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inn [45]
2 years ago
10

A 675-pound trailer is sitting on an exit ramp inclined at 40o on Highway 35. How much vertical force is required to keep the tr

ailer from rolling back down the exit ramp?
Physics
1 answer:
Phantasy [73]2 years ago
8 0

Answer:

433.88 lbs

Explanation:

We are given that mass of trailer =675 pound

Angle=40^{\circ}

We have to find the vertical force required to keep the trailer from rolling back down the exit ramp.

The force required to keep  the trailer from rolling back down the exit ramp=wsin\theta

Where w=675 pounds and \theta=40^{\circ}

Substitute the values then we get

F=675 sin40^{\circ}

sin40^{\circ}=0.642787

Substitute the value then we get

F=675\times 0.642787=433.88 lbs

Hence, the force required to keep the trailer from rolling back down the exits ramp=433.88 lbs

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Compare the components that make up the windsurfer, his board, and his surroundings.
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Answer:

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if a torque of 55.0 N/m is required and the largest force that can be exerted by you is 135 N what is th e length of the lever a
Whitepunk [10]

Answer:

r=0.41m

Explanation:

Torque is defined as the cross product between the position vector ( the lever arm vector connecting the origin to the point of force application) and the force vector.

\tau=r\times F

Due to the definition of cross product, the magnitude of the torque is given by:

\tau=rFsin\theta

Where \theta is the angle between the force and lever arm vectors. So, the length of the lever arm (r) is minimun when sin\theta is equal to one, solving for r:

r=\frac{\tau}{F}\\r=\frac{55\frac{N}{m}}{135N}\\r=0.41m

7 0
2 years ago
Consider the two moving boxcars in Example 5. Car 1 has a mass of m1 = 65000 kg and a velocity of v01 = +0.80 m/s. Car 2 has a m
Amiraneli [1.4K]

Answer:

1.034m/s

Explanation:

We define the two moments to develop the problem. The first before the collision will be determined by the center of velocity mass, while the second by the momentum preservation. Our values are given by,

m_1 = 65000kg\\v_1 = 0.8m/s\\m_2 = 92000kg\\v_2 = 1.2m/s

<em>Part A)</em> We apply the center of mass for velocity in this case, the equation is given by,

V_{cm} = \frac{m_1v_1+m_2v_2}{m_1+m_2}

Substituting,

V_{cm} = \frac{(65000*0.8)+(92000*1.2)}{92000+65000}

V_{cm} = 1.034m/s

Part B)

For the Part B we need to apply conserving momentum equation, this formula is given by,

m_1v_1+m_2v_2 = (m_1+m_2)v_f

Where here v_f is the velocity after the collision.

v_f = \frac{m_1v_1+m_2v_2}{m_1+m_2}

v_f = \frac{(65000*0.8)+(92000*1.2)}{92000+65000}

v_f = 1.034m/s

8 0
2 years ago
Given a die, would it be more likely to get a single 6 in six rolls, at least two 6s in twelve rolls, or at least one-hundred 6s
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Answer:

Explanation:

In first case we are interested in one time 6 in six rolls

Thus probability = number of chances required/Total chances

= 1/6

Similarly in the second case probability = 2/12 = 1/6

In the same way in last case probability = 100/600 = 1/6

The probability is the same . Thus all the cases has equal chances  

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2 years ago
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