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2 years ago

If an object is accelerating to the right , then net force on the object must be directed towards the right ! True Or False ?

2 answers:

8 0

Not so fast.

I think you're using 'accelerating' to mean 'speeding up', but you really need
to be more careful with it. "Acceleration" means ANY change in speed OR
direction.

If an object's speed to the left is decreasing, or its speed to the right is
increasing, then the net force on the object must be directed towards
the right.

If an object is moving with constant speed in a circular path, then it's
constantly accelerating, because its direction is constantly changing.
The force on it is always directed towards the center of the circle, so
there's one point on the path where the force is directed straight to the right.

Serggg [28]2 years ago

6 0

This is correct. Just think about it. If you push an object to the right then it will accelerate to the right.

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Genrish500 [490]

Answer:

3.0 A Is the correct option

Explanation:

I = V / R

I = 12/ 4

= 3

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2 years ago

A balloon of hydrogen is put into to pressure chamber. The initial pressure and volume of hydrogen is 1 atm and .5 cm3. The pres

Sedbober [7]

Answer:

2.5 cm³

Explanation:

Use Boyle's Law

1×5=2×V

5=2V

V=5/2=2.5 cm³

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2 years ago

Nancy needs to do a scientific investigation for her class. She is interested in pottery, plants, and basketball. Which of the f

JulijaS [17]

I believe that the answer is A hope this helped.

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2 years ago

Read 2 more answers

The power radiated by the sun is 3.90 1026 W. The earth orbits the sun in a nearly circular orbit of radius 1.50 1011 m. The ear

WITCHER [35]

Answer:

3234.2 W

Explanation:

Since intensity I = Power/Area. The intensity of the light from the sun, I = power radiated by sun/area of sphere of radius, r = 1.5 × 10¹¹ m.

So, I = 3.9 10²⁶W/4π(1.5 × 10¹¹ m)² = 2.069 × 10³ W/m².

Now, the power radiated on the patch of area 0.570 m² at the equator is

P = Icos27/A = 2.069 × 10³ W/m² cos27/0.570 m² = 1843.49/0.570 = 3234.2 W

6 0

2 years ago

An object falls from rest on a high tower and takes 5.0 s to hit the ground. Calculate the object's position from the top of the

Lena [83]

Answer:

After 1 sec = 4.9 m

After 2 sec = 19.6 m

After 3 sec = 44.1 m

After 4 sec = 78.4 m

After 5 sec = 122.5 m

Explanation:

After 1 sec:

u=0m/s t=1 s a=9.8m/s²

s = ut + (1/2)at²

=0(1) + (1/2)(9.8)(1²) = 4.9m

After 2 sec:

u=0m/s t=2 s a=9.8m/s²

s = ut + (1/2)at²

=0(2) + (1/2)(9.8)(2²) = 19.6m

After 3 sec:

u=0m/s t=3 s a=9.8m/s²

s = ut + (1/2)at²

=0(3) + (1/2)(9.8)(3²) = 44.1m

After 4 sec:

u=0m/s t=4 s a=9.8m/s²

s = ut + (1/2)at²

=0(4) + (1/2)(9.8)(4²) = 78.4m

After 5 sec:

u=0m/s t=5 s a=9.8m/s²

s = ut + (1/2)at²

=0(5) + (1/2)(9.8)(5²) = 122.5m

7 0

2 years ago