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2 years ago

A biker starts moving from rest with a constant acceleration of 2.0 m/s^2. What is his velocity after 5.0 s?

Physics

1 answer:

Ber [7]2 years ago

5 0

a = v-u/t

2 = v - 0 / 5

2 = v/5

v = 10 m /s

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Galileo started modern science because he started using ? Rather than speculation

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By using the telescope

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2 years ago

an object moving at 10. km/hr has a kinetic energy of 10. J. what is the kinetic energy of the same object if it is moving at 20

Schach [20]

Kinetic energy is related to velocity by:
KE = (1/2)mv^2

solve for mass m
10 = (1/2)m(10)^2
10 = (1/2)m(100)
10= 50m
10/50 = m
1/5 = m

at 20 km/hr

KE = (1/2)(1/5)(20)^2
KE = (1/10)(400)
KE = 40 J

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2 years ago

Two light bulbs are wired in series and one bulb burns out (opens.) Technician A says that the other bulb will still work. Techn

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Answer:

Neither technician

Explanation:

Neither technician is correct.

two bulbs are connected in series one bulb burn out

If one bulb in the series burns out then the circuit will break.

In a series circuit same current passes each resistor.

so, both the technician is incorrect bulb B will not work and current will not increase in the other bulb.

6 0

3 years ago

Luke stands on the edge of a roof throws a ball downward. It strikes the ground with 100J of kinetic energy. Luke now throws ano

attashe74 [19]

A. Less than 100 this the answer

3 0

2 years ago

A thundercloud has an electric charge of 48.8 C near the top of the cloud and –41.7 C near the bottom of the cloud. The magnitud

IceJOKER [234]

Answer: 1.51 km

Explanation:

Coulomb's Law: The electrostatic force between two charge particles Q: and Q2 is directly proportional to product of magnitude of charges and inversely proportional to square of separation distance between them.

Or, $\vec{F}=k \frac{Q_{1} Q_{2}}{r^{2}}$

Where Q1 and Q2 are magnitude of two charges and r is distance between them:

Given:

Q1 = Charge near top of cloud = 48.8 C

Q2 = Charge near the bottom of cloud = -41.7 C

Force between charge at top and bottom of cloud (i.e. between Q: and Q2) (F) = 7.98 x 10^6N

k = 8.99 x 109Nm^2/C^2

So,

$\begin{aligned}&7.98 \times 10^{6}=\left(8.99 \times 10^{9} \mathrm{Nm}^{2} / \mathrm{C}^{2}\right) \frac{48.8 \mathrm{C} \times 41.7 \mathrm{C}}{\mathrm{r}^{2}} \\&r=\sqrt{\frac{1.8294 \times 10^{13}}{7.98 \times 10^{6}}}=1.514 \times 10^{3} \mathrm{~m}=1.51 \mathrm{~km}\end{aligned}$

Therefore, the separation between the two charges (r) = 1.51 km

3 0

1 year ago