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vovikov84 [41]
2 years ago
8

A force is applied to a block sliding along a surface (Figure 2). The magnitude of the force is 15 N, and the horizontal compone

nt of the force is 4.5 N. At what angle (in degrees) above the horizontal is the force directed?

Physics
2 answers:
patriot [66]2 years ago
4 0

If I have done my math correctly, your answer is 67.5

denpristay [2]2 years ago
3 0

Explanation:

It is given that,

The magnitude of force applied force on the block is, F = 15 N

The horizontal component of the force, F_x=4.5\ N

We need to find the angle (in degrees) above the horizontal is the force directed. The horizontal component of the force is given by :

F_x=F\ cos\theta

4.5=15\ cos\theta    

\theta=cos^{-1}(\dfrac{4.5}{15})

\theta=72.54^{\circ}

So, the angle above the horizontal force and the surface is 72.54 degrees. Hence, this is the required solution.

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A balloon drifts 140m toward the west in 45s ; then the wind suddenly changes and the balloon flies 90m toward the east in the n
Bogdan [553]

Answer: 140 m

Explanation:

Let's begin by stating clear that motiont is the change of position of a body at a certain time. So, during this motion, the balloon will have a trajectory and a displacement, being both different:

The<u> trajectory</u> is <u>the path followed by the body, the distance it travelled</u> (is a scalar quantity).  

The displacement is <u>the distance in a straight line between the initial and final position</u> (is a vector quantity).  

So, according to this, the distance the balloon traveled during the first 45 s (its trajectory) is 140 m.

But, if we talk about displacement, we have to draw a straight line between the initial position of the balloon (point 0) to its final position (point 90 m).  Being its displacement 95 m.

8 0
1 year ago
A rabbit is trying to cross the street. Its velocity v as a function of time t is given in the graph below where
Amiraneli [1.4K]

Answer:

2.5

Explanation:

8 0
2 years ago
If you accidentally touch the "hot" wire connected to the 120 V line, how much current will pass through your body?
Sav [38]
<h2>Complete Question:</h2>

You are on an aluminum ladder that is standing on the ground, trying to fix an electrical connection with a metal screwdriver having a metal handle. Your body is wet because you are sweating from the exertion; therefore, it has a resistance of 1.60 kΩ .

(a) If you accidentally touch the "hot" wire connected to the 120 V line, how much current will pass through your body?

(b) How much electrical power is delivered to your body?

<h2>Answer:</h2>

(a) 0.075A

(b) 9W

<h2>Explanation:</h2>

The voltage (V) passing across or supplied to a body is directly proportional to the current (I) flowing through the body as stated by Ohm's law. i.e

V ∝ I

=> V = I x R                 ----------------------(i)

Where;

R = constant of proportionality called resistance of the body

(a) As stated in the question;

The body is wet and thus will conduct electricity and has the following;

V = voltage supplied = 120V

R = resistance of the wet body = 1.60kΩ = 1.6 x 1000Ω = 1600Ω

Substitute these values into equation(i) as follows;

120 = I x 1600

Solve for I;

I = \frac{120}{1600}

I = 0.075A

Therefore the amount of current that will pass through your body is 0.075A

(b) Electrical power(P), which is commonly measured in Watts(W), delivered to a body is the product of the current(I) and voltage (V) supplied to the body. i.e

P = I x V           ---------------------(ii)

Where;

I = 0.075A   [as calculated above]

V = 120V     [given in the question]

Substitute these values into equation (ii) as follows;

P = 0.075 x 120

P = 9W

Therefore, the electric power delivered to your body is 9W

7 0
1 year ago
The wheels of the locomotive push back on the tracks with a constant net force of 7.50 × 105 N, so the tracks push forward on th
Rasek [7]

Answer:

The freight train would take 542.265 second to increase the speed of the train from rest to 80.0 kilometers per hour.

Explanation:

Statement is incomplete. Complete description is presented below:

<em>A freight train has a mass of </em>1.83\times 10^{7}\,kg<em>. The wheels of the locomotive push back on the tracks with a constant net force of </em>7.50\times 10^{5}\,N<em>, so the tracks push forward on the locomotive with a force of the same magnitude. Ignore aerodynamics and friction on the other wheels of the train. How long, in seconds, would it take to increase the speed of the train from rest to 80.0 kilometers per hour?</em>

If locomotive have a constant net force (F), measured in newtons, then acceleration (a), measured in meters per square second, must be constant and can be found by the following expression:

a = \frac{F}{m} (1)

Where m is the mass of the freight train, measured in kilograms.

If we know that F = 7.50\times 10^{5}\,N and m = 1.83\times 10^{7}\,kg, then the acceleration experimented by the train is:

a = \frac{7.50\times 10^{5}\,N}{1.83\times 10^{7}\,kg}

a = 4.098\times 10^{-2}\,\frac{m}{s^{2}}

Now, the time taken to accelerate the freight train from rest (t), measured in seconds, is determined by the following formula:

t = \frac{v-v_{o}}{a} (2)

Where:

v - Final speed of the train, measured in meters per second.

v_{o} - Initial speed of the train, measured in meters per second.

If we know that a = 4.098\times 10^{-2}\,\frac{m}{s^{2}}, v_{o} = 0\,\frac{m}{s} and v = 22.222\,\frac{m}{s}, the time taken by the freight train is:

t = \frac{22.222\,\frac{m}{s}-0\,\frac{m}{s}  }{4.098\times 10^{-2}\,\frac{m}{s^{2}} }

t = 542.265\,s

The freight train would take 542.265 second to increase the speed of the train from rest to 80.0 kilometers per hour.

6 0
1 year ago
A rock is rolling down a hill. At position 1, it’s velocity is 2.0 m/s. Twelve seconds later, as it passes position 2, it’s velo
mr Goodwill [35]

Answer

Hi,

correct answer is {D} 3.5 m/s²

Explanation

Acceleration is the rate of change of velocity with time. Acceleration can occur when a moving body is speeding up, slowing down or changing direction.

Acceleration is calculated by the equation =change in velocity/change in time

a= {velocity final-velocity initial}/(change in time)

a=v-u/Δt

The units for acceleration is meters per second square m/s²

In this example, initial velocity =2.0m/s⇒u

Final velocity=44.0m/s⇒v

Time taken for change in velocity=12 s⇒Δt

a= (44-2)/12  = 42/12

3.5 m/s²

Best Wishes!

5 0
2 years ago
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