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3 years ago

A truck travels 1430 m uphill along a road that makes a constant angle of 5.76◦ with the horizontal.

Physics

1 answer:

Alex73 [517]3 years ago

5 0

Answer:

The horizontal component of displacement is d' = 1422.7 m

Explanation:

Given data,

The distance covered by the truck, d = 1430 m

The angle formed with the horizontal, Ф = 5.76°

The displacement is a vector quantity.

The horizontal component of displacement is given by,

d' = d cos Ф

= 1430 cos 5.76°

= 1422.7 m

Hence, the horizontal component of displacement is d' = 1422.7 m

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The attraction will vary directly with the separation between the charges.

Burka [1]

No it won't. It'll vary inversely as the square of the separation.

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3 years ago

a spherical mirror is cut in half horizontally will an image be formed by the bottom half of the mirror how

monitta

Answer:

Explanation:

the spherical mirror can form an image even if it is cut in half horizontally , but the image formed may be blurred.

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7 0

3 years ago

A 175-kg roller coaster car starts from rest at the top of an 18.0-m hill and rolls down the hill, then up a second hill that ha

Anni [7]

Answer:

The work done by non-conservative forces on the car from the top of the first hill to the top of the second hill is 6574.75 joules.

Explanation:

By Principle of Energy Conservation and Work-Energy Theorem we present the equations that describe the situation of the roller coaster car on each top of the hill. Let consider that bottom has a height of zero meters.

From top of the first hill to the bottom

$m\cdot g \cdot h_{1} = \frac{1}{2}\cdot m\cdot v_{1}^{2} +W_{1, loss}$ (1)

From the bottom to the top of the second hill

$\frac{1}{2}\cdot m\cdot v_{1}^{2} = m\cdot g \cdot h_{2} + \frac{1}{2}\cdot m \cdot v_{2}^{2}+W_{2,loss}$ (2)

Where:

$m$ - Mass of the roller coaster car, in kilograms.

$v_{1}$ - Speed of the roller coaster car at the bottom between the two hills, in meters per second.

$g$ - Gravitational acceleration, in meters per square second.

$h_{1}$ - Height of the first top of the hill with respect to the bottom, in meters.

$W_{1, loss}$ - Work done by non-conservative forces on the car between the top of the first hill and the bottom, in joules.

$v_{2}$ - Speed of the roller coaster car at the top of the second hill, in meters per seconds.

$h_{2}$ - Height of the second top of the hill with respect to the bottom, in meters.

$W_{2, loss}$ - Work done by non-conservative forces on the car bewteen the bottom between the two hills and the top of the second hill, in joules.

By using (1) and (2), we reduce the system of equation into a sole expression:

$m\cdot g\cdot h_{1} = m\cdot g\cdot h_{2} + \frac{1}{2}\cdot m \cdot v_{2}^{2} + W_{loss}$ (3)

Where $W_{loss}$ is the work done by non-conservative forces on the car from the top of the first hill to the top of the second hill, in joules.

If we know that $m = 175\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ , $h_{1} = 18\,m$ , $h_{2} = 8\,m$ and $v_{2} = 11\,\frac{m}{s}$ , then the work done by non-conservative force is:

$W_{loss} = m\cdot\left[ g\cdot \left(h_{1}-h_{2}\right)-\frac{1}{2}\cdot v_{2}^{2} \right]$

$W_{loss} = 6574.75\,J$

The work done by non-conservative forces on the car from the top of the first hill to the top of the second hill is 6574.75 joules.

8 0

2 years ago

A textbook of mass 2.02 kg rests on a frictionless, horizontal surface. A cord attached to the book passes over a pulley whose d

Anika [276]

Answer:

a. 7.38 N b. 40.87 N c. 0.113 kg-m²

Explanation:

7 0

3 years ago

Guys No one's answering my question so sad! Once again I'm asking the same question –Here

nexus9112 [7]

For the front glass of the car to get wet, $V_c \geq 10 \ m/s$ .

The given parameters:

<em>Speed of the car, = Vc</em>
<em>Speed of the rain, = 10 m/s</em>

The relative velocity of the car with respect to the falling rain is calculated as;

$V_{C/R} = V_C- V_R$

If the speed of the car equals the speed of the rain, the rain will fall behind the car.
If the speed of the rain is greater than speed of the car, the rain will fall far in front of the car.
If the speed of the car is greater than speed of the rain, the rain will fall on the car.

Thus, for the front glass of the car to get wet, $V_c \geq 10 \ m/s$ .

Learn more about relative velocity here: brainly.com/question/17228388

8 0

3 years ago