Ask question

Ask question

All categories

3 years ago

11

The increase in volume of a material due to a temerature increased is called?

1 answer:

kakasveta [241]3 years ago

7 0

I believe the correct answer is expansion.

You might be interested in

An object traveling at a constant speed but with a changing direction is accelerating.

prohojiy [21]

Strange as it may seem, that's true. (choice 'a'.)

"Acceleration" doesn't mean "speeding up". It means ANY change in
the speed or direction of motion. So a car with the brakes applied
and slowing down, and a point on the rim of a bicycle wheel that's
turning at a constant rate, are both accelerating.

6 0

3 years ago

Read 2 more answers

Reaction rates are affected by reactant concentrations and temperature. this is accounted for by the ________.

Black_prince [1.1K]

<span>
Reaction rates are affected by reactant concentrations and temperature. this is accounted for by the c</span>ollision model.

-Hope this helps.

7 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

3 years ago

Use this table of a school bus during morning pickups to calculate its average speed between 0 h and 2.340 h.

Gelneren [198K]

The average speed between 0 h and 2.340 h is 6.97 Km/h

Average speed is defined as the total distance travelled divided by the total time taken to cover the distance.

$Average \: speed = \frac{total \: distance}{total \: time} \\ \\$

With the above formula, we can obtain the average speed between 0 h and 2.340 h as illustrated below:

Total time = 2.340 – 0 = 2.340 h
Total distance = 16.3 – 0 = 16.3 Km
Average speed =?

$Average \: speed = \frac{total \: distance}{total \: time} \\ \\Average \: speed = \frac{16.3}{2.340} \\ \\ Average \: speed = 6.97 \: Km/h \\ \\$

Learn more about average speed: brainly.com/question/24884027

8 0

2 years ago

E. weight = mass x gravitational field strength

allsm [11]

Answer:

e. weight = mass x gravitational field strength

7 0

2 years ago