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3 years ago

How much force is required to accelerate a 2 kg mass at 3 m/s^2?*

Physics

1 answer:

alexgriva [62]3 years ago

5 0

Answer: 6N

Explanation:

F = MA due to Newton's second law

M = 2

A = 3

plug in to get

F = 2 * 3 = 6N

You might be interested in

If a snail travels at 5 m/s, how far will it travel in 90 seconds?

labwork [276]

Answer:

in 90 seconds It'll travel 450m At constant speed of 5m/s

8 0

3 years ago

Read 2 more answers

If the 78.0 kg astronaut were in a spacecraft 6R from the center of the earth, what would the astronaut's weight be on earth? 76

den301095 [7]

(a) 764.4 N

The weight of the astronaut on Earth is given by:

$F=mg$

where

m is the astronaut's mass

g is the acceleration due to gravity

Here we have

m = 78.0 kg

g = 9.8 m/s^2 at the Earth's surface

So the weight of the astronaut is

$F=(78.0)(9.8)=764.4 N$

(b) 21.1 N

The spacecraft is located at a distance of

$r=6R$

from the center of Earth.

The acceleration due to gravity at a generic distance r from the Earth's center is

$g=\frac{GM}{r^2}$

where G is the gravitational constant and M is the Earth's mass.

We know that at a distance of r = R (at the Earth's surface) the value of g is 9.8 m/s^2, so we can write:

$GM=9.8R^2$ (1)

the acceleration due to gravity at r=6R instead will be

$g'=\frac{GM}{(6R)^2}$

And substituting (1) into this formula,

$g'=\frac{9.8R^2}{36R^2}=0.27 m/s^2$

So the weight of the astronaut at the spacecratf location is

$F'=mg'=(78.0 kg)(0.27 m/s^2)=21.1 N$

6 0

4 years ago

An elastic circular bar is fixed at one end and attached to a rubber grommet at the other end. The grommet functions as a torsio

Artist 52 [7]

Answer:

2.1 rad(anticlockwise).

Explanation:

So, we are given the following data or parameters or information in the question above:

=> "The torsional stiffness of the spring support is k = 50 N m/rad. "

=> "If a concentrated torque of mag- nitude Ta = 500 Nm is applied in the center of the bar"

=> "L = 300 mm Assume a shear modu- lus G = 10 kN/mm2 and polar monnent of inertia J = 2000 mln"

Hence;

G × J = 10 kN/mm2 × 2000 mln = 20 Nm^2.

Also, L/2 = 300 mm /2 = 0.15 m (converted to metre).

==> 0.15/20 (V - w) + θ = 0.

==> 0.15/20 (V - w ) = -θ.

Where V = k = 50 N m/rad

w = 183.3 θ.

Therefore, w + Vθ = 500 Nm.

==> 183.3 + 50 θ = 500 Nm.

= 6.3

Anticlockwise,

θ = 2.1 rad.

4 0

3 years ago

The velocity time graph of a car shown below a) Calculate the magnitude of displacement of the car in 40 seconds. b) During whic

Gekata [30.6K]

Answer:

a) 0 metres

b) From time 0 s to 10 s , the car was accelerated. Its velocity accelerated from 0m/s to 20 m/s

c) 20 m/s

Explanation:

a) <em>Formula of displacement= velocity x time</em>

time=40 s

velocity =0 m/s

∴ displacement= 0 x 40 = 0 m

Magnitude of displacement is 0 m

b) The increase in velocity shows that there has been acceleration.

c) The average velocity of the car is = $\frac{0+40}{2\\}$ {initial velocity + final velocity}

= $\frac{40}{2}$

=20

Therefore, the magnitude of the average velocity of the car is 20 m/s

3 0

3 years ago

Two men, Joel and Jerry, each pushes an object that are identical on a horizontal frictionless floor starting from rest. Joel an

Nataliya [291]

Answer:

The work done by Joel is greater than the work done by Jerry.

Explanation:

Let suppose that forces are parallel or antiparallel to the direction of motion. Given that Joel and Jerry exert constant forces on the object, the definition of work can be simplified as:

$W = F\cdot \Delta s$

Where:

$W$ - Work, measured in joules.

$F$ - Force exerted on the object, measured in newtons.

$\Delta s$ - Travelled distance by the object, measured in meters.

During the first 10 minutes, the net work exerted on the object is zero. That is:

$W_{net} = W_{Joel} - W_{Jerry}$

$W_{net} = F\cdot \Delta s - F\cdot \Delta s$

$W_{net} = (F-F)\cdot \Delta s$

$W_{net} = 0\cdot \Delta s$

$W_{net} = 0\,J$

In exchange, the net work in the next 5 minutes is the work done by Joel on the object:

$W_{net} = W_{Joel}$

$W_{net} = F\cdot \Delta s$

Hence, the work done by Joel is greater than the work done by Jerry.

7 0

4 years ago