12.8 A
Substitute these values into equation 2
t = 6.8×10⁰/12.8
t = 8.85×10⁻³ minutes
(b) n = Q/(1.602×10⁻¹⁹)................. Equation 3
Where n = number of electrons.
Given: Q = 6.8×10⁰ C
Substitute into equation 2
n = 6.8×10⁰/1.602×10⁻¹⁹
n = 4.24×10¹⁹ electrons
(a) The time taken by the charge to flow from the drier will be [tex]\d8.85[/tex]×[tex]\d10^{-3}[/tex]minutes
(b) Amount of the electrons in the charge will be [tex]\d4.24[/tex]×[tex]\d10^{19}[/tex] electrons
What will be the time of the charge and number of the electrons in the charge ?As we know Q = IT
Where Q = quantity of charge, i = current, T = time.
From the above equation
T= Q/I.
Given: Q = [tex]6.8\times\d10^{0}[/tex] C, i = 12.8 A
Substitute these values
T= [tex]6.8[/tex]×[tex]\d10^{0}[/tex] /12.8
T = [tex]8.85[/tex]×[tex]\d10^{-3}[/tex] minutes
Now the number of the electrons present in the charge will be
n = Q/( [tex]1.602[/tex]×[tex]\d10^{-19}[/tex])
Where n = number of electrons.
Given: Q = [tex]6.8\times\d10^{0}[/tex] C
Substitute Value of Q
n = [tex]6.8\times\d10^{0}[/tex]/ [tex]1.602\times\d10^{-19}[/tex]
n = [tex]4.24\times\d10^{19}[/tex] electrons
Thus
(a)The time taken by the charge to flow from the drier will be [tex]\d8.85[/tex]×[tex]\d10^{-3}[/tex]minutes
(b)Amount of the electrons in the charge will be [tex]\d4.24[/tex]×[tex]\d10^{19}[/tex] electrons
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the region where a magnet force is strongest is at the
Answer:
Magnetic poles
Explanation:
Magnetic poles are the regions where the magnet's force is strongest, one is the north pole and one is the south pole.
When grip strength increases:
a. action potential voltage increases.
b. action potential frequency decreases.
c. action potential frequency increases.
d. action potential frequency increases.
e. the number of active motor units increases.
Answer:
e. the number of active motor units increases.
Explanation:
There is a direct relationship between the number of active motor units and the grip strength in a given scenario. For example, increase in the grip strength leads to increase in the number of active motor units. In the other-hand, the decrease in grip strength leads to the decrease in the number of active motor units.
A 3.0-kilogram object initially at rest explodes and splits into three fragments. One fragment has a mass of 0.50 kg and flies off along the negative x axis at a speed of 2.8 m/s, and another has a mass of 1.3 kg and flies off along the negative y axis at a speed of 1.5 m/s.
Required:
What are the speed and direction of the third fragment?
Answer:
The speed and direction of the third fragment are 2 meters per second and 54.316º, respectively.
Explanation:
The object is not under the influence of any external force, meaning that Principle of Momentum Conservation to calculation of the velocity of the third fragment:
[tex](m_{1}+m_{2}+m_{3})\cdot \vec {v}_{o} = m_{1}\cdot \vec v_{1} + m_{2}\cdot \vec v_{2} + m_{3}\cdot \vec v_{3}[/tex] (1)
Where:
[tex]m_{1}[/tex], [tex]m_{2}[/tex], [tex]m_{3}[/tex] - Masses of the first, second and third fragments, in kilograms.
[tex]\vec v_{o}[/tex] - Initial velocity of the object, in meters per second.
[tex]\vec v_{1}[/tex], [tex]\vec v_{2}[/tex], [tex]\vec v_{3}[/tex] - Velocities of the first, second and third fragments, in meters per second.
If we know that [tex]m_{1} = 0.5\,kg[/tex], [tex]m_{2} = 1.3\,kg[/tex], [tex]m_{3} = 1.2\,kg[/tex], [tex]\vec v_{o} = (0,0)\,\left[\frac{m}{s} \right][/tex], [tex]\vec v_{1} = \left(-2.8, 0\right)\,\left[\frac{m}{s} \right][/tex] and [tex]\vec v_{2} = \left(0,-1.5\right)\,\left[\frac{m}{s} \right][/tex], the velocity of the third fragment is:
[tex](-1.4,0) + (0,-1.95) + 1.2\cdot \vec v_{3} = (0,0)[/tex]
[tex]1.2\cdot \vec v_{3} = (1.4,1.95)[/tex]
[tex]\vec v_{3} = (1.167, 1.625)\,\left[\frac{m}{s} \right][/tex]
The speed of the third fragment is the magnitude of the result found above:
[tex]v_{3} = 2\,\frac{m}{s}[/tex]
And the direction of the third fragment is:
[tex]\theta_{3} = \tan^{-1} \left(\frac{1.625}{1.167}\right)[/tex]
[tex]\theta_{3} \approx 54.316^{\circ}[/tex]
The speed and direction of the third fragment are 2 meters per second and 54.316º, respectively.
a place where two bones come together is known as an
Answer:
a place where two bones come together is known as a join
Answer:
Hey mate.....
Explanation:
This is ur answer....
JointsJoints – A place in the body where bones come together. Non-Moveable Joints (sometimes called fixed or fibrous) – A place in the body where two or more bones come together but do not move.
Hope it helps!
mark me brainliest plz....
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A force of 10 N is applied at right angles to the handle of a spanner, 0.5 m from the centre of a nut. The
moment on the nut is:
20 Nm
50 Nm
5 Nm
Explanation:
the movement of the nut is 20Nm
1 point
3. A 75 N box requires 250 J of work to move to a shelf. How high is the
shelf?
O 3.3 m
3.25m
0 3.9m
O 3.6m
A string is wound tightly around a wheel. When the end of the string is pulled through a distance of 10 cm, the wheel rotates through 5 revolutions. What is the radius of the wheel
Answer:
0.318cm
Explanation:
The computation of the radius of the wheel is shown below:
As we know that
One revolution of the wheel = 2πr
Here r denotes the radius of the wheel
Now
5 revolutions of the wheel would be
= 2πr × 5
= 10πr
So,
10πr = 10cm
Thus r = 1 ÷ π
= 0.318cm
The radius of the wheel will be r=0.318 cm
What will be the radius of the wheel?The radius is defined as the distance between the center of circle and the outer layer of the circle.
It is given in the question that
Revolutions of the wheel = 5
The length of the string =10 cm
Now the radius will be calculated as
One revolution of the wheel
= [tex]2\pi r[/tex]
Now 5 revolutions of the wheel would be
= [tex]2\pi r\times 5=10\pi\ r[/tex]
Since the string has the length of 10 cm then
[tex]10\pi\ r=10[/tex]
[tex]r=\dfrac{10}{10\pi}[/tex]
[tex]r=0.318\ cm[/tex]
Thus the radius of the wheel will be r=0.318 cm
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Diffraction occurs for all types of waves, including sound waves.
a. True
b. False
Answer:
a. True
Explanation:
Sound are mechanical waves that are highly dependent on matter for their propagation and transmission.
Sound travels faster through solids than it does through either liquids or gases. A student could verify this statement by measuring the time required for sound to travel a set distance through a solid, a liquid, and a gas.
Mathematically, the speed of a sound is given by the formula:
[tex] Speed = wavelength * frequency [/tex]
Generally, the frequency of a sound wave determines the pitch of the sound that would be heard.
Diffraction occurs for all types of waves, including sound waves.
In high air pressure the molecules are
A-Warm and moving fast
b-Close together and moving slowly
c-far apart and moving slowly
d-hot and moving rapidly
Q1what is pinhole camera?
A force of 12 N changes the momentum of a toy car from 3kgm/s t0 10kgm/s. Calculate the time the force took to produce this change in momentum.
Answer:
Time = 0.58 seconds
Explanation:
Given the following data;
Initial momentum = 3 kgm/s
Final momentum = 10 kgm/s
Force = 12 N
To find the time required for the change in momentum;
First of all, we would determine the change in momentum.
[tex] Change \; in \; momentum = final \; momentum - initial \; momentum [/tex]
[tex] Change \; in \; momentum = 10 - 3 [/tex]
Change in momentum = 7 kgm/s
Now, we can find the time required;
Note: the impulse of an object is equal to the change in momentum experienced by the object.
Mathematically, impulse (change in momentum) is given by the formula;
[tex] Impulse = force * time [/tex]
Making "time" the subject of formula, we have;
[tex] Time = \frac {impulse}{force} [/tex]
Substituting into the formula, we have;
[tex] Time = \frac {7}{12} [/tex]
Time = 0.58 seconds
1. A horizontal force of 50 N is applied to push a desk 40 m across a
warehouse floor. How much work is done?
2000 J
1000 J
3000 J
2001 J
3. Materials that lets electricity to pass
Answer:
materials that allow electricity to pass through them are called conductors, some examples of conductors are many metals, such as copper, iron and steel.
A falling 0.60 kg object experiences a frictional force due to air resistance of 1.5 N. What is the object's acceleration?
Answer:
7.5 m/s².
Explanation:
From the question given above, the following data were:
Mass (m) of object = 0.6 Kg
Force of friction (Fբ) = 1.5 N
Acceleration (a) =?
Next, we shall determine the force of gravity on the object. This can be obtained as follow:
Mass (m) of object = 0.6 Kg
Acceleration due to gravity (g) = 10 m/s²
Force of gravity (F₉) =?
F₉ = mg
F₉ = 0.6 × 10
F₉ = 6 N
Next, we shall determine the net force acting on the object. This can be obtained as follow:
Force of friction (Fբ) = 1.5 N
Force of gravity (F₉) = 6 N
Net force (Fₙ) =?
Fₙ = F₉ – Fբ
Fₙ = 6 – 1.5
Fₙ = 4.5 N
Finally, we shall determine the acceleration of the object. This can be obtained as follow:
Mass (m) of object = 0.6 Kg
Net force (Fₙ) = 4.5 N
Acceleration (a) of object =?
Fₙ = ma
4.5 = 0.6 × a
Divide both side by 0.6
a = 4.5 / 0.6
a = 7.5 m/s²
Therefore, the acceleration of the object is 7.5 m/s²
A small town has decided to forego the use of electrical power and send energy through town via mechanical waves on ropes. They use rope with a mass per length of 1.50 kg/m under 6000 N tension. If they are limited to a wave amplitude of 0.500 m, what must be the frequency of waves necessary to transmit power at the average rate of 2.00 kW
Answer:
the required frequency of waves is 2.066 Hz
Explanation:
Given the data in the question;
μ = 1.50 kg/m
T = 6000 N
Amplitude A = 0.500 m
P = 2.00 kW = 2000 W
we know that, the average power transmit through the rope can be expressed as;
p = [tex]\frac{1}{2}[/tex]vμω²A²
p = [tex]\frac{1}{2}[/tex]√(T/μ)μω²A²
so we solve for ω
ω² = 2P / √(T/μ)μA²
we substitute
ω² = 2(2000) / √(6000/1.5)(1.5)(0.500)²
ω² = 4000 / 23.71708
ω² = 168.65
(2πf)² = ω²
so
(2πf)² = 168.65
4π²f² = 168.65
f² = 168.65 / 4π²
f² = 4.27195
f = √4.27195
f = 2.066 Hz
Therefore, the required frequency of waves is 2.066 Hz