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3 years ago

14

A car is traveling 20 meters per second and is brought to rest by applying brakes over a period of 4 seconds. What is the averag

e acceleration over this time interval

1 answer:

seraphim [82]3 years ago

8 0

It would be 0-20/4 so it’d be -5 meters per second

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You are driving your 1700 kg car at 21 m/s down a hill with a 5.0∘ slope when a deer suddenly jumps out onto the roadway. You sl

podryga [215]

Answer:

d = 27.7 m

Explanation:

Here the car is driving on the inclined plane

So here we can say that work done by the gravity and work done by friction is equal to change in kinetic energy of the system

So here we can write it as

$mgsin\theta \times d - F_f \times d = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$

now we have

m = 1700 kg

$v_f = 0$

$v_i = 21 m/s$

$F_f = 1.5 \times 10^4 N$

$(1700)(9.81)sin5 (d) - (1.5\times 10^4)d = 0 - \frac{1}{2}(1700)(21^2)$

$1453.5 d - (1.5\times 10^4)d = -374850$

$d = 27.7 m$

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3 years ago

A proton moves with a speed of 1.17 105 m/s through Earth's magnetic field, which has a value of 50.0 µT at a particular locatio

vlabodo [156]

a) Southward you need to apply right hand rule. If you close your hand to the east, your thumb will indicate south.

b) Given the equation for Magnetic Force

$F= qVB$

Replacing

$F= (1.16*10^{-19})(1.17*10^5)(50*10^{-6})$

$F=9.36*10^{-19}$

c) Given the second Newton's Law by

$F_g = 1.67*10^{-27}*9.81$

$F_g = 1.64*10^{-26}$

Given the electric force by,

$F_e = 1.6*10^{-19}*1.5*10^2$

$F_e = 2.4*10^{-17}N$

$F=9.36*10^{-19}N$

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3 years ago

State a situation in which force is applied on a body, but no work is done

LenaWriter [7]

If you press on your arm force is applied work done is if it moves.
One answer could be if I was to press my hand on a table.
Have a great day!

8 0

3 years ago

Read 2 more answers

Which of the following is known as the following cycle because it's reservoir is a rock.

inessss [21]

Letra a obrigadoaijahahavqvqgqgqg

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3 years ago

A boat moves through the water with two forces acting on it. One is a 1,800-N forward push by the water on the propeller, and th

baherus [9]

(a)The acceleration of the 1,400-kg boat will be 0.425 m/sec²

(b) If it starts from rest, the distance through which the boat moves in 20.0 s will be 85 m.

(c)Velocity at the end of that time will be 8.5 m/sec.

<h3>What is velocity?</h3>

The change of distance with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.

Given data;

Forwad force acting on the boat,v₁ = 1,800-N

Resistive force acting on the boat,v₂ = 1,200-N

The acceleration of the boat,a

Mass of boat,m = 1,400-kg

Initial velocity of boat,u= 0 m/sec

Distance travelled by boat,S = ?

Time for the boat travels,t = 20.0 s

Final velocity,V = ? m/sec

The net force on the boat;

F = F₁ - F₂

F = 1800 N - 1200 N

F = 600 N

From the defination of force;

F= ma

a = F / m

a = 600 N / 1400 kg

a = 0.425 m/sec²

b)

The distance through which the boat moves is 20.0 s;

$\rm x_f = x_ 0 + v_0 t + \frac{1}{2} at^2 \\\\ x_f = \frac{1}{2}at^2 \\\\ x_f = 0+0 + \frac{1}{2} at^2 \\\\ x_f = \frac{1}{2}\times 0.425 \times (20 )^ 2 \\\\ x_f = 85 \ m$

c)

The velocity at the end of that time is found as;

$\rm v_f = v_ 0 + at \\\\ v_f =- 0+ at \\\\ v_f = 8.5$ m/sec

Hence the acceleration of the boat, the distance through which the boat moves in 20.0 s, and velocity at the end of that time will be 0.425 m/sec²,85 m, and 8.5 m/sec.

To learn more about the velocity, refer to the link: brainly.com/question/862972

#SPJ1

8 0

2 years ago