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3 years ago

A 1500kg car, moving at a speed of 20m/s comes to a halt. How much work was done by the brakes?

Physics

1 answer:

victus00 [196]3 years ago

6 0

The work done by the brakes is $-3.0\cdot 10^5 J$

Explanation:

According to the work-energy theorem, the work done by the brakes on the car is equal to the change in kinetic energy of the car. Therefore:

$W=K_f - K_i = \frac{1}{2}mv^2-\frac{1}{2}mu^2$

where

W is the work done

m is the mass of the car

v is the final speed

u is the initial speed

For the car in this problem, we have:

m = 1500 kg

u = 20 m/s

v = 0 (the car comes to a halt)

Substituting, we find the work done:

$W=0-\frac{1}{2}(1500)(20)^2=-3.0\cdot 10^5 J$

And the work is negative because the brakes apply a force in the opposite direction to the motion of the car.

Learn more about work:

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Q 6: A body throws a ball vertically up. It returns to the ground after 5 seconds. Find

Rasek [7]

Answer:

Taking gravity to be 9.8m/s2, The velocity is 24.5m/s2.

Taking gravity to be 10m/s2, The velocity is 25m/s2.

Explanation:

According the first formula of motion under the influence of gravity for upward motion, v=u-gt, where v=final velocity, u=initial velocity, and t= time taken.

Here the time taken for the ball to reach the maximum point is half of 5, which is 2.5 seconds.

And v is 0, since at the maximum point gravity slows down the velocity to 0.

Finding the initial velocity,

v=u-gt

0=u-10(2.5)

u=10(2.5)

u=25m/s

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2 years ago

How do you identify the number of valence electrons in an element using periodic table

frosja888 [35]

You may look at what group they are in
Group
1A=Group 1
2A = Group 2
3A = Group 13
4A= Group 14
5A=Group 15
6A=Group 16
7A=Group 17

The #A tells you how many valence electrons there are by the # before A. Such as Chlorine, which is in 7A, so therefore has 7 valence electrons.

5 0

2 years ago

Read 2 more answers

A long solenoid that has 1 200 turns uniformly distributed over a length of 0.420 m produces a magnetic field of magnitude 1.00

Fantom [35]

Answer:

<h2>The current required winding is $2.65*10^-^2 mA$ </h2>

Explanation:

We can use the expression B=μ₀*n*I-------1 for the magnetic field that enters a coil and

n= N/L (number of turns per unit length)

Given data

The number of turns n= 1200 turns

length L= 0.42 m

magnetic field B= 1*10^-4 T

μ₀= $4\pi*10^-^7 T.m/A$

Applying the equation B=μ₀*n*I

I= B/μ₀*n

I= B*L/μ₀*n

$I= \frac{1*10^-^4*0.42}{4\pi*10^-^7*1.2*10^3 }$

$I= 2.65*10^-^2 mA$

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3 years ago

a. Find the magnitude of the gravitational force (in N) between a planet with mass 6.50 ✕ 1024 kg and its moon, with mass 2.45 ✕

Vanyuwa [196]

Answer:

a. $1.70\times 10^{20} N$ b. $6.94\times 10^{-3}m/s^2$ c. $2.62\times 10^{-5} m/s^2$

Explanation:

Use Gravitational Force:

$F= G\frac{Mm}{r^2}$

Use values given to have the results.

$g=\frac{F}{m}$

Use m as the mass of the Moon. Using values of a. and m you have the answer.

$g = \frac{F}{M}$

Use M mass of the planet. Use value of a and Mass of Planet.

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3 years ago

Earth is 149.6 million meters from the Sun and takes 365 days to make one complete revolution around the Sun. Mars is 227.9 mill

luda_lava [24]

Answer:

$\frac{a_{r,earth}}{a_{r,mars}} = 2.325$

Explanation:

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