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2 years ago

What is the time required for an object starting from rest.

Physics

1 answer:

sashaice [31]2 years ago

6 0

Time required for what?

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What is the density of iron if 5.0 cm3 has a mass of 39.5g

weqwewe [10]

Answer : $\rho = 7.9\ g/cm^3$

Explanation :

It is given that

Mass of iron, m = 39.5 g

Volume of iron, $V = 5\ cm^3$

So, density is :

density, $\rho =\dfrac{mass}{volume}$

$\rho =\dfrac{39.5g}{5cm^3}$

$\rho = 7.9\ g/cm^3$

7 0

4 years ago

Convert 525 pounds to kilograms using dimensional analysis setup

kobusy [5.1K]

You can do it I believe in you

6 0

4 years ago

A wave has a wavelength of 14 m and a frequency of 24 Hz, what is the speed of the wave?

lapo4ka [179]

Answer:

Explanation: Speed = Wavelength x Wave Frequency. In this equation, wavelength is measured in meters and frequency is measured in hertz (Hz), .

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3 years ago

Starting from zero, an electric current is established in a circuit made of a battery of emf E, a resistor of resistance R and a

igor_vitrenko [27]

Answer:

time constant will decrease and steady state current will decrease on increasing the resistance

Explanation:

As we know that the EMF of cell is E which is used to connected across a resistor and an inductor.

So we will have

$E - iR - L\frac{di}{dt} = 0$

here we know that

$i = \frac{E}{R}(1 - e^{-Rt/L})$

now here we have

$\tau = \frac{L}{R}$

so if we increase the value of resistance of the wire then the time constant will decrease

and hence it will take less time to reach near the steady state value

also the steady state current will be smaller in that case

8 0

3 years ago

the acceleration of a rocket traveling upward is given by a=(6+.02s)m/s^s, where s is in meters. Determine the rocket's velocity

NikAS [45]

Answer:

a) velocity v = 322.5m/s

b) time t = 19.27s

Explanation:

Note that;

ads = vdv

where

a is acceleration

s is distance

v is velocity

Given;

a = 6 + 0.02s

so,

$\int\limits^s_0 {a} \, ds = \int\limits^v_0 {v} \, dv\\ \int\limits^s_0 {6+0.02s} \, ds = \int\limits^v_0 {v} \, dv\\ 6s + \frac{0.02s^{2} }{2} = \frac{1}{2} v^{2} \\v = \sqrt{12s + 0.02s^{2} } .....................1 \\\\\\$

Remember that

$v = \frac{ds}{dt} \\\frac{ds}{v} = dt\\\int\limits^s_0 {\frac{ds}{\sqrt{12s+0.02s^{2} } } } \, ds = \int\limits^t_0 {} \, dt \\t= (5\sqrt{2} ) ln \frac{| [s + 300 + \sqrt{(s^{2} + 600s)} ] |}{300} .......2$

substituting s = 2km =2000m, into equation 1

v = 322.5m/s

substituting s = 2000m into equation 2

t = 19.27s

8 0

4 years ago