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Katarina [22]
1 year ago
12

How much net force is required to accelerate a 2000 kg car at 3.00 m/s^2

Physics
1 answer:
andrezito [222]1 year ago
6 0

The net force required to accelerate a car is 6000 N.

Force is defined as the product of the mass and acceleration of the body. Force is used to changing the velocity that is to accelerate an object or a body of a particular mass. The unit of Force is Newton or kg m/s^2.

The formula used to calculate the net force is :

F = ma

where, F = Force

m = mass = 2000 kg

a = acceleration = 3.00 m/s^2

∴ F = 2000*3

F = 6000 N

Thus, to accelerate the car at 3.00 m/s^2 of mass 2000 kg net force required is 6000 N.

To learn more about force,

brainly.com/question/1046166

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A fox is 7 meters from the base of a tree that is 15 m tall. A nest of owls is at the top of the tree. One of the owlets tosses
mamaluj [8]
Refer to the diagram shown below.

In this analysis, wind resistance is ignored, and g = 9.8 m/s².

The meat falls with zero vertical velocity, therefore the time, t, before the meat hits the ground is
\frac{1}{2}*(9.8 \, \frac{m}{s^{2}})*(t \, s)^{2} = (15 \, m) \\ t= \sqrt{ \frac{15}{4.9} }= 1.75 \, s

If the fox catches the meat before it hits the ground, then the fox should travel a horizontal distance d in the same time that the meat travels a horizontal distance (7 -d).
The meat travels a distance of 
7 - d = (1.2 m/s)*(1.75 s) = 2.1 m
or
d = 4.9 m

Let v =  velocity of the fox when it catches the meat.
If the acceleration of the fox is a m/s², then
v = 1.75a
Also,
d= \frac{1}{2} *(a \,  \frac{m}{s^{2}} )*(1.75 \, s)^{2} =  \frac{1}{2}( \frac{v}{1.75})^{2}*(1.75^{2})  \\ 4.9 = 0.875v^{2} \\ v^{2} = 5.6 \\ v = 2.366 \, m/s

Answer:  2.37 m/s  (nearest hundredth)



5 0
3 years ago
What is the name of the unit you use to measure force?
Anettt [7]

Answer:

that would be newtons.

Explanation:

I hope this helps

8 0
2 years ago
Which types of heat transfer involve heat flow from hot objects to colder objects? A. convection, conduction, and radiation B. c
Soloha48 [4]
B is definitely the answer. Please let it be
4 0
3 years ago
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You are working on a laboratory device that includes a small sphere with a large electric charge Q. Because of this charged sphe
madam [21]

Answer:

the only effect it has is to create more induced charge at the closest points, but the net face remains zero, so it has no effect on the flow.

Explanation:

We can answer this exercise using Gauss's law

      Ф = ∫ e . dA = q_{int} / ε₀

field flow is directly proportionate to the charge found inside it, therefore if we place a Gaussian surface outside the plastic spherical shell.  the flow must be zero since the charge of the sphere is equal  induced in the shell, for which the net charge is zero. we see with this analysis that this shell meets the requirement to block the elective field

From the same Gaussian law it follows that if the sphere is not in the center, the only effect it has is to create more induced charge at the closest points, but the net face remains zero, so it has no effect on the flow , so no matter where the sphere is, the total induced charge is always equal to the charge on the sphere.

5 0
3 years ago
A roundabout in a fairground requires an input power of 2.5 kW when operating at a constant angular velocity of 0.47 rad s–1 . (
natita [175]

Answer:

Explanation:

Since the roundabout is rotating with uniform velocity ,

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b )

When power is switched off , it will decelerate because of frictional torque .

5 0
3 years ago
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