3 years ago

What is the effect that distance has on gravity?

Physics

1 answer:

IgorLugansk [536]3 years ago

3 0

The forces of gravity between two objects are smaller when the objects are farther apart.

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The battleship and enemy ships 1 and 2 lie along a straight line. Neglect air friction. battleship 1 2 Consider the motion of th

DaniilM [7]

Answer:

$\frac{t_1}{t_2} = \frac{sin\theta_1}{sin\theta_2}$

Explanation:

The vertical component of the initial velocities are

$v_v = v_0sin\theta$

If we ignore air resistance, and let g = -9.81 m/s2. The the time it takes for the projectiles to travel, vertically speaking, can be calculated in the following motion equation

$v_vt - gt^2/2 = s = 0$

$t(v_v - gt/2) = 0$

$v_v - gt/2 = 0$

$t = 2v_v/g = 2v_0sin\theta/g$

So the ratio of the times of the flights is

$t_1 / t_2 = \frac{2v_0sin\theta_1/g}{2v_0sin\theta_2/g} = \frac{sin\theta_1}{sin\theta_2}$

8 0

3 years ago

Increasing the amount of current that flows through a wire ______ the strength of an electromagnet

Anestetic [448]

Electromagnet is in form of solenoid

and the magnetic field due to solenoid is given as

$B = \mu_0 n i$

here

i = current in the loop

so when we increase the current in electromagnet the magnetic field of the solenoid will increase

this will increase the strength of the electromagnet

so the answer would be

<em>INCREASE</em>

7 0

3 years ago

Read 2 more answers

Pls help! True or False?

jasenka [17]

Answer:

True

Explanation:

i hope it helps

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7 0

2 years ago

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Imagine a glass box that is completely sealed and in that box is a cactus in a small pot which statement is most accurate?

Dennis_Churaev [7]

I THINK C BECAUSE IF IT IS A GLASS BOX HOW DID A CACTUS GET IN AND NOTHING CAN GET IN OR OUT OF THE BOX SO THERE IS NO CACTUS IN THE BOX

6 0

3 years ago

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Confirm if this is correct or not. If it isn't correct, please correct it.

kow [346]

Answer:

d = 421.83 m

Explanation:

It is given that,

Height, h = 396.9 m

Horizontal speed, v = 46.87 m/s

We need to find the distance traveled by the ball horizontally. Let t is the time taken by the ball. Using second equation of motion for vertical direction. So,

$396.9=0\times t+\dfrac{1}{2}\times 9.8 t^2\\\\t=9\ s$

Now d is the distance covered by the cannonball. So,

$d=vt\\\\d=46.87\times 9\\\\d=421.83\ m$

Hence, this is the required solution.

3 0

3 years ago