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3 years ago

A 2500-N net force acting on a 880-kg car accelerates it at a rate of ______ m/s/s

Physics

1 answer:

Bond [772]3 years ago

8 0

Answer:

a = 2.84 m/s²

Explanation:

Given that,

Net force, F = 2500 N

Mass of the car, m = 880 kg

We need to find the acceleration of the car. Net force is given by :

F = ma

$a=\dfrac{F}{m}\\\\a=\dfrac{2500\ N}{880\ m/s^2}\\\\a=2.84\ m/s^2$

So, the acceleration of the car is 2.84 m/s².

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The cable holding a 2125 kg elevator has a maximum strength of 21,750 N. What is the maximum upward acceleration the cable can g

vivado [14]

Answer:

10.23m/s^2

Explanation:

GIven data

mass of elevator = 2125 kg

Force= 21,750 N

Required

The maximum acceleration upward

F= ma

a= F/m

a=21,750/2125

a= 10.23m/s^2

Hence the acceleration is 10.23m/s^2

4 0

3 years ago

A brick and a feather fall to the earth at their respective terminal velocities. Which objectexperiences the greater force of ai

Karo-lina-s [1.5K]

Answer:

Under the reasonable assumption that the brick has more mass than the feather, the brick experiences a greater force of air friction.

Explanation:

Objects at terminal velocity, only under the influence of gravity, have maximized their speed and have an acceleration of zero. Thus, neither object is accelerating.

Recall Newton's second law: $\sum {\vec {F}}=m \vec {a}$

Since acceleration for each object is zero, the sum of the force acting on each of those objects must also be zero.

Since the only forces acting on the objects are gravity and the force of air friction, in order to zero out, the force of air friction must be equal in magnitude and opposite in direction to the force of gravity.

Recall that near the surface of the earth, $F_{gravity}=mg$ , so the Force of Gravity acting on an object is directly proportional to the object's mass. (A similar argument could be made even if this were not taking place on the surface of the earth, so long as the objects were the same distance from the object providing gravitational influence).

If the masses of the objects are different, the object with the greater mass will experience a larger force of gravity, and hence a larger force of air friction at terminal velocity.

Under the reasonable assumption that the brick has more mass than the feather, the brick experiences a greater force of air friction.

4 0

2 years ago

(b) A car of mass 3000 kg travels at a constant velocity of 5.0 m/s.

Tamiku [17]

Answer:

16.7 s

Explanation:

T= Vf - Vo a= F

a m

4,500 / 3000 = 1.5 (a)

30 - 5 / 1.5(a) = 16.7 s

4 0

3 years ago

calculate the mass of displaced water when a piece of 30cm Iceberg with surface area 1000 cm^2 floats on water density of ice is

sweet [91]

Vi = As * h = 1000 * 30 = 30,000 cm^3 = Vol. of the ice.

Vb = (Di/Dw) * Vi = (0.9/1.0) * 30,000 = 27,000 cm^3 = Vol. below surface - Vol. of water displaced.

27,000cm^3 * 1g/cm^3 = 27,000 grams = 27 kg = Mass of water displaced.

3 0

3 years ago

A boxer punches a sheet of paper in mid air, and brings it from rest up to a speed of 25 m/s in 0.05 s. if the mass of the paper

zepelin [54]

Ok, so you've got to figure out a force F and you have the speed in which the boxer punches on determinate time and the mass of the sheet of paper.

So based on the formula that says that the Force is equal to the mass multiplied by the acceleration => F=ma.
You look at it and see that you only have mass which is measured on KG so there is no problem.
then you have the acceleration which is measured on meters and is defined by: a = Δv/Δt
So now you can replace the velocity and the time you have there
⇒ a 25m/s / 0.05s
you have computing that ⇒ 50m because the seconds were cancelled out.
and then you plug the meters into the force equation.
F=(0.005kg)(50)
F=0.25N
so the boxer will have a force of 0.25 Newton's.

6 0

3 years ago

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