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3 years ago

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A man of mass 50kg ascends a flight of stairs 5m high in 5seconds. If acceleration due to gravity is 10ms-2 the power expended i

s

2 answers:

Bad White [126]3 years ago

4 0

Answer:

-12

Explanation:

Mariana [72]3 years ago

3 0

$P = \frac{W}{t} = \frac{mgh}{t} = \frac{50*10*5}{5} = 500W$

So therefore the answer is, 500W

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A small, 3 kg weight is moved from 5 m from the ground to 8 m. What is the change in potential energy?

Ulleksa [173]

Answer: 88.2 J

Explanation: PE, OR Potential Energy, equals to mass gravity and height.

mass - 3kg

height - 3 because (8-5)

gravity is 9.8kg

so 3 times 3 times 9.8 = 88.2J

4 0

3 years ago

If the volume is held constant, what happens to the pressure of a gas as temperature is decreased? Explain.

Lady bird [3.3K]

Answer:Decreases

Explanation:

Given

Volume is held constant that is it is a isochoric process.

We know that

PV=nRT

as n,V& R are constant therefore only variables are

P & T

so $\frac{P_1}{T_1}=\frac{P_2}{T_2}$

$\frac{P_1}{P_2}=\frac{T_1}{T_2}$

As $T_1$ is decreasing therefore Pressure must also decrease so that ratio remains constant.

6 0

3 years ago

The tension in cable da has a magnitude of tda=6.27 lb. find the cartesian components of tension tda, which is directed from d t

Oliga [24]

Complete Question

The Complete Question is attached below

We have that the Cartesian components of tension $T_{da}$ is

$T_{DA}=-4.433i-3.49j+2.735k$

From the Question we are told that

$M_{da}=6.27 lb\\\\w=9.50ft\\\\d=6.60ft\\\\h=4.50ft$

$\vec {DA}=-4.7i-3.7j+2.9k)ft$

$\vec {DB}=-1.9i-3.7j+1.9k)ft\\\\\vec {DC}=-1.9i+5.8j-1.6k)ft$

Generally the equation for $T_{DA}$ is mathematically given as

$T_{DA}=\phi_{DA}* M_{da}$

Where

$\phi_{DA}=\frac{-4.7i-3.7j+2.9k}{(-4.7)^2+(-3.7)^2+(2.9)^2}\\\\\phi_{DA}=\frac{-4.7i-3.7j+2.9k}{6.65}$

Therefore

$T_{DA}=\phi_{DA}* M_{da}$

$T_{DA}=\frac{-4.7i-3.7j+2.9k}{6.65}* 6.27$

$T_{DA}=-4.433i-3.49j+2.735k$

For more information on this visit

brainly.com/question/20746649?referrer=searchResults

5 0

3 years ago

I was driving along at 20 m/s, trying to change a CD and not watching where I was going. When I looked up, I found myself 45 m f

cricket20 [7]

Answer:

$a=4.44\frac{m}{s^2}$

Explanation:

First we have to find the time required for train to travel 60 meters and impact the car, this is an uniform linear motion:

$t=\frac{d}{v}\\\\t=\frac{60m}{30\frac{m}{s}}=2s$

The reaction time of the driver before starting to accelerate was 0.50 seconds. So, remaining time for driver is 1.5 seconds.

Now, we have to calculate the distance traveled for the driver in this 0.5 seconds before he start to accelerate. Again, is an uniform linear motion:

$d=vt\\d=20\frac{m}{s}(0.5s)=10m$

The driver cover 10 meters in this 0.5 seconds. So, the remaining distance to be cover in 1.5 seconds by the driver are 35 meters. We calculate the minimum acceleration required by the car in order to cross the tracks before the train arrive, Since this is an uniformly accelerated motion, we use the following equation:

$d=v_0t+\frac{1}{2}at^2\\a=\frac{2(d-v_0t)}{t^2}\\a=\frac{2(35m-20\frac{m}{s}*1.5s}{(1.5s)^2}\\a=4.44\frac{m}{s^2}$

7 0

3 years ago

Inside most ball-point pens is a small spring that compresses as the pen is pressed against the paper. If a force of 0.1 N compr

AnnZ [28]

Answer:

20 N/m

Explanation:

From the question,

The ball-point pen obays hook's law.

From hook's law,

F = ke............................ Equation 1

Where F = Force, k = spring constant, e = compression.

Make k the subject of the equation

k = F/e........................ Equation 2

Given: F = 0.1 N, e = 0.005 m.

Substitute these values into equation 2

k = 0.1/0.005

k = 20 N/m.

Hence the spring constant of the tiny spring is 20 N/m

8 0

3 years ago