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Oksi-84 [34.3K]
4 years ago
10

A bicyclist goes from a speed of 5 m/s to a speed of 10 m/s. What happens to her kinetic energy?

Physics
1 answer:
serg [7]4 years ago
4 0

Answer:

C

Explanation:

The kinetic energy of an object can be calculated via the following equation:

k =  \frac{1}{2} m {u}^{2}

where:

K : kinetic energy

m: mass of the object

v: its velocity

when the speed changes, the kinetic energy changes as well. Lets suppose that it doubles, as on the problem stated. Then K is increased by a factor of four. in other words...

newk =  \frac{1}{2} m {(newu)}^{2}  =  \frac{1}{2} m {(2u)}^{2}   \\  =  \frac{1}{2} m(4 {u}^{2} ) = 4( \frac{1}{2} m {u}^{2} ) = 4k

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A baseball leaves a bat with a horizontal velocity of 20 m/s. In a time of 0.25 s, How far will it have traveled horizontally?
Maurinko [17]

Distance traveled by the ball is given by

distance = speed \times time

here we know that

speed = 20 m/s

times = 0.25 s

now we have

distance = 20 \times 0.25

distance = 5 m

so ball will travel 5 m distance in the given interval of time

6 0
3 years ago
One of your summer lunar space camp activities is to launch a 1130 kg1130 kg rocket from the surface of the Moon. You are a seri
maxonik [38]

Answer:

∆U = 2.296×10^10Joules

Explanation:

Gravitational potential energy is defined as the energy possessed by an object under the influence of gravity due to its virtue of position.

Potential energy U = Fr where;

F is the force of attraction between the masses of the moon and the rocket.

r is the radius or height of the object.

From Newton's law of universal gravitation, F = GMm/r²

Potential energy U = (-GMm/r²)×r

Potential energy U = -GMm/r

The force is negative because the objects act upward.

M is the mass of the rocket

m is the mass of the moon

Gravitational potential energy possessed by the rocket

U1 = -GMm/r1

r1 is the altitude covered by the rocket

Gravitational potential energy possessed by the Moon

U2 = -GMm/(r2+r1)

r2 is the radius of the moon

Change in gravitational potential energy ∆U = U2-U1

∆U = -GMm/(r2+r1)-(-GMm/r1)

∆U = -GMm/(r2+r1) + GMm/r1

∆U = -GMm{1/(r2+r1)-1/r1}

Given

G = 6.67×10^-11m³/kgs²

M = 1130kg

m = 7.36×10²²kg

r1 = 215km = 215,000m

r2 = 1740km = 1,740,000m

∆U = -6.67×10^-11× 7.36×10²² × 1130{1/(215,000+1,740,000)-1/215000}

∆U= -55.47×10¹⁴{1/1955000-1/215000}

∆U = -55.47×10¹⁴{5.12×10^-7 - 4.65×10^-6}

∆U = -284×10^7 + 257.94×10^8

∆U = 22,954,000,000Joules

∆U = 2.296×10^10Joules

8 0
4 years ago
a plane travels 204 km, northeast in 15.0 minutes. It also increases elevation by 1.6 km, upward in the same amount of time. Wha
LUCKY_DIMON [66]

Answer:

230 m/s northeast, 1.8 m/s up

Explanation:

204 kilometres = 204000 metres

15.0 minutes = 900 seconds

Velocity = Distance / Time

= 204000 / 900

= 230 m/s northeast (to 2 sf.)

1.6km = 1600 metres

Velocity = 1600 / 900

= 1.8 m/s up (to 2 sf.)

Read more on Brainly.com - brainly.com/question/13863590#readmore

5 0
3 years ago
Read 2 more answers
What is the mass of an object if a 30 N force makes it accelerate at 6 m/s2
jasenka [17]

Answer:

5 kg

Explanation:

Acceleration = 6 m/s^2

Force = 30 N

Force = mass * acceleration

mass = force / acceleration

mass = 30 / 6

mass = 5 kg

4 0
3 years ago
A small sphere with mass m is attached to a massless rod of length L that is pivoted at the top, forming a simple pendulum. The
USPshnik [31]

Answer:

a) see attached, a = g sin θ

b)

c)   v = √(2gL (1-cos θ))

Explanation:

In the attached we can see the forces on the sphere, which are the attention of the bar that is perpendicular to the movement and the weight of the sphere that is vertical at all times. To solve this problem, a reference system is created with one axis parallel to the bar and the other perpendicular to the rod, the weight of decomposing in this reference system and the linear acceleration is given by

          Wₓ = m a

          W sin θ = m a

          a = g sin θ

b) The diagram is the same, the only thing that changes is the angle that is less

                θ' = 9/2  θ

             

c) At this point the weight and the force of the bar are in the same line of action, so that at linear acceleration it is zero, even when the pendulum has velocity v, so it follows its path.

The easiest way to find linear speed is to use conservation of energy

Highest point

            Em₀ = mg h = mg L (1-cos tea)

Lowest point

          Emf = K = ½ m v²

          Em₀ = Emf

          g L (1-cos θ) = v² / 2

              v = √(2gL (1-cos θ))

4 0
4 years ago
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