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2 years ago

Hi! I really need help on these questions, it would be great if you can answer them!

Physics

2 answers:

Nikitich [7]2 years ago

6 0

1: Equilibrium

[] When two forces are equal in opposite directions, they are in a state of equilibrium

2: Friction

[] It can be hard to sometimes slide a paper across a table, roll a car across the ground, etc etc. This is because of a force called friction which resists motion

3: Displacement

[] Work is only "done" when the object or surroundings are having movement, or in motion, and this creates displacement

4: Force

[] A force is a push/pull in any direction, this is a simple definition so not much can be added.

5: Inertia

[] Inertia is a force that likes to remain unchanged, or without movement

6: Effort

[] "Meg is holding up weight without movement. She hasn't done any work since it did not move, but is tired. Why?" Effort! She still put in effort even though no "work" happened

7: Work

[] Force times displacement is the equation for work

8: Motion

[] Look back to #3, when there is motion we can tell work is done

Have a nice day!

I hope this is what you are looking for, but if not - comment! I will edit and update my answer accordingly. (ノ^∇^)

- Heather

Ilya [14]2 years ago

5 0

Equilibrium, friction, displacement, force, inertia

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I need to choose a theme for my physics assignment My experiment is finding g

Kobotan [32]

<h3>Question:</h3>

How to find g (acceleration due to gravity)

<h3>Solution:</h3>

We know,

Acceleration due to gravity (g)

$= \frac{GM}{ {R}^{2} }$

where, G = Gravitational constant

$= 6.67 \times {10}^{11} N {m}^{2}/k {g}^{2} \\$

M = Mass of the earth

$= 6 \times {10}^{24} \: kg$

R = Radius of the earth

$= 6.4 \times {10}^{6} m$

Putting these values of G, M and R in the above formula, we get

$g \: = \: \frac{6.67 \times {10}^{11} N {m}^{2}/k {g}^{2} \times \: 6 \times {10}^{24} \: kg }{(6.4 \times {10}^{6}m {)}^{2} } \\ = 9.8m/ {s}^{2}$

So, the value of acceleration due to gravity is

$9.8m/s ^{2}$

Hope it helps.

Do comment if you have any query.

5 0

2 years ago

The amount of work done against friction to slide a box in a straight line across a uniform, horizontal floor depends most on th

Ivenika [448]

Amount of force that is applied to the box

7 0

2 years ago

Read 2 more answers

a trolley of mass 20kg was originally at rest on a smoth horizontal surface. bu how much will it accelerate if a pulling force o

DochEvi [55]

Answer:

just divide 22 N by 20 kg to get the acceleration in m/s2

Explanation:

I hope this is right-

6 0

3 years ago

A 500 kg sack of coal falls vertically onto a 2000 kg railroad flatcar which was initially moving horizontally at 3 m/s. no exte

Zinaida [17]

Since there are no external forces, including friction, act on the flatcar. after the sack rests on the flatcar, we would assume that momentum is conserved. This means that

total momentum of car before collision = total momentum of car after collision.

Recall,

momentum = mass x velocity

From the information given,

mass of car before collision = 2000

velocity of car before collision = 3

Thus,

total momentum of car before collision = 2000 x 3 = 6000

Also,

mass of sack = 500

mass of car and sack after collision = 500 + 2000 = 2500

velocity after collision = v

momentum after collision = 2500 x v = 2500v

Since momentum is conserved, then

6000 = 2500v

v = 6000/2500

v = 2.4

the speed of the flatcar is 2.4 m/s

6 0

10 months ago

During a very quick stop, a car decelerates at 6.8 m/s^2. Assume the forward motion of the car corresponds to a positive directi

geniusboy [140]

Answer:

-24.28571 rad/s²

29.57239 revolutions

3.91176 seconds

52.026478 m

Explanation:

$a_t$ = Tangential acceleration = -6.8 m/s²

r = Radius of wheel = 0.28

$\omega_i$ = Initial angular velocity = 95 rad/s

$\theta$ = Angle of rotation

$\omega_f$ = Final angular velocity

t = Time taken

Angular acceleration is given by

$\alpha=\frac{a_t}{t}\\\Rightarrow \alpha=\frac{-6.8}{0.28}\\\Rightarrow \alpha=-24.28571\ rad/s^2$

The angular acceleration is -24.28571 rad/s²

$\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \theta=\frac{\omega_f^2-\omega_i^2^2}{2\alpha}\\\Rightarrow \theta=\frac{0^2-95^2}{2\times -24.28571}\\\Rightarrow \theta=185.80885\ rad=185.80885\times \frac{1}{2\pi}\\\Rightarrow \theta=29.57239\ rev$

The number of revolutions is 29.57239

$\omega_f=\omega_i+\alpha t\\\Rightarrow t=\frac{\omega_f-\omega_i}{\alpha}\\\Rightarrow t=\frac{0-95}{-24.28571}\\\Rightarrow t=3.91176\ s$

The time it takes for the car to stop is 3.91176 seconds

Linear distance

$s=r\theta\\\Rightarrow s=0.28\times 185.80885\\\Rightarrow s=52.026478\ m$

The distance the car travels is 52.026478 m

8 0

3 years ago