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soldier1979 [14.2K]
3 years ago
12

If John ran a 5-kilometer marathon, how long was that in meters? A. 50 m. B. 5,000 m. C. 500 m. D. 1000 m.

Physics
2 answers:
Vladimir79 [104]3 years ago
8 0
The answer is B good lessons
andreyandreev [35.5K]3 years ago
4 0
1 km = 1000 M . . 5 km = 5000 M The fact that John ran it is not involved anywhere in the math.
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The bodies in this universe attract one another name the scientist who propounded this statement​
murzikaleks [220]

Answer:

It was proposed by Isaac Newton

Explanation:

The law of universal attraction of expression

          F = G \ \frac{m_1m_2}{ r^2}G m1m2 / r ^ 2

where G is a constant, m₁ and m₂ are the masses of the bodies and r the distance between them.

It was proposed by Isaac Newton

With this law Newton explained that the force that pulls the moon towards the earth is the same as that which attracts an apple towards the earth

8 0
2 years ago
NEED HELP
ch4aika [34]

Answer: C.

Explanation:

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8 0
2 years ago
Scott travels north 5 miles, then goes west 3 miles, and then goes south for 2 miles.
Yuki888 [10]
Scott traveled 10 miles
5 0
2 years ago
NEED HELP ASAP
Dafna11 [192]

Answers:

a) -2.54 m/s

b) -2351.25 J

Explanation:

This problem can be solved by the <u>Conservation of Momentum principle</u>, which establishes that the initial momentum p_{o} must be equal to the final momentum p_{f}:  

p_{o}=p_{f} (1)  

Where:  

p_{o}=m_{1} V_{o} + m_{2} U_{o} (2)  

p_{f}=(m_{1} + m_{2}) V_{f} (3)

m_{1}=110 kg is the mass of the first football player

V{o}=-7 m/s is the velocity of the first football player (to the south)

m_{2}=75 kg  is the mass of the second football player

U_{o}=4 m/s is the velocity of the second football player (to the north)

V_{f} is the final velocity of both football players

With this in mind, let's begin with the answers:

a) Velocity of the players just after the tackle

Substituting (2) and (3) in (1):

m_{1} V_{o} + m_{2} U_{o}=(m_{1} + m_{2}) V_{f} (4)  

Isolating V_{f}:

V_{f}=\frac{m_{1} V_{o} + m_{2} U_{o}}{m_{1} + m_{2}} (5)

V_{f}=\frac{(110 kg)(-7 m/s) + (75 kg) (4 m/s)}{110 kg + 75 kg} (6)

V_{f}=-2.54 m/s (7) The negative sign indicates the direction of the final velocity, to the south

b) Decrease in kinetic energy of the 110kg player

The change in Kinetic energy \Delta K is defined as:

\Delta K=\frac{1}{2} m_{1}V_{f}^{2} - \frac{1}{2} m_{1}V_{o}^{2} (8)

Simplifying:

\Delta K=\frac{1}{2} m_{1}(V_{f}^{2} - V_{o}^{2}) (9)

\Delta K=\frac{1}{2} 110 kg((-2.5 m/s)^{2} - (-7 m/s)^{2}) (10)

Finally:

\Delta K=-2351.25 J (10) Where the minus sign indicates the player's kinetic energy has decreased due to the perfectly inelastic collision

6 0
3 years ago
If the man has initial velocity of 0.4 m/s, acceleration is 0.343 m/s^2, a time of 5.8 seconds, and experimental final velocity
agasfer [191]

Answer:

The answer is going to be C.

Explanation:

Trust me. Im an expert in physics

6 0
3 years ago
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