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1 year ago

If you accelerate from rest at 12m/s^2 for 5 seconds, what is your finally velocity?

Physics

1 answer:

jeka941 year ago

3 0

Answer:

$v=60m/s$

Explanation:

Use the Kinematic Equation:

$v=v_o+at$

Plug in what is given and solve

$v=0+(12)(5)$

$v=60m/s$

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The different atoms have different quantized energy levels

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3 years ago

Two cars A and B are moving with velocities 20 m/s and 15 m/s in the direction east and west respectively. If

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Answer:

Distance between them is 4,200 meters.

Explanation:

Consinder car A:

${ \bf{distance = speed \times time }}$

substitute:

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Consider car B:

$distance = 15 \times (2 \times 60) \\ = 1800 \: m$

since these cars move in opposite directions, distance between them is their summation:

$distance \: between = { \sum(distance \: of \: each \: car)} \\ = 2400 + 1800 \\ = 4200 \: m$

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2 years ago

If railroad tracks were built over the boundary of two plates what would happen to the railroad tracks as the plates moved

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The railroad tracks will move with the plate boundaries

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3 years ago

Aconstant current of 3 Afor 4 hours is required to charge an automotive battery. If the terminal voltage is V, where t is in hou

kirza4 [7]

Answer:

(a) 43.2 kC

(b) 0.012V kWh

(c) 0.108V cents

Explanation:

Given:

i = current flow = 3 A
t = time interval for which the current flow = $4\ h = 4\times 3600\ s = 14400\ s$
V = terminal voltage of the battery
R = rate of energy = 9 cents/kWh

Assume:

Q = charge transported as a result of charging
E = energy expended
C = cost of charging

Part (a):

We know that the charge flow rate is the electric current flow through a wire.

$\therefore i = \dfrac{Q}{t}\\\Rightarrow Q =it\\\Rightarrow Q = 3\times 14400\\\Rightarrow Q = 43200\ C\\\Rightarrow Q = 43.200\ kC\\$

Hence, 43.2 kC of charge is transported as a result of charging.

Part (b):

We know the electrical energy dissipated due to current flow across a voltage drop for a time interval is given by:

$E = Vit\\\Rightarrow E = V\times 3\times 4\\\Rightarrow E = 12V\ Wh\\\Rightarrow E = 0.012V\ kWh\\$

Hence, 0.012V kWh is expended in charging the battery.

Part (c):

We know that the energy cost is equal to the product of energy expended and the rate of energy.

$\therefore \textrm{Cost}=\textrm{Energy}\times \textrm{Rate}\\\Rightarrow C = ER\\\Rightarrow C = 0.012V\times 9\\\Rightarrow C =0.108V\ cents$

Hence, 0.108V cents is the charging cost of the battery.

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3 years ago

Challenge: Some but not all of the gasoline used by car's enegine is transformed into kinetic energy. Where might some of the en

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Most of the excess energy is released as waste heat into the air surrounding the engine. Small amounts of excess energy are also released as sound energy, and as electrical energy generated by the alternator in a car's engine.

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2 years ago