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4 years ago

If a sprinter accelerates from rest to 12 m/s north , what is their change in velocity ?

Physics

2 answers:

Elden [556K]4 years ago

7 0

12 is the answer gjbxrhcdcvhbcbj

natima [27]4 years ago

5 0

Answer: It would be 12 m/s.

Explanation: It would be this because If you go from rest to sprint it would be 12 m/s. Also, I did this the other day.

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Consider an experimental setup where charged particles (electrons or protons) are first accelerated by an electric field and the

MissTica

Answer:

V = 10581.59 V

Explanation:

To solve this problem, we have to take into the relation between the kinetic energy of an electron and the energy generated by the potential difference of an electric field. This relation is given by

$E_{k} = E_{V}$

$\frac{1}{2}mv^{2} = eV$

where m is the mass of the electron, v is the velocity of the electron, e is the electric charge and V is the potential difference that accelerated the electron.

By taking V from the last expression we have:

$V = \frac{mv^{2}}{2e} = \frac{(9.1*10^{-31} )(6.1*10^{7})^{2}}{2*(-1.6*10^{-19})}} = 10581.59 V$

where we have taken that

me = 9.1*10^(-31) Kg

e = -1.6*10^(-19) C

v = 6.1*10^(7)

I hope this is useful for you

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6 0

4 years ago

A 77 kg student traveling in a car with a constant velocity has a kinetic energy of 1.7 104 J. What is the speedometer reading o

AVprozaik [17]

ANSWER

75.65 km/h

EXPLANATION

Given:

• The student's mass, m = 77 kg

• The kinetic energy of the student in the car, KE = 1.7 x 10⁴ J

Find:

• The speed read in the speedometer of the car, which is the speed of the student, v (in km/h)

The kinetic energy of an object with mass m, traveling at a speed v, is,

$KE=\frac{1}{2}mv^2$

Solving for v,

$v=\sqrt{\frac{2KE}{m}}$

Replace the known values and solve,

$v=\sqrt{\frac{2\cdot1.7\cdot10^4J}{77kg}}\approx21.013m/s$

Note that because the kinetic energy is given in Joules - which is equivalent to kg*m²/s², the speed we found is in m/s. Now, knowing that there are 3600 seconds in 1 hour and that 1 km is equivalent to 1000 m, we can convert this to km/s,

$v=21.013\frac{m}{s}\cdot\frac{3600s}{1h}\cdot\frac{1km}{1000m}\approx75.65km/h$

Hence, the speedometer reading of the car is 75.65 km/h, rounded to the nearest hundredth.

7 0

2 years ago

The electric power consumed by a circuit with one light bulb is 24 W. The voltage of the battery is 3 V. What is the current in

IceJOKER [234]

Answer:

8 ampere

Explanation:

P= V x I

I = P/V

I= 24 /3

I = 8 A

5 0

3 years ago

PLEASE HELP!! I'll mark brainliest for correct answer.

Tju [1.3M]

1. Car A
2. Car C
3. Car A
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5. 750 J
6. 750 J
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9. B

6 0

3 years ago

How much does a 0.15 kg baseball weigh on earth?

Gnesinka [82]

1.472 N

to get weight you multiply an object's mass in kilograms with the acceleration of gravity(9.81m/s) :)

7 0

3 years ago