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vazorg [7]
3 years ago
9

Jason applies a force of 4.00 newtons to a sled at an angle 62.0 degrees from the ground. What is the component of force effecti

ve in pulling the sled horizontally along the ground?
Physics
2 answers:
Andreas93 [3]3 years ago
7 0
Answer: 1.88 N

Explanation:

Data:

Force = 4.00N
angle = 62°
horizontal force = ?

Solution:

The trigonometric ratio that relates horizontal - leg to hypotenuse is the cosine.

That ratio is:

                        horizontal - leg
cos(angle) = -------------------------
                          hypotenuse

So, applied to the force, that is:

                             horizontal force
cos (angle) = -----------------------------------
                                 total force

So, clearing the horizontal component you get:
                                         
horizontal force = force * cos (angle)

Substitute the data given:

horizontal force = 4.00N * cos(62°) = 4.00N * 0.4695 = 1.88 N

Answer: 1.88N


Alex73 [517]3 years ago
4 0
The horizontal force will be equivalent to the total force multiplied by the cosine of the angle between them, that is, since the angle is 62.0 degrees from the ground, we multiply the force by cos 62.
So the total force will be 4cos62 = 1.88 Newtons of force.
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Soloha48 [4]

Answer:

1400 units of momentum.

Explanation:

Using the formula p=mv. We can get the momentum using 70*20 =1400 units of momentum

6 0
3 years ago
Please help!!! A river has a constant current of 3 km per hour. If a motorboat, capable of maintaining a constant speed of 20km
PilotLPTM [1.2K]
Construct a vector diagram. It will be a right-angled triangle. One vector (the hypotenuse) represents the heading of the boat, one represents the current and one represents the resultant speed of the boat, which I'll call x. Their magnitudes are 20, 3 and x. Let the required angle = theta. We have: 

<span>theta = arcsin(3/20) = approx. 8.63° </span>

<span>The boat should head against the current in a direction approx. 8.63° to the line connecting the dock with the point opposite, or approx. 81.37° to the shore line. </span>

<span>x = sqrt(20^2 - 3^2) </span>
<span>= sqrt(400 - 9) </span>
<span>= sqrt 391 </span>

<span>The boat's crossing time = </span>
<span>0.5 km/(sqrt 391 km/hr) </span>
<span>= (0.5/sqrt 391) hr </span>
<span>= approx. 0.025 hr </span>
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4 0
3 years ago
The flow of electrons through a circuit is measured in which of the following units? A. electrical pressure B. amperes C. volts
damaskus [11]

The total quantity of electrons that have flowed through a circuit is a
quantity of charge, measured in Coulombs, or in Ampere-seconds.

The <em><u>rate</u></em> of flow of electrons, or more accurately the rate of flow of
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8 0
3 years ago
n outer space, a constant net force with a magnitude of 140 N is exerted on a 32.5 kg probe initially at rest. A) What accelerat
musickatia [10]

Answer:

a) 4.31 m/s²

b) 215.5 m

Explanation:

a) According to Newton's first law of motion

The net force applied to particular mass produced acceleration, a, according to

F = ma

F = 140 N

m = 32.5 kg

a = ?

140 = 32.5 × a

a = 140/32.5 = 4.31 m/s²

b) Using the equations of motion, we can obtain the distance travelled by the object in t = 10 s

u = initial velocity of the probe = 0 m/s (since it was initially at rest)

a = 4.31 m/s²

t = 10 s

s = distance travelled = ?

s = ut + at²/2

s = 0 + (4.31×10²)/2 = 215.5 m

7 0
3 years ago
A shot is fired at an angle of 60 degree horizontal with Kinetic energy E. If air resistance is ignored, the K.E at the top of t
Lapatulllka [165]
I'm not sure what "60 degree horizontal" means.

I'm going to assume that it means a direction aimed 60 degrees
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Now, I'll answer the question that I have invented.

When the shot is fired with speed of 'S' in that direction,
the horizontal component of its velocity is    S cos(60)  =  0.5 S ,
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-- 0.75 of its kinetic energy is due to its vertical velocity.
That much of its KE gets used up by climbing against gravity.

-- 0.25 of its kinetic energy is due to its horizontal velocity.
That doesn't change. 

-- So at the top of its trajectory, its KE is 0.25 of what it had originally. 

That's  E/4 .
3 0
2 years ago
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