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2 years ago

What are the two variables that affect the strength of gravity?

Physics

1 answer:

Dovator [93]2 years ago

4 0

Answer:

mass or size, distance

Explanation:

using newton's law of of gravity the only values that aren't constants are distance and mass since an objects mass applies to its gravity along with the distance

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A 0.40 kg object is attached to a spring with force constant 157 N/m so that the object is allowed to move on a horizontal frict

nirvana33 [79]

Answer:

Part a)

$F = 26.7 N$

Part b)

$a = 66.7 m/s^2$

Explanation:

Part a)

Force on the object due to spring force is given as

$F = kx$

here we know that

$k = 157 N/m$

$x = 0.17 m$

so we have

$F = 157\times 0.17$

$F = 26.7 N$

Part b)

Acceleration of the object is given as

$a = \frac{F}{m}$

$a = \frac{26.7}{0.40}$

$a = 66.7 m/s^2$

8 0

3 years ago

How many days does the average male live?

Marysya12 [62]

Answer:

An average male lives 27,375 Days

4 0

4 years ago

In what way could a random mutation provide an organism with<br> an advantage?

andreyandreev [35.5K]

Answer:

They are called beneficial mutations. They lead to new versions of proteins that help organisms adapt to changes in their environment. Beneficial mutations are essential for evolution to occur. They increase an organism's changes of surviving or reproducing, so they are likely to become more common over time.

Explanation:

7 0

3 years ago

A car accelerates from rest at 3m/s2 along a straight road how far has the car traveled after 4s?

murzikaleks [220]

Answer:

24 m

Explanation:

$x = 1/2at^{2} + v_{0}t + x_{0}\\a = 3 m/s^{2}\\v_{0} = 0$

7 0

3 years ago

A ball is thrown upward with a speed of 28.2 m/s.A. What is its maximum height?B. How long is the ball in the air?C. When does t

Ede4ka [16]

Answer:

(A) The maximum height of the ball is 40.57 m

(B) Time spent by the ball on air is 5.76 s

Explanation:

Given;

initial velocity of the ball, u = 28.2 m/s

(A) The maximum height

At maximum height, the final velocity, v = 0

v² = u² -2gh

u² = 2gh

$h = \frac{u^2}{2g}\\\\h = \frac{(28.2)^2}{2*9.8}\\\\h = 40.57 \ m$

(B) Time spent by the ball on air

Time of flight = Time to reach maximum height + time to hit ground.

Time to reach maximum height = time to hit ground.

Time to reach maximum height is given by;

v = u - gt

u = gt

$t = \frac{u}{g}$

Time of flight, T = 2t

$T = \frac{2u}{g}\\\\ T = \frac{2*28.2}{9.8}\\\\ T = 5.76 \ s$

As the ball moves upwards, the speed drops, then the height of the ball when the speed drops to 12m/s will be calculated by applying the equation below.

v² = u² - 2gh

12² = 28.2² - 2(9.8)h

12² - 28.2² = - 2(9.8)h

-651.24 = -19.6h

h = 651.24 / 19.6

h = 33.23 m

Thus, at 33.23 m the speed will be 12 m/s

6 0

3 years ago