How can you identify a moveable pulley?

If a neutral metal comb is held near an object with a negative charge, the Como become charged by

LenKa [72]

The local charging of the comb is due to induction. The negative object's charge pushes electrons in the comb away from the side close to the charged object causing that part to be positively charged. Note that the comb's net charge is still zero provided it doesn't touch the object. When you move the comb away its electrons will redistribute.

4 0

3 years ago

Read 2 more answers

The distance between two planets is 1600 km. How much time would the light

Snowcat [4.5K]

Answer:

5.33*10^-3 seconds

Explanation:

c = d/t

c = speed of light constant (3.0*10^5 km/s)

d = distance (1600 km)

t = ?

3.0*10^5 = 1600/t

t = 1600/3.0*10^5

t = 5.33*10^-3 seconds

I hope this helped! :)

6 0

3 years ago

A 10-turn coil of wire having a diameter of 1.0 cm and a resistance of 0.50 Ω is in a 1.0 mT magnetic field, with the coil orien

n200080 [17]

Answer:

The voltage across the capacitor is 1.57 V.

Explanation:

Given that,

Number of turns = 10

Diameter = 1.0 cm

Resistance = 0.50 Ω

Capacitor = 1.0μ F

Magnetic field = 1.0 mT

We need to calculate the flux

Using formula of flux

$\phi=NBA$

Put the value into the formula

$\phi=10\times1.0\times10^{-3}\times\pi\times(0.5\times10^{-2})^2$

$\phi=7.85\times10^{-7}\ Tm^2$

We need to calculate the induced emf

Using formula of induced emf

$\epsilon=\dfrac{d\phi}{dt}$

Put the value into the formula

$\epsilon=\dfrac{7.85\times10^{-7}}{dt}$

Put the value of emf from ohm's law

$\epsilon =IR$

$IR=\dfrac{7.85\times10^{-7}}{dt}$

$Idt=\dfrac{7.85\times10^{-7}}{R}$

$Idt=\dfrac{7.85\times10^{-7}}{0.50}$

$Idt=0.00000157=1.57\times10^{-6}\ C$

We know that,

$Idt=dq$

$dq=1.57\times10^{-6}\ C$

We need to calculate the voltage across the capacitor

Using formula of charge

$dq=C dV$

$dV=\dfrac{dq}{C}$

Put the value into the formula

$dV=\dfrac{1.57\times10^{-6}}{1.0\times10^{-6}}$

$dV=1.57\ V$

Hence, The voltage across the capacitor is 1.57 V.

5 0

3 years ago

A frog jumps for 4.0 seconds at a maximum horizontal distance of 0.8m. what is its velocity along the road?

Lerok [7]

Answer:

The frog's horizontal velocity is 0.2 m/s.

Explanation:

To solve this problem, we must first remember what velocity is and how we solve for it. Velocity can be solved for using the formula x/t, where x represents horizontal distance and t represents time (in seconds), that it takes to travel this distance. If we plug in the given numbers for these variables and solve, we get the following:

v = x/t

v = 0.8m/4s

v = 0.2 m/s

Therefore, the correct answer is 0.2 m/s. We can verify that these units are correct because the formula calls for distance divided by time, so meters per second is a sensible answer.

Hope this helps!

3 0

2 years ago

rickey approaches third base. He dives head first, hitting the ground at 6.75 m/s and reaching the base at 5.91 m/s in 2.5 secon

Gekata [30.6K]

Answer:

15.825 m

Explanation:

t = Time taken = 2.5 s

u = Initial velocity = 6.75 m/s

v = Final velocity = 5.91 m/s

s = Displacement

a = Acceleration

Equation of motion

$v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow a=\frac{5.91-6.75}{2.5}\\\Rightarrow a=-0.336\ m/s^2$

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{5.91^2-6.75^2}{2\times -0.336}\\\Rightarrow s=15.825\ m$

The distance Rickey slides across the ground before touching the base is 15.825 m

4 0

3 years ago