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2 years ago

8

A ball took 0.45s to hit the ground 0.72m from the table. What was the horizontal velocity of the ball as it rolled off the tabl

e?

1 answer:

dusya [7]2 years ago

8 0

the ball hits the ground a very high velocity

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Suppose the foreman had released the box from rest at a height of 0.25 m above the ground. What would the crate's speed be when

Arturiano [62]

Answer:

v = 2.21 m/s

Explanation:

The foreman had released the box from rest at a height of 0.25 m above the ground.

We need to find the speed of the crate when it reaches the bottom of the ramp. Let v is the velocity at the bottom of the ramp. It can be calculated using conservation of energy as follows :

$mgh=\dfrac{1}{2}mv^2\\\\v=\sqrt{2gh} \\\\v=\sqrt{2\times 9.8\times 0.25} \\\\v=2.21\ m/s$

So, its velocity at the bottom of the ramp is 2.21 m/s.

4 0

2 years ago

Maverick and goose are flying a training mission in their F-14. They are flying at an altitude of 1500 m and are traveling at 68

den301095 [7]

Answer:

The bomb will remain in air for <u>17.5 s</u> before hitting the ground.

Explanation:

Given:

Initial vertical height is, $y_0=1500\ m$

Initial horizontal velocity is, $u_x=688\ m/s$

Initial vertical velocity is, $u_y=0(\textrm{Horizontal velocity only initially)}$

Let the time taken by the bomb to reach the ground be 't'.

So, consider the equation of motion of the bomb in the vertical direction.

The displacement of the bomb vertically is $S=y-y_0=0-1500=-1500\ m$

Acceleration in the vertical direction is due to gravity, $g=-9.8\ m/s^2$

Therefore, the displacement of the bomb is given as:

$S=u_yt+\frac{1}{2}gt^2\\-1500=0-\frac{1}{2}(9.8)(t^2)\\1500=4.9t^2\\t^2=\frac{1500}{4.9}\\t=\sqrt{\frac{1500}{4.9}}=17.5\ s$

So, the bomb will remain in air for 17.5 s before hitting the ground.

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2 years ago

If the water vapor content of air remains constant, lowering air temperature causes _____.

Molodets [167]

<em>ANSWER</em>

<u>An increase in relative humidity</u>

<em><u>Could you mark me brainliest plz?</u></em>

8 0

2 years ago

Read 2 more answers

(a) What is the sum of the following four vectors in unit-vector notation? For that sum, what are (b) the magnitude, (c) the ang

avanturin [10]

6m at 0.9 radian means 6m at $51.57^0$

Since positive angles are counter clockwise

6m at $51.57^0$ can be written as 6*cos $51.57^0$ i+6*sin $51.57^0$ j = 3.73i+4.70j

5m at $-75^0$ can be written as 5*cos $(-75)^0$ i+5*sin $(-75)^0$ j = 1.294i-4.83j

4m at 1.2 radian means 4m at $68.75^0$

4m at $68.75^0$ can be written as 4*cos $68.75^0$ i+4*sin $68.75^0$ j = 1.45i+3.73j

6m at $-210^0$ can be written as 6*cos $(-210)^0$ i+6*sin $(-210)^0$ j = -5.20i-3j

a) So sum of the vector = 1.274i+0.6j

b) Magnitude = $\sqrt{1.274^2+0.6^2} = 1.98 m$

c) Angle, $tan\theta = \frac{0.6}{1.274} \\ \\ \theta = 25.22^0$

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2 years ago

Imagine you want to get 1 kcal of energy from a cow. How much energy would the cow need to get from plants? Why?

ZanzabumX [31]

1000 kcal because you only get 10% of the energy of the thing you eat

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2 years ago