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Brums [2.3K]
2 years ago
5

Kelly is riding a bicycle, moves with an initial velocity of 5 m/s. Ten seconds later, she is moving at 15 m/s. What is her acce

leration?
Physics
1 answer:
sladkih [1.3K]2 years ago
3 0

Answer:

her acceleration is 1 m/sec

Explanation:

The following information is given in the question

The initial velocity is 5 m/s

After 10 seconds, she would be moved at 15 m/s

We need to find the acceleration

As we know that

Acceleration = Change in speed ÷ time

Acceleration = (15 - 5) ÷ (10)

= 1 m/sec

Hence, her acceleration is 1 m/sec

The same would be considered  

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A hill is 132 m long and makes an angle of 12.0 degrees with the horizontal. As a 54 kg jogger runs up the hill, how much work d
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Answer:

14523.55J

Explanation:

The work done by the jogger against gravity is given by the following equation;

W=mgh.................(1)

where m is the mass, g is acceleration due to gravity taken as 9.8m/s^2 and h is the height of the hill.

Since the length of the hill is 132m and it is inclined at 12 degrees to the horizontal, the height is thus given as follows;

h=132sin12^o\\h=27.44m

Substituting this into equation (1) with all other necessary parameters, we obtain the following;

W=54*9.8*27.44\\W=14523.55J

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3 years ago
According to the picture, how would the gravitational potential energy of the car change if its weight was doubled? A) The gravi
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Answer is A

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A wheel initially spinning at wo = 50.0 rad/s comes to a halt in 20.0 seconds. Determine the constant angular acceleration and t
svetlana [45]

Answer:

(I). The angular acceleration and number of revolution are -2.5 rad/s² and 500 rad.

(II). The torque is 84.87 N-m.

Explanation:

Given that,

Initial spinning = 50.0 rad/s

Time = 20.0

Distance = 2.5 m

Mass of pole = 4 kg

Angle = 60°

We need to calculate the angular acceleration

Using formula of angular velocity

\omega=-\alpha t

\alpha=-\dfrac{\omega}{t}

\alpha=-\dfrac{50.0}{20.0}

\alpha=-2.5\ rad/s^2

The angular acceleration is -2.5 rad/s²

We need to calculate the number of revolution

Using angular equation of motion

\theta=\omega_{0}t+\dfrac{1}{2}\alpha t

Put the value into the formula

\theta=50\times20-\dfrac{1}{2}\times2.5\times20^2

\theta=500\ rad

The number of revolution is 500 rad.

(II). We need to calculate the torque

Using formula of torque

\vec{\tau}=\vec{r}\times\vec{f}

\tau=r\times f\sin\theta

Put the value into the formula

\tau=2.5\times4\times 9.8\sin60

\tau=84.87\ N-m

Hence, (I). The angular acceleration and number of revolution are -2.5 rad/s² and 500 rad.

(II). The torque is 84.87 N-m.

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3 years ago
Whats the difference between speed and velocity of an object
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The difference between speed and velocity is that the speed is a scalar quantity which means that you can say that this object has a speed of x m/s but you don't have to define its direction
while the velocity is a vector quantity which means that you have to express the velocity by which it moves in x,y and z directions and its norm is the speed
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Introduction to Symbolic Answers, Part B, Enter the expression 2cos2(θ)−1, where θ is the lowercase Greek letter theta.
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2cos2(o)-1    is the answer
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