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Sedaia [141]
3 years ago
5

A force of 50N is applied to the end of a lever to lift a rock that is 350N . what is the mechanical advantage of a lever?

Physics
1 answer:
Ostrovityanka [42]3 years ago
5 0

Answer:

MA  = 7

Explanation:

Mechanical Advantage = Load/Effort

From the question,

Load  = 350N

Effort applied = 50N

Mechanical Advantage = 350N/50N

Mechanical Advantage = 7

Hence the mechanical advantage is 7

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A 1.0-kg ball has a velocity of 12 m/s downward just before it strikes the ground and bounces up with a velocity of 12 m/s upwar
Nezavi [6.7K]

Answer:

The change in momentum of the ball is 24 kg-m/s  

Explanation:

It is given that,

Mass of the ball, m = 1 kg

Initial velocity of the ball, u = -12 m/s (in downwards)

Final velocity of the ball, v = +12 m/s (in upward)

We need to find the change in momentum of the ball.

Initial momentum of the ball, p_i=mu=1\ kg\times (-12\ m/s)=-12\ kg-m/s

Final momentum of the ball, p_f=mv=1\ kg\times (12\ m/s)=12\ kg-m/s

Change in momentum of the ball, \Delta p=p_f-p_i

\Delta p=12-(-12)=24\ kg-m/s

So, the change in momentum of the ball is 24 kg-m/s. Hence, this is the required solution.

3 0
3 years ago
Rocks A and B are located at the same height on top of a hill. The mass of rock A is twice the mass of rock B. How does the pote
Ronch [10]

Potential energy is defined by formula

U = mgh

here

m = mass

g = acceleration due to gravity

h = height

Now here two different stones are located at same height

while mass of stone A is twice that of stone B

so here we can say potential energy of A is

U_a = (2m)gh

Similarly potential energy of B is

U_b = mgh

now if we take the ratio of two energy

\frac{U_a}{U_b} = 2

so we can say potential energy of stone A is two times the potential energy of B

7 0
3 years ago
Calculate the lateral displacement of a ray of light incident at an angle of 30o on a parallel – sided block of glass of thickne
OLga [1]
The lateral displacement is I don’t know tbh I think 16.8
6 0
3 years ago
Atomic math challenge will give brainly and thanks
nevsk [136]

Answer:

1. Hydrogen

Atomic # = 1

Atomic Mass = 1.00794  ( If you round it it's 1.008 )

# of protons = 1

# of neutrons = none

# of electrons = 1

8 0
3 years ago
Read 2 more answers
A parallel-plate capacitor stores charge Q. The capacitor is then disconnected from its voltage source, and the space between th
Stells [14]

Answer:

The relationship between the initial stored energy PE_{i} and the stored energy after the dielectric is inserted PE_{f} is:

c) PE_{f} =0.5\ PE_{i}

Explanation:

A parallel plate capacitor with C_{o} that is connected to a voltage source V_{o} holds a charge of Q_{o} =C_{o} V_{o}. Then we disconnect the voltage source and keep the charge Q_{o} constant . If we insert a dielectric of \kappa=2 between the plates while we keep the charge constant, we found that the potential decreases as:

                                                     V=\frac{V_{o}}{\kappa}

The capacitance is modified as:

                                              C=\frac{Q}{V} =\kappa\frac{Q_{o}}{V_{o}}=\kappa\ C_{o}

The stored energy without the dielectric is

                                               PE_{i}=\frac{1}{2}\frac{Q_{o}^{2}}{C_{o}}=\frac{1}{2}C_{o}V_{o}^{2}

The stored energy after the dielectric is inserted is:

                                               PE_{f}=\frac{1}{2}\frac{Q^{2}}{C}=\frac{1}{2}CV^{2}

If we replace in the above equation the values of V and C we get that

                                         PE_{f}=\frac{1}{2}\kappa\ C_{o}(\frac{V_{o}}{\kappa})^{2}=\frac{1}{\kappa}(\frac{1}{2}C_{o}V_{o}^{2})

                                                   PE_{f} =\frac{PE_{i}}{\kappa}

Finally

                                                  PE_{f} =0.5\ PE_{i}

                                               

                                     

5 0
3 years ago
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