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g100num [7]
3 years ago
10

A woman places her briefcase on the backseat of her car. As she drives to work, the car negotiates an unbanked curve in the road

that can be regarded as an arc of a circle of radius 54.5 m. While on the curve, the speed of the car is 12.4 m/s at the instant the briefcase starts to slide across the backseat toward the side of the car. What force causes the centripetal acceleration of the briefcase when it is stationary relative to the car? Under what condition does the briefcase begin to move relative to the car?
Physics
1 answer:
PSYCHO15rus [73]3 years ago
6 0

Answer:

The force that cause the centripetal acceleration is the static friction, and the briefcase begin to move if the force of centripetal acceleration exceeds the force due to the static friction

Explanation:

Given data:

r = radius = 54.5 m

v = speed of the car = 12.4 m/s

The force due to the static friction that is exerted on the curve in the road is what causes the centripetal acceleration. If the force due to centripetal acceleration is greater than the force of static friction, then the briefcase will begin to roll.

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A bowling ball traveling with constant speed hits the pins at the end of a bowling lane 16.5 m long. The bowler hears the sound
pshichka [43]

Let , speed of bowling ball is v .

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Now ,

Time taken by sound to hear after release :

2.79=\dfrac{16.5}{V}+0.048\\\\2.79-0.048=\dfrac{16.5}{V}\\\\2.742=\dfrac{16.5}{V}\\\\V=\dfrac{16.5}{2.742}\\\\V=6.02\ m/s

Hence , this is the required solution .

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An arrow is launched upward with an initial speed of 100 meters per second (m/s). The equations above describe the constant-acce
ankoles [38]

Answer:

d=510.2m

t=10.2s

Explanation:

The formulas for accelerated motion are:

v=v_0+at\\x=x_0+v_0t+\frac{at^2}{2}

From them we can get v^2=v_0^2+2ad.

We have:

v-v_0=at\\t=\frac{v-v_0}{a}

And substitute:

x=x_0+v_0(\frac{v-v_0}{a})+\frac{a}{2}(\frac{v-v_0}{a})^2\\x-x_0=\frac{v_0(v-v_0)}{a}+\frac{(v-v_0)^2}{2a}

We multiply both sides by 2a, and continue:

2a(x-x_0)=2v_0(v-v_0)+(v-v_0)^2=2v_0v-2v_0^2+v^2+v_0^2-2vv_0=v^2-v_0^2

Being d the displacement x-x_0, we have v^2=v_0^2+2ad

For our exercise, we will write this as:

d=\frac{v^2-v_0^2}{2a}

And taking upwards direction positive and imposing final velocity 0m/s (for maximum height), we have:

d=\frac{-v_0^2}{2a}=\frac{-(100m/s)^2}{2(-9,8m/s^2)}=510.2m

For the time we use:

t=\frac{v-v_0}{a}=\frac{-v_0}{a}=\frac{-(100m/s)}{(-9.8m/s^2)}=10.2s

6 0
3 years ago
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steposvetlana [31]

Answer:

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