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3 years ago

A 2 kg car accelerated from 10 m's to 20 m/s using a force of 3000N. How quickly did it

Physics

1 answer:

elena-14-01-66 [18.8K]3 years ago

7 0

Answer:

the car accelerates in

$\frac{1}{150} \: or \: 0.006 \: second$

Explanation:

here's the solution : -

we know,

=》

$force = mass \times acceleration$

=》

$acceleration = \frac{force}{mass}$

=》

$a = \frac{3000}{2}$

=》

$a = 1500$

so, acceleration = 1500 m/s^2

now,

=》

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

here, a = acceleration, v = final velocity,

u = initial velocity, t = time taken.

So,

=》

$1500 = \frac{20 - 10}{t}$

=》

$1500 = \frac{10}{t}$

=》

$t = \frac{10}{1500}$

=》

$t = 0.006 \: sec$

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What are two limitations of using a marble to model an atom?

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Answer:

Explanation:

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3 years ago

A tennis racquet swung with an angular velocity of 12 rad/s strikes a motionless ball at a distance of 0.5 m from the axis of ro

Anestetic [448]

Answer:

The linear velocity of the racquet at the point of contact with the ball is 6 m/s.

Explanation:

Given;

angular velocity of the racquet, ω = 12 rad/s

distance of strike, r = 0.5 m

The linear velocity of the racquet at the point of contact is given by;

V = ωr

V = (12)(0.5)

V = 6 m/s

Therefore, linear velocity of the racquet at the point of contact with the ball is 6 m/s.

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3 years ago

A water-skier is moving at a speed of 14.3 m/s. When she skis in the same direction as a traveling wave, she springs upward ever

kiruha [24]

Answer:

a) 1.95 m/s

b) 5.56 m

Explanation:

Given that:

Velocity of the skier $(V_s)$ = 14.3 m/s

For the skier moving in the direction of the wave, we have:

Period (T) = 0.450 s

Relative velocity (V) of the skier in regard with the wave = $(V_s - V_w)$

where:

$V_s$ = velocity of the skier

$V_w$ = velocity of the wave

The wavelength $(\lambda)$ can be written as:

$\lambda = (V_s-V_w)T$

$\lambda = (V_s-V_w) 0.450m$ ---------------> Equation (1)

For the skier moving opposite in the direction of the wave, we have:

Period (T) = 0.342 s

Relative velocity (V) of the skier in regard with the wave = $(V_s + V_w)$

The wavelength $(\lambda)$ can be written as:

$\lambda = (V_s+V_w)T$

$\lambda = (V_s+V_w) 0.342m$ ------------------> Equation 2

Equating equation (1) and equation (2) and substituting $V_s$ = 14.3 m/s ; we have:

$(V_s-V_w) 0.450m = (V_s-V_w) 0.342m$

$0.450m(V_s)-0.450m(V_w) = 0.342m(V_s)+0.342m(V_w)$

Collecting the like terms; we have:

$0.450m(V_s) - 0.342m(V_s) = 0.342m(V_w)+0.450m(V_w)$

$(V_s)(0.450m - 0.342m) = (V_w)0.342m+0.450m$

$14.3m/s(0.450m - 0.342m) = (V_w)0.342m+0.450m$

$14.3m/s(0.108m = (V_w)0.792m$

$1.5444m^2/s = (V_w)0.792m$

$(V_w) = \frac{1.5444m^2/s}{ 0.792m}$

$(V_w) = 1.95 m/s$

The Wavelength of the wave can be calculated using : $( \lambda }) = (V_s-V_w) 0.450m$

$({\lambda}) = (14.3 m/s -1.95 m/s)(0.450)$

$(\lambda) = (12.35)0.450m$

$(\lambda)= 5.5575 m$

λ ≅ 5.56 m

5 0

3 years ago

a smaller cannon ball leaves a cannon much faster than a larger, heavier cannon ball fired at the same time. which one of newton

Sever21 [200]

Answer:

that is his first law

Explanation:

8 0

4 years ago

Two workers pull horizontally on a heavy box but one pulls twice as hard as the other. The larger pull is directed at 25.0 west

icang [17]

1) Call F1 the larger force and F1x and F1y its its x-and-y- components.respectively.

I will use the complementary angle: 90 - 25 = 65 to work with the normal convention.

=> cos(65) = F1x / F1 => F1x = - F1*cos(65) (I choose negative as the west direction)

=> sin(65) = F1y / F1 => F1y = F1*sin(65) (I choose positive the north direction)

2) Call F2 the shorter force and F2x and F2y its components

=> cos(x) = F2x / F2 => F2x = F2*cos(x)

=> sin(x) = F2y / F2=> F2y = F2*sin(x)

3) You know that:

- F1 = 2F2
- The net force in the y direction is 430 N
- The net force in the x direction is 0

a) F1x + F2x = 0

=> -F1*cos(65) + F2*sin(x) = 0

=> F1*cos(65) = F2 sin(x) => sin(x) = [F1/F2] cos(65)

Remember F1 = 2F2 => F1/F2 = 2 => sin(x) = 2 cos(65) = 0.84524

=> x = arcsin(0.84524) = 57.7

b) F1y + F2y = 430 =>

F1 sin(65) + F2*sin(57.7) = 430 =>

0.9060F1 + 0.84524F2 430

F1 = 2F2 => 0.9060*2F2 + 0.84524F2 = 430 => 1.7512F2 = 430

=> F2 = 430 / 1.7512 = 245.54 N

=> F1 = 2*245.54 =491.1N

There you have the two forces.

The angle of the shorter force is 57.7 measured from the east to the north (this is north of east), which would be 90 - 57.7 = 32.3 degrees east of north..

Then the shorter force is 245.5 N at 32.3 degrees east of north

And the larger force is 491.1 N at 25.0 degrees west of north.

3 0

4 years ago