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2 years ago

Calculate the force of gravity between two objects of masses 1300 kg and 7800 kg, which are 0.23 m apart.

Physics

1 answer:

uranmaximum [27]2 years ago

6 0

Answer:

F = Gm1m2/r^2 where G = 6.67x10^-11, m1 =1300, m2 = 7800, r = 0.23m

F = 6.67x10^-11 *1300*7800/(0.23)^2 = 0.0127852N

Explanation:

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A 5.09 × 1014-hertz electromagnetic wave is traveling through a transparent medium. The main factor that determines the speed of

sergiy2304 [10]

We are given an electromagnetic wave with a frequency of 5.09 x 10^14 Hz and travelling through a transparent medium. If the medium was vacuum, the speed of the wave would be equal to the speed of light. Otherwise, the main factor that would determine the speed of the wave is its wavelength.

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2 years ago

A thin beam of light enters a thick plastic sheet from air at an angle of 32.0° with the normal and continues in the sheet at an

Cloud [144]

Answer:

1.36

Explanation:

$n_{air}$ = Index of refraction of air = 1

$n_{plastic}$ = Index of refraction of plastic = ?

i = angle of incidence in air = 32.0° deg

r = angle of refraction in plastic = 23.0° deg

Using Snell's law

$n_{air}$ Sini = $n_{plastic}$ Sinr

(1) SIn32.0 = $n_{plastic}$ Sin23.0

$n_{plastic}$ = 1.36

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2 years ago

A 75-g bullet is fired from a rifle having a barrel 0.540 m long. Choose the origin to be at the location where the bullet begin

Mashutka [201]

The given question is incomplete. The complete question is as follows.

A 75-g bullet is fired from a rifle having a barrel 0.540 m long. Choose the origin to be at the location where the bullet begins to move. Then the force (in newtons) exerted by the expanding gas on the bullet is $14,000 + 10,000x − 26,000x^{2}$ , where x is in meters. Determine the work done by the gas on the bullet as the bullet travels the length of the barrel.

Explanation:

We will calculate the work done as follows.

W = $\int_{0}^{0.54} F dx$

= $\int_{0}^{0.54} (14,000 + 10,000x - 26,000x^{2}) dx$

= $[14000x + 5000x^{2} - 8666.7x^{3}]^{0.54}_{0}$

= 7560 + 1458 - 1364.69

= 7653.31 J

or, = 7.65 kJ (as 1 kJ = 1000 J)

Thus, we can conclude that the work done by the gas on the bullet as the bullet travels the length of the barrel is 7.65 kJ.

5 0

2 years ago

When a car drives along a "washboard" road, the regular bumps cause the wheels to oscillate on the springs. (What actually oscil

marishachu [46]

Answer:

a) 40,000 N/m

b) f = 6.37 Hz

c) v = 4,8 m/s

Explanation:

part a)

First in order to estimate the spring constant k, we need to know the expression or formula to use in this case:

k = ΔF / Δx

Where:

ΔF: force that the men puts in the car, in this case, the weight.

Δx: the sinking of the car, which is 2 cm or 0.02 m.

With this data, and knowing that there are four mens, replace the data in the above formula:

W = 80 * 10 = 800 N

This is the weight for 1 man, so the 4 men together would be:

W = 800 * 4 = 3200 N

So, replacing this data in the formula:

k = 3200 / 0.02 = 160,000 N/m

This means that one spring will be:

k' = 160,000 / 4 = 40,000 N/m

b) An axle and two wheels has a mass of 50 kg, so we can assume they have a parallel connection to the car. If this is true, then:

k^n = 2k

To get the frequency, we need to know the angular speed of the car with the following expression:

wo = √k^n / M

M: mass of the wheel and axle, which is 50 kg

k = 40,000 N/m

Replacing the data:

wo = √2 * 40,000 / 50 = 40 rad/s

And the frequency:

f = wo/2π

f = 40 / 2π = 6.37 Hz

c) finally for the speed, we have the time and the distance, so:

V = x * t

The only way to hit bumps at this frequency, is covering the gaps of bumping, about 6 times per second so:

x: distance of 80 cm or 0.8 m

V = 0.8 * 6 =

V = 4.8 m/s

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2 years ago

A 25kg child sits on one end of a 2m see saw. How far from the pivot point should a rock of 50kg be placed on the other side of

ivann1987 [24]

Answer:

a rock of 50kg should be placed =drock=0.5m from the pivot point of see saw

Explanation:

τchild=τrock

Use the equation for torque in this equation.

(F)child(d)child)=(F)rock(d)rock)

The force of each object will be equal to the force of gravity.

(m)childg(d)child)=(m)rockg(d)rock)

Gravity can be canceled from each side of the equation. for simplicity.

(m)child(d)child)=(m)rock(d)rock)

Now we can use the mass of the rock and the mass of the child. The total length of the seesaw is two meters, and the child sits at one end. The child's distance from the center of the seesaw will be one meter.

(25kg)(1m)=(50kg)drock

Solve for the distance between the rock and the center of the seesaw.

drock=25kg⋅m50kg

drock=0.5m

6 0

2 years ago