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2 years ago

8

A ball, which has a mass of 1.25 kg, is thrown straight up from the top of a building 225 meters tall with a velocity of 52.0 m/

s. what will be the momentum of this ball just as it reaches the ground

1 answer:

Elena-2011 [213]2 years ago

6 0

First we will find the speed of the ball just before it will hit the floor

so in order to find the speed of the cart we will first use energy conservation

$KE_i + PE_i = KE_f + PE_f$

$\frac{1}{2}mv_i^2 + mgh = \frac{1}{2}mv_f^2 + 0$

$\frac{1}{2}(1.25)(52)^2 + 1.25(9.8)(225) = \frac{1}{2}(1.25)v_f^2$

So by solving above equation we will have

$v_f = 84.3 m/s$

now in order to find the momentum we can use

$P = mv$

$P = 1.25 \times 84.3$

$P = 105.4 kg m/s$

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FromTheMoon [43]

Answer:

The weight would be greater in the Mariana because the radius of the earth is lower.

Explanation:

We will make a comparison through constants and equations to see which one is more viable.

Using the force of gravity

$F = \frac{Gm}{R^2}$

F = Gravity force

G= Gravitational constant

m= mass

R is the radius of the earth, $R_E = 6384$ km and $R_M =6370$ km

Density $(\rho)$ is equal in both places,

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-Gravity Force at bottom of Mariana trench, $F_M=\frac{Gm}{R_M^2}$

Making the relation,

$\frac{F_E}{F_M}= \frac{\frac{Gm}{R_E^2}}{\frac{Gm}{R_M^2}}$

$\frac{F_E}{F_M}= (\frac{R_M}{R_E})^2$

For the Ecuador,

$F_E= F_M * \frac{R_M^2}{R_E^2}$

If we take $F_M * R_M^2$ as a constant X then

$F_E= \frac{X}{R_E^2}$

So the gravity force of the place is inversely proportion of the radius

Same for the Mariana trench,

$FM= \frac{X}{R_M^2}$

Then, the weight would be greater in the Mariana because the radius of the earth is lower.

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2 years ago

When you skid to a stop on your bike, you can significantly heat the small patch of tire that rubs against the road surface. Sup

Wittaler [7]

Answer:

$W_f = 148.17J$

Explanation:

During the exchange of applied force, thermal energy is generated by the friction that exists between the ground and the tire.

Said force according to the statement is the reaction of half the force on the rear tire. In this way the normal force acted is,

$N=\frac{mg}{2} = \frac{90*9.8}{2} = 441N$

The work done is given by the friction force and the distance traveled,

$W_f = fd = \mu_k Nd$

Where $\mu_k [/ tex] is the coefficient of kinetic frictionN is the normal force previously found d is the distance traveled,Replacing,[tex]W_f = (0.80)(441)(0.42)$

The thermal energy released through the work done is,

$W_f = 148.17J$

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2 years ago

A charge Q is uniformly spread over one surface of a very large nonconducting square elastic sheet having sides of length d. At

GuDViN [60]

Answer:

E/4

Explanation:

The formula for electric field of a very large (essentially infinitely large) plane of charge is given by:

E = σ/(2ε₀)

Where;

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Formula to calculate σ is;

σ = Q/A

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Putting Q/d² for σ in the electric field equation to obtain;

E = Q/(2ε₀d²)

Now, we can see that E is inversely proportional to the square of d i.e.

E ∝ 1/d²

The electric field at P has some magnitude E. We now double the side length of the sheet to 2L while keeping the same amount of charge Q distributed over the sheet.

From the relationship of E with d, the magnitude of electric field at P will now have a quarter of its original magnitude which is;

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We know that,

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m × a = 2m × a'

a = 2 × a'

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