[tex]\purple{ \longrightarrow \bf{h_m = \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2g} }} [/tex]
Projectile is thrown with a velocity = v Angle of projection = θVelocity of projectile at a height half of the maximum height covered be [tex] \sf{v_0}[/tex][tex]\qquad[/tex]______________________________
Then –
[tex]\qquad[/tex] [tex]\pink{ \longrightarrow \bf{ \dfrac{h_m}{2} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }}[/tex]
[tex]\qquad[/tex] [tex] \longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2g} \times \dfrac{1}{2} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }[/tex]
[tex]\qquad[/tex][tex]\longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{4g} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }[/tex]
[tex]\qquad[/tex][tex]\longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2} = {v_0}^{2} \: {sin}^{2} \theta }[/tex]
[tex]\qquad[/tex][tex] \longrightarrow \sf{ \dfrac{ {v}^{2} }{2} = {v_0}^{2} }[/tex]
[tex]\qquad[/tex][tex] \longrightarrow \bf{v_0 = \sqrt{ \dfrac{ {v}^{2} }{2} } = \dfrac{v}{ \sqrt{2} } }[/tex]
Now, the vertical component of velocity of projectile at the height half of [tex] \sf{h_m}[/tex] will be –[tex]\qquad[/tex] [tex]\longrightarrow \bf{v_{(y)}=v_0 \: sin \theta }[/tex]
[tex]\qquad[/tex] [tex] \longrightarrow \bf{v_{(y)} = \dfrac{v}{ \sqrt{2} } \: sin \theta = \dfrac{v \: sin \: \theta}{ \sqrt{2} } }[/tex]
Therefore, the vertical component of velocity of projectile at this height will be–
☀️[tex]\qquad[/tex][tex] \pink {\bf{ \dfrac{v \: sin \: \theta}{ \sqrt{2} }} }[/tex]
Answer:
A projectile is thrown with velocity v at an angle θ with horizontal. When the projectile is at a height equal to half of the maximum height, the vertical component of the velocity of projectile is v sintheta / √2
Which branch of science deals with the study of the structures shown here?
FISH
AMPHIBIAN
REPTILE
BIRD?
Answer:
Vertebrate zoology
Explanation:
Have a great day!
which is the hottest and coldest planet in solar system ?
Answer:
hottest plant: vensus, close to the sun
coldest plant: Neptune, farthest from the sun
Explanation:
What ions are produced from acids and bases?
Answer:
Give person above me brainliest
Explanation:
A projectile with an initial velocity of 48 feet per second is launched from a building 190 feet tall. The path of the projectile is modeled using the equation h(t) = â€""16t2 48t 190. Approximately when will the projectile hit the ground? 1. 5 seconds 3. 2 seconds 5. 3 seconds 6. 2 seconds.
The time for the projectile to hit the ground from the given position is 1.5 s.
The given parameters:
Initial velocity of the projectile, u = 48 ft/sHeight, h = 190 ftModel of the projectile path, h(t) = –16t² + 48t + 190.The time for the projectile to hit the ground is calculated as follows:
when the projectile hits the ground, the final velocity = 0
[tex]v_f = 0\\\\\frac{dh(t)}{dt} = v_f\\\\\frac{dh(t)}{dt} = -32t + 48\\\\-32t+48 = 0\\\\32t = 48\\\\t = \frac{48}{32} \\\\t = 1.5 \ s[/tex]
Thus, the time for the projectile to hit the ground from the given position is 1.5 s.
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what is the cost of monthly (30 days) electric bill of ana if her city's cost of electricity is 0.05$ per kwh and she uses three refrigerators running in 600-watt power rating and open 24 hours
Answer:
1kW = 1000W
600W = 0.6kW
Cost of electric bill = 0.6kWh × 24 × 30 × $0.05
= $21.60
A car accelerates at a constant rate of 3 m/s2 for 5 seconds. If it reaches a velocity of 27 m/s, what was its initial velocity?
A
15 m/s
B
42 m/s
C
12 m/s
D
2.1 m/s
The initial velocity of a car that accelerates at a constant rate of 3m/s² for 5 seconds is 12m/s.
CALCULATE INITIAL VELOCITY:
The initial velocity of the car can be calculated by using one of the equation of motion as follows:
V = u + at
Where;
V = final velocity (m/s)u = initial velocity (m/s)a = acceleration due to gravity (m/s²)t = time (s)According to this question, a car accelerates at a constant rate of 3 m/s² for 5 seconds. If it reaches a velocity of 27 m/s, its initial velocity is calculated as follows:
u = v - at
u = 27 - 3(5)
u = 27 - 15
u = 12m/s.
Therefore, the initial velocity of a car that accelerates at a constant rate of 3m/s² for 5 seconds is 12m/s.
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the impulse-momentum relationship is a direct result of
Newton's second law.
__________________
o0o0o0o0o0o0o0o0o0
The impulse-momentum relationship is a direct result of Newton's second law of motion.
What is the impulse-momentum?The impulse-momentum theorem states that the impulse applied to an object is equal to the change in its momentum. It proves that the change in momentum of an object depends not only on the amount of force applied but also on the duration of force applied.
Newton's second law states that the force acting on an object is equal to the rate of change of its momentum.
This means that a force applied to an object will cause a change in its momentum. The impulse-momentum relationship describes the relationship between the force applied to an object and the resulting change in its momentum.
The impulse-momentum relationship states that the impulse acting on an object is equal to the change in its momentum.
Impulse is defined as the force applied to an object over a period of time, while momentum is the product of an object's mass and velocity. Therefore, the impulse-momentum relationship can be expressed as:
Impulse = Change in momentum
This relationship is important in understanding the behavior of objects in motion, particularly in collisions or other situations where forces act over a period of time.
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Which statement best explains the environmental drawback of using geothermal energy over energy generated by fossil fuels?
More cost-effective over time
Less pollution than fossil fuel plants
More energy plant locations available
Changes to regional landscapes
Answer:
Die Antwort lautet: Weniger Umweltverschmutzung als Kraftwerke mit fossilen Brennstoffen
Explanation:
hoffe das hat geholfen!! :))
Answer:
Changes to regional landscapes
Explanation:
It would be D because this is a cost or drawback and geothermal energy production would cause changes to the landscape, unlike how fossil fuels would. I also got this right on the quiz.
Hope this helps!
A 2.0 kg block rests on a level surface. The coefficient of static friction is, and the coefficient of kinetic friction is A horizontal force, X, is applied to the block. As X is increased, the block begins moving. Describe how the force of friction varies as X increases from the moment the block is at rest to when it begins moving. Indicate how you could determine the force of friction at each value of X―before the block starts moving, at the point it starts moving, and after it is moving. Show your work.
ps. I had to change F to X because of brainly.
By Newton's second law, the net force acting on the block in the vertical direction is
∑ F [ver] = n - mg = 0
where n = magnitude of normal force and mg = weight of the block. It follows that n = mg.
When the block is at rest, the applied force X will not be enough to move the box until it can overcome the maximum mag. of static friction. If µ[s] is the coefficient of static friction, then the maximum mag. of the frictional force is
f = µ[s] n = µ[s] mg
The net horizontal force would be
∑ F [hor] = X - µ[s] mg = 0
so a minimum force of X = µ[s] mg is required to get the block moving. Any mag. smaller than this and the block stays at rest/in equilibrium.
Once the mag. of X exceeds µ[s] mg, the block will begin to move. At that point, if the coefficient of kinetic friction is µ[k], then the net force on the block is
∑ F [hor] = X - µ[k] mg = 0
so a minimum force of X = µ[k] mg would be needed to keep the block moving at constant speed, or otherwise X = µ[k] mg + ma if the block is accelerating with mag. a.
The principles here are captured in the attached plot.
The magnetic field 0. 02 m from a wire is 0. 1 T. What is the magnitude of the magnetic field 0. 01 m from the same wire? 0. 01 T 0. 05 T 0. 1 T 0. 2 T.
The magnitude of the magnetic field 0.01 m from the same wire is 0.2 T.
Given to us:
Magnetic field, [tex]B_1 = 0.1\ T[/tex]
Radius of wire, [tex]R_1 = 0.02\ m[/tex]
To find out the magnitude of the magnetic field 0. 01 m from the same wire, we need to find out current first. we will use the formula,
[tex]B = \dfrac{\mu_oI }{2\pi R},\\\rn\\where,\\B= magnetic\ field\\\mu_o = 4\pi\times 10^{-7} m\cdot kg\cdot s^{-2} A^{-2}\ is\ the\ magnetic\ constant\\I= current\\R= radius\ of\ the\ wire[/tex]
Putting the values,
[tex]B_1 = \dfrac{\mu_oI }{2\pi R_1},\\\rn\\\\0.1= \dfrac{4\times \pi \times 10^{-7}\times I}{2\times \pi\times0.02}\\\\I=10,000\ A[/tex]
Now, for [tex]B_2[/tex]
[tex]B_2 = \dfrac{\mu_oI }{2\pi R_2},\\\rn\\\\B_2= \dfrac{4\times \pi \times 10^{-7}\times 10,000}{2\times \pi\times0.01}\\\\B_2= 0.2\ T[/tex]
Hence, the magnitude of the magnetic field 0. 01 m from the same wire is 0.2 T.
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how did you identify the layer that belonged next to the cambrian layer?
Answer:
Morphology and phylogenetics revealed by fossils. Perhaps the strongest evidence to support the Cambrian evolutionary explosion of animal forms is the first clear appearance, in the Early Cambrian, of skeletal fossils representing members of many marine bilaterian animal phyla
a body of mass 10 kg is moving with velocity of 10ms-1.a force acts for 5 seconds to reduce its velocity to 2ms-1.find the momentum of the body before and after application of the force on it
Answer:
momentum before: 100kgm/s
momentum after: 20kgm/s
Explanation:
initial velocity: 10
final velocity: 2
p=mv
before: 10kg×10m/s
after:10kg×2m/s
I'm not 100% sure btw
Under the Big Top elephant. Ella [2500 kg]. is attracted to Phant, the 3,000 kg elephant. They are separated by 8 m. What is the gravitational attraction between them? G=6.67×10^-11 (-11 is an exponent)
Hi there!
We can use the same equation for Gravitational Force:
[tex]\large\boxed{F_g = G\frac{m_1m_2}{r^2}}[/tex]
Fg = force due to gravity (N)
G = gravitational constant
m1,m2 = masses of objects (kg)
r = distance between objects (m)
Plug in the values provided:
[tex]F_g = (6.67*10^{-11})\frac{(2500)(3000)}{8^2} = \large\boxed{7.814 * 10^{-6}N }[/tex]
[tex]\huge\bf\underline{\underline{\pink{A}\orange{N}\blue{S}\green{W}\red{E}\purple{R:-}}}[/tex]
Here we've been given,
Universal gravitational constant (G) = [tex] \sf{6.67 \times {10}^{ - 11} }[/tex]Mass of object 1 (m1) = 2500 kg Mass of object 2 (m2) = 3000 kg Distance between two objects (r) = 8 mWe have to find the gravitational attraction force (Fg) = ?
The standard formula to solve is given by,
[tex]:\implies\tt{F_g = g \frac{m_1m_2}{ {r}^{2} } } [/tex]
[tex]:\implies\tt{F_g = 6.67 \times {10}^{ - 11} \times \frac{(2500 )(3000)}{ {8}^{2} } }[/tex]
[tex]:\implies\tt{F_g = 6.67 \times {10}^{ - 11} \times \frac{7500000}{64} }[/tex]
[tex]:\implies\tt{F_g = 7.814 \times {10}^{ - 6} }[/tex]
Gravitational force of attraction is 7.814 × 10^-6 N.do forces that act a distance come in pairs?
Answer:
Forces come in pairs.
Explanation:
Distances are between two points. The distance from A to B is equal but opposite the distance from B to A.
. a. Calculate the work done while lifting 300 kg of wate through a vertical height of 6 m. (Assume g = 10 m a =
Answer:
potential energy = mgh = 300 × 10 × 6m = 18000 joule or 18 kilo joule.
Explanation:
It takes a car 4 seconds to slow down from 22 m/s to 6 m/s , what was its acceleration?
To Find The Acceleration of A Car Slowing Down:
The change in speed is the final speed minus the initial speed. To find acceleration, divide the change in velocity by the length of time during which the velocity changed.
Question:
The US Navy is developing a railgun that uses magnetic field repulsion. The railgun can propel a 981 N projectile at seven times the speed of sound. How much energy is stored in this magnetic field? Assume a complete conversion of energy and a speed of sound of 340 m/s.
Answer choices:
A. 119 MJ
B. 283 MJ
C. 578 MJ
D. 2780 MJ
The relation of the kinetic energy allows to find the correct result for the energy stored in the electromagnet is:
B) 283 MJ
Kinetic energy is the energy due to the movement of bodies.
K = ½ m v²
Where K is the kinetic energy, m is the mass and v the spped.
They indicate that the weight of the bodye is W = 981 N and its final velocity is v = 7 [tex]v_s[/tex].
W = m g
Since the projectile starts from rest, its initial velocity is zero, therefore the change in energy is
ΔK = [tex]K_f - K_o = K_f[/tex]
we substitute
ΔK = ½ 981 / 9.8 (7 340) ²
ΔK = 2.835 10⁸ J
They indicate that all the energy of the electromagnet is transformed into the energy of the projectile,
Em = K
When reviewing the results, the correct one is:
B) 283 MJ
In conclusion, using the relationship of kinetic energy we can find the correct result for the energy stored in the electromagnet is:
B) 283 MJ
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the smallest division value of electronic balance
Answer:
0.1g to 0.0000001g hope it helps uu
A wave has a frequency of 2700 Hz and λ of 6m. Calculate the speed of the wave.pls help
Answer:
16,200 m/sExplanation:
The speed of the wave given only it's frequency and wavelength can be found by using the formula
[tex]c = f \times \lambda[/tex]
where
c is the velocity of the wave in m/s
[tex] \lambda[/tex] is the wavelength in m
f is the frequency in Hz
From the question
c = 2700 × 6 = 16,200
We have the final answer as
16,200 m/sHope this helps you
what does the slope of the curve on a velocity vs. time graph represent?
Answer:
the slope of velocity-time graph represent an object acceleration
if vector b is added to vector a under what conditions does the resultant vector has magnitude a+b?
Explanation:
hope it's useful for you knows
what is the energy equivalent of an object with a mass of 2.5 kg?
Answer:
2,500g is the answer yes
what is value of 2.37 in sl units
A box slides across a rough surface, eventually coming to rest.
Part A
a. Use the work-energy principle (Khan Academy:Work and the work-Energy Principle and the definition of work to explain why the box comes to rest.
b. Explain how the motion of the box is consistent with the Law of Conservation of Energy.
Answer:
he was a random act like you to be the first time I see is a great day and night to get a very happy to see you soon I hope that the world is not the same thing to say about this
a) When a box slides across a rough surface, eventually coming to rest, its kinetic energy is used during work done against frictional resistance force.
b) This work done is stored in the box as a potential energy. Thus the motion of the box is consistent with the Law of Conservation of Energy.
What is work-Energy Principle?According to the work-energy theorem, the work done by the net force acting on a body equals the change in kinetic energy.
It can simply be written as:
Work done = initial kinetic energy - final kinetic energy
The work energy theorem equation is the one presented above.
Now when a box slides across a rough surface, eventually coming to rest, frictional force comes into play. This force is opposite to the direction of motion of the box. Hence, the kinetic energy of the box is used during work done against frictional resistance force and stored in the box as a potential energy and thus Law of Conservation of Energy followed.
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Before a rifle is fired, the linear momentum of the bullet-rifle system is zero.
After the rifle is fired:
I. kinetic energy of the system is zero
II. linear momentum of the system is zero
Hi there!
II. Linear momentum of the system is zero.
This is an example of a RECOIL collision. With the Law of Conservation of Momentum, momentum remains constant before and after the collision.
Thus, the total momentum would also be equivalent to zero after the collision.
A 0.15 kg baseball moving at 20 m/s is stopped by a player in 0.010 s. What is the average force of the ball?
Answer: 300N
Explanation:
Impulse= Mass * Velocity
F.T = M * V
F= MV/T
F= (0.15*20)/ 0.01
F= 300N
what is the energy (in j) of a photon required to excite an electron from n = 2 to n = 8 in a he⁺ ion? submit an answer to three signficant figures.
Answer:
Approximately [tex]5.11 \times 10^{-19}\; {\rm J}[/tex].
Explanation:
Since the result needs to be accurate to three significant figures, keep at least four significant figures in the calculations.
Look up the Rydberg constant for hydrogen: [tex]R_{\text{H}} \approx 1.0968\times 10^{7}\; {\rm m^{-1}[/tex].
Look up the speed of light in vacuum: [tex]c \approx 2.9979 \times 10^{8}\; {\rm m \cdot s^{-1}}[/tex].
Look up Planck's constant: [tex]h \approx 6.6261 \times 10^{-34}\; {\rm J \cdot s}[/tex].
Apply the Rydberg formula to find the wavelength [tex]\lambda[/tex] (in vacuum) of the photon in question:
[tex]\begin{aligned}\frac{1}{\lambda} &= R_{\text{H}} \, \left(\frac{1}{{n_{1}}^{2}} - \frac{1}{{n_{2}}^{2}}\right)\end{aligned}[/tex].
The frequency of that photon would be:
[tex]\begin{aligned}f &= \frac{c}{\lambda}\end{aligned}[/tex].
Combine this expression with the Rydberg formula to find the frequency of this photon:
[tex]\begin{aligned}f &= \frac{c}{\lambda} \\ &= c\, \left(\frac{1}{\lambda}\right) \\ &= c\, \left(R_{\text{H}}\, \left(\frac{1}{{n_{1}}^{2}} - \frac{1}{{n_{2}}^{2}}\right)\right) \\ &\approx (2.9979 \times 10^{8}\; {\rm m \cdot s^{-1}}) \\ &\quad \times (1.0968 \times 10^{7}\; {\rm m^{-1}}) \times \left(\frac{1}{2^{2}} - \frac{1}{8^{2}}\right)\\ &\approx 7.7065 \times 10^{14}\; {\rm s^{-1}} \end{aligned}[/tex].
Apply the Einstein-Planck equation to find the energy of this photon:
[tex]\begin{aligned}E &= h\, f \\ &\approx (6.6261 \times 10^{-34}\; {\rm J \cdot s}) \times (7.7065 \times 10^{14}\; {\rm s^{-1}) \\ &\approx 5.11 \times 10^{-19}\; {\rm J}\end{aligned}[/tex].
(Rounded to three significant figures.)
Which are examples of perfectly inelastic collisions? Check all that apply.
a baseball bouncing off a bat
bumper cars bumping off of each other
a cue ball hitting an eight ball and stopping
a plane landing on an aircraft carrier
rain sticking to a window
two train cars coupling together
Answer:
D. a plane landing on an aircraft carrier
E. rain sticking to a window
F. two train cars coupling together
Explanation:
Yoo what it do I don’t know the answer tho
Answer:
You should try your best to answer the question.
Tim is pushing a heavy box across the floor. He is using 300N of force and can accelerate at 2m/s/s. What is the mass of the box?
-298 kg
-600 kg
-0.006 kg
-150 kg
Using Newton's second law
[tex]\\ \sf\Rrightarrow F=ma[/tex]
[tex]\\ \sf\Rrightarrow 300=2m[/tex]
[tex]\\ \sf\Rrightarrow m=150kg[/tex]
Hey there!
The formula of “mass” in physics is:
m = F/a
Whereas “f” is ‘force’, “a” is ‘acceleration’, & “m” is your ‘mass’ of course.
mass = 300 Net force/2 acceleration
300 Net force/2 acceleration = m
mass = 150
Therefore, your answer is: 150 kg
Good luck on your assignment and enjoy your day!
~Amphitrite1040:)