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2 years ago

the mass of a car is 1000 kg. how much force would be required to accelerate the car at a rate of 3 m/s^2

Physics

2 answers:

Olenka [21]2 years ago

5 0

Answer:y?

Explanation:

GalinKa [24]2 years ago

4 0

Answer:

<h3>The answer is 3000 N</h3>

Explanation:

The force acting on an object given the mass and acceleration we use the formula

<h3>force = mass × acceleration</h3>

From the question

mass = 1000 kg

acceleration = 3 m/s²

So we have

force = 1000 × 3

We have the final answer as

Hope this helps you

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A mass m attached to a spring of constant k is oscillating on a frictionless surface. A second mass of mass m is dropped on top

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Answer:

E. The period of oscillation increases.

Explanation:

The period of oscillation is:

T = 2π√(m/k)

Frequency is the inverse of period (f = 1/T), so as period increases, frequency decreases.

Increasing the mass will increase the period and decrease the frequency.

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Answer:

Skeletal system, muscular system

Explanation:

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2 years ago

Read 2 more answers

A balloon is rising vertically upwards at a velocity of 10m/s. When it is at a height of 45m from the ground, a parachute bails

harina [27]

(a) 30.9 m

Let's analyze the motion of the parachutist. Its vertical position above the ground is given by

$y=h+ut+\frac{1}{2}gt^2$

where

h = 45 m is the initial height

u = 10 m/s is the initial velocity (upward)

t is the time

g = -9.8 m/s^2 is the acceleration of gravity (downward)

Substituting t=3 s , we find the height of the parachutist when it opens the parachute:

$y=45 m+(10 m/s)(3 s)+\frac{1}{2}(-9.8 m/s^2)(3 s)^2=30.9 m$

(b) 44.1 m

Here we have to find first the height of the balloon 3 seconds after the parachutist has jumped off from it. The vertical position of the balloon is given by

y = h + ut

where

h = 45 m is the initial height

u = 10 m/s is the initial velocity (upward)

t is the time

Substituting t = 3 s, we find

y = 45 m + (10 m/s)(3 s) = 75 m

So the distance between the balloon and the parachutist after 3 s is

d = 75 m - 30.9 m = 44.1 m

(c) 8.2 m/s downward

The velocity of the parachutist at the moment he opens the parachute is:

v = u +gt

where

u = 10 m/s is the initial velocity (upward)

t is the time

g = -9.8 m/s^2 is the acceleration of gravity (downward)

Substituting t = 3 s,

v = 10 m/s + (-9.8 m/s^2)(3 s)= -19.4 m/s

where the negative sign means it is downward

After t=3 s, the parachutist open the parachute and it starts moving with a deceleration of

a =+5 m/s^2

where we put a positive sign since this time the acceleration is upward.

The total distance he still has to cover till the ground is

d = 30.9 m

So we can find the final velocity by using

$v^2-u^2 = 2ad$

where this time we have u = 19.4 m/s as initial velocity. Taking the downward direction as positive, the deceleration must be considered as negative:

a = -5 m/s^2

Solving for v,

$v=\sqrt{u^2 +2ad}=\sqrt{(19.4 m/s)^2+2(-5 m/s^2)(30.9 m)}=8.2 m/s$

(d) 5.24 s

We can find the duration of the second part of the motion of the parachutist (after he has opened the parachute) by using

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

where

a = -5 m/s^2 is the deceleration

v = 8.2 m/s is the final velocity

u = 19.4 m/s is the initial velocity

t is the time

Solving for t, we find

$t=\frac{v-u}{a}=\frac{8.2 m/s-19.4 m/s}{-5 m/s^2}=2.24 s$

And added to the 3 seconds between the instant of the jump and the moment he opens the parachute, the total time is

t = 3 s + 2.24 s = 5.24 s

8 0

2 years ago

A steel ball with mass m is suspended from the ceiling at the bottom end of a light, 17.0-m-long rope. The ball swings back and

iogann1982 [59]

Answer:

1. 18.25 m/s

2. 0 m/s

Explanation:

1.So the centripetal acceleration of the ball at this lowest point must be, taking gravity into account

$a_c = \frac{T - mg}{m} = \frac{3mg - mg}{m} = 2g$

The speed at this point would then be

$v^2 = a_c r = 2gr = 2*9.8*17 = 333.2$

$v = \sqrt{333.2} = 18.25 m/s$

2. Similarly, if T = mg, then the centripetal acceleration must be

$a_c = \frac{T - mg}{m} = \frac{mg - mg}{m} = 0$

As the ball has no centripetal acceleration, its speed must also be 0 as well.

6 0

3 years ago