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2 years ago

If the velocity of an object changes from 15 m/s during a time interval of 4s what is the acceleration of the object

1 answer:

6 0

The acceleration of an object is defined as change in velocity divided by change in time, that is,
Acceleration = change in velocity / change in time
In the question given above,
Change in velocity = 15
Change in time = 4
Acceleration = 15 /4 = 3.75
Therefore, acceleration = 3.8 m/s^2.

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Please help.. urgent Which statement is equivalent to Newton's first law? a. 15,300 N b. 1.20*10^3 N c. 2,030 N d. 1,560 N

taurus [48]

According to Newton laws of motion,
F = m*a
Here, m = 1,560 Kg
a = 1.30 m/s²

Substitute their values,
F = 1,560 * 1.30
F = 2028 N ~ 2030 N [ Closest value ]

In short, Your Answer would be Option C

Hope this helps!

6 0

3 years ago

I wish to use a step up transformer to turn an initial RMS AC voltage of 100 V into a final RMS AC voltage of 200 V. What is the

zhuklara [117]

Answer:

1:2

Explanation:

It is given that,

Initial RMS AC voltage is 100 V and final RMS AC voltage is 200 V.

We need to find the ratio of the number of turns in the primary to the secondary for step up transformer.

For a transformer, $\dfrac{V_1}{V_2}=\dfrac{N_1}{N_2}$

So,

$\dfrac{N_1}{N_2}=\dfrac{100}{200}\\\\\dfrac{N_1}{N_2}=\dfrac{1}{2}$

So, the ratio of the number of turns in the primary to the secondary is 1:2.

4 0

3 years ago

The traffic on the freeway is moving at a constant speed of 24 m/sm/s. What distance does the traffic travel while the car is mo

ExtremeBDS [4]

Incomplete question as there is so much information is missing.The complete question is here

A car sits on an entrance ramp to a freeway, waiting for a break in the traffic. Then the driver accelerates with constant acceleration along the ramp and onto the freeway. The car starts from rest, moves in a straight line, and has a speed of 24 m/s (54 mi/h) when it reaches the end of the 120-m-long ramp. The traffic on the freeway is moving at a constant speed of 24 m/s. What distance does the traffic travel while the car is moving the length of the ramp?

Answer:

Distance traveled=240 m

Explanation:

Given data

Initial velocity of car v₀=0 m/s

Final velocity of car vf=24 m/s

Distance traveled by car S=120 m

To find

Distance does the traffic travel

Solution

To find the distance first we need to find time, for time first we need acceleration

$(V_{f})^{2}=(V_{o})^{2}+2aS\\ So\\a=\frac{(V_{f})^{2}-(V_{o})^{2} }{2S}\\ a=\frac{(24m/s)^{2}-(0m/s)^{2} }{2(120)}\\a=2.4 m/s^{2}$

As we find acceleration.Now we need to find time

$V_{f}=V_{i}+at\\t=\frac{V_{f}-V_{i}}{a}\\t=\frac{(24m/s)-(0m/s)}{(2.4m/s^{2} )}\\t=10s$

Now for distance

$Distance=velocity*time\\Distance=(24m/s)*(10s)\\Distance=240m$

7 0

3 years ago

Three equal 1.55-μC point charges are placed at the corners of an equilateral triangle whose sides are 0.500 m long. What is the

kati45 [8]

Answer:

0.12959085 J

Explanation:

k = Coulomb constant = $8.99\times 10^{9}\ Nm^2/C^2$

q = Charge = 1.55 μC

d = Distance between charge = 0.5 m

Electric potential energy is given by

$U=k\dfrac{q^2}{d}$

In this system with three charges which are equidistant from each other

$U=k\dfrac{q^2}{d}+k\dfrac{q^2}{d}+k\dfrac{q^2}{d}$

$\\\Rightarrow U=k\dfrac{3q^2}{d}\\\Rightarrow U=8.99\times 10^9\times \dfrac{3\times (1.55\times 10^{-6})^2}{0.5}\\\Rightarrow U=0.12959085\ J$

The potential energy of the system is 0.12959085 J

6 0

3 years ago

Difference between sold and liquid

Tcecarenko [31]

Explanation:

Solids have closely packed particles and vibrate about a fixed position, they also have a fixed volume.

liquid have close particles but which are able to move with a bit of kinetic energy, for this reason they have no fixed volume but take the volume of the container or vessel

4 0

2 years ago