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castortr0y [4]
3 years ago
13

What landforms are found on the Moon? (Select all that apply.)

Physics
1 answer:
Novay_Z [31]3 years ago
3 0

Answer: I think it’s craters

Explanation:

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You are pushing a 30-kg block on a rough floor in a direction that is parallel to the floor. The block moves with a uniform 2 m/
kykrilka [37]

Answer:dd

Explanation:

I dunno

3 0
3 years ago
The acceleration of a block attached to a spring is given by a=−(0.324m/s2)cos([2.50rad/s]t) a = − ( 0.324 m / s 2 ) c o s ( [ 2
allsm [11]

Answer:

Looks like you have:

a = -.324 cos 2.5 t

In this case   ω^2 A = .324

ω = 2.5

f = ω / (2 * pi) = 2.5 / 6.28 = .40 / sec

5 0
2 years ago
A car starts from rest and accelerates uniformly at a rate of 2.0 meter per second squared for 4.0 seconds. During this time int
vagabundo [1.1K]

Answer:

16

Explanation:

\frac{1}{2}  \times 2 \times  {4}^{2}  = 16

7 0
3 years ago
A motorcycle traveling at a speed of 44.0 mi/h needs a minimum of 44.0 ft to stop. If the same motorcycle is traveling 79.0 mi/h
Tasya [4]

Answer:

141.78 ft

Explanation:

When speed, u = 44mi/h, minimum stopping distance, s = 44 ft = 0.00833 mi.

Calculating the acceleration using one of Newton's equations of motion:

v^2 = u^2 + 2as\\\\v = 0 mi/h\\\\u = 44 mi/h\\\\s = 0.00833 mi\\\\=> 0^2 = 44^2 + 2 * a * 0.00833\\\\=> 1936 = -0.01666a\\\\a = -116206.48 mi/h^2 or -14.43 m/s^2

Note: The negative sign denotes deceleration.

When speed, v = 79mi/h, the acceleration is equal to when it is 44mi/h i.e. -116206.48 mi/h^2

Hence, we can find the minimum stopping distance using:

v^2 = u^2 + 2as\\\\v = 0 mi/h\\\\u = 79 mi/h\\\\a = -116206.48 mi/h\\\\=> 0^2 = 79^2 + (2 * -116206.48 * s)\\\\6241 = 232412.96s\\\\s = \frac{6241}{232412.96} \\\\s = 0.0268531 mi = 141.78 ft

The minimum stopping distance is 141.78 ft.

4 0
3 years ago
A farsighted boy has a near point at 2.3 m and requires eyeglasses to correct his vision. Corrective lenses are available in inc
tino4ka555 [31]

Answer:

P = 3.5 D

Explanation:

As we know that convex lens is to be used to make the near point of eye to be correct

So we will have

\frac{1}{d_i} + \frac{1}{d_o} = \frac{1}{f}

here we have

d_i = 2.3 m = 230 cm

d_o = 25 cm

now plug in all values into the formula

-\frac{1}{230} + \frac{1}{25} = \frac{1}{f}

f = 28 cm

now for power of lens

P = \frac{1}{f}

P = \frac{1}{0.28} = 3.5 D

so the power in dioptre is

P = 3.5 D

5 0
3 years ago
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