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2 years ago

A load of bricks is lifted to the second floor of a building. How do work and power relate to this job?

Physics

2 answers:

kow [346]2 years ago

6 0

The faster the job is done, the greater the power. Just took the quiz on e2020 got 100%

Nady [450]2 years ago

5 0

The faster the job is done, the greater the power

You might be interested in

An incident ray of light strikes a diamond at an angle of

lys-0071 [83]

Answer:

The angle of refraction is option b: 17°.

Explanation:

We can find the angle of refraction by using Snell's law:

$n_{1}sin(\theta_{1}) = n_{2}sin(\theta_{2})$

Where:

n₁: is the index of refraction of the medium 1 (air) = 1.0003

n₂: is the index of refraction of the medium 2 (diamond) = 2.42

θ₁: is the angle of incidence = 45°

θ₂: is the angle of refraction =?

Hence, the angle of refraction is:

$sin(\theta_{2}) = \frac{n_{1}sin(\theta_{1})}{n_{2}} = \frac{1.0003*sin(45)}{2.42} = 0.2922$

$\theta_{2} = 17 ^{\circ}$

Therefore, the correct option is b: 17°.

I hope it helps you!

4 0

3 years ago

Read 2 more answers

A particle moves along a straight line and its position at time t is given by

Alisiya [41]

(A) Use interval notation to indicate the time interval or union of time intervals when the particle is moving forward and backward

the particle is speeding up

if a(t) >0, it means 12t-48>0, and t >4, I = ]4, infinity[

the particle is speeding down

if a(t) < 0, it means 12t-48>0, and t < 4, I = ] -infinity, 4[

when the particle is moving forward and backward

that is depending of the sign of v(t)

moving forward if v(t)> 0

moving backward if v(t)< 0

but v(t) = 6t² - 48 t +90, this quadratic equation has exactly two solutions, t =1, and t =3

so after taking care the sign of 6t² - 48 t +90,

v(t)> o when t is in I = ] –inf, 1[ U ]4, inf[,the particle is moving forward

v(t) < o when t is in I = ]1, 4[,the particle is moving backward

3 0

3 years ago

A block is held at rest against a wall by a force of magnitude F exerted at an angle theta from the horizontal, as shown in the

wel

Answers:

B.) $F cos\theta=F_{n}$

C.) $F sin\theta=F_{g} \pm F_{f}$

Explanation:

The image attached shows the way the force $F$ is acting on the block. Now, if we draw a free body diagram of the situation and write the equations for the Net Force in X and Y, we will have the following:

Net Force in X:

$-F_{n}+F cos\theta=0$ (1)

Where:

$F_{n}$ is the Normal force

$F$ is the magnitude of the force exerted on the block

$\theta$ is the angle

Net Force in Y:

$F sin\theta \pm F_{f}-F_{g}=0$ (2)

Where:

$F_{f}$ is the Friction force (it is expresed with the $\pm$ sign because this force may be up or down, we cannot know because the block is at rest)

$F_{g}$ is the gravity force

Rewrittin (1):

$F cos\theta=F_{n}$ (3) This is according to option B

Rewritting (2):

$F sin\theta=F_{g}\pm F_{f}$ (3) This is according to option C

3 0

2 years ago

Two forces act in opposite directions on a wood block. What will happen if the forces are unbalanced?

worty [1.4K]

Answer:

Explanation:

If two unequal forces act in opposite directions, one larger force must cancel out the smaller force,

leaving the net force to be some number in one direction.

Take for example a game of tug-of-war; there are two OPPOSITE forces (groups of people) acting on the rope, one force is pulling with a force in the negative direction while the other force is pulling in the positive direction.

If the forces on the rope were unequal, then the stronger force (group) will pull everything in their direction.

The same will happen on two unequal forces of opposite directions acting on a wooden block. Therefore, since the resultant force will have a non-zero magnitude and direction, there will be a change in the block's motion and position.

7 0

2 years ago

A ball is thrown into the air with a vertical velocity of 50 m/s and a horizontal

daser333 [38]

$\mathfrak{\huge{\pink{\underline{\underline{AnSwEr:-}}}}}$

Actually Welcome to the Concept of the Projectile Motion.

Since, here given that, vertical velocity= 50m/s

we know that u*sin(theta) = vertical velocity

so the time taken to reach the maximum height or the time of Ascent is equal to

T = Usin(theta) ÷ g, here g = 9.8 m/s^2

so we get as,

T = 50/9.8

T = 5.10 seconds

thus the time taken to reach max height is 5.10 seconds.

5 0

2 years ago