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den301095 [7]
3 years ago
5

A car moving south speeds up from 10 m/s to 40 m/s in 15 seconds. What is the car’s acceleration? A. 2 m/s2 B.15 m/s2 C.30 m/s2

D.50 m/s2
Physics
2 answers:
Ainat [17]3 years ago
6 0
A. 2 m/s^2

40-10=30
30÷15 =2
Black_prince [1.1K]3 years ago
5 0

Answer:

Acceleration, a=2\ m/s^2

Explanation:

It is given that,

A car moving south speeds up from 10 m/s to 40 m/s in 15 seconds.

It means,

Initial velocity of the car, u = 10 m/s

Final velocity of the car, v = 40 m/s

Time taken, t = 15 seconds

We have to find the acceleration of the car. Mathematically, acceleration of the car is given by taking difference in velocities divided by time taken i.e.

a=\dfrac{v-u}{t}

a=\dfrac{40\ m/s-10\ m/s}{15\ s}

a=2\ m/s^2

Hence, this is the required solution.

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Darya [45]
N= energy efficiency pout means output and pin means input the reason this would show efficiency is because your output should be greater then your input and because depending on how small your number is after your division will tell you how efficient it is you want a big number.
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3 years ago
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Dos masas de 8kg es tan unidas en el extremo de una varilla de aluminio de 400mm de longitud. La varilla está sostenida en su pa
enyata [817]

Answer:

The maximum frequency of revolution is 3.6 Hz.

Explanation:

Given that,

Mass = 8 kg

Distance = 400 mm

Tension = 800 N

We need to calculate the velocity

Using centripetal force

F=\dfrac{mv^2}{r}

Where, F= tension

m = mass

v= velocity

r = radius of circle

Put the value into the formula

800=\dfrac{8\times v^2}{200\times10^{-3}}

v^2=\sqrt{\dfrac{800\times200\times10^{-3}}{8}}

v=4.47\ m/s

We need to calculate the maximum frequency of revolution

Using formula of frequency

f=\dfrac{v}{2\pi r}

Put the value into the formula

f=\dfrac{4.47}{2\pi\times200\times10^{-3}}

f=3.6\ Hz

Hence, The maximum frequency of revolution is 3.6 Hz.

7 0
2 years ago
A boat is floating in a small pond. the boat then sinks so that it is completely submerged. what happens to the level of the pon
lukranit [14]
A boat is floating in a small pond. the boat then sinks so that it is completely submerged. what happens to the level of the pond?
It increases!
4 0
3 years ago
To practice Problem-Solving Strategy 30.1: Inductors in Circuits. A circuit has a 1 V battery connected in series with a switch.
ki77a [65]

Answer:

0.0133A

Explanation:

Since we have two sections, for the Inductor region there would be a current i_1. In the case of resistance 2, it will cross a current i_2

Defined this we proceed to obtain our equations,

For i_1,

\frac{di_1}{dt}+i_1R_1 = V

I_1 = \frac{V}{R_1} (1-e^{-\frac{R_1t}{L}})

For i_2,

I_2R_2 =V

I_2 = \frac{V}{R_2}

The current in the entire battery is equivalent to,

i_t = I_1+I_2

i_t = \frac{V}{R_2}+\frac{V}{R_1} (1-e^{-\frac{R_1t}{L}})

Our values are,

V=1V

R_1 = 95\Omega

L= 1.5*10^{-2}H

R_2 =360\Omega

Replacing in the current for t= 0.4m/s

i=\frac{1}{360}+\frac{1}{95}(1-e^{-\frac{95*0.4}{1.5*10^{-2}}})

i= 0.0133A

i_1 = 0.01052A

3 0
3 years ago
When a mass of 0.350 kg is attached to a vertical spring and lowered slowly, the spring stretches 12.0 cm. The mass is now displ
Ray Of Light [21]

Answer:

period of oscillations is 0.695 second

Explanation:

given data

mass m = 0.350 kg

spring stretches x = 12 cm = 0.12 m

to find out

period of oscillations

solution

we know here that force

force = k × x   .........1

so force = mg =  0.35 (9.8)  = 3.43 N

3.43 = k × 0.12

k = 28.58 N/m

so period of oscillations is

period of oscillations = 2π × \sqrt{\frac{m}{k} }   ................2

put here value

period of oscillations = 2π × \sqrt{\frac{0.35}{28.58} }  

period of oscillations = 0.6953

so period of oscillations is 0.695 second

4 0
3 years ago
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