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kolbaska11 [484]
3 years ago
6

You are driving on the highway at a speed of 40 m/s (which is over the speed limit) when you notice a cop in front of you. To av

oid a ticket, you press on the brake and slow to a speed of 30 m/s over the course of 5 seconds. What is the acceleration of the car? SHOW WORK OR GET REPORTED(have 5 accounts)
What is your car's initial velocity?

What is your car's final velocity?

How long does it take the car to slow down?

Write the equation you will use to solve this problem.

What is the acceleration of your vehicle?
+ 2.0 m/s^2
- 2.0 m/s^2
+ 8.0 m/s^2
- 6.0 m/s^2
Physics
1 answer:
ladessa [460]3 years ago
3 0

Answer:

What is the acceleration of your vehicle?

+ 2.0 m/s^2 ✔

- 2.0 m/s^2

+ 8.0 m/s^2

- 6.0 m/s^2

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How far did a frog jump if he travels at a rate of 2.1 m/s for 10 seconds?
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21 m

Explanation:

The motion of the frog is a uniform motion (constant speed), therefore we can find the distance travelled by using

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The frog in this problem has a speed of

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3 years ago
Two wooden crates rest on top of one another. The smaller top crate has a mass of m1 = 24 kg and the larger bottom crate has a m
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Answer:

The sum of all forces for the two objects with force of friction F and tension T are:

(i) m₁a₁ = F

(ii) m₂a₂ = T - F

1) no sliding infers: a₁ = a₂= a

The two equations become:

m₂a = T - m₁a

Solving for a:

a = T / (m₁+m₂) = 2.1 m/s²

2) Using equation(i):

F = m₁a = 51.1 N

3) The maximum friction is given by:

F = μsm₁g

Using equation(i) to find a₁ = a₂ = a:

a₁ = μs*g

Using equation(ii)

T = m₁μsg + m₂μsg = (m₁ + m₂)μsg = 851.6 N

4) The kinetic friction is given by: F = μkm₁g

Using equation (i) and the kinetic friction:

a₁ = μkg = 6.1 m/s²

5) Using equation(ii) and the kinetic friction:

m₂a₂ = T - μkm₁g

a₂ = (T - μkm₁g)/m₂ = 12.1 m/s²

4 0
3 years ago
What is the total energy of a particle with a rest mass of 1 gram moving with half the speed of light? 1 eV = 1.6 x 10^-19 J. An
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Answer:

6.5 x 10^32 eV

Explanation:

mass of particle, mo = 1 g = 0.001 kg

velocity of particle, v = half of velocity of light = c / 2

c = 3 x 10^8 m/s

Energy associated to the particle

E = γ mo c^2

E=\frac{m_{0}c^}2}{\sqrt{1-\frac{v^{2}}{c^{2}}}}

E=\frac{m_{0}c^}2}{\sqrt{1-\frac{c^{2}}{4c^{2}}}}

E=\frac{2m_{0}c^}2}{\sqrt{3}}

E=\frac{2\times0.001\times9\times10^{16}}{1.732}

E=1.04\times10^{14}J

Convert Joule into eV

1 eV = 1.6 x 10^-19 J

So, E=\frac{1.04\times10^{14}}{1.6\times10^{-19}}=6.5\times10^{32}eV

4 0
2 years ago
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