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Dafna11 [192]
3 years ago
6

4. A force of 6.0 Newtons is applied to a block at rest on a horizontal frictionless surface over a 7.0 meter span. How much ene

rgy is gained by the block?
Physics
1 answer:
Nina [5.8K]3 years ago
7 0

Answer:

42 Joules

Explanation:

Given that a force of 6.0 Newtons is applied to a block at rest on a horizontal frictionless surface over a 7.0 meter span. How much energy is gained by the block?

The parameters given are:

Distance covered = 7.0 m

The force = 6.0 m

The formula for work done is the product of force and distance. That is,

Work done = force × distance

Where work done is the same as energy.

Energy = force × distance

Energy = 6 × 7

Energy = 42 Joules.

Therefore, the energy gained by the block is 42 Joules.

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How many x-ray photons per second are created by an x-ray tube that produces a flux of x rays having a power of 1.00 W
ArbitrLikvidat [17]

Complete question:

How many x-ray photons per second are created by an x-ray tube that produces a flux of x rays having a power of 1.00 W. Assume the average energy per photon in 78.0 keV.

Answer:

The number of x-ray photons per second created by the x-ray tube is 8.01 x 10¹³ photons/sec

Explanation:

Given;

power of the flux produced, P = 1 W = 1 J/s

energy per photon, E = 78 keV

Convert the energy per photon to J

E = 78 x 10³ x 1.6 x 10⁻¹⁹ = 1.248 x 10⁻¹⁴ J / photon

let the number of photons = n

n(1.248 x 10⁻¹⁴ J / photon) = 1 J/s

n = \frac{1 \ J/s}{1.248 \times 10^{-14}\  J/photon } = 8.01 \times 10^{13} \ photons/s

Therefore, the number of x-ray photons per second created by the x-ray tube is 8.01 x 10¹³ photons/sec

8 0
3 years ago
Gravity anything with mass as gravity we know the earth has gravity because you and I are standing on earth and not floating off
Paraphin [41]

Answer:

F=mg

Explanation:

Close to Earth's surface, the force of gravity that pulls an object towards the ground is

F=mg (2)

where

m is the mass of the object

g is the acceleration due to gravity, which is 9.81 m/s^2 close to Earth's surface

This is an approximation of the general formula of gravity valid only close to Earth's surface. The more general formula is

F=G\frac{Mm}{r^2} (1)

where

G is the gravitational constant

M is the Earth's mass

m is the object's mass

r is the distance of the object from Earth's center

At the Earth's surface,

r = R (Earth's radius), and by calling the following factor

g=\frac{GM}{R^2}

we see that eq.(1) becomes eq.(2).

8 0
3 years ago
Can someone help me w this ?
lukranit [14]

Answer:

i can't sorry

Explanation:

I didn't really pay attention in that class

7 0
2 years ago
A wire has a cross sectional area of 4.00 mm2 and is stretched by 0.100 mm by a certain force. How far will a wire of the same m
Nina [5.8K]

Answer: 0.05\ mm

Explanation:

Given

Cross-sectional area of wire A_1=4\ mm^2

Extension of wire \delta l=0.1\ mm

Extension in a wire is given by

\Rightarrow \delta l=\dfrac{FL}{AE}

where, E=\text{Youngs modulus}

\Rightarrow \delta_1=\dfrac{FL}{A_1E}\quad \ldots(i)

for same force, length and material

\Rightarrow \delta_2=\dfrac{FL}{A_2E}\quad \ldots(ii)

Divide (i) and (ii)

\Rightarrow \dfrac{0.1}{\delta_2}=\dfrac{A_2}{A_1}\\\\\Rightarrow \delta_2=0.1\times \dfrac{4}{8}\\\\\Rightarrow \delta_2=0.05\ mm

5 0
3 years ago
Which statement is true about a planet’s orbital motion?
lana66690 [7]

Answer:

Orbital motion results when the object’s forward motion is balanced by a second object’s gravitational pull.

Explanation:

The gravitational force is responsible for the orbital motion of the planet, satellite, artificial satellite, and other heavenly bodies in outer space.

When an object is applied with a velocity that is equal to the velocity of the orbit at that location, the body continues to move forward. And, this motion is balanced by the gravitational pull of the second object.

The orbiting body experience a centripetal force that is equal to the gravitational force of the second object towards the body.

The velocity of the orbit is given by the relation,

                                    V = \sqrt{\frac{GM}{R + h} }

Where

                   V - velocity of the orbit at a height h from the surface

                    R - Radius of the second object

                    G - Gravitational constant

                    h - height from the surface

The body will be in orbital motion when its kinetic motion is balanced by gravitational force.

                         1/2 mV^{2} = GMm/R

Hence, the orbital motion results when the object’s forward motion is balanced by a second object’s gravitational pull.

3 0
3 years ago
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