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3 years ago

Which of the following is a force acting at a distance

Physics

2 answers:

wariber [46]3 years ago

8 0

D. gravitational force. is your answer hope this helps

OLga [1]3 years ago

6 0

Answer:

D Gravitational Force

Explanation:

Gravitational Force act's at a certain distance also depending on it's mass

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What two factors keep Earth in orbit and around the sun and the moon in orbit around Earth?

Alex_Xolod [135]

D because bits makes the most sense logically

8 0

4 years ago

Read 2 more answers

When starting a foot race, a 70.0kg sprinter exerts an average force of 650 N backward on the ground for 0.800 s. How far does h

melomori [17]

Answer:

Answer is option b) 2.97m

Explanation:

With the relationship between the force exerted by the runner and the mass that it has, I can determine the acceleration it will have:

F= m × a ⇒ a= (650 kg ×(m/s^2)) / (70kg)= 9.286 (m/s^2)

With the acceleration that prints the force exerted and the time I can determine the distance traveled in the interval:

Distance= (1/2) × a × t^2 = (1/2) × 9.286 (m/s^2) × ((0.8s)^2)= 2.97m

8 0

3 years ago

The intensity of light from a star (its brightness) is the power it outputs divided by the surface area over which it’s spread:

kow [346]

Answer:

$\frac{d_{1}}{d_{2}}=0.36$

Explanation:

1. We can find the temperature of each star using the Wien's Law. This law is given by:

$\lambda_{max}=\frac{b}{T}=\frac{2.9x10^{-3}[mK]}{T[K]}$ (1)

So, the temperature of the first and the second star will be:

$T_{1}=3866.7 K$

$T_{2}=6444.4 K$

Now the relation between the absolute luminosity and apparent brightness is given:

$L=l\cdot 4\pi r^{2}$ (2)

Where:

L is the absolute luminosity
l is the apparent brightness
r is the distance from us in light years

Now, we know that two stars have the same apparent brightness, in other words l₁ = l₂

If we use the equation (2) we have:

$\frac{L_{1}}{4\pi r_{1}^2}=\frac{L_{2}}{4\pi r_{2}^2}$

So the relative distance between both stars will be:

$\left(\frac{d_{1}}{d_{2}}\right)^{2}=\frac{L_{1}}{L_{2}}$ (3)

The Boltzmann Law says, $L=A\sigma T^{4}$ (4)

σ is the Boltzmann constant
A is the area
T is the temperature
L is the absolute luminosity

Let's put (4) in (3) for each star.

$\left(\frac{d_{1}}{d_{2}}\right)^{2}=\frac{A_{1}\sigma T_{1}^{4}}{A_{2}\sigma T_{2}^{4}}$

As we know both stars have the same size we can canceled out the areas.

$\left(\frac{d_{1}}{d_{2}}\right)^{2}=\frac{T_{1}^{4}}{T_{2}^{4}}$

$\frac{d_{1}}{d_{2}}=\sqrt{\frac{T_{1}^{4}}{T_{2}^{4}}}$

$\frac{d_{1}}{d_{2}}=0.36$

I hope it helps!

5 0

3 years ago

A constant net force that has a magnitude of 135.5 N is exerted on a 26.7 kg object that is initially not moving. a. Draw a free

jek_recluse [69]

Explanation:

Given that,

Net force = 135.5 N

Mass = 26.7 kg

(a). We need to draw a free body diagram

A force is exerted on a object.

(b). We need to calculate the acceleration

Using formula of acceleration

$a =\dfrac{F}{m}$

$a=\dfrac{135.5}{26.7}$

$a=5.07\ m/s^2$

(c). We need to draw the sketch a position vs. time graph for the object.

Using equation of motion

$s=ut+\dfrac{1}{2}at^2$

Put the value into the formula

$s=0+\dfrac{1}{2}\times5.07\times t^2$

$s=2.535 t^2$

(d). We need to draw the sketch a velocity vs. time graph for the object.

Using equation of motion

$v = u+at$

Put the value into the formula

$v=0+5.07t$

$v =5.07 t$

(e). We need to calculate the distance travel in 8.82 s

Using equation of motion

$s=ut+\dfrac{1}{2}at^2$

$s=0+\dfrac{1}{2}\times5.07\times(8.82)^2$

$s=197.20\ m$

Hence, This is the required solution.

8 0

3 years ago

Harmonics problem. A square wave of frequency f contains harmonics (sine waves) at f, 3f, 5f, 7f, ... . Suppose a system respond

ira [324]

Answer:

Explanation:

A square of frequency of consists of the infinite sum of sine waves, whose frequencies are the odd multiples of the main frequency f i.e f, 3f,5f, 7f, ... . Given that the range of frequencies, to which the system responds is 20-40 kHz, for a square wave of frequency 10kHz we need to look for the harmonics whose frequencies are in the systems' respond range, which are the harmonics of 20, 30 and 40 kHz

4 0

3 years ago