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Artemon [7]
3 years ago
8

Q:Why does a sheet of sandpaper become warm when you rub it against a wooden board?

Physics
2 answers:
otez555 [7]3 years ago
7 0

Answer: D and friction

Explanation:

PLz brainliest

avanturin [10]3 years ago
4 0

Answer:

D. The mechanical energy of the moving sandpaper is converted to thermal energy

Explanation:

The friction between wood and sandpaper moves quickly and can warm up... the friction causing the heat.

You might be interested in
A piano wire with mass 2.60g and length 84.0 cm is stretched with a tension of 25.0 N. A wave with frequency 120.0 Hz and amplit
likoan [24]

Answer:

Power will be 0.2023 watt

And when amplitude is halved then power will be 0.0505 watt

Explanation:

We have given mass of the Piano wire m = 2.60 gram = 0.0026 kg

Length of wire l = 84 cm = 0.84 m

So mass density \mu =\frac{m}{l}=\frac{0.0026}{0.84}=0.0031kg/m

Tension in the wire T = 25 N

Frequency f = 120 Hz

So angular frequency \omega =2\pi f=2\times 3.14\times 120=753.6rad/sec

And amplitude A = 1.6 mm = 0.0016 m

We have to find the generated power

Power is given by P=\frac{1}{2}\sqrt{\mu T}\omega ^2A^2=\frac{1}{2}\times \sqrt{0.0031\times 25}\times 753.6^2\times 0.0016^2=0.2023watt

From the relation we can see that power P\ \propto\ A^2

So if amplitude is halved then power will be \frac{1}{4} times

So power will be equal to \frac{0.2023}{2}=0.0505watt

4 0
3 years ago
Two identical small metal spheres with q1 > 0 and |q1| > |q2| attract each other with a force of magnitude 72.1 mN when se
Brrunno [24]

1) +2.19\mu C

The electrostatic force between two charges is given by

F=k\frac{q_1 q_2}{r^2} (1)

where

k is the Coulomb's constant

q1, q2 are the two charges

r is the separation between the charges

When the two spheres are brought in contact with each other, the charge equally redistribute among the two spheres, such that each sphere will have a charge of

\frac{Q}{2}

where Q is the total charge between the two spheres.

So we can actually rewrite the force as

F=k\frac{(\frac{Q}{2})^2}{r^2}

And since we know that

r = 1.41 m (distance between the spheres)

F= 21.63 mN = 0.02163 N

(the sign is positive since the charges repel each other)

We can solve the equation for Q:

Q=2\sqrt{\frac{Fr^2}{k}}=2\sqrt{\frac{(0.02163)(1.41)^2}{8.98755\cdot 10^9}}}=4.37\cdot 10^{-6} C

So, the final charge on the sphere on the right is

\frac{Q}{2}=\frac{4.37\cdot 10^{-6} C}{2}=2.19\cdot 10^{-6}C=+2.19\mu C

2) q_1 = +6.70 \mu C

Now we know the total charge initially on the two spheres. Moreover, at the beginning we know that

F = -72.1 mN = -0.0721 N (we put a negative sign since the force is attractive, which means that the charges have opposite signs)

r = 1.41 m is the separation between the charges

And also,

q_2 = Q-q_1

So we can rewrite eq.(1) as

F=k \frac{q_1 (Q-q_1)}{r^2}

Solving for q1,

Fr^2=k (q_1 Q-q_1^2})\\kq_1^2 -kQ q_1 +Fr^2 = 0

Since Q=4.37\cdot 10^{-6} C, we can substituting all numbers into the equation:

8.98755\cdot 10^9 q_1^2 -3.93\cdot 10^4 q_1 -0.141 = 0

which gives two solutions:

q_1 = 6.70\cdot 10^{-6} C\\q_2 = -2.34\cdot 10^{-6} C

Which correspond to the values of the two charges. Therefore, the initial charge q1 on the first sphere is

q_1 = +6.70 \mu C

8 0
4 years ago
Two forces act on a 6.00-kg object. One of the forces is 10.0 N. If the object accelerates at 2.00 m/s2
liubo4ka [24]

Given :

Two forces act on a 6.00-kg object. One of the forces is 10.0 N.

Acceleration of object 2 m/s².

To Find :

The greatest possible magnitude of the other force.\

Solution :

Let, other force is f.

So, net force, F = 10 + f.

Now, acceleration is given by :

a=\dfrac{F}{mass}\\\\a= \dfrac{10+f}{6}\\\\\dfrac{10+f}{6}=2\\\\f = 12 - 10\\\\f = 2 \ N

Therefore, the greatest possible magnitude of the other force is 2 N.

Hence, this is the required solution.

7 0
3 years ago
What is the mechanical advantage of a wedge that is 2 inches at its widest part and has a sloped side with a length of 10 inches
Delicious77 [7]
The correct answer would be 5 hope this helps.
7 0
3 years ago
Read 2 more answers
In a given volume of whole blood, there are more white blood cells than red blood cells.
navik [9.2K]

Answer:

its false there are fewer white blood cells

Explanation:

8 0
3 years ago
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