All categories

3 years ago

I NEED HELP ASAP IM ON TIMER !!!!

Physics

1 answer:

Angelina_Jolie [31]3 years ago

8 0

Answer:

a.) 20 km east

b.) 50 km west

c.) 80 km west

Explanation:

As given,

Railway station - A B C

Distance(km) - 0 30 60

Starts from A , train reaches B be in 15 minutes

Starts from A , train reaches C be in 30 minutes

a.)

If A is origin

As train covers 30km = 15 minutes

⇒ 15 minutes = 30 km

1 minute = $\frac{30}{15} = 2$ km

⇒ 10 minutes = 2×10 = 20 km

∴ we get

The position of the train moving from station A to station B after 10 minutes = 20 km east

b.)

If B is origin then , the position of the train moving from station A to station B after 10 minutes = 30 + 20 = 50 km west

( Because B is the origin , so firstly train go from B to A in 15 minutes ( 30 km ) then A to B in 10 minutes (20 km) )

c.)

If C is the origin then , the position of the train moving from station A to station B after 10 minutes = 60 + 20 = 80 km west

( Because C is the origin , so firstly train go from C to A in 30 minutes ( 60 km ) then A to B in 10 minutes (20 km) )

You might be interested in

A planet is 10 light years away from Earth. What speed would you need to go for a trip to the planet and back to take only 5 yea

viva [34]

Answer:

a. speed, v = 0.97 c

b. time, t' = 20.56 years

Given:

t' = 5 years

distance of the planet from the earth, d = 10 light years = 10 c

Solution:

(a) Distance travelled in a round trip, d' = 2d = 20 c = L'

Now, using Length contraction formula of relativity theory:

$L'' = L'\sqrt{1 - \frac{v^{2}}{c^{2}}}$ (1)

time taken = 5 years

We know that :

time = $\frac{distance}{speed}$

5 = $\frac{L''}{v}$ (2)

Dividing eqn (1) by v on both the sides and substituting eqn (2) in eqn (1):

$\frac{L'\sqrt{1 - \frac{v^{2}}{c^{2}}}}{v} = 5$

$\frac{20'\sqrt{1 - \frac{v^{2}}{c^{2}}}}{v} = 5$

Squaring both the sides and Solving above eqution, we get:

v = 0.97 c

(b) Time observed from Earth:

Using time dilation:

$t'' = \frac{t'}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}$

$t'' = \frac{5}{\sqrt{1 - \frac{(0.97c)^{2}}{c^{2}}}}$

Solving the above eqn:

t'' = 20.56 years

4 0

3 years ago

A flat disk of radius 0.50 m is oriented so that the plane of the disk makes an angle of 30 degrees with a uniform electric fiel

nexus9112 [7]

Answer:

The electric flux is $280\ \rm N.m^2/C$

Explanation:

Given:

Radius of the disc R=0.50 m
Angle made by disk with the horizontal $\theta=30^\circ$
Magnitude of the electric Field $E=713.0\ \rm N/C$

The flux of the Electric Field E due to the are dA in space can be found out by using Gauss Law which is as follows

$\phi=\int E.dA$

where

$\phi$ is the total Electric Flux
E is the Electric Field
dA is the Area through which the electric flux is to be calculated.

Now according to question we have

$=EA\cos\theta \\=713\times 3.14\times 0.5^2 \times \cos60^\circ\\=280\ \rm N.m^2/C$

Hence the electric flux is calculated.

8 0

3 years ago

6 A test of a driver's perception/reaction time is being conducted on a special testing track with level, wet pavement and a dri

mylen [45]

Answer:

a. 10.5 s b. 6.6 s

Explanation:

a. The driver's perception/reaction time before drinking.

To find the driver's perception time before drinking, we first find his deceleration from

v² = u² + 2as where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 0 m/s (since he stops), a = deceleration of driver and s = distance moved by driver = 385 ft = 385 × 0.3048 m = 117.35 m

So, a = v² - u²/2s

substituting the values of the variables into the equation, we have

a = v² - u²/2s

a = (0 m/s)² - (22.35 m/s)²/2(117.35 m)

a = - 499.52 m²/s²/234.7 m

a = -2.13 m/s²

Using a = (v - u)/t where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 0 m/s (since he stops), a = deceleration of driver = -2.13 m/s² and t = reaction time

So, t = (v - u)/a

Substituting the values of the variables into the equation, we have

t = (0 m/s - 22.35 m/s)/-2.13 m/s²

t = - 22.35 m/s/-2.13 m/s²

t = 10.5 s

b. The driver's perception/reaction time after drinking.

To find the driver's perception time after drinking, we first find his deceleration from

v² = u² + 2as where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 30 mi/h = 30 × 1609 m/3600 s = 13.41 m/s, a = deceleration of driver and s = distance moved by driver = 385 ft = 385 × 0.3048 m = 117.35 m

So, a = v² - u²/2s

substituting the values of the variables into the equation, we have

a = v² - u²/2s

a = (13.41 m/s)² - (22.35 m/s)²/2(117.35 m)

a = 179.83 m²/s² - 499.52 m²/s²/234.7 m

a = -319.69 m²/s² ÷ 234.7 m

a = -1.36 m/s²

Using a = (v - u)/t where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 30 mi/h = 30 × 1609 m/3600 s = 13.41 m/s, a = deceleration of driver = -1.36 m/s² and t = reaction time

So, t = (v - u)/a

Substituting the values of the variables into the equation, we have

t = (13.41 m/s - 22.35 m/s)/-1.36 m/s²

t = - 8.94 m/s/-1.36 m/s²

t = 6.6 s

4 0

3 years ago

For each of the following, describe the relationship as directly or inversely proportional:

Maksim231197 [3]

The relationships can best be described as follows:
As frequency increases, wavelength decreases. The greater the energy, the larger the frequency and the shorter (smaller) the wavelength. 

a) wavelength vs. frequency = inversely proportional
b) wavelength vs. energy = inversely proportional
c) frequency vs. energy = directly proportional

Hope this answers the questions. Have a nice day. Feel free to ask more questions.

7 0

3 years ago

A 0.0600-kilogram ball traveling at 60.0 meters

Vikki [24]

The momentum of ball is given by:

$\Delta Q=mv \\ \Delta Q=6*10^{-2}*60 \\ \Delta Q=3.6kg*m/s$

Since both have the same momentum, we have:

$\Delta Q=MV \\ 3.6=10^{-2}V \\ \boxed {V=360m/s}$

Number 3

If you notice any mistake in my english, please let me know, because i am not native.

7 0

3 years ago