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UNO [17]
1 year ago
11

A stone of weight 10N falls from the top of a 250m high cliff. a) Calculate how much work is done by the force of gravity in pul

ling the stone to the foot of the cliff. b) How much energy is transferred to the stone?​
Physics
1 answer:
tiny-mole [99]1 year ago
4 0

Answer:

work done = ( force × displacement)

(a)The force acting on the block is it's self weight and displacement is equal to height of the tower.

work done by gravity = (250 × 10) = 2500 joule

(b) The work done by gravity 2500 joule is transferred to the object in the form of it's kinetic energy.

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allsm [11]
Really long we’ll not long but far in distance
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2 years ago
A compact car has a mass of 1380 kg . Assume that the car has one spring on each wheel, that the springs are identical, and that
astraxan [27]

Answer:

A) k=34867.3384\ N.m^{-1}

B) \omega'\approx84\ Hz

Explanation:

Given:

mass of car, m=1380\ kg

A)

frequency of spring oscillation, f=1.6\ Hz

We knkow the formula for spring oscillation frequency:

\omega=2\pi.f

\Rightarrow \sqrt{\frac{k_{eq}}{m} } =2\pi.f

\sqrt{\frac{k_{eq}}{1380} } =2\times \pi\times 1.6

k_{eq}=139469.3537\ N.m^{-1}

Now as we know that the springs are in parallel and their stiffness constant gets added up in parallel.

<u>So, the stiffness of each spring is (as they are identical):</u>

k=\frac{k_{eq}}{4}

k=\frac{139469.3537}{4}

k=34867.3384\ N.m^{-1}

B)

given that 4 passengers of mass 70 kg each are in the car, then the oscillation frequency:

\omega'=\sqrt{\frac{k_{eq}}{(m+70\times 4)} }

\omega'=\sqrt{\frac{139469.3537}{(1380+280)} }

\omega'\approx84\ Hz

7 0
3 years ago
The lowest point in Death Valley is 85 m below sea level. The summit of nearby Mt. Whitney has an elevation of 4420 m.What is th
mario62 [17]

Answer:

\Delta E=2.87\times 10^6\ J

Explanation:

It is given that,

Depth of Death valley is 85 m below sea level, h_i=-85\ m

The summit of nearby Mt. Whitney has an elevation of 4420 m, h_f=4420\ m

Mass of the hiker, m = 65 kg

We need to find the change in potential energy. It is given by :

\Delta E=mg(h_f-h_i)

\Delta E=65\times 9.8(4420-(-85))

\Delta E=2869685\ J

or

\Delta E=2.87\times 10^6\ J

So, the change in potential energy of the hiker is 2.87\times 10^6\ J. Hence, this is the required solution.

5 0
3 years ago
A girl is sledding down a slope that is inclined at 30º with respect to the horizontal. The wind is aiding the motion by providi
OleMash [197]

Answer:

The sled required 9.96 s to travel down the slope.

Explanation:

Please, see the figure for a description of the problem. In red are the x and y-components of the gravity force (Fg). Since the y-component of Fg (Fgy) is of equal magnitude as Fn but in the opposite direction, both forces get canceled.

Then, the forces that cause the acceleration of the sled are the force of the wind (Fw), the friction force (Ff) and the x-component of the gravity force (Fgx).

The sum of all these forces make the sled move. Finding the resulting force will allow us to find the acceleration of the sled and, with it, we can find the time the sled travel.

The magnitude of the friction force is calculated as follows:

Ff = μ · Fn

where :

μ = coefficient of kinetic friction

Fn =  normal force

The normal force has the same magnitude as the y-component of the gravity force:

Fgy = Fg · cos 30º = m · g · cos 30º

Where

m = mass

g = acceleration due to gravity

Then:

Fgy = m · g · cos 30º = 87.7 kg · 9.8 m/s² · cos 30º

Fgy = 744 N

Then, the magnitude of Fn is also 744 N and the friction force will be:

Ff = μ · Fn = 0.151 · 744 N = 112 N

The x-component of Fg, Fgx, is calculated as follows:

Fgx = Fg · sin 30º = m·g · sin 30º = 87.7 kg · 9.8 m/s² · sin 30º = 430 N

The resulting force, Fr, will be the sum of all these forces:

Fw + Fgx - Ff = Fr

(Notice that forces are vectors and the direction of the friction force is opposite to the other forces, then, it has to be of opposite sign).

Fr = 161 N + 430 N - 112 N = 479 N

With this resulting force, we can calculate the acceleration of the sled:

F = m·a

where:

F = force

m = mass of the object

a = acceleration

Then:

F/m = a

a = 479N/87.7 kg = 5.46 m/s²

The equation for the position of an accelerated object moving in a straight line is as follows:

x = x0 + v0 · t + 1/2 · a · t²

where:

x = position at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

Since the sled starts from rest and the origin of the reference system is located where the sled starts sliding, x0 and v0 = 0.

x = 1/2· a ·t²

Let´s find the time at which the position of the sled is 271 m:

271 m = 1/2 · 5.46 m/s² · t²

2 · 271 m / 5.46 m/s² = t²

<u>t = 9.96 s </u>

The sled required almost 10 s to travel down the slope.

8 0
3 years ago
Imagine two billiard balls on a pool table. Ball A has a mass of 7 kilograms and ball B has a mass of 2 kilograms. The initial v
olga_2 [115]
A) the final velocity = 66/9 m/s.
b) The total momentum before and after collision is the same because energy is destroyed or made.
Thanks brainly. <span />
7 0
3 years ago
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