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3 years ago

A ball with a mass of 0.7 kg is thrown straight upward, flies up to its maximum height, and

Physics

1 answer:

vichka [17]3 years ago

7 0

Answer:

250.05 J

Explanation:

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

Mass (m) = 0.7 kg

Kinetic energy (KE) = 250.1 J

Potential energy (PE) =?

Next, we shall determine the velocity of the ball. This can be obtained as follow:

Mass (m) = 0.7 kg

Kinetic energy (KE) = 250.1 J

Velocity (v) =?

KE = ½mv²

250.1 = ½ × 0.7 × v²

250.1 = 0.35 × v²

Divide both side by 0.35

v² = 250.1 / 0.35

Take the square root of both side

v = √(250.1 / 0.35)

v = 26.73 m/s

Next, we shall determine the maximum height reached by the ball. This can be obtained as follow:

Initial velocity (u) = 26.73 m/s

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

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

Maximum height (h) =.?

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

0² = 26.73² – (2 × 9.8 × h)

0 = 714.4929 – 19.6h

Collect like terms:

0 – 714.4929 = – 19.6h

– 714.4929 = – 19.6h

Divide both side by –19.6

h = –714.4929 / –19.6

h = 36.45 m

Finally, we shall determine the potential energy of the ball at the maximum height. This can be obtained as follow:

Mass (m) = 0.7 kg

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

Maximum height (h) = 36.45 m

Potential energy (PE) =?

PE = mgh

PE = 0.7 × 9.8 × 36.45

PE = 250.05 J

Thus, the potential energy at the maximum height is 250.05 J.

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What % of an object’s mass is above the water line if the object’s density is 0.82g/ml?

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To develop this problem we will apply the Archimedes model. As well as the definitions of Weight based on mass and acceleration. The first in turn will be considered under the relationship of Density and Volume. From the values given we have to:

$\rho_w =1 g/mL \rightarrow \text{Water Density}$

$\rho_o = 0.82g/mL \rightarrow \text{Object density}$

Since it is in equilibrium, the weight of the object will have a reaction from the water, which will cause the sum of forces between the two objects to be zero, therefore

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$F_w = F_o$

$m_w g = m_og$

$\rho_w V_w g = \rho_o V_o g$

The value of gravity is canceled because it is a constant

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$\frac{V_w}{V_o} = \frac{0.82}{1}$

$\frac{V_w}{V_o} = 0.82$

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3 years ago

Calculate the momentum for a 0:2 kg rifle bullet traveling 300m/a

Ksenya-84 [330]

Given data

mass (m) = 0.2 Kg ,

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4 years ago

A cannon is fired from a castle wall at some unknown height above the ground. The cannonball leaves the cannon with speed 30.0m/

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Answer:

Part a)

$t = 3.85 s$

Part b)

$h = 72.67 m$

Part C)

$v_x = 25.98 m/s$

$v_y = 0$

Part d)

In horizontal direction velocity will remain constant

$v_x = 30 cos30 = 25.98 m/s$

in vertical direction we have

$v_y = -22.77 m/s$

Explanation:

Part a)

Horizontal speed of the cannon

$v = 30.0 m/s$

angle of projection

$\theta = 30^o$

now we have

horizontal speed = $v_x = vcos30 = 30 cos30 =25.98 m/s$

vertical speed = $v_y = vsin30 = 30 sin30 = 15 m/s$

now the time taken by it to cover the distance 100 m from the wall

$x = v_x t$

$100 = 25.98 t$

$t = 3.85 s$

Part b)

Since it hits the ground in the same time

so the height of the castle is given as

$h = \frac{1}{2}gt^2$

$h = \frac{1}{2}(9.81)(3.85^2)$

$h = 72.67 m$

Part C)

At highest point of the projection

the vertical component of the velocity will become zero

so we will have

$v_x = 25.98 m/s$

$v_y = 0$

Part d)

In horizontal direction velocity will remain constant

so we have

$v_x = 30 cos30 = 25.98 m/s$

in vertical direction we have

$v_y = v_i + at$

$v_y = 15 - 9.81(3.85)$

$v_y = -22.77 m/s$

Part e)

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