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Sidana [21]
3 years ago
13

You are in a planet where the acceleration due to gravity is known to be 3.28 m/s^2. You drop a ball and record that the ball ta

kes 2.40 s to reach the ground. How fast is the ball moving just as it lands? Neglect air resistance.
A.) 1.12 m/s

B.) 7.87 m/s

C.) 3.55 m/s

D.) 2.81 m/s
Physics
1 answer:
notsponge [240]3 years ago
5 0
B) 7.87 m/s

The gravitational pull is the rate of change of velocity which is the acceleration. Formula for acceleration is;
a =  \frac{final \: velocity - initial \: velocity}{time \: taken}

Given:

• Initial velocity = 0m/s; I dropped the ball, and didn't throw it, so it was at rest firstly
• Time taken = 2.40s
• Acceleration = 3.28m/s^2

We're require to find the final velocity, at which the ball hit the ground with. Ignoring air resistance, keep in mind that the velocity of an object increases as it comes closer to the ground.

3.28  =  \frac{final \: velocity - 0}{2.40}
3.28 \times 2.40 = final \: velocity
final \: velocity = 7.872 \frac{m}{ {s} }

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Answer:

(a) t = 22.9 s

(b) α= - 0.467 rad/s²

Explanation:

The uniformly accelerated circular movement, is a circular path movement in which the angular acceleration is constant.

We apply the equations of circular motion uniformly accelerated :

ωf ²=  ω₀² + 2*α*θ  Formula (1)

ωf= ω₀ + α*t Formula (2)

Where:

θ : angle that the body has rotated in a given time interval (rad)

α : angular acceleration (rad/s²)

t : time interval (s)

ω₀ : initial angular speed ( rad/s)

ωf : final angular speed ( rad/s)

Data

θ =  19.5 revolutions  : angular displacement of each wheel or angle that the  wheel has rotated in a given time interval

ω₀= 10.7 rad/s :  initial angular speed of the Wheel ( rad/s)

ωf = 0 : final angular speed  of the Whee( rad/s)

Calculating of the angular acceleration (α )

We replace data in the fómula (1),considering that 1 revolution is equal to 2π radians :

ωf ²=  ω₀² + 2*α*θ

(0 )²=  (10.7)² + 2*α*(19.5*2*π )

0= 114.49 + (245.04)*α

-114.49 =  (245.04)*α

α= (-114.49) /(245.04)

α= -114.49 /(245.04)

α= -0.467 rad/s²

Time does it take for the bike to come to rest

We replace data in the formula (2)

ωf = ω₀ + α*t

0 =  10.7 + -0.467*t

-10.7  = - 0.467*t    we multiply by (-1) both sides of the equation :

10.7  = 0.467*t  

t = 10.7 / 0.467

t = 22.9 s

3 0
3 years ago
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Answer:

The maximum speed that the truck can have and still be stopped by the 100m road is the speed that it can go and be stopped at exactly 100m. Since there is no friction, this problem is similar to a projectile problem. You can think of the problem as being a ball tossed into the air except here you know the highest point and you are looking for the initial velocity needed to reach that point. Also, in this problem, because there is an incline, the value of the acceleration due to gravity is not simply g; it is the component of gravity acting parallel to the incline. Since we are working parallel to the plane, also keep in mind that the highest point is given in the problem as 100m. Solving for the initial velocity needed to have the truck stop after 100m, you should find that the maximum velocity the truck can have and be stopped by the road is 18.5 m/s.

Explanation:

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Answer:

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