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sveticcg [70]
3 years ago
7

Barry has a 10 kg object. He exerts a horizontal force which causes it to accelerate at 5 m/s2, moving it a distance of 20 m, ho

rizontally. Calculate the work done by the force.
Physics
1 answer:
kodGreya [7K]3 years ago
8 0

Answer:

Work done by the force is 1000 Joules

Explanation:

The relation ship between the variables is as follows:

(Work W) = (Force F) x (Displacement d) = (Mass m) x (Accelleration a) x (Displacement d)

so:

W = m\cdot a\cdot d = 10 kg \cdot 5 \frac{m}{s^2} \cdot 20 m = 1000 J


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Calculate the air pressure in the pressurized tank, if h1 = 0. 18 m, h2 = 0. 2m and h3 = 0. 25m. The density of the mercury, wat
fomenos

The air pressure in the pressurized tank will be 24014.88 N/m²,196.2 N/m²,2084.625  N/m².

<h3 /><h3>What is pressure?</h3>

The force applied perpendicular to the surface of an item per unit area across which that force is spread is known as pressure.

It is denoted by P. The pressure relative to the ambient pressure is known as gauge pressure.

Pressure is found as the product of the density,acceleraton due to gravity and the height.

P₁=ρ₁gh₁

P₁=13,600 kg/m³×9.81 (m/s²)×0.18 m

P₁=24014.88 N/m²

P₂=ρ₂gh₂

P₂= 1000 kg/m³×9.81 (m/s²)×00.2 m

P₂=196.2 N/m²

P₃=ρ₃gh₃

P₃=850 kg/m³×9.81 (m/s²)×0.25

P₃=2084.625  N/m²

Hence,the air pressure in the pressurized tank will be 24014.88 N/m²,196.2 N/m²,2084.625  N/m².

To learn more about the pressure refer to the link;

brainly.com/question/356585

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If a neutral metal comb is held near an object with a negative charge, the Como become charged by
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The local charging of the comb is due to induction. The negative object's charge pushes electrons in the comb away from the side close to the charged object causing that part to be positively charged. Note that the comb's net charge is still zero provided it doesn't touch the object. When you move the comb away its electrons will redistribute.
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TRUE or FALSE: Most Electromagnetic waves are blocked by the atmosphere.<br> True<br> False
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An object islaunched at velocity of 20 m/s in a direction making an angle of 25 degree upward with the horizontal. what is the m
Aneli [31]

Answer:

\displaystyle y_m=3.65m

Explanation:

<u>Motion in The Plane</u>

When an object is launched in free air with some angle respect to the horizontal, it describes a known parabolic path, comes to a maximum height and finally drops back to the ground level at a certain distance from the launching place.

The movement is split into two components: the horizontal component with constant speed and the vertical component with variable speed, modified by the acceleration of gravity. If we are given the values of v_o and \theta\\ as the initial speed and angle, then we have

\displaystyle v_x=v_o\ cos\theta

\displaystyle v_y=v_o\ sin\theta-gt

\displaystyle x=v_o\ cos\theta\ t

\displaystyle y=v_o\ sin\theta\ t -\frac{gt^2}{2}

If we want to know the maximum height reached by the object, we find the value of t when v_y becomes zero, because the object stops going up and starts going down

\displaystyle v_y=o\Rightarrow v_o\ sin\theta =gt

Solving for t

\displaystyle t=\frac{v_o\ sin\theta }{g}

Then we replace that value into y, to find the maximum height

\displaystyle y_m=v_o\ sin\theta \ \frac{v_o\ sin\theta }{g}-\frac{g}{2}\left (\frac{v_o\ sin\theta }{g}\right )^2

Operating and simplifying

\displaystyle y_m=\frac{v_o^2\ sin^2\theta }{2g}

We have

\displaystyle v_o=20\ m/s,\ \theta=25^o

The maximum height is

\displaystyle y_m=\frac{(20)^2(sin25^o)^2}{2(9.8)}=\frac{71.44}{19.6}

\displaystyle y_m=3.65m

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3 years ago
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