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JulsSmile [24]
3 years ago
5

If we rub a rod and a fabric made of the same material, would the rod become charged? Please explain.

Physics
1 answer:
vfiekz [6]3 years ago
3 0
If you rub a glass rod with a cloth, the charge produced on the glass rod is called positive, whereas when a plastic rod is rubbed with a cloth, the charge produced on the plastic rod is negative.
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Which skateboarder has greater momentum?
Verizon [17]

Answer:

skateboard b

Explanation:

p=mv

skateboard a

p=(60kg)(1.5m/s)=90kg*m/s

skateboard b

p=(50kg)(2m/s)=100kg*m/s

5 0
3 years ago
Water is flowing in a pipe with a circular cross section but with varying cross-sectional area, and at all points the water comp
slamgirl [31]

(a) 5.66 m/s

The flow rate of the water in the pipe is given by

Q=Av

where

Q is the flow rate

A is the cross-sectional area of the pipe

v is the speed of the water

Here we have

Q=1.20 m^3/s

the radius of the pipe is

r = 0.260 m

So the cross-sectional area is

A=\pi r^2 = \pi (0.260 m)^2=0.212 m^2

So we can re-arrange the equation to find the speed of the water:

v=\frac{Q}{A}=\frac{1.20 m^3/s}{0.212 m^2}=5.66 m/s

(b) 0.326 m

The flow rate along the pipe is conserved, so we can write:

Q_1 = Q_2\\A_1 v_1 = A_2 v_2

where we have

A_1 = 0.212 m^2\\v_1 = 5.66 m/s\\v_2 = 3.60 m/s

and where A_2 is the cross-sectional area of the pipe at the second point.

Solving for A2,

A_2 = \frac{A_1 v_1}{v_2}=\frac{(0.212 m^2)(5.66 m/s)}{3.60 m/s}=0.333 m^2

And finally we can find the radius of the pipe at that point:

A_2 = \pi r_2^2\\r_2 = \sqrt{\frac{A_2}{\pi}}=\sqrt{\frac{0.333 m^2}{\pi}}=0.326 m

6 0
3 years ago
Which of the following is an example of a physical change?
Stels [109]

Answer:

1 question is- I believe 4. Melting Question 2 is Density

Explanation:

THOUGHT ABOUT IT!!!

7 0
4 years ago
A 0.55-μF capacitor is connected to a 3.5-V battery. How much charge is on each plate of the capacitor?
yan [13]

Answer:

1.925 μC

Explanation:

Charge: This can be defined as the product of the capacitance of a capacitor and the voltage. The S.I unit of charge is Coulombs (C)

The formula for the charge stored in a capacitor is given as,

Q = CV ................... Equation 1

Where Q = charge, C = Capacitor, V = Voltage.

Note: 1 μF  = 10⁻⁶  F

Given: C = 0.55 μF = 0.55×10⁻⁶ F, V = 3.5 V.

Substitute into equation 1

Q = 0.55×10⁻⁶×3.5

Q = 1.925×10⁻⁶ C.

Q = 1.925 μC

Hence the charge on the plate = 1.925 μC

6 0
4 years ago
Read 2 more answers
A magnetic field is perpendicular to the plane of a flat coil. Since the magnitude of the field is increasing, an emf will be in
Blababa [14]

Answer:

D) Reduce the time interval during which the magnitude of the field increases.

6 0
3 years ago
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