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denis-greek [22]

2 years ago

15

A person weighing 600 Newtons gets on an elevator. The elevator lifts the person 6 meters in 10 seconds. How much power was used

?

1 answer:

Svetlanka [38]2 years ago

4 0

Answer:

The power used is 360 Watts.

Explanation:

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PLEASE HELP!!! 25 pts!!

mina [271]

Answer:

The slope of the graph is what you need. That tells you the speed not the velocity. In order to find the velocity you would also need to know the direction of the motion.

3 0

3 years ago

Read 2 more answers

What formula is used to find an objects acceleration

ololo11 [35]

Acceleration = velocity / time.

7 0

3 years ago

A(n) 0.2 kg object is swung in a vertical circular path on a string 0.1 m long. The acceleration of gravity is 9.8 m/s2 . If a c

Leya [2.2K]

Answer:

$T=83.37N$

Explanation:

Since the object is under a circular motion, according to Newton's second law, when the object is at the top of the circle we have:

$\sum F_y: T-mg=F_c$

Where $F_c$ is the centripetal force and is given by:

$F_c=ma_c=m\frac{v^2}{r}$

Replacing and solving for T:

$T=m\frac{v^2}{r}+mg\\T=0.2kg\frac{(6.38\frac{m}{s})^2}{0.1m}+0.2kg(9.8\frac{m}{s^2})\\T=83.37N$

8 0

3 years ago

A plane has an airspeed of 142 m/s. A 30.0 m/s wind is blowing southward at the same time as the plane is flying. What must be t

const2013 [10]

Answer:

$\theta=12.19^{\circ}$

Explanation:

Given that

The speed of the airplane ,v= 142 m/s

The speed of the air ,u = 30 m/s

Lets take angle make by airplane from east direction towards north direction is θ .

Now by using diagram ,we can say that

$sin\theta =\dfrac{u}{v}$

Now by putting the values in the above equation we get

$sin\theta =\dfrac{30}{142}$

$sin\theta=0.21$

$\theta=12.19^{\circ}$

Therefore the angle will be 12.19° .

4 0

3 years ago

A spherical submersible 1.84 m in radius, armed with multiple cameras, descends under water in a region of the Atlantic Ocean kn

Drupady [299]

Answer:

23932242.5 Pa

Explanation:

$P_a$ = Atmospheric pressure = $1.013\times 10^5\ Pa$

$P_w$ = Pressure of seawater

$\rho$ = Density of sea water = $1.025\times 10^3\ kg/m^3$

h = Depth of shipwreck = $2.37\times 10^3\ m$

g = Acceleration due to gravity = 9.81 m/s²

The absolute pressure is given by

$P_{ab}=P_a+P_w\\\Rightarrow P_{ab}=1.013\times 10^5+1.025\times 10^3\times 9.81\times 2.37\times 10^3\\\Rightarrow P_{ab}=23932242.5\ Pa$

The absolute pressure at the depth of the shipwreck is 23932242.5 Pa

5 0

3 years ago