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garik1379 [7]
3 years ago
13

[Please help fast! Offering 100 points if it works!}

Physics
2 answers:
Alex17521 [72]3 years ago
8 0

(8.3 min)/(1 AU) = (T)/(1.52 AU)

(8.3 min)x(1.52 AU) = (T x 1 AU)

T = (8.3 min x 1.52 AU) / (1 AU)

T = 12.62 minutes

solmaris [256]3 years ago
3 0

Answer:

d=1.49×1011m

Explanation:

Velocity is defined as the rate of travel, and can be found using the distance formula.

velocity=distancetime

Rearranging this formula we can solve for distance given velocity and time of travel.

d=vt

We are given velocity and time, and so can solve for distance, but if we plug in the values given;

d=(3.00×108m/s)(8.3minutes)

We can see that the units do not match up. Since seconds are the SI unit for time, we will need to convert 8.3 minutes to seconds.

t=(8.3minutes)(60seconds/minute)=(498s)

Now our units work out and we can solve for distance.

= 15.85

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In 1993, Cuban athlete Javier Sotomayor set the world record for the high jump. The gravitational potential energy associated wi
Katarina [22]

Answer:2.541

Explanation:

Well , Potential Energy = mgh

m=mass = 82

g=acceleration of gravity=9.80m/s^2

h=what we are looking for

PE=mgh

PE/(mg) = h

Substitute in the values:

1970/(82 x 9.8) = h 2.541

5 0
3 years ago
A 1090 kg car moving +11.0 m/s
stiks02 [169]

The mass of the second car is 1434.21 kg

<u>Explanation:</u>

Using law of conservation of momentum,

          m_{1} u_{1}+m_{2} u_{2}=\left(m_{1}+m_{2}\right) v

Given:

m_{1} = 1090 kg

u_{1} = 11 m/s

u_{2} = 0

v = 4.75 m/s

We need to find m_{2}

When substituting the given values in the above equation, we get

(1090 \times 11)+\left(m_{2} \times 0\right)=\left(1090+m_{2}\right) 4.75

11990=5177.5+4.75 m_{2}

4.75 m_{2}=11990-5177.5

4.75 m_{2}=6812.5

m_{2}=\frac{6812.5}{4.75}=1434.21 \mathrm{kg}

6 0
3 years ago
A. Telephone signals are often transmitted over long distances by microwaves. What is the frequency of microwave radiation with
zzz [600]

(a) 10 GHz is the frequency of microwave radiation.

(b) 0.167 ms is required by the microwave to travel between two mountains.

Answer:

Explanation:

(a). 1 MHz is the frequency of microwave radiation.

(b)  0.167 ms is required by the microwave to travel between two mountains.

Answer:

Explanation:

a. Frequency is the measure of number of times a same thing will be repeated in a given time interval for a given time. And wavelength is the measure of distance between two successive crests or troughs. So wavelength and frequency are inversely proportional to each other. And velocity of light is the proportionality constant.

So frequency of microwave radiation = Speed of light/Wavelength of radiation

Frequency = \frac{3*10^{8} }{3*10^{-2} }

Frequency = 10^{8+2} = 10^{10}=10 GHz

So 10 GHz is the frequency of microwave radiation.

b). As microwave is a part of light waves, so it will be experiencing the speed of light.

As the speed is 3*10^{8} m/s and the distance between the two mountains is given as 50 km, then time can be calculated as

Time = Distance/Velocity

Time = \frac{50*10^{3} m}{3*10^{8} }=16.67*10^{3-8}=16.67*10^{-5}

So time = 0.167 ms.

Thus, 0.167 ms is required by the microwave to travel between two mountains.

6 0
3 years ago
A uniform electric field exists in the region between two oppositely charged plane parallel plates. a proton is released from re
dedylja [7]

According to Newton's second law

E.e = a * mp  ..... (1)

where

E is the magnitude of the electric field; e = 1.6 * 10^-19 is the elementary charge; mp = 1.67*10^-27 kg is the proton mass; a is the acceleration.

So, the distance

l = at^2/2 .......(2)

The proton accelerated

a = 2l / t^2 ...........(3)

From equations (1) and (3)

E= 32.51 V/m

Electric field

The physical field that surrounds electrically charged particles and exerts a force on all other charged particles in the field, either attracting or repelling them, is known as an electric field (also known as an E-field).  It can also refer to a system of charged particles' physical field. Electric charges and time-varying electric currents are the building blocks of electric fields. The electromagnetic field, one of the four fundamental interactions (also known as forces) of nature, manifests itself in both electric and magnetic fields.

To learn more about an electric field refer here:

brainly.com/question/15800304

#SPJ4

5 0
2 years ago
A ball filled with an unknown material starts from rest at the top of a 2 m high incline that makes a 28o with respect to the ho
Lady_Fox [76]

Answer:

<u>Searching in google I found the total mass and the radius of the ball (m = 1.5 kg and r = 10 cm) which are needed to solve the problem!</u>  

The ball rotates 6.78 revolutions.

     

Explanation:

<u>Searching in google I found the total mass and the radius of the ball (m = 1.5 kg and r = 10 cm) which are needed to solve the problem!</u>        

At the bottom the ball has the following angular speed:

\omega_{f} = \frac{v_{f}}{r} = \frac{4.9 m/s}{0.10 m} = 49 rad/s

Now, we need to find the distance traveled by the ball (L) by using θ=28° and h(height) = 2 m:

sin(\theta) = \frac{h}{L} \rightarrow L = \frac{h}{sin(\theta)} = \frac{2 m}{sin(28)} = 4.26 m

To find the revolutions we need the time, which can be found using the following equation:                

v_{f} = v_{0} + at  

t = \frac{v_{f} - v_{0}}{a} (1)

So first, we need to find the acceleration:

v_{f}^{2} = v_{0}^{2} + 2aL \rightarrow a = \frac{v_{f}^{2} - v_{0}^{2}}{2L}    (2)  

By entering equation (2) into (1) we have:

t = \frac{v_{f} - v_{0}}{\frac{v_{f}^{2} - v_{0}^{2}}{2L}}

Since it starts from rest (v₀ = 0):  

t = \frac{2L}{v_{f}} = \frac{2*4.26 m}{4.9 m/s} = 1.74 s

Finally, we can find the revolutions:  

\theta_{f} = \frac{1}{2} \omega_{f}*t = \frac{1}{2}*49 rad/s*1.74 s = 42.63 rad*\frac{1 rev}{2\pi rad} = 6.78 rev

Therefore, the ball rotates 6.78 revolutions.

I hope it helps you!                                                                                                                                                                                          

3 0
3 years ago
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