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Explain two ways friction can be reduced

BlackZzzverrR [31]

Friction can be reduced by making a surface smooth. It can also be reduced by applying lubricants such as oil and grease

6 0

3 years ago

A student removes a 10.5 kg stereo amplifier from a shelf that is 1.82 m high. The amplifier is lowered at a constant speed to a

Nonamiya [84]

Answer:

(a) the work done by the student is 110.1 J

(b) The gravitational force that acts on the amplifier is 102.9 N

Explanation:

Given;

mass of the amplifier, m = 10.5 kg

initial position of the amplifier, x₀ = 1.82 m

final position of the amplifier, x₁ =0.75 m

The dispalcement of the amplifier Δx = x₁ - x₀ = 1.82 m - 0.75 m = 1.07 m

(b) The gravitational force that acts on the amplifier;

F = mg

F = 10.5 x 9.8

F = 102.9 N

(a) the work done by the student is calculated as;

W = FΔx

W = 102.9 x 1.07

W = 110.1 J

7 0

3 years ago

At a certain distance from the center of the Earth, a 0.4-kg object has a weight of 2.0 N. (a) Find this distance. (b) If the ob

Alika [10]

Answer:

a) The distance of the object from the center of the Earth is 8.92x10⁶ m.

b) The initial acceleration of the object is 5 m/s².

Explanation:

a) The distance can be found using the equation of gravitational force:

$F = \frac{GMm}{r^{2}}$

Where:

G: is the gravitational constant = 6.67x10⁻¹¹ Nm²/kg²

M: is the Earth's mass = 5.97x10²⁴ kg

m: is the object's mass = 0.4 kg

F: is the force or the weight = 2.0 N

r: is the distance =?

The distance is:

$r = \sqrt{\frac{GMm}{F}} = \sqrt{\frac{6.67 \cdot 10^{-11} Nm^{2}/kg^{2}*5.97 \cdot 10^{24} kg*0.4 kg}{2.0 N}} = 8.92 \cdot 10^{6} m$

Hence, the distance of the object from the center of the Earth is 8.92x10⁶ m.

b) The initial acceleration of the object can be calculated knowing the weight:

$W = ma$

Where:

W: is the weight = 2 N

a: is the initial acceleration =?

$a = \frac{W}{m} = \frac{2 N}{0.4 kg} = 5 m/s^{2}$

Therefore, the initial acceleration of the object is 5 m/s².

I hope it helps you!

4 0

3 years ago

A satellite is in a circular Earth orbit of radius r. The area A enclosed by the orbit depends on r2 because A = πr2. Determine

Salsk061 [2.6K]

Answer:

a. $T=r^{3/2}$

b. $K=\frac{1}{r}$

c. $v=\frac{1}{\sqrt{r}}$

d. $v=\sqrt{r}$

Explanation:

To make analysis about the satellite circular earth the depends or r and A

$T^2=\frac{4\pi}{GM}*r^3$

$K=\frac{GM*m}{2*r}$

a.

$T^2=r^3$

$T=r^{3/2}$

b.

$K=\frac{GM*m}{2}*\frac{1}{r}$

$K=\frac{1}{r}$

c.

$K=\frac{1}{2}*m*v^2$

$v^2=\frac{2*K}{m}=\frac{2*Gm*m}{2*m*r}$

$v=\frac{1}{\sqrt{r}}$

d.

$v=\sqrt{2*GMr}$

$v=\sqrt{r}$

3 0

3 years ago

A 1.0 kg ball falls from rest a distance of 15 m. what was its change in potential energy?

patriot [66]

Answer:

147 J

Explanation:

PE = mgh

PE = (1.0)(9.8)(15)

PE = 147 joules

7 0

3 years ago

Please Help Me!!!!!!!​

Please Help Me!!!!!!!