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3 years ago

Plz help me with this activity :(

Physics

1 answer:

Marrrta [24]3 years ago

8 0

As mass increases, the potential energy also increases, I hope that helped :)

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A sphere has a surface uniformly charged with 3.30 C. At what distance from its center is the potential 4.50 MV

Bogdan [553]

Answer:

r = 6.6 x 10³ m = 6600 m

Explanation:

The potential at a distance from a charged sphere can be given as follows:

$V = \frac{kq}{r}\\\\r = \frac{kq}{V}$

where,

r = distance = ?

k = Colomb Constant = 9 x 10⁹ Nm²/C²

q = charge on sphere = 3.3 C

V = potential = 4.5 MV = 4.5 x 10⁶ V

Therefore,

$r = \frac{(9\ x\ 10^9\ Nm^2/C^2)(3.3\ C)}{4.5\ x\ 10^6\ V}$

6 0

3 years ago

What is the frequency of a tuning fork that vibrates 518 times in 3.50 s?

Serga [27]

Given:

The number of vibrations of the tuning fork is,

$51\text{8 times}$

in,

$t=3.50\text{ s}$

To find:

The frequency of the fork

Explanation:

The frequency is the number of vibrations in 1 second

The frequency for the given fork is,

$\begin{gathered} \frac{The\text{ total vibrations}}{The\text{ time}} \\ =\frac{518}{3.50} \\ =148\text{ Hz} \end{gathered}$

Hence, the frequency is 148 Hz.

7 0

1 year ago

in 10 minutes a heart can beat 700 times at this rate in how many minutes will a heart beat 140 times at what rate can a heart b

Andru [333]

Answer: Heart can beat 140 times in = 2 minutes

3 0

3 years ago

During a very quick stop, a car decelerates at 7.00 m/s2 . (a) What is the angular acceleration of its 0.280-m-radius tires, ass

love history [14]

Answer:

-25 rad/s²

29 times

3.8 seconds

50.54 m

26.6 m/s

Explanation:

$\omega_f$ = Final angular velocity

$\omega_i$ = Initial angular velocity

$\alpha$ = Angular acceleration

$\theta$ = Angle of rotation

a = Acceleration = -7 m/s² (negative because of deceleration)

r = Radius of wheel = 0.28 m

Angular acceleration is given by

$\alpha=\frac{a}{r}\\\Rightarrow \alpha=\frac{-7}{0.28}\\\Rightarrow \alpha=-25\ rad/s^$

Angular acceleration of the wheel is -25 rad/s²

$\omega_f=\omega_i+\alpha t\\\Rightarrow t=\frac{\omega_f-\omega_i}{\alpha}\\\Rightarrow t=\frac{0-95}{-25}\\\Rightarrow t=3.8\ s$

It took 3.8 seconds for the car to stop

$\theta=\omega_it+\frac{1}{2}\alpha t^2\\\Rightarrow \theta=95\times 3.8+\frac{1}{2}\times -25\times 3.8^2\\\Rightarrow \theta=180.5\ rad$

$\frac{180.5}{2\pi}=28.72746\ rev$

The wheel rotated 29 times.

The

Initial velocity

$u=r\omega_i\\\Rightarrow u=0.28\times 95\\\Rightarrow u=26.6\ m/s$

The car’s initial velocity is 26.6 m/s

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=26.6\times 3.8+\frac{1}{2}\times -7\times 3.8^2\\\Rightarrow s=50.54\ m$

Distance covered in the time is 50.54 m

If the distance is found in meters we get 50.54 m which is very low. Considering the initial velocity is 95.76 km/h. Coming to a stop in that distance is very low. So, the values do not seem reasonable. At least a 100 m is required to stop from that speed.

3 0

3 years ago

What is m=35g, v=17cm³

shepuryov [24]

The density is $2.06 g/cm^3$