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Fantom [35]
1 year ago
12

Help pls it’s about physics I need all of them

Physics
1 answer:
damaskus [11]1 year ago
4 0

1) Amy runs 2 miles south, then takes a turn and runs 3 miles north the displacement is 1 mile North.

2) Distance is 800m and Displacement will be zero.

3) Distance is 90 feet, and Displacement is 30 feet.

4) Distance is 4 miles and Displacement is 0.

Displacement is the measurement of "how far an object is out of place," whereas distance refers to "how much ground an object has covered during its motion."

The spacing between two specified points is represented by the one-dimensional quantity of displacement (symbolized as d or s), commonly known as length or distance. The meter serves as the standard displacement unit in the International System of Units (SI) (m). Usually, displacement is defined or measured along a straight line.

1) Here, the person goes 2 miles south and then 3 miles north.

3-2 = 1 mile north

So, Displacement is 1 mile North.

2) It is given that Jermaine runs 2 laps around the track.

The perimeter of the track on which he runs = 400m

Here as it is given laps, which means the track is circular in shape.

Distance covered in 1 lap

So total distance covered in 2 laps= 2 × 400 = 800 m

Displacement will be zero as the initial and the final position is the same. So, no positional change happens.

To know more about Displacement refer to:  brainly.com/question/2109763

#SPJ9

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A cosmic ray muon with mass mμ = 1.88 ✕ 10−28 kg impacting the Earth's atmosphere slows down in proportion to the amount of matt
anyanavicka [17]

Answer:

a. the magnitude of the force experienced by the muon is 2.55 × 10⁻¹⁹N

b.  this force compare to the weight of the muon; the force is 1.38 × 10⁸ greater than muon

Explanation:

F= ma

v²=u² -2aS

(1.56 ✕ 10⁶)²=(2.40 ✕ 10⁶)²-2a(1220)

a=1.36×10⁹m/s²

recall

F=ma

F = 1.88 ✕ 10⁻²⁸ kg × 1.36×10⁹m/s²

F= 2.55 × 10⁻¹⁹N

the magnitude of the force experienced by the muon is 2.55 × 10⁻¹⁹N

b.  this force compare to the weight of the muon

F/mg= 2.55 × 10⁻¹⁹/ (1.88 ✕ 10⁻²⁸ × 9.8)

= 1.38 × 10⁸

6 0
3 years ago
A wind turbine has a total input power of 2 500 kW.
Sonbull [250]

Answer:

Output power = 500 KW

Explanation:

Given the following data;

Efficiency = 20%

Input power = 2500 KW

To find the output power;

Efficiency = \frac {Out-put \; power}{In-put \; power} * 100

Substituting into the equation, we have;

LET Output power = OP

20 = \frac {OP}{2500} * 100

Cross-multiplying, we have;

20 * 2500 = OP * 100

50000 = OP * 100

OP = \frac {50000}{100}

Output power = 500 KW

6 0
3 years ago
A baseball is batted. It's a long fly ball. 4 seconds later the ball reaches the outfield 100 meters away and returns to the hei
Vitek1552 [10]

Answer:

25 m/s

Explanation:

First we should define the variables

T=4

Dx = 100

ay=-9.8

ax=0

We can use formula 1 from the BIG 5

x=(v+v0)t/2

By plugging in our variables we can get 100=4(v+v0)/2

Which is 50=v+v0

v=v0 since horizontal acceleration always equals zero

so 2v0 = 50

v0 = 25

8 0
2 years ago
A laser with a power of 1.0 mW has a beam radius of 1.0 mm. What is the peak value of the electric field in that beam
Fynjy0 [20]

Answer:

The peak value of the electric field is 489.64 V/m

Explanation:

Given;

power of the laser, P = 1.0 mW = 1 x 10⁻³ W

Radius of the beam, R = 1.0 mm = 1 x 10⁻³ m

Area of the beam = πr² = π(1 x 10⁻³ )² = 3.142 x 10⁻⁶ m²

The average intensity of the light = P / A

The average intensity of the light = ( 1 x 10⁻³) / (3.142 x 10⁻⁶)

The average intensity of the light = 318.27 W/m²

The peak value of the electric field is given by;

E_o = \sqrt{\frac{2I_{avg}}{c\epsilon_o}}\\\\E_o = \sqrt{\frac{2(318.27)}{(3*10^8)(8.85*10^{-12})}}\\\\E_o = 489.64 \ V/m

Therefore, the peak value of the electric field is 489.64 V/m.

4 0
3 years ago
An electron in the beam of a cathod-ray tube is accelerated by a potential difference of 2.12 kV . Then it passes through a regi
son4ous [18]

Answer:

B=9.1397*10^-4 Tesla

Explanation:

To find the velocity first we put kinetic energy og electron is equal to potential energy of electron

K.E=P.E

\frac{1}{2}*m*v^{2}  =e*V

where :

m is the mass of electron

v is the velocity

V is the potential difference

v=\sqrt{\frac{2*e*V}{m} }    eq 1

Radius of electron moving in magnetic field is given by:

R=\frac{m*v}{q*B}       eq 2

where:

m is the mass of electron

v is the velocity

q=e=charge of electron

B is the magnitude of magnetic field

Put v from eq 1 into eq 2

R=\frac{m*\sqrt{\frac{2*e*V}{m} } }{e B}

B=\sqrt{\frac{2*m*V}{e*R^{2} } }

B=\sqrt{\frac{2*(9.31*10^{-31})*(2.12*10^{3})  }{(1.60*10^{-19})*(0.170)^{2}  } }

B=9.1397*10^-4 Tesla

3 0
3 years ago
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