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otez555 [7]
2 years ago
13

What is the gravitational force of attraction between a planet and a 17-kilogram mass that is falling freely toward the surface

of the planet at 8.8 meters per second squared?a. 150 Nb. 8.8 Nc. 1.9 Nd. 0.52 N
Physics
1 answer:
PolarNik [594]2 years ago
5 0

Answer:

a. 150 N

Explanation:

Gravitational Force: This is the force that act on a body under gravity.

The gravitational force always attract every object on or near the earth's surface. The earth therefore, exerts an attractive force on every object on or near it.

The S.I unit of gravitational force is Newton(N).

Mathematically, gravitational force of attraction is expressed as

(i) F = GmM/r² ........................ Equation 1 ( when it involves two object of different masses on the earth)

(ii) F = mg ............................... Equation 2 ( when it involves one mass and the gravitational field).

Given: m = 17 kg, g = 8.8 m/s²

Substituting into equation 2,

F = 17(8.8)

F = 149.6 N

F ≈ 150 N.

Thus the gravitational force = 150 N

The correct option is a. 150 N

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A bullet with a mass m b = 11.9 mb=11.9 g is fired into a block of wood at velocity v b = 261 m/s. vb=261 m/s. The block is atta
fredd [130]

Answer:

0.372 kg

Explanation:

The collision between the bullet and the block is inelastic, so only the total momentum of the system is conserved. So we can write:

mu=(M+m)v (1)

where

m=11.9 g = 11.9\cdot 10^{-3}kg is the mass of the bullet

u=261 m/s is the initial velocity of the bullet

M is the mass of the block

v is the velocity at which the bullet and the block travels after the collision

We also know that the block is attached to a spring, and that the surface over which the block slides after the collision is frictionless. This means that the energy is conserved: so, the total kinetic energy of the block+bullet system just after the collision will entirely convert into elastic potential energy of the spring when the system comes to rest. So we can write

\frac{1}{2}(M+m)v^2 = \frac{1}{2}kx^2 (2)

where

k = 205 N/m is the spring constant

x = 35.0 cm = 0.35 m is the compression of the spring

From eq(1) we get

v=\frac{mu}{M+m}

And substituting into eq(2), we can solve to find the mass of the block:

(M+m) \frac{(mu)^2}{(M+m)^2}=kx^2\\\frac{(mu)^2}{M+m}=kx^2\\M+m=\frac{(mu)^2}{kx^2}\\M=\frac{(mu)^2}{kx^2}-m=\frac{(11.9\cdot 10^{-3}\cdot 261)^2}{(205)(0.35)^2}-11.9\cdot 10^{-3}=0.372 kg

4 0
3 years ago
Why electrons flow in a wire when connected to a battery?
vesna_86 [32]

Answer:

The metal atoms in the wire can't move, but their outer electrons can. The force pushes those electrons and they move to further parts of the wire, trying to reach the other end. As the electrons move away, new electrons flow into the wire through the battery to take their place.

Explanation:

6 0
3 years ago
A student is given three wires that are made from different materials, but each wire has the same physical dimensions. For a giv
Ymorist [56]

Answer:

Use the ammeter to measure the current that flows through each wire, because a larger current that flows through the wire corresponds to a smaller resistivity

Explanation:

Since they are connected to a constant voltage power source, the potential difference does not change. The potential difference is proportional to the product of the current and the resistance and, the resistance opposes the flow of electric current. It is clear to see that a large current that flows through the current means there is a lesser resistance to the flow of current at constant potential difference across the circuit.

6 0
3 years ago
Ryan applied a force of 10N and moved a book 30 cm in the direction of the force. How much was the workdone by Ryan?​
Xelga [282]
<h2><u>Question</u><u>:</u><u>-</u></h2>

Ryan applied a force of 10N and moved a book 30 cm in the direction of the force. How much was the work done by Ryan?

<h2><u>Answer:</u><u>-</u></h2>

<h3>Given,</h3>

=> Force applied by Ryan = 10N

=> Distance covered by the book after applying force = 30 cm

<h3>And,</h3>

30 cm = 0.3 m (distance)

<h3>So,</h3>

=> Work done = Force × Distance

=> 10 × 0.3

=> 3 Joules

\small \boxed{work \: done \:  by \: Ryan \:  = 3 \: Joules}

4 0
2 years ago
What pulls all objects in the universe, including the moon and earth and the su
Salsk061 [2.6K]
It's called gravity, it attract the sun toward the gravitational pull making everything circulate. I don't really know how to explain it though.
6 0
3 years ago
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