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3 years ago

How much work should be done to lift a 5kg brick to the height of 12m

Physics

2 answers:

Goshia [24]3 years ago

8 0

Answer: 588 joules

Explanation:

Work is done when force is applied on an object over a distance ( whether vertical or horizontal). It is measured in joules.

Thus, Workdone = Force X distance

- Vertical distance to be moved by the brick = 12 metres

- Mass of box = 5kg

- Acceleration due to gravity when box was lifted represented by g is a constant with value of 9.8m/s^2

Now, recall that Force = Mass x acceleration due to gravity

i.e Force = 5kg x 9.8m/s^2

Force = 49 Newton

So, Workdone = Force X Distance

Workdone = 49 Newton X 12 metres

Workdone = 588 joules

Thus, 588 joules of work was done.

zhenek [66]3 years ago

8 0

Answer:

588 j

Explanation:

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An 85-kg man plans to tow a 109 000-kg airplane along a runway by pulling horizontally on a cable attached to it. Suppose that h

Lelu [443]

Answer:

The greatest acceleration the man can give the airplane is 0.0059 m/s².

Explanation:

Given that,

Mass of man = 85 kg

Mass of airplane = 109000 kg

Distance = 9.08

Coefficient of static friction = 0.77

We need to calculate the greatest friction force

Using formula of friction

$F=\mu mg$

Where, m = mass of man

g = acceleration due to gravity

Put the value into the formula

$F = 0.77\times85\times9.8$

$F= 641.41\ N$

We need to calculate the acceleration

Using formula of newton's second law

$F = ma$

$a=\dfrac{F}{m}$

Put the value into the formula

$a=\dfrac{ 641.41}{109000}$

$a=0.0059\ m/s^2$

Hence, The greatest acceleration the man can give the airplane is 0.0059 m/s².

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3 years ago

Why is the moon's umbra much smaller during a solar eclipse?

Stolb23 [73]

Answer:

the sun is larger than the moon so it creates a shadow

7 0

2 years ago

A ball is dropped from a rooftop 60m high. <br> How long is the ball in the air?

alina1380 [7]

Answer: 3.49 s

Explanation:

We can solve this problem with the following equation of motion:

$y=y_{o}+V_{o}t-\frac{1}{2}gt^{2}$ (1)

Where:

$y=0 m$ is the final height of the ball

$y_{o}=60 m$ is the initial height of the ball

$V_{o}=0 m/s$ is the initial velocity (the ball was dropped)

$g=9.8 m/s^{2}$ is the acceleratio due gravity

$t$ is the time

Isolating $t$ :

$t=\sqrt{\frac{2 y_{o}}{g}}$ (2)

$t=\sqrt{\frac{2 (60 m)}{9.8 m/s^{2}}}$ (3)

Finally we find the time the ball is in the air:

$t=3.49 s$ (4)

7 0

3 years ago

Water enters a student's house 10.0 m above the ground through a pipe with a cross section area of 1.00 x 10-4m2 at ground. Insi

dezoksy [38]

Answer:

(a). V₁ = 10m/s (velocity inside the house), V₂ = 5m/s (velocity at ground level)

(b). P₂ = 236500 Pa

Explanation:

This is quite straight-forward so let us begin by defining the terms given.

Given that;

The cross-section area inside the student's house A₁ = 0.50 0.50 x 10-4m2.

Let us make the velocity of water inside the house be V₁

such that the Volume of water entering the per second is = A₁V₁

Therefore, in 90sec:

45 L = 90 A₁V₁

V₁ = 45 * 10⁻³m³ / 90*0.5*10⁻⁴

V₁ = 10m/s (velocity of water inside the house)

From the continuity equation we have that;

A₁V₁ = A₂V₂

0.5*10⁻⁴ * 10 = 1*10⁻⁴ V₂

V₂ = 5m/s (velocity at ground level)

(b). We are told to calculate the water pressure in the pipeline at the ground level.

Using Bernoulli's equation;

P₁ + pgh₁ + 1/2PV₁² (inside) = P₂ + pgh₂ + 1/2PV₂² (ground level)

1.01*10⁵ + 1000*9.8*10 + 1/2*1000*(10)² = P₂ + 0 + 1/2*1000*(5)²

P₂ (pressure) = 1.01*10⁵Pa

Therefore we have;

101000 + 98000 + 50000 = P₂ + 12500

P₂ = 236500 Pa

cheers I hope this helped !!

3 0

2 years ago

A sealed container holding 0.0255 L of an ideal gas at 0.981 atm and 65 ∘ C is placed into a refrigerator and cooled to 41 ∘ C w

user100 [1]

Answer:

0.911 atm

Explanation:

In this problem, there is no change in volume of the gas, since the container is sealed.

Therefore, we can apply Gay-Lussac's law, which states that:

"For a fixed mass of an ideal gas kept at constant volume, the pressure of the gas is proportional to its absolute temperature"

Mathematically:

$p\propto T$

where

p is the gas pressure

T is the absolute temperature

For a gas undergoing a transformation, the law can be rewritten as:

$\frac{p_1}{T_1}=\frac{p_2}{T_2}$

where in this problem:

$p_1=0.981 atm$ is the initial pressure of the gas

$T_1=65^{\circ}+273=338 K$ is the initial absolute temperature of the gas

$T_2=41^{\circ}+273=314 K$ is the final temperature of the gas

Solving for p2, we find the final pressure of the gas:

$p_2=\frac{p_1 T_2}{T_1}=\frac{(0.981)(314)}{338}=0.911 atm$

3 0

3 years ago