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3 years ago

A boy and his dog are running. The dog has a mass of 20kg and the boy has a mass of 45kg.

Physics

1 answer:

Ad libitum [116K]3 years ago

4 0

Answer:

Explanation:

mass of boy =m1=45 kg

mass of dog =m2=20kg

the speed of boy=3 m/s

the speed of dog =3 m/s

K.E of boy

K.E=1/2mv²

K.E=1/2(45)(3)²

K.E=202.5 J

K.E of dog

K.E=1/2mv²

K.E=1/2(20)(3)²

K.E=90 J

now the new speed of dog is 6 m/s

therefore new K.E of dog is

K.E=1/2(20)(6)²

K.E=360 j

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$W = \frac{N}{g} = \frac{100N}{9.8m/s^{2}} = \frac{100}{9.8} = 10.2kg$

Then;
Draw an arrow at the base of the ladder pointing towards the wall with a value of 30N, to show the frictional force.

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3 years ago

A tank is full of water. Find the work W required to pump the water out of the spout. (Use 9.8 m/s2 for g. Use 1000 kg/m3 as the

Sergio039 [100]

Answer:

W = 1.06 MJ

Explanation:

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- Make a differential volume of water in the tank with thickness dx. We see as we traverse up or down the differential volume of water the side length is always constant, hence, its always 8.

- As for the width of the part w we see that it varies as we move up and down the differential element. We will draw a rectangle whose base axis is x and vertical axis is y. we will find the equation of the slant line that comes out to be y = 0.5*x. And the width spans towards both of the sides its going to be 2*y = x.

- Now develop and expression of Force required:

F = p*V*g

F = 1000*(2*0.5*x*8*dx)*g

F = 78480*x*dx

- Now, the work done is given by:

W = F.s

- Where, s is the distance from top of hose to the differential volume:

s = (5 - x)

- We have the work as follows:

dW = 78400*x*(5-x)dx

- Now integrate the following express from 0 to 3 till the tank is empty:

W = 78400*(2.5*x^2 - (1/3)*x^3)

W = 78400*(2.5*3^2 - (1/3)*3^3)

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3 years ago

In order to do work, the force vector must be

gregori [183]

As we know that total work done by a force is given by

$W = F.d$

$W = Fdcos\theta$

so it is product of force and displacement along same direction

as we can write it as

$W = (Fcos\theta)(d)$

so it must be the product of force and displacement in same directions so correct answer must be

B. in the same direction as the displacement vector and the motion.

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3 years ago

Read 2 more answers

8. What is the momentum of a 120-kg professional fullback running across the line at 11.2 m/s?

Alchen [17]

Answer:

134r kgm^-1 or 1344 kg /m

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p is momentum, m is mass in kg and v is velocity in ms−1

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3 years ago

A ball is hit with a paddle, causing it to travel straight upward. It takes 2.90 s for the ball to reach its maximum height afte

zmey [24]

Answer:

A. 28.42 m/s

B. 41.21 m

Explanation:

From the question given above, the following data were obtained:

Time (t) to reach the maximum height = 2.90 s

Initial velocity (u) =?

Maximum height (h) =?

A. Determination of the initial velocity of the ball.

Time (t) to reach the maximum height = 2.90 s

Final velocity (v) = 0 m/s (at maximum height)

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Initial velocity (u) =?

v = u – gt (since the ball is going against gravity)

0 = u – (9.8 × 2.9)

0 = u – 28.42

Collect like terms

0 + 28.42 = u

u = 28.42 m/s

Thus, the initial velocity of the ball is 28.42 m/s

B. Determination of the maximum height reached by the ball.

Final velocity (v) = 0 m/s (at maximum height)

Acceleration due to gravity (g) = 9.8 m/s²

Initial velocity (u) = 28.42 m/s

Maximum height (h) =?

v² = u² – 2gh (since the ball is going against gravity)

0² = 28.42² – (2 × 9.8 × h)

0 = 807.6964 – 19.6h

Collect like terms

0 – 807.6964 = – 19.6h

– 807.6964 = – 19.6h

Divide both side by – 19.6

h = – 807.6964 / – 19.6

h = 41.21 m

Thus, the maximum height reached by the ball is 41.21 m

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3 years ago