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2 years ago

Find acceleration. Will give brainliest!

Physics

2 answers:

Art [367]2 years ago

8 0

Answer:

$\boxed{\mathrm{4.5 \: kmph/s \: or \: 1.25 \: m/s^2 }}$

Explanation:

$\displaystyle \mathrm{acceleration = \frac{change \: in \: velocity}{time \: taken}}$

$\displaystyle \mathrm{a = \frac{v - u}{t}}$

$\displaystyle \mathrm{v=final \: velocity}\\\displaystyle \mathrm{u=initial \: velocity}$

$\displaystyle \mathrm{a = \frac{45- 0}{10}}$

$\displaystyle \mathrm{a = \frac{45}{10}}$

$\displaystyle \mathrm{a = 4.5}$

$\mathrm{4.5 \: kmph/s = 1.25 \: m/s^2 }$

zalisa [80]2 years ago

7 0

Answer:

16200 km/s

270 km/min

4.5 km/h

Explanation:

Acceleration Formula: Average Acceleration = Δv/Δt (change in velocity over change in time)

Simply plug in our known variables and solve:

a = (45.0 - 0)/10

a = 45.0/10

a = 4.5 km/h

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A car starting from rest accelerated at a rate of 0 .5 m/s up to 2 km what would be the final velocity of the car and how much t

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From equation of motion v^2 = u^2 +2aS

Hence, the final velocity is 40 m/s.

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2 years ago

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An 1100-kg car traveling at 27.0 m/s starts to slow down and comes to a complete stop in 578 m. What is the magnitude of the ave

Trava [24]

Answer:

693.685 N

Explanation:

t = Time taken

u = Initial velocity = 27 m/s

v = Final velocity

s = Displacement = 578 m

a = Acceleration

m = Mass

Equation of motion

$v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-27^2}{2\times 578}\\\Rightarrow a=-0.63062\ m/s^2$

Force

F = ma

$F=1100\times -0.63062\\\Rightarrow F=-693.6851\ N$

Magnitude of braking force required to stop the car is 693.685 N

7 0

3 years ago

Jack (mass 59.0 kg ) is sliding due east with speed 8.00 m/s on the surface of a frozen pond. He collides with Jill (mass 47.0 k

Phantasy [73]

Answer:

Part(A): The magnitude of Jill's final velocity is $\bf{6.59~m/s}$ .

Part(B): The direction is $\bf{42.7^{0}}$ south to east.

Explanation:

Given:

Mass of Jack, $m_{1} = 59.0~Kg$

Mass of Jill, $m_{2} = 47..0~Kg$

Initial velocity of Jack, $v_{1i} = 8.00~m/s$

Initial velocity of Jill, $v_{2i} = 0$

Final velocity of Jack, $v_{1f} 5.00~m/s$

The final angle made by Jack after collision, $\alpha = 34.0^{0}$

Consider that the final velocity of Jill be $v_{2f}$ and it makes an angle of $\beta$ with respect to east, as shown in the figure.

Conservation of momentum of the system along east direction is given by

$~~~~&& m_{1}v_{1i} + m_{2}v_{2i} = m_{1}v_{1f} \cos \alpha + m_{2}v_{2f}^{x}\\&or,& v_{2f}^{x} = \dfrac{m_{1}(v_{1i} - v_{1f} \cos \alpha)}{m_{2}}$

where, $v_{2f}^{x}$ is the component of Jill's final velocity along east. The direction of this component will be along east.

Substituting the value, we have

$v_{2f}^{x} &=& \dfrac{(59.0~Kg)(8.00~m/s - 5.00 \cos 34.0^{0}~m/s)}{47.0~Kg}\\~~~~~&=& 4.84~m/s$

Conservation of momentum of the system along north direction is given by

$~~~~&& v_{2f}^{y} + v_{1f} \sin \alpha = 0\\&or,& v_{2f}^{y} = - v_{1f} \sin \alpha = (8.00~m/s) \sin 34^{0} = 4.47~m/s$

where, $v_{2f}^{y}$ is the component of Jill's final velocity along north. The direction of this component will be along the opposite to north.

Part(A):

The magnitude of the final velocity of Jill is given by

$v_{2f} &=& \sqrt{(v_{2f}^{x})^{2} + (v_{2f}^{y})^{2}}\\~~~~~&=& 6.59~m/s$

Part(B):

The direction is given by

$\beta &=& \tan^{-1}(\dfrac{4.47~m/s}{4.84~m/s})\\~~~~&=& 42.7^{0}$

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3 years ago

A 20 gram bullet exits the rifle at a speed of 985 m/s. The gun barrel is 0.8 m long. What

borishaifa [10]

Answer:

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Explanation:

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1 year ago

What two gases in the atmosphere do animals and plants need to breathe and grow?

kenny6666 [7]

Answer: Explanation:

Plants and animals each produce the gases that the other needs to live. Plants need carbon dioxide—people and other animals exhale carbon dioxide as a waste product. People and other animals need oxygen—plants produce oxygen during an important process called photosynthesis, which turns the sun's energy into nutrients.

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3 years ago

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