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3 years ago

A car can accelerate from rest to 28 m/s and travels 280meters. How long does this acceleration take?

Physics

1 answer:

Reika [66]3 years ago

4 0

Answer:

$a=1.4m/s^2$

Explanation:

From the question we are told that

Velocity v=28m/s

Distance d=280

Generally the Newtons equation for motion is mathematically given as

Acceleration

$v^2=2as$

$a=\frac{V^2}{2s}$

$a=\frac{28^2}{2*280}$

$a=1.4m/s^2$

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In a certain clock, a pendulum of length L1 has a period T1 = 0.95s. The length of the pendulum

gulaghasi [49]

Answer:

Ratio of length will be $\frac{L_2}{L_1}=1.108$

Explanation:

We have given time period of the pendulum when length is $L_1$ is $T_1=0.95sec$

And when length is $L_2$ time period $T_2=1sec$

We know that time period is given by

$T=2\pi \sqrt{\frac{L}{g}}$

So $0.95=2\pi \sqrt{\frac{L_1}{g}}$ ----eqn 1

And $1=2\pi \sqrt{\frac{L_2}{g}}$ -------eqn 2

Dividing eqn 2 by eqn 1

$\frac{1}{0.95}=\sqrt{\frac{L_2}{L_1}}$

Squaring both side

$\frac{L_2}{L_1}=1.108$

8 0

3 years ago

Two carts connected by a 0.05 m spring hit a wall, compressing the spring to 0.02 m. The spring constant k is

vivado [14]

Answer:

0.045 J

Explanation:

From the question,

The elastic potential energy stored in a spring is given as,

E = 1/2ke²...................... Equation 1

Where E = elastic potential energy, k = spring constant, e = compression.

Given: k = 100 N/m, e = 0.05-0.02 = 0.03 m

Substitute these values into equation 1

E = 1/2(100)(0.03²)

E = 50(9×10⁻⁴)

E = 0.045 J

Hence the right option is 0.045 J

4 0

2 years ago

What is evaporation? Condensation? Drag the terms on the left to the appropriate blanks on the right to complete the sentences.

krok68 [10]

Answer:

Evaporation is the physical change in which a substance converts from its liquid state to its gaseous state. Condensation is the physical change in which a substance converts from its gaseous state to its liquid state.

Explanation:

Evaporation and condensation are opposite processes to each other. Evaporation changes a liquid to a gas and condensation is the reverse.

7 0

3 years ago

A sinusoidally oscillating current I ( t ) with an amplitude of 9.55 A and a frequency of 359 cycles per second is carried by a

UNO [17]

Answer:

$P_{avg} = 6.283*10^{-9} \ W$

Explanation:

Given that;

I₀ = 9.55 A

f = 359 cycles/s

b = 72.2 cm

c = 32.5 cm

a = 80.2 cm

Using the formula;

$\phi = \frac{\mu_o Ic }{2 \pi} In (\frac{b+a}{b})$

where;

$E= \frac{d \phi}{dt}$

$E = \frac{\mu_o}{2 \pi}c In (\frac{b+a}{a}) I_o \omega cos \omega t$

$E_{rms} = \frac { {\frac{\mu_o \ c}{2 \pi} In (\frac{b+a}{a}) I_o (2 \pi f)}}{\sqrt{2}}$

Replacing our values into above equation; we have:

$E_{rms} = \frac { {\frac{4 \pi*10^{-7}*0.325}{2 \pi} In (\frac{72.2+80.2}{80.2}) *9.55 (2 \pi *359)}}{\sqrt{2}}$

$E_{rms} = \frac {8.98909588*10^{-4} }{\sqrt{2}}$

$E_{rms} = 6.356*10^{-4} \ V$

Then the $P_{avg$ is calculated as:

$P_{avg} = \frac{E^2}{R}$

$P_{avg} = \frac{(6.356*10^{-4})^2}{64.3}$

$P_{avg} = 6.283*10^{-9} \ W$

6 0

3 years ago

Suppose you observe two stars and you know they have the same luminosity. If one star is twice as far away as the other, the mor

rosijanka [135]

Answer:

The farther star will appear 4 times fainter than the star that is near to the observer.

Explanation:

Since it is given that the luminosity of the 2 stars is same thus they radiate the same energy per unit time

Consider a spherical wave front of energy 'E' that leaves both the stars (Both radiate 'E' as they have same luminosity)

This Energy is spread over the whole surface area of sphere Thus when the wave front is at a distance 'r' the energy per unit surface area is given by

$e_{1}=\frac{E}{4\pi r^{2}}$

For the star that is twice away from the earth the distance is '2r' thus we will receive an energy given by

$e_{2}=\frac{E}{4\pi (2r)^{2}}=\frac{E}{8\pi r^{2}}=\frac{e_{1}}{4}$

Hence we sense it as 4 times fainter than the nearer star.

5 0

2 years ago