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STALIN [3.7K]
3 years ago
6

The universal law of gravitation states that the force of attraction between two objects depends on which quantities?

Physics
1 answer:
xxMikexx [17]3 years ago
7 0

Answer:depends on the masses of the objects and the distance between them

Explanation:

According to Newton's law of universal gravitation,the force of attraction between two objects depends on the masses of the objects and the distance between them

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Plz help me
Reil [10]

Answer:

Yes

Explanation:

When an object has more mass it takes more gravity to keep it down therefore producing friction which in return reduces the amount of kinetic energy created. A change in an object's speed has an greater effect on its kinetic energy. than a change in its mass has, because kinetic energy is proportional to.

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What is the kinetic energy of a 9.0 kg steelhead if its speed is 16 m/s?
Vesnalui [34]

<u>We are given:</u>

Mass of the Steelhead(m) = 9 kg

Velocity of the Steelhead(v) = 16 m/s

<u>Calculating the Kinetic Energy:</u>

KE = 1/2mv²

replacing the variables

KE = 1/2 * 9 * (16)²

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3 years ago
1. What type of wave is sound? Does sound need a medium (substance) to travel through?
denpristay [2]

Answer:

1. What type of wave is sound?

2.Does sound need a medium to travel through?

3. What are the properties of sound waves?

Explanation:

1. The type of waves sound are is mechanical waves.


2. Sound needs a solid, liquid or gas (material medium) to travel through.


3. I believe they are wavelength, amplitude, frequency, time period and velocity. 


I apologize if it is incorrect-

I hope it helps! Have a great day!

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6 0
2 years ago
A 3.9 g dart is fired into a block of wood with a mass of 24.6 g. The wood block is initially at rest on a 1.5 m tall post. Afte
Galina-37 [17]

Answer:

46.48m/s

Explanation:

The problem is a combination of the principle of conservation of linear momentum and projectile motion.

The principle of conservation of linear momentum states that in a closed system, the total momentum of colliding bodies before impact is equal to the total momentum after impact. The masses stated in the problem experienced an inelastic collision. In an inelastic collision, the bodies involved stick together after the collision and move with a common velocity.

For two bodies of masses m_1 and m_2 moving with velocities u_1 and u_2 before impact, if they experience inelastic collision, the conservation of their momenta is as stated in equation (1);

m_1u_1+m_2u_2=(m_1+m_2)v..................(1)

were v is their common velocity after impact. If the second mass m_2 was at rest before the impact, then its initial velocity u_2=0m/s. therefore m_2u_2=0. Equation (1) then becomes;

m_1u_1=(m_1+m_2)v..............(2)

In the problem stated, the second mass taken as the mass of the wooden block was at rest before the impact and the collision was inelastic since both the wood and the dart stuck together and moved with a common velocity after the impact. Therefore we can use equation (2) for the problem.

Given;

m_1=3.9g=0.0039kg\\u_1=?\\m_2=24.6g=0.0246kg\\v=?

Substituting these values into (2), we get the following;

0.0039*u_1=(0.0039+0.024)v\\0.0039u_1=0.0285v.........(3)

Their common v velocity after impact now makes both the wooden block and the dart (as a single body) to fall vertically through a height h of 1.5m over a range R of 3.5m as stated by the problem; hence by the principle of projectile motion for a body projected horizontally, the following relationship holds;

R= vt............(4)

were t is the time taken to fall through the height h. To obtain t we use the second equation of free fall under gravity;

h=\frac{1}{2}gt^2...........(5)

were g is acceleration due to gravity taken as 9.8m/s^2. Therefore;

1.5=\frac{1}{2}*9.8*t^2\\1.5=4.9t^2\\t^2=\frac{1.5}{4.9}=0.306\\t=\sqrt{0.306} =0.55s

We then substitute R and t into equation (4) to obtain v.

3.5=v*0.55\\v=\frac{3.5}{0.55}\\v=6.36m/s

We now further substitute this value of v into (3) to obtain u_1;

u_1=\frac{0.0285v}{0.0039}\\\\u_1=\frac{0.0285*6.36}{0.0039}\\\\u_1=\frac{0.18126}{0.0039}\\\\u_1=46.48m/s

4 0
3 years ago
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