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2 years ago

4. Shirley Bored sits in her seat in the English classroom. The force of gravity on Earth pulls

Physics

1 answer:

Sedbober [7]2 years ago

4 0

Answer:

600 N upward

Explanation:

According to Newton's third law, for every action their is an equal and opposite reaction. The equal and opposite reaction to a downward force of magnitude 600 N is an upward force of magnitude 600 N

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1) A boy drags a wooden crate with a mass of 20 kg, a distance of 12 m, across a rough level floor at a constant speed of 1.5 m/

mojhsa [17]

Answer: a) 49.560 and 21.13 b) i) 50 N, ii) 196 N iii) 196 N iv) 47.685 N

c) i) 594.72 ii) 0 iii) 0 iv) 0

d) 594.72

Explanation: question a)

The force is inclined at an angle of 25° to the horizontal

The horizontal component of force = 50 cos 25° = 49.560 N

The vertical component of force = 50 sin 30°= 21.130N

Question b)

i) according to the question applied force is 50 N

ii) if g = 9.8m/s², w=mg where m = mass of object = 20kg hence weight = 20* 9.8 = 196 N

iii) the normal force is the force the floor exerts on the body as a result of the weight of the object.

Normal reaction R = W = mg, we already deduced that w = mg, hence R = 196 N.

iv) according to newton's laws of motion

F - Fr = ma

F = applied force = horizontal component of force = 49.560 N.

We need to get the acceleration (a) by using Newton laws of motion before we can be able to compute the frictional force..

The body started from rest hence initial velocity u = 0

Final velocity v = 1.5m/s distance covered (s) = 12m

v ² = u² + 2as

But u = 0

v² = 2as

1.5² = 2(a) * 12

2.25 = 24a

a = 2.25/24 = 0.09735m/s²

From F - Fr = ma

49.560 - Fr = 20 * 0.09735

49.560 - Fr = 1.875

Fr = 49.560 - 1.875

Fr = 47.685 N

Question c)

i) The applied force = 49.560 N, distance covered = 12m

Work done = force * distance

Work done = 49.560 * 12

Work done = 594.72 J

ii) the weight of the object does not make the object move a distance, hence work done = 0 ( since distance covered is 0)

iii) the normal force is the same thing as the weight and they did not cover any distance hence work done is zero.

iv) the frictional force does not cover any distance, hence work done is zero.

Question d)

The total work done = work done by applied force + work done by weight + work done by normal reaction + work done by frictional force.

Total work done = 594.72 + 0 + 0 + 0 = 594.72 J

8 0

2 years ago

When you throw a ball upward, its kinetic energy and its potential energy . When the ball reaches maximum height, its kinetic en

emmasim [6.3K]

It has zero kinetic energy, more potential energy

8 0

3 years ago

Read 2 more answers

Do mirrors reflect light

777dan777 [17]

Answer:

Yes, Mirror are a surface that reflects light more perfectly than ordinary objects.

Explanation:

8 0

2 years ago

The electric flux through a spherical surface is 1.4 ✕ 105 N · m2/C. What is the net charge (in C) enclosed by the surface?

Anit [1.1K]

Answer:

The value is $Q_{net} = 1.239 *10^{-6} \ C$

Explanation:

From the question we are told that

The electric flux is $\Phi = 1.4*10^{5} \ N\cdot m^2/C$

Generally the net charge is mathematically represented as

$Q_{net} = \Phi * \epsilon_o$

Here $\epsilon_o$ is the permetivity of free space with value

$\epsilon_o = 8.85*10^{-12} \ \ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2$

$Q_{net} = 1.4*10^5 * 8.85*10^{-12}$

=> $Q_{net} = 1.239 *10^{-6} \ C$

8 0

3 years ago

By how much does the volume of an aluminum cube 4.00 cm on an edge increase when the cube is heated from 19.0°C to 67.0°C? The l

Genrish500 [490]

Answer:

The volume of an aluminum cube is 0.212 cm³.

Explanation:

Given that,

Edge of cube = 4.00 cm

Initial temperature = 19.0°C

Final temperature = 67.0°C

linear expansion coefficient $\alpha=23.0\times10^{-6}/C^{\circ}$

We need to calculate the volume expansion coefficient

Using formula of volume expansion coefficient

$\beta=3\alpha$

Put the value into the formula

$\beta=3\times23.0\times10^{-6}$

$\beta=0.000069=69\times10^{-6}/C^{\circ}$

We need to calculate the volume

$V= a^3$

$V=4^3$

$V=64\ cm^3$

The change temperature of the cube is

$\Delta T=T_{f}-T_{i}$

Put the value into the formula

$\Delta T=67-19 = 48^{\circ}C$

We need to calculate the increases volume

Using formula of increases volume

$\Delta V=V\beta\Delta T$

Put the value into the formula

$\Delta V=64\times69\times10^{-6}\times48$

$\Delta V=0.212\ cm^3$

Hence, The volume of an aluminum cube is 0.212 cm³.

5 0

3 years ago