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3 years ago

You are on a 4 m high ladder and throw a ball upwards at 12 m/s. It lands on the ground below the ladder.

Physics

1 answer:

Gelneren [198K]3 years ago

7 0

a) 2.75 s

The vertical position of the ball at time t is given by the equation

$y= h+ut-\frac{1}{2}gt^2$

where

h = 4 m is the initial height of the ball

u = 12 m/s is the initial velocity of the ball (upward)

g = 9.8 m/s^2 is the acceleration of gravity (downward)

We can find the time t at which the ball reaches the ground by substituting y=0 into the equation:

$0 = 4 + 12t - 4.9 t^2$

This is a second-order equation. By solving it for t, we find:

t = -0.30 s

t = 2.75 s

The first solution is negative, so we discard it; the second solution, t = 2.75 s, is the one we are looking for.

b) -15.0 m/s (downward)

The final velocity of the ball can be calculated by using the equation:

$v=u-gt$

where

u = 12 m/s is the initial (upward) velocity

g = 9.8 m/s^2 is the acceleration of gravity (downward)

t is the time

By subsisuting t = 2.75 s, we find the velocity of the ball as it reaches the ground:

$v=12 -(9.8)(2.75)=-15.0 m/s$

And the negative sign means the direction is downward.

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Answer:

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Explanation:

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3 years ago

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If you were capable of converting mass to energy with 100%, efficiency, how much mass would you need to produce 3.5x10^12 Joules

Alexeev081 [22]

Answer:

a) 3.9 x 10⁻⁵ kg

Explanation:

The amount of mass required to produce the energy can be given by Einstein's formula:

$E = mc^2\\\\m = \frac{E}{c^2}$

where,

m = mass required = ?

E = Energy produced = 3.5 x 10¹² J

c = speed of light = 3 x 10⁸ m/s

Therefore,

$m = \frac{3.5\ x\ 10^{12}\ J}{(3\ x\ 10^8\ m/s)^2} \\\\m = 3.9\ x\ 10^{-5}\ kg$

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2 years ago

A 1.05 kg block slides with a speed of 0.865 m/s on a frictionless horizontal surface until it encounters a spring with a force

djyliett [7]

Answer:

a) U = 0 J

k = 0.393 J

E = 0.393 J

b) U = 0.0229J

k = 0.370 J

E = 0.393 J

c) U = 0.0914 J

k = 0.302 J

E = 0.393 J

d) U = 0.206 J

k = 0.187 J

E = 0.393 J

e) U = 0.366 J

k = 0.027 J

E = 0.393 J

Explanation:

Hi there!

The equations of kinetic energy and elastic potential energy are as follows:

k = 1/2 · m · v²

U = 1/2 · ks · x²

Where:

m = mass of the block.

v = velocity.

ks = spring constant.

x = displacement of the string.

a) When the spring is not compressed, the spring potential energy will be zero:

U = 1/2 · ks · x²

U = 1/2 · 457 N/m · (0 cm)²

U = 0 J

The kinetic energy of the block will be:

k = 1/2 · m · v²

k = 1/2 · 1.05 kg · (0.865 m/s)²

k = 0.393 J

The mechanical energy will be:

E = k + U = 0.393 J + 0 J = 0.393 J

This energy will be conserved, i.e., it will remain constant because there is no work done by friction nor by any other dissipative force (like air resistance). This means that the kinetic energy will be converted only into spring potential energy (there is no thermal energy due to friction, for example).

b) The spring potential energy will be:

U = 1/2 · 457 N/m · (0.01 m)²

U = 0.0229 J

Since the mechanical energy has to remain constant, we can use the equation of mechanical energy to obtain the kinetic energy:

E = k + U

0.393 J = k + 0.0229 J

0.393 J - 0.0229 J = k

k = 0.370 J

c) The procedure is now the same. Let´s calculate the spring potential energy with x = 0.02 m.

U = 1/2 · 457 N/m · (0.02 m)²

U = 0.0914 J

Using the equation of mechanical energy:

E = k + U

0.393 J = k + 0.0914 J

k = 0.393 J - 0.0914 J = 0.302 J

d) U = 1/2 · 457 N/m · (0.03 m)²

U = 0.206 J

E = 0.393 J

k = E - U = 0.393 J - 0.206 J

k = 0.187 J

e) U = 1/2 · 457 N/m · (0.04 m)²

U = 0.366 J

E = 0.393 J

k = E - U = 0.393 J - 0.366 J = 0.027 J.

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3 years ago

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3 years ago

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How many ohms of resistance must be present in a circuit that has 240 volts and a current of 15 amps?

PIT_PIT [208]

Answer:

16 ohms

Explanation:

V= I ⋅ R

where, V is the net potential difference in the circuit, I is the current in the circuit and R is the net resistance of the circuit.

In this case, V = 240 volts, I = 15 amperes.

240 = 15 ⋅ R

⇒ R = 240/ 15 = 16 ohms

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3 years ago

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