Which of the diagrams is the correct model for the element beryllium? A) B) C) D)

In the equation for the gravitational force between two objects, which quantity must be squared?

Alex787 [66]

Answer:

The distance between the two objects must be squared.

Explanation:

Gravitational force always act between two objects that have mass. The gravitational force is a weak force and attractive in nature.

The force of pull depends on the masses of the two objects and the distance between them.

The formula to calculate gravitational force between two objects having masses 'm' and 'M' and separated by a distance 'd' is given as:

$F_g=\frac{GmM}{d^2}$

Where, 'G' is called the universal gravitational constant and its value is equal to $6.674\times10^{-11}\ m^3 kg^{-1} s^{-2}$ .

Now, from the above formula, it is clear that, the force of gravitation is inversely proportional to the square of the distance between the two objects.

Thus, the quantity that must be squared in the equation of gravitational force between two objects is the distance 'd'.

7 0

3 years ago

What change occurs to the mass of an object when an unbalanced force is applied to it

strojnjashka [21]

The mass of an object always stays the same since it is really just the amount of matter in an object so no matter the force applied, as long as the object does not lose or gain matter, the object stays the same

5 0

1 year ago

A train travels due north in a straight line with a constant speed of 100 m/s. Another train leaves a station 2,881 m away trave

damaskus [11]

Answer:

The trains will collide at a distance 1660 m from the station

Explanation:

Let the train traveling due north with a constant speed of 100 m/s be Train A.

Let the train traveling due south with a constant speed of 136 m/s be Train B.

From the question, Train B leaves a station 2,881 m away (that is 2,881 m away from Train A position).

Hence, the two trains would have traveled a total distance of 2,881 m by the time they collide.

∴ If train A has covered a distance $x$ m by the time of collision, then train B would have traveled $(2881 - x)$ m.

Also,

At the position where the trains will collide, the two trains must have traveled for equal time, t.

That is, At the point of collision,

$t_{A} = t_{B}$

$t_{A}$ is the time spent by train A

$t_{B}$ is the time spent by train B

From,

$Velocity = \frac{Distance }{Time }\\$

$Time = \frac{Distance}{Velocity}$

Since the time spent by the two trains is equal,

Then,

$\frac{Distance_{A} }{Velocity_{A} } = \frac{Distance_{B} }{Velocity_{B} }$

${Distance_{A} = x$ m

${Distance_{B} = 2881 - x$ m

${Velocity_{A} = 100$ m/s

${Velocity_{B} = 136$ m/s

Hence,

$\frac{x}{100} = \frac{2881 - x}{136}$

$136(x) = 100(2881 - x)\\136x = 288100 - 100x\\136x + 100x = 288100\\236x = 288100\\x = \frac{288100}{236} \\x = 1220.76m\\$

$x$ ≅ 1,221 m

This is the distance covered by train A by the time of collision.

Hence, Train B would have covered (2881 - 1221)m = 1660 m

Train B would have covered 1660 m by the time of collision

Since it is train B that leaves a station,

∴ The trains will collide at a distance 1660 m from the station.

7 0

3 years ago

A light bulb emits light uniformly in all directions. The average emitted power is 150.0 W. At a distance of 5 m from the bulb,

beks73 [17]

Answer:

a) 0.477 W/m²

b) 13.407 N/C

c) 18.96 N/C

Explanation:

P = Power = 150 W

r = Distance = 5 m

ε₀ = Permittivity of space = 8.854×10⁻¹² F/m

a) Average intensity

$\bar{S}=\frac{P}{A}\\\Rightarrow \bar{S}=\frac{150}{4\pi r^2}\\\Rightarrow \bar{S}=\frac{150}{4\pi 5^2}\\\Rightarrow \bar{S}=0.477\ W/m^2$

∴ Average intensity is 0.477 W/m²

b) Rms value

$\bar{S}=c\epsilon_0E_{rms}^2\\\Rightarrow E_{rms}=\sqrt{\frac{\bar{S}}{c\epsilon_0}}\\\Rightarrow E_{rms}=\sqrt{\frac{\bar{0.477}}{3\times 10^8\times 8.854\times 10^{-12}}}\\\Rightarrow E_{rms}=13.407\ N/C$

∴ Rms value of the electric field is 13.407 N/C

c) Peak value

$E_0=\sqrt 2E_{rms}\\\Rightarrow E_0=\sqrt 2\times 13.407\\\Rightarrow E_0=18.96\ N/C$

∴ Peak value of the electric field is 18.96 N/C

8 0

3 years ago

Why-does-the-energy-of-a-sound-wave-decrease-over-time

KengaRu [80]

The sound wave never actually loses energy, the bulk of energy it had at the beginning of the sound, is spread out into a sphere, making it decrease in amplitude and frequency.

7 0

3 years ago