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4 years ago

Which energy resources are found above the Earth's surface

Physics

1 answer:

never [62]4 years ago

6 0

Answer: the sun

Explanation:

The sun's radiant energy reaches the earth's surface either directly through radiation, indirectly through convection, or it can move "across" or "through" objects or materials on the surface via conduction. Let's look more closely at each case. We've probably experienced the feeling of "warmth" of the sun on our skin on a sunny day. Light energy from the sun is reaching us across space and down through the atmosphere through radiation. A dark colored vehicle in the sun quickly becomes warm (or hot!) to the touch because of radiation. The light energy from the sun heats the air in the earth's atmosphere, and this drives convection and transfers thermal energy around. It is possible that we've felt a "hot breeze" on our skin on sunny days. The thermal energy in the air will be carried to objects in its path, and it will warm them.

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Charge q is accelerated starting from rest up to speed v through the potential difference V. What speed will charge q have after

adoni [48]

Answer: v = $1.19 * 10^{6} m/s$

Explanation: q = magnitude of electronic charge = $1.609 * 10^{-19} c$

mass of an electronic charge = $9.10 * 10^{-31} kg$

V= potential difference = 4V

v = velocity of electron

by using the work- energy theorem which states that the kinetic energy of the the electron must equal the work done use in accelerating the electron.

kinetic energy = $\frac{mv^{2} }{2}$ , potential energy = qV

hence, $\frac{mv^{2} }{2} = qV$

$\frac{9.10 *10^{-31} * v^{2} }{2} = 1.609 * 10^{-16} * 4\\\\\\\\9.10*10^{-31} * v^{2} = 2 * 1.609 *10^{-16} * 4\\\\\\9.10 *10^{-31} * v^{2} = 1.287 *10^{-15} \\\\v^{2} = \frac{1.287 *10^{-15} }{9.10 *10^{31} } \\\\v^{2} = 1.414*10^{15} \\\\v = \sqrt{1.414*10^{15} } \\\\v = 1.19 * 10^{6} m/s$

7 0

3 years ago

A spring stretches by 0.0177 m when a 2.82-kg object is suspended from its end. How much mass should be attached to this spring

Artemon [7]

The mass attached to the spring must be 0.72 kg

Explanation:

The frequency of vibration of a spring-mass system is given by:

$f=\frac{1}{2\pi} \sqrt{\frac{k}{m}}$ (1)

where

k is the spring constant

m is the mass attached to the spring

We can find the spring constant by using Hookes' law:

$F=kx$

where

F is the force applied on the spring

x is the stretching of the spring

When a mass of m = 2.82 kg is applied to the spring, the force applied is the weight of the mass, so we have

$mg=kx$

and using $g=9.8 m/s^2$ and $x=0.0177 m$ , we find

$k=\frac{mg}{x}=\frac{(2.82)(9.8)}{0.0177}=1561.3 N/m$

Now we want the frequency of vibration to be

f = 7.42 Hz

So we can rearrange eq.(1) to find the mass m that we need to attach to the spring:

$m=\frac{k}{(2\pi f)^2}=\frac{1561.3}{(2\pi (7.42))^2}=0.72 kg$

#LearnwithBrainly

6 0

3 years ago

Two insulated wires, each 2.64 m long, are taped together to form a two-wire unit that is 2.64 m long. One wire carries a curren

nikklg [1K]

Answer:

$4.77\ \text{A}$

Explanation:

F = Magnetic force = 4.11 N

$I_n$ = Net current

$I_2$ = Current in one of the wires = 7.68 A

B = Magnetic field = 0.59 T

$\theta$ = Angle between current and magnetic field = $65^{\circ}$

$l$ = Length of wires = 2.64 m

$I$ = Current in the other wire

Magnetic force is given by

$F=I_nlB\sin\theta\\\Rightarrow I_n=\dfrac{F}{lB\sin\theta}\\\Rightarrow I_n=\dfrac{4.11}{2.64\times 0.59 \sin65^{\circ}}\\\Rightarrow I_n=2.91\ \text{A}$

Net current is given by

$I_n=I_2-I\\\Rightarrow I=I_2-I_n\\\Rightarrow I=7.68-2.91\\\Rightarrow I=4.77\ \text{A}$

The current I is $4.77\ \text{A}$ .

8 0

3 years ago

A 90-kg astronaut is stranded in space at a point 6.0 m from his spaceship, and he needs to get back in 4.0 min to control the s

Stella [2.4K]

Momentum = 0.5 * 4 = 2
to conclude the man’s velocity after he throws the piece of equipment, divide this number by the man’s mass.

v = 2/90

This is about 0.0222 m/s. To know if he can move 6 meters at velocity in 4minutes, use the following equation.

d = v * t, t = 4 * 60 = 240 s
d = 2/90 * 240 = 5⅓ meters.

This is ⅔ of a meter from the spaceship. To know the velocity that he must have to move 6 meter, use the same equation.

6 = v * 240
v = 6/240
This is about 0.00416 m/s.
His final momentum = 90 * 6/240 = 2.25

To know the velocity of the package, divide this number by the mass of the package.
v = 2.25/0.5 = 4.5 m/s

8 0

3 years ago

Calculate the change internal energy (Î´e) for a system that is giving off 45.0 kj of heat and is performing 855 j of work on th

adelina 88 [10]

The change in internal energy of a system is given by (second law of thermodynamics)
$\Delta U = Q + W$
where Q is the heat absorbed by the system and W is the work done on the system.

In order to correctly evaluate the internal energy change, we must be careful with the signs of Q and W:
Q positive -> Q absorbed by the system
Q negative -> Q released by the system
W positive -> W done on the system by the surroundings
W negative -> W done by the system on the surroundings

In our problem, the heat released by the system is $Q=-45 kJ=-45000 J$ (with negative sign since it is released by the system), and the work done is $W=-855 J$ still with negative sign because it is performed by the system on the surrounding, so the change in internal energy is
$\Delta U = Q +W=-45000 J - 855 J=-45855 J$

3 0

3 years ago

Which energy resources are found above the Earth's surface​

Which energy resources are found above the Earth's surface