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andre [41]
2 years ago
7

A footballer kicks a ball from rest. The foot is in contact with the ball for 0.30s and the final velocity of the ball is 15ms-1

.What is the average acceleration of the ball?​
Physics
1 answer:
Ne4ueva [31]2 years ago
7 0

Answer:

<h2>50m/s^2</h2>

Explanation:

Step one:

given data

initial velocity u= 0m/s since the ball is at rest

time of contact t= 0.3s

final velocity v=15m/s

Required

acceleration a

from the first law of motion

v=u+at

substitute our given data

15=0+a*0.3

15=0.3a

divide both sides by 0.3

a=15/0.3

a=50m/s

<u>The average acceleration is 50m/s^2</u>

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Triss [41]

When we jump from the truck and accelerate towards the earth surface, the earth also accelerates towards us but it's acceleration is very negligible.

To find the answer, we need to know about the acceleration of earth due to the gravitational attraction.

<h3>What's the gravitational force between the earth and a person?</h3>
  • Gravitational attraction force is GMm/r² between the earth and a person.
  • M= mass of the earth

m= mass of the person

r= separation between them.

<h3>What's the acceleration of the earth towards the person when he jumps from a truck?</h3>
  • According to Newton's second law, Force = M×acceleration
  • Acceleration= Force / M
  • Here, Force = GMm/r²,

so acceleration of earth= Gm/r²

  • As this acceleration is very small, so we can't notice it.

Thus, we can conclude that the earth also accelerates towards us.

Learn more about the gravitational force here:

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7 0
2 years ago
Martha was leaning out of the window on the second floor of her house and speaking to Steve. Suddenly, her glasses slipped from
DiKsa [7]

Answer:

s = 23.72 m

v = 21.56 m/s²  

Explanation:

given

time to reach the ground (t) = 2.2 second

we know that

a) s = u t + 0.5 g t²

   u = 0 m/s

   g = 9.8 m/s²

   s = 0 + 0.5 × 9.8 × 2.2²

  s = 23.72 m

b) impact velocity

      v = √(2gh)

      v = √(2× 9.8 × 23.72)    

      v =  √464.912

      v = 21.56 m/s²  

6 0
3 years ago
4:list one food chain that is part of the food web.
Sophie [7]
4. Grass - Caterpillar - Hedgehog - Fox
5. Caterpillar, Rabbit, Mouse.
6. Cougar and Fox.
7. Bacteria
8. The bird, hedgehog, Fox and cougar would be effected since the Hedgehogs and birds would soon die out due to the loss of their food. Once they die out, the cougar and Fox would have no predators left to eat.
6 0
3 years ago
А A pool of water of refractive index
babymother [125]

Answer:

Apparent depth = 45 cm

Explanation:

The refractive index of water in a pool, n = 4/3

Real depth, d = 60 cm

We need to find its apparent  depth when viewed vertically through  air.​ The ratio of real depth to the apparent depth is equal to the refractive index of the material. Let the apparent depth is d'. So,

n=\dfrac{d}{d'}\\\\d'=\dfrac{d}{n}\\\\d'=\dfrac{60}{\dfrac{4}{3}}\\\\d'=45\ cm

So, the apparent depth is 45 cm.

3 0
2 years ago
Calculate the linear acceleration (in m/s2) of a car, the 0.310 m radius tires of which have an angular acceleration of 15.0 rad
love history [14]

Answer:

a) The linear acceleration of the car is 4.65\,\frac{m}{s^{2}}, b) The tires did 7.46 revolutions in 2.50 seconds from rest.

Explanation:

a) A tire experiments a general plane motion, which is the sum of rotation and translation. The linear acceleration experimented by the car corresponds to the linear acceleration at the center of the tire with respect to the point of contact between tire and ground, whose magnitude is described by the following formula measured in meters per square second:

\| \vec a \| = \sqrt{a_{r}^{2} + a_{t}^{2}}

Where:

a_{r} - Magnitude of the radial acceleration, measured in meters per square second.

a_{t} - Magnitude of the tangent acceleration, measured in meters per square second.

Let suppose that tire is moving on a horizontal ground, since radius of curvature is too big, then radial acceleration tends to be zero. So that:

\| \vec a \| = a_{t}

\| \vec a \| = r \cdot \alpha

Where:

\alpha - Angular acceleration, measured in radians per square second.

r - Radius of rotation (Radius of a tire), measured in meters.

Given that \alpha = 15\,\frac{rad}{s^{2}} and r = 0.31\,m. The linear acceleration experimented by the car is:

\| \vec a \| = (0.31\,m)\cdot \left(15\,\frac{rad}{s^{2}} \right)

\| \vec a \| = 4.65\,\frac{m}{s^{2}}

The linear acceleration of the car is 4.65\,\frac{m}{s^{2}}.

b) Assuming that angular acceleration is constant, the following kinematic equation is used:

\theta = \theta_{o} + \omega_{o}\cdot t + \frac{1}{2}\cdot \alpha \cdot t^{2}

Where:

\theta - Final angular position, measured in radians.

\theta_{o} - Initial angular position, measured in radians.

\omega_{o} - Initial angular speed, measured in radians per second.

\alpha - Angular acceleration, measured in radians per square second.

t - Time, measured in seconds.

If \theta_{o} = 0\,rad, \omega_{o} = 0\,\frac{rad}{s}, \alpha = 15\,\frac{rad}{s^{2}}, the final angular position is:

\theta = 0\,rad + \left(0\,\frac{rad}{s}\right)\cdot (2.50\,s) + \frac{1}{2}\cdot \left(15\,\frac{rad}{s^{2}}\right)\cdot (2.50\,s)^{2}

\theta = 46.875\,rad

Let convert this outcome into revolutions: (1 revolution is equal to 2π radians)

\theta = 7.46\,rev

The tires did 7.46 revolutions in 2.50 seconds from rest.

3 0
3 years ago
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