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4 years ago

HELPPPPP!!!!!

Physics

1 answer:

leva [86]4 years ago

3 0

Answer:

Option C

Maximum potential energy is at point R.

Explanation:

Potential energy is a product of mass, acceleration due to gravity and height ie

PE=mgh where PE is the potential energy, m is mass of an object, g is acceleration due to gravity whose value is normally taken as 9.81 and h is height. Since at point R we have the maximum height, the potential energy will be highest at this point.

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At NASA's Zero Gravity Research Facility in Cleveland, Ohio, experimental payloads fall freely from rest in an evacuated vertica

Diano4ka-milaya [45]

Answer:

(a). Energy is 64,680 J

(b) velocity is 51.43m/s

Explanation:

(a).

The potential energy $P$ of the payload of mass $m$ is at a vertical distance $h$ is

$P =mgh$ .

Therefore, for the payload of mass $m = 50kg$ at a vertical distance of $h = 132 m$ , the potential energy is

$P = (50kg)(9.8m/s^2)(132m)$

$\boxed{P = 64,680J}$

(b).

When the payload reaches the bottom of the shaft, all of its potential energy is converted into its kinetic energy; therefore,

$mgh= \dfrac{1}{2}mv^2$

$v= \sqrt{2gh}$

$v = \sqrt{2*9.8*135}$

$\boxed{v = 51.43m/s}$

(c).

The velocity in mph is

$\dfrac{51.43m}{s} * \dfrac{3600s}{hr} * \dfrac{1mile}{1609.34m}$

$\boxed{v= 115.0mph}$

5 0

4 years ago

These two pls :)))) ill mark brainliest :)

Anna71 [15]

Answer:

Bowling Ball: weight on Earth = 49 N

Textbook: Mass = 2 kg; weight on the moon = 3.2 N

Large dog: weight on Earth = 490 N; weight on the moon = 80 N

Law of Universal Gravitation: $F_{G}=\frac{Gm_{1}m_{2}}{r^{2}}$

$F_{G}$ = gravitational force (Newtons/N)

G = gravitational constant, 6.67430 × 10¹¹ $\frac{N*m^{2}}{kg^{2}}$

m₁ and m₂ = masses of two objects (kilograms/kg)

r² = square of distance between centers of the two objects (meters/m)

Have a fantastic day!

4 0

3 years ago

Two basketballs of equal mass are rolling toward each other at constant velocities. The first basketball (B1) has a velocity of

slamgirl [31]

$v'_2 = \frac{2m_1}{m_1+m_2} (4.3) - \frac{m_1-m_2}{m_1+m_2} (4.3)\\\\v'_1 = \frac{m_1-m_2}{m_1+m_2} (4.3) + \frac{2m_2}{m_1+m_2} (4.3)$

Explanation:

Velocity of B₁ = 4.3m/s

Velocity of B₂ = -4.3m/s

For perfectly elastic collision:, momentum is conserved

$m_1v_1 + m_2v_2 = m_1v'_1 + m_2v'_2$

where,

m₁ = mass of Ball 1

m₂ = mass of Ball 2

v₁ = initial velocity of Ball 1

v₂ = initial velocity of ball 2

v'₁ = final velocity of ball 1

v'₂ = final velocity of ball 2

The final velocity of the balls after head on elastic collision would be

$v'_2 = \frac{2m_1}{m_1+m_2} v_1 - \frac{m_1-m_2}{m_1+m_2} v_2\\\\v'_1 = \frac{m_1-m_2}{m_1+m_2} v_1 + \frac{2m_2}{m_1+m_2} v_2$

Substituting the velocities in the equation

$v'_2 = \frac{2m_1}{m_1+m_2} (4.3) - \frac{m_1-m_2}{m_1+m_2} (4.3)\\\\v'_1 = \frac{m_1-m_2}{m_1+m_2} (4.3) + \frac{2m_2}{m_1+m_2} (4.3)$

If the masses of the ball is known then substitute the value in the above equation to get the final velocity of the ball.

5 0

3 years ago

A student pushed a box 32.0 m across a smooth, horizontal floor using a constant force of 124 N. If the force was applied for 8.

Brilliant_brown [7]

The power developed is 500 W ( to the nearest Watt)

Power(P) is the rate at which work is done. Work done (W) is the product of the force applied on the object and the displacement (s) made by the point of application of the force.

$P = \frac{W}{t}$

$W= F*s$

Therefore,

$P=\frac{F*s}{t}$

Substitute the given values of force , displacement and time

$F = 124 N,s = 32.0 m,t = 8.0 s$

$P =\frac{W*s}{t} =\frac{124N*22.0s}{8.0s} =496 W$

Thus the Power can be rounded off to the nearest value of 500 W

3 0

4 years ago

Read 2 more answers

A pot on the stove contains 200 g of water at 20°C. An unknown mass of ice that is originally at −10°C is placed in an identical

Mumz [18]

Answer:

a) The mass of the ice is smaller than the mass of the water

b) The ice reaches first 80°C ,

Explanation:

Since the heat Q that should be provided to ice

Q = sensible heat to equilibrium temperature (as ice) + latent heat + sensible heat until final temperature ( as water)

m ice * c ice * ( T equil -T initial ) + m ice* L + m ice* c water * ( T final - T equil)

and the heat Q that should be provided to water is

Q= m water * c water * ( T final - T equil )

since the rate of heat addition q = constant and the time t taken to reach the final temperature is the same , then the heat absorbed Q=q*t is the same for both, therefore

m water * c water * ( T final - T equil ) = m ice* [c ice *( T equil -T initial ) + L + c water * ( T final - T equil)]

m water/ m ice = [c ice * ( T equil -T initial ) + L + c water * ( T final - T equil)]/ [ c water * ( T final - T equil)]

m water/ m ice = [c ice * ( T equil -T initial ) + L ]/[c water * ( T final - T equil) ] + 1

since [c ice * ( T equil -T initial ) + L ]/[c water * ( T final - T equil) ] >0 , then

m water/ m ice > 1

m water > m ice

so the mass of ice is smaller that the mass of water

b) Since the heat Q that should be provided to the ice, starting from 55°C mass would be

Q ice= m ice * c water * ( T final2 - T final1 )

and for the water mass

Q water = m water * c water * ( T final2 - T final1 )

dividing both equations

Q water / Q ice = m water / m ice >1

thus

Q water > Q ice

since the heat addition rate is constant

Q water = q* t water and Q ice=q* t ice

therefore

q* t water > q* t ice

t water > t ice

so the time that takes to reach 80°C is higher for water , thus the ice mass reaches it first.

5 0

4 years ago