If 80 joules of work were necessary to move a 5 newton box, how far was the box moved?

Answer:

The presence of dwarf galaxies around the Milky Way supports what picture that our galaxy was formed by a coming together or combination of smaller systems

Answer:

Average net force = 0.62 N

Explanation:

We are given;

Mass; m = 1.7 kg

Initial velocity; u = 12.5 m/s

Final velocity; v = 25 m/s

time; t = 25 seconds

Now, we are told that the final velocity was 30° west of North. So, resolving this velocity along the horizontal gives;

v = 25 cos 30°

Now, using Newton's first equation of motion gives;

v = u + at

Where a is acceleration

Plugging in the relevant values gives;

25 cos 30° = 12.5 + 25a

21.6506 - 12.5 = 25a

a = (21.6506 - 12.5)/25

a = 0.3660 m/s²

Now, magnitude of the average net force would be; F = ma

F = 1.7 × 0.366

F ≈ 0.62 N

Answer:

Energy dissipated = 13.453 Joules

Explanation:

In order to solve this problem, we first compute the gravitational potential energy the child has, and then find the kinetic energy at the lowest position.

The gravitational potential energy (relative to lowest position) is found as follows:

G.P.E = mass * gravity * height

Where,  Height = 2 - 2 * Cos(34°)

Height = 0.3193 meters

G.P.E = 30 * 9.8 * 0.3193

G.P.E = 93.874 J

Kinetic energy:

K.E = 0.5 * mass * velocity^2

K.E = 0.5 * 30 * 2.31547^2

K.E = 80.421 J

Energy dissipated = G.P.E - K.E

Energy dissipated = 93.874 - 80.421

Energy dissipated = 13.453 J

The change in temperature of this cube of aluminum is equal to: B. 10°C

Given the following data:

• Edge length, L = 20 cm.

• Density of Aluminum = 2.7 g/cm

• Specific heat capacity (C) of aluminum = 0.217 Cal/g°С

• Internal energy = 47000 calories.

To find the change in temperature of this cube of aluminum:

First of all, we would determine the volume of this cube of aluminum.

Next, we calculate the mass of this cube of aluminum:

Mass = 21,600 grams.

Now, we can find the change in temperature of this cube of aluminum:

Mathematically, the quantity of heat energy is given by the formula;

Where:

• Q represents the quantity of heat energy.

• m represents the mass of an object.

• c is the specific heat capacity.

• ∅ is the change in temperature.

Substituting the parameters into the formula, we have;

Change in temperature = 10°C

Read more: brainly.com/question/18877825

Answer:

10 °C

Explanation:

From the question given above, the following data were obtained:

Egde length (L) of aluminum = 20 cm

Density of Aluminum = 2.7 g/cm³

Specific heat capacity (C) of aluminum = 0.217 cal/ g°С

Heat (Q) energy = 47000 cal

Change in Temperature (ΔT) =?

Next, we shall determine the volume of the aluminum. This can be obtained as follow:

Egde length (L) of aluminum = 20 cm

Volume (V) of aluminum =?

V = L³

V = 20³

V = 8000 cm³

Thus, the volume of the aluminum is 8000 cm³

Next, we shall determine the mass of the aluminum. This can be obtained as follow:

Density of Aluminum = 2.7 g/cm³

Volume of Aluminum = 8000 cm³

Mass of aluminum =.?

Density = mass/volume

2.7 = mass /8000

Cross multiply

Mass of aluminum = 2.7 × 8000

Mass of Aluminum = 21600 g

Finally, we shall determine the change in temperature of the aluminum as follow:

Specific heat capacity (C) of aluminum = 0.217 Cal/g°С

Heat (Q) energy = 47000 Cal

Mass (M) of Aluminum = 21600 g

Change in Temperature (ΔT) =?

Q = MCΔT

47000 = 21600 × 0.217 × ΔT

47000 = 4687.2 × ΔT

Divide both side by 4687.2

ΔT = 47000 / 4687.2

ΔT = 10 °C

Therefore, the increase in the temperature of the aluminum is 10 °C.

Answer: wavelength =3.52m

Explanation:

,λ=c/μ

where c=speed of the light,λ=wave length, μ=frequncy

c=3x10^8m/s

And

μ=83.5/MHz =85.3x10^6Hz==85.3x10^6Hz=

=85.3x10^6s-1

λ=c/μ

=3x10^8m/s/85.3x10^6s-1

=3.51699883

=3.52m