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Brilliant_brown [7]
3 years ago
8

Which of the following are true about centripetal force? Check all that apply. A. Without centripetal force, an object cannot ac

celerate. B. Friction can be a centripetal force, such as when it keeps a car on the road going around a curve. C. Gravity can be a centripetal force, such as when it pulls a satellite in its orbit. D. It's also called an outward force.
Physics
2 answers:
klemol [59]3 years ago
5 0
Centripital force is a center seeking force, meaning it is always directed to the center point of rotation.

A is vague, but true. Assuming there is no friction and an object is already in centripetal motion; without centripital force, there will be no acceleration. The centripital force is responsible for an acceleration which doesn't affect the magnitude of the object's velocity, but rather the change in direction of the velocity.

B is true. Friction keeps the car from sliding due to the tendency for the car to want to maintain a straight path. Thus it is responsible for allowing the car to execute a turn by causing a center seeking force to the center of the point of the curve.

C is true. Newton's law of gravitation states that there is an attraction force between two masses. If a mass is undergoing an orbit around another mass, the gravitation force of attraction is responsible for maintaining the object's orbit.

D. This is false, as stated above and exemplified, centripital force is a center seeking force.
Katyanochek1 [597]3 years ago
3 0

B. Friction can be a centripetal force, such as when it keeps a car on the road going around a curve.

C. Gravity can be a centripetal force, such as when it pulls a satellite in its orbit.

Explanation:

The centripetal force is any force that keeps an object moving in circular motion, "pulling" the object towards the centre of the circular trajectory.

Several forces can act as centripetal force. Examples are:

- friction: when a car is going around the curve, is moving by circular motion. The force that keeps the car in circular motion is, in fact, the friction between the tires and the road.

- Gravity: when a satellite moves around the Earth, it is moving by circular motion. The force that keeps the satellite in circular motion is the gravitational attraction between the Earth and the satellite, that pushes the satellite towards the Earth.

The other two options are not correct because:

A) An object can also accelerate if there is no centripetal force (for example, a car speeding up on a straight road is accelerating, but there is no centripetal force since there is no circular motion

D) Centripetal force is not an outward force, since it pushes the object inwards (towards the centre of the trajectory).

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A distracted driver is driving towards a turn where the edge of the road leads into a 75.0 m cliff. The velocity of the vehicle
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As long as the car is on the road, it moves with a constant speed of 80km/h.

As soon as the car starts to fall down the cliff, it follows a parabolic motion. It means that it still moves with constant speed along the x axis, but it also starts to move along the y axis, with constant acceleration (i.e. the acceleration due to gravity).

The good thing about parabolic motions is that the two motions along the x and y axes are completely separable.

So, first of all, we need to know how long it takes for an object to fall for 75m. The equation of a constantly accelerated motion is

s=s_0+v_0t+\dfrac{1}{2}at^2

Where s_0 is the initial position, v_0 is the initial speed, and a is the constant rate of acceleration. In our case, we start from an initial height of 75m, an initial (vertical!) speed of zero, and our acceleration is -g. So, our equation becomes

s=75-\dfrac{g}{2}t^2

And we want to solve for the time when s=0 (i.e. we want to know how long will it take for the object to reach the ground). We have

0=75-\dfrac{g}{2}t^2 \iff 75=\dfrac{g}{2}t^2 \iff \dfrac{2\cdot75}{g}=t^2 \iff t=\sqrt{\dfrac{150}{g}}

(I'm discarding the negative solution because it wouldn't make sense)

Now that we've used the vertical motion to find out the falling time, we can go back to the horizontal motion. We know that the car moves for a certain amount of time at a certain speed. So, we simply have to plug our values in the s=vt equation, to get

s=80\sqrt\dfrac{150}{g}}

This is how far from the base of the cliff the vehicle lands.

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6. An earthquake releases two types of traveling seismic waves, called transverse and longitudinal waves. The average speed of t
Ymorist [56]

Answer:

The distance away the center of the earthquake is 1083.24 km.

Explanation:

Given that,

Speed of transverse wave = 9.1\ km/s

Speed of longitudinal wave = 5.7 km/s

Time = 71 sec

We need to calculate the distance of transverse wave

Using formula of distance

d=v\times t

d=9.1\times t....(I)

The distance of longitudinal wave

d=5.7\times (t+71)....(II)

From the first equation

t=\dfrac{d}{9.1}

Put the value of t in equation (II)

d =5.7\times(\dfrac{d}{9.1}+71)

\dfrac{9.1d-5.7d}{9.1}=71\times5.7

d0.3736=404.7

d =1083.24\ km

Hence, The distance away the center of the earthquake is 1083.24 km.

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