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expeople1 [14]
2 years ago
3

A school bus leaves school and heads east for 2 miles before making its first stop. It then turns left and heads north 3 miles b

efore making another stop. Find the distance and displacement of the school bus after completing its first two stops

Physics
1 answer:
Gala2k [10]2 years ago
7 0

Answer:

The answer is below in the picture.

BRAINLIEST IF YOU WILL

Explanation:

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50 J of work was performed in 20 seconds. How much power was used to do this task?
yuradex [85]
Power=work/time
power=50/20
50/20=2.5
Therefore A. 2.5 W
7 0
2 years ago
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You are using a lightweight rope to pull a sled along level ground. The sled weighs 485 N, the coefficient of kinetic friction b
Bogdan [553]

Answer:

N=459.01N

Explanation:

According to Newton's first law:

N+F_y-W=0

The component of the force on the y-axis can be obtained through the Pythagorean Theorem. This is because the components are the cathetus of a right triangle and its hypotenuse is the magnitude of the force:

sin12^\circ=\frac{F_y}{F}\\F_y=Fsin12^\circ

Replacing and solving for N:

N=W-Fsin12^\circ\\N=485N-(125N)sin12^\circ\\N=459.01N

5 0
2 years ago
A skydiver finds that she speeds up when she holds her arms close to her body. What does this do?
ArbitrLikvidat [17]

D. Reduces the force of air resistance

7 0
2 years ago
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A 250 GeV beam of protons is fired over a distance of 1 km. If the initial size of the wave packet is 1 mm, find its final size
Margarita [4]

Answer:

The final size is approximately equal to the initial size due to a very small relative increase of 1.055\times 10^{- 7} in its size

Solution:

As per the question:

The energy of the proton beam, E = 250 GeV =250\times 10^{9}\times 1.6\times 10^{- 19} = 4\times 10^{- 8} J

Distance covered by photon, d = 1 km = 1000 m

Mass of proton, m_{p} = 1.67\times 10^{- 27} kg

The initial size of the wave packet, \Delta t_{o} = 1 mm = 1\times 10^{- 3} m

Now,

This is relativistic in nature

The rest mass energy associated with the proton is given by:

E = m_{p}c^{2}

E = 1.67\times 10^{- 27}\times (3\times 10^{8})^{2} = 1.503\times 10^{- 10} J

This energy of proton is \simeq 250 GeV

Thus the speed of the proton, v\simeq c

Now, the time taken to cover 1 km = 1000 m of the distance:

T = \frac{1000}{v}

T = \frac{1000}{c} = \frac{1000}{3\times 10^{8}} = 3.34\times 10^{- 6} s

Now, in accordance to the dispersion factor;

\frac{\delta t_{o}}{\Delta t_{o}} = \frac{ht_{o}}{2\pi m_{p}\Delta t_{o}^{2}}

\frac{\delta t_{o}}{\Delta t_{o}} = \frac{6.626\times 10^{- 34}\times 3.34\times 10^{- 6}}{2\pi 1.67\times 10^{- 27}\times (10^{- 3})^{2} = 1.055\times 10^{- 7}

Thus the increase in wave packet's width is relatively quite small.

Hence, we can say that:

\Delta t_{o} = \Delta t

where

\Delta t = final width

3 0
2 years ago
Two swift canaries fly toward each other, each moving at 15.0 m/s relative to the ground, each warbling a note of frequency 1750
lesya692 [45]

Answer:

Part a)

f = 1911.5 Hz

Part b)

\lambda = 0.186 m

Explanation:

Here the source and observer both are moving towards each other

so we know that the apparent frequency is given as

f' = f_0 (\frac{v + v_o}{v - v_s})

here we know that

f_0 = 1750 Hz

v_o = 15 m/s

v_s = 15 m/s

now we will have

f' = (1750)(\frac{340 + 15}{340 - 15})

f' = 1911.5 Hz

Part b)

Apparent wavelength is given by the formula

\lambda = \frac{v_{relative}}{f_{app}}

here we will have

\lambda = \frac{340 + 15}{1911.5}

\lambda = 0.186 m

3 0
2 years ago
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