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3 years ago

5

A bullet is shot downward out of a high building with an initial speed of 10 m/s. How fast is the bullet traveling after 5 secon

ds (by m/s, neglect air friction, g = 9.8 m/s)?
a.) -10
b.) 9.8
c.)-59
d.)-49
e.)-9.8

1 answer:

serious [3.7K]3 years ago

6 0

Answer:

-59

Explanation:

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Without being provided a list of items, I would have to generally say that everything around you is matter. There are a few exceptions to this list, but a general rule of thumb is anything you can touch, taste, smell or hold would be considered matter. Sound, light, time (Dr. Who may disagree) and heat would be considered non-matter items.

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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

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Answer:

(a) 43.2 kC

(b) 0.012V kWh

(c) 0.108V cents

Explanation:

<u>Given:</u>

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

<u>Assume:</u>

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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4 years ago

A 111 ‑turn circular coil of radius 2.11 cm and negligible resistance is immersed in a uniform magnetic field that is perpendicu

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Answer:

0.0061 J

Explanation:

Parameters given:

Number of turns, N = 111

Radius of turn, r = 2.11 cm = 0.0211 m

Resistance, R = 14.1 ohms

Time taken, t = 0.125 s

Initial magnetic field, Bin = 0.669 T

Final magnetic field, Bfin = 0 T

The energy dissipated in the resistor is given as:

E = P * t

Where P = Power dissipated in the resistor

Power, P, is given as:

P = V² / R

Hence, energy will be:

E = (V² * t) / R

To find the induced voltage (EMF), V:

EMF = [-(Bfin - Bin) * N * A] / t

A is Area of coil

EMF = [-(0 - 0.669) * 111 * pi * 0.0211²] / 0.125

EMF = 0.83 V

Hence, the energy dissipated will be:

E = (0.83² * 0.125) / 14.1

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3 years ago

The Indianapolis speedway consists of a 2.5 mile track having four turns, each 0.25 mile long and banked at 9 12'

blsea [12.9K]

Answer: Your question is missing below is the question

Question : What is the no-friction needed speed (in m/s ) for these turns?

answer:

20.1 m/s

Explanation:

2.5 mile track

number of turns = 4

length of each turn = 0.25 mile

banked at 9 12'

<u>Determine the no-friction needed speed </u>

First step : calculate the value of R

2πR / 4 = πR / 2

note : πR / 2 = 0.25 mile

∴ R = ( 0.25 * 2 ) / π

= 0.159 mile ≈ 256 m

Finally no-friction needed speed

tan θ = v^2 / gR

∴ v^2 = gR * tan θ

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