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3 years ago

A Car accelerates from 4 m/s to 16 m/s in 4 seconds. What is the car's

Physics

2 answers:

klemol [59]3 years ago

5 0

Answer:

4 m/s^2

that is the right acceleretion a car

irakobra [83]3 years ago

5 0

Answer:

I believe that the answer is 3 m/s^2

Explanation:

since acceleration is final velocity minus initial velocity divided by time, we can plug in known values and solve.

a=(f-i)/t

a=(16-4)/4

a=12/4

a=3

3m/s^2 is the acceleration

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If the volume is held constant, what happens to the pressure of a gas as temperature is decreased? Explain.

Lady bird [3.3K]

Answer:Decreases

Explanation:

Given

Volume is held constant that is it is a isochoric process.

We know that

PV=nRT

as n,V& R are constant therefore only variables are

P & T

so $\frac{P_1}{T_1}=\frac{P_2}{T_2}$

$\frac{P_1}{P_2}=\frac{T_1}{T_2}$

As $T_1$ is decreasing therefore Pressure must also decrease so that ratio remains constant.

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3 years ago

Which letter represents the location of the resister in this diagram? A B C D

ololo11 [35]

Answer:

The letter B is the letter that represents the location of the resister in the diagram.

Explanation:

I hope it helped!Please mark brainliest and have a wonderful night! and day!!!!

5 0

3 years ago

Read 2 more answers

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.

4 0

3 years ago

A point charge of 4.0 µC is placed at a distance of 0.10 m from a hard rubber rod with an electric field of 1.0 × 103 . What is

schepotkina [342]

4 for the first one and 5.2 for the second one.

3 0

3 years ago

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A source charge of 3 µC generates an electric field of 2.86 × 105 N/C at the location of a test charge. Using k = 8.99 × 109N.m^

Nataliya [291]

Variables:

Source charge, Q = 3 micro C = 3 * 10^ - 6 C

E = electric field = 2.86 * 10 ^5 N/C

K = 8.99 * 10^9 N * m^2 / C

d = distance = ?

Formula:

E = K * Q / (d^2) => d^2 = K * Q / E

=> d^2 = 8.99 * 10^9 N * m^2 / C * 3 * 10^ -6 C / (2.86 * 10^ 5 N/C)

d^2 = 9.43 * 10 ^ -2 m^2

=> d = 3.07 * 10^-1 m

Answer: 0.307 m

Note: it is a long distance due to the Electric field is very low

7 0

3 years ago

Read 2 more answers