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Papessa [141]
2 years ago
15

How much work must be done on a 750 kg car to slow it down from 1.0 x 10 km/h to 2 50.0 km/h?

Physics
1 answer:
Papessa [141]2 years ago
8 0

Answer:

I got the answar ...............................

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4. A student wants to determine what type of cereal his classmates like best. He buys 3 boxes of his most favorite puffed rice c
Ann [662]
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The student's conclusion shows experimental bias
Just because most of nis classmates like puffed cereal ,it is inappropriate to conclude puffed cereal is better than oats cereal
4 0
3 years ago
What is the magnitude of the angular momentum (in kgm2/s) of a 40 g golf ball flying through the air and spinning at 4300 rpm af
sladkih [1.3K]

Answer:

L=0.0045\ kg-m^2/s

Explanation:

Given that,

The mass of a golf ball, m = 40 g = 0.04 kg

Its angular velocity, \omega=4300\ rpm=450.29\ rad/s

The radius of the sphere is 2.5 cm or 0.025 m

We need to find the magnitude of the angular momentum of the ball. It is given by the formula as follows:

L=I\omega

Where I is moment of inertia

For sphere, I=\dfrac{2}{5}mr^2

L=\dfrac{2}{5}mr^2\omega\\\\L=\dfrac{2}{5}\times 0.04\times (0.025)^2\times 450.29\\\\L=0.0045\ kg-m^2/s

So, the magnitude of the angular momentum of the sphere is 0.0045\ kg-m^2/s.

4 0
3 years ago
A car drives around a curve with a radius of 42 m at a velocity of 24 m/s. What is the centripetal acceleration of the car?
defon
Hope this helps you.

8 0
3 years ago
Read 2 more answers
10. 1. Why is Newton's law of gravitation called universal law ?​
Eduardwww [97]

Newtons law of gravitation is called the universal law of gravitation because it is applicable for all masses at all distances, independent of the medium.

6 0
2 years ago
Read 2 more answers
An unstable particle at rest breaks up into two fragments of unequal mass. The mass of the lighter fragment is equal to 2.90 ✕ 1
motikmotik

Answer:

The speed of the heavier fragment is 0.335c.

Explanation:

Given that,

Mass of the lighter fragment M_{l}=2.90\times10^{-28}\ kg

Mass of the heavier fragment M_{h}=1.62\times10^{-27}\ Kg

Speed of lighter fragment = 0.893c

We need to calculate the speed of the heavier fragment

Let v is the speed of the second fragment after decay

Using conservation of relativistic momentum

0=\drac{m_{1}v_{1}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}-\drac{m_{2}v_{2}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}

\drac{m_{1}v_{1}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}=\drac{m_{2}v_{2}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}

\dfrac{2.90\times10^{-28}\times0.893c}{\sqrt{1-(0.893)^2}}=\dfrac{1.62\times10^{-27}v_{2}}{\sqrt{1-\dfrac{v_{2}^2}{c^2}}}

\dfrac{v_{2}}{\sqrt{1-\dfrac{v_{2}^2}{c^2}}}=\dfrac{2.90\times10^{-28}\times0.893c}{1.62\times10^{-27}\times0.45}

\dfrac{v_{2}}{\sqrt{1-\dfrac{v_{2}^2}{c^2}}}=0.355c

\dfrac{v_{2}}{1-\dfrac{v_{2}^{2}}{c^2}}=(0.355c)^2

\dfrac{1-\dfrac{v_{2}^2}{c^2}}{v_{2}^2}=\dfrac{1}{(0.355c)}

\dfrac{1}{v_{2}^2}-\dfrac{1}{c^2}=\dfrac{1}{(0.355c)^2}

\dfrac{1}{v_{2}^2}=\dfrac{1}{c^2}+\dfrac{1}{0.126c^2}

\dfrac{1}{v_{2}^2}=\dfrac{1}{c^2}(1+\dfrac{1}{0.126})

\dfrac{1}{v_{2}^2}=\dfrac{8.93}{c^2}

v_{2}^2=\dfrac{c^2}{8.93}

v_{2}=0.335c

Hence, The speed of the heavier fragment is 0.335c.

7 0
3 years ago
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