All categories

4 years ago

A car accelerates uniformly from rest at 2.2 m / s^2 for 3.0 s.Calculate the speed of the car at time t = 3.0 s

Physics

1 answer:

Alexxx [7]4 years ago

4 0

Answer:

6.6m/s

Explanation:

we know that

v = u + at

= 0+ 2.2*3

=6.6m/s

You might be interested in

A turntable that spins at a constant 80.0 rpmrpm takes 3.50 ss to reach this angular speed after it is turned on. Find its angul

Veronika [31]

Answer:

The angular acceleration is 2.39 rad/s².

The number of degrees it rotates is 841.68 degrees.

Explanation:

Given:

Initial angular speed (ω₀) = 0 rad/s

Final angular speed in rpm (N) = 80.0 rpm

Time taken (t) = 3.50 s

First, let us determine the final angular speed in radians per second.

We know that,

$\omega=\frac{2\pi N}{60}\ rad/s$

Plug in the values and find the final angular speed, 'ω'. This gives,

$\omega=\frac{2\pi\times 80.0}{60}=8.38\ rad/s$

Now, using equation of motion for rotational motion, we have:

$\omega=\omega_0+\alpha t\\\\\alpha\to angular\ acceleration$

Plug in the given values and solve for α. This gives,

$8.38=0+\alpha \times 3.50\\\\\alpha=\frac{8.38}{3.50}=2.39\ rad/s^2$

Therefore, the angular acceleration is 2.39 rad/s².

Now, again using rotational equation of motion relating angular displacement, we have:

$\omega^2=\omega_0^2+2\alpha\theta\\\\\theta=\frac{\omega^2-\omega_0^2}{2\alpha }$

Plug in the given values and solve for 'θ'. This gives,

$\theta=\frac{(8.38)^2-0}{2\times 2.39}\\\\\theta=\frac{70.2244}{4.78}=14.69\ rad$

Convert radians to degrees using the conversion factor. This gives,

π radians = 180°

So, 1 radian =( 180 ÷ π ) degrees

Therefore, $14.69\ rad=14.69\times (\frac{180}{\pi})=841.68^\circ$

So, the number of degrees it rotates is 841.68 degrees.

3 0

3 years ago

Why does a tail appear when a comet nears the sun?

ahrayia [7]

It heats up. the ice transforms directly from a solid to a vapor, releasing dust particles. the solar wind sweeps the material and it forms what to appears to be a tail.

3 0

4 years ago

A car moves at speed v across a bridge made in the shape of a circular arc of radius r. (a) Find an expression for the normal fo

inna [77]

Answer:

(a) FN = m (g - $\frac{v^{2} }{r}$ )

(b) vmin = 17.146 m/s

Explanation:

The radius of the arc is

r = 30m

The normal force acting on the car form the highest point is

FN = m (g - $\frac{v^{2} }{r}$ )

If the normal force become 0 we have

m (g - $\frac{v^{2} }{r}$ ) = 0

g - $\frac{v^{2} }{r}$ = 0

This way, when FN = 0, then v = vmin, so

g - $\frac{vmin^{2} }{r}$ = 0

vmin = $\sqrt[.]{g*r}$ = $\sqrt[.]{9.8 m/s^{2} * 30m } = 17.146 m/s$

4 0

4 years ago

Of the planets with atmospheres, which is the warmest? a. Venusb. Earthc. Marsd. Jupiter

tangare [24]

The Atmosphere in Jupiter is full of gases that move at high speeds in giant eddies. Its atmosphere consists mostly of gases such as hydrogen that generate a temperature fluctuation of around 128K.

On Earth, due to the protection of the Ozone Layer and the presence of Nitrogen and Oxygen, the temperature fluctuates by an average of 300K.

In the case of Mars, its atmosphere is thin, mostly composed of Carbon Dioxide and Diatomic Nitrogen, which allow a temperature oscillation of 210K.

In contrast, the atmosphere of Venus is thick and is composed of carbon dioxide that does not allow the sun's rays to escape, generating an extreme 'greenhouse effect' with temperatures ranging from 737K,

Correct Answer is A.

7 0

4 years ago

NASA has asked your team of rocket scientists about the feasibility of a new satellite launcher that will save rocket fuel. NASA

kkurt [141]

Answer:

The answer is " $q=0.0945\,C$ ".

Explanation:

Its minimum velocity energy is provided whenever the satellite(charge 4 q) becomes 15 m far below the square center generated by the electrode (charge q).

$U_i=\frac{1}{4\pi\epsilon_0} \times \frac{4\times4q^2}{\sqrt{(15)^2+(5/\sqrt2)^2}}$

It's ultimate energy capacity whenever the satellite is now in the middle of the electric squares:

$U_f=\frac{1}{4\pi\epsilon_0}\ \times \frac{4\times4q^2}{( \frac{5}{\sqrt{2}})}$

Potential energy shifts:

$= U_f -U_i \\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{\sqrt{(15)^2+( \frac{5}{\sqrt{2})^2)}}\right ) \\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ 15 +( \frac{5}{2})}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ (\frac{30+5}{2})}}\right )\\\\$

$=\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ (\frac{35}{2})}}\right )\\\\=\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{17.5}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{ 24.74- 5 }{87.5}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{ 19.74- 5 }{87.5}}\right )\\\\ =\frac{4q^2}{\pi\epsilon_0}\left ( 0.2256 }\right )\\\\= \frac{0.28 \times q^2}{ \epsilon_0}\\\\=q^2\times31.35 \times10^9\,J$

Now that's the energy necessary to lift a satellite of 100 kg to 300 km across the surface of the earth.

$=\frac{GMm}{R}-\frac{GMm}{R+h} \\\\=(6.67\times10^{-11}\times6.0\times10^{24}\times100)\left(\frac{1}{6400\times1000}-\frac{1}{6700\times1000} \right ) \\\\ =(6.67\times10^{-11}\times6.0\times10^{26})\left(\frac{1}{64\times10^{5}}-\frac{1}{67\times10^{5}} \right ) \\\\=(6.67\times6.0\times10^{15})\left(\frac{67 \times 10^{5} - 64 \times 10^{5} }{ 4,228 \times10^{5}} \right ) \\\\$

$=( 40.02\times10^{15})\left(\frac{3 \times 10^{5}}{ 4,228 \times10^{5}} \right ) \\\\ =40.02 \times10^{15} \times 0.0007 \\\\$

$\\\\ =0.02799\times10^{10}\,J \\\\= q^2\times31.35\times10^{9} \\\\ =0.02799\times10^{10} \\\\q=0.0945\,C$

This satellite is transmitted by it system at a height of 300 km and not in orbit, any other mechanism is required to bring the satellite into space.

6 0

3 years ago