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2 years ago

What is the color of lava at its hottest? Help asap

Physics

1 answer:

DENIUS [597]2 years ago

5 0

Yellow !!!!!!!!!!!!!!!!

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Which statement about cellulose is true?

zlopas [31]

Is there any answers? Or is it asking you to choose?

6 0

3 years ago

Read 2 more answers

When point charges q = +8.4 uC and q2 = +5.6 uC are brought near each other, each experiences a repulsive force of magnitude 0.6

Stella [2.4K]

Answer:

it is separated by 80 cm distance

Explanation:

As per Coulombs law we know that force between two point charges is given by

$F = \frac{kq_1q_2}{r^2}$

here we know that

$q_1 = +8.4\mu C$

$q_2 = +5.6 \mu C$

force between two charges is given as

$F = 0.66 N$

now we have

$F = \frac{kq_1q_2}{r^2}$

$0.66 = \frac{(9\times 10^9)(8.4 \mu C)(5.6 \mu C)}{r^2}$

$r = 0.8 m$

so it is separated by 80 cm distance

3 0

3 years ago

Read 2 more answers

(b) A car of mass 3000 kg travels at a constant velocity of 5.0 m/s.

Tamiku [17]

Answer:

16.7 s

Explanation:

T= <u>Vf - Vo</u> a= <u>F</u>

a m

4,500 / 3000 = 1.5 (a)

30 - 5 / 1.5(a) = 16.7 s

4 0

3 years ago

S.I. Units Ampere is use for what?

lawyer [7]

The ampere (symbol: A) is the SI base unit of electric current equal to one coulomb per second.

The ampere is that constant current which, if maintained in two straight parallel conductors of infinite length, of negligible circular cross section, and placed 1 meter apart in vacuum, would produce between these conductors a force equal to 2 times 10–7 newton per meter of length.

Electric current is the time rate of change or displacement of electric charge.

One ampere represents the rate of 1 coulomb of charge per second.

The ampere is defined first (it is a base unit, along with the meter, the second, and the kilogram), without reference to the quantity of charge.

The unit of charge, the coulomb, is defined to be the amount of charge displaced by a one ampere current in the time of one second.

This is your answer friend. Hope it helps you.

3 0

3 years ago

a ball of mass 100g moving at a velocity of 100m/s collides with another ball of mass 400g moving at 50m/s in same direction, if

klio [65]

Answer:

Velocity of the two balls after collision: $60\; \rm m \cdot s^{-1}$ .

$100\; \rm J$ of kinetic energy would be lost.

Explanation:

<h3>Velocity</h3>

Because the question asked about energy, convert all units to standard units to keep the calculation simple:

Mass of the first ball: $100\; \rm g = 0.1\; \rm kg$ .
Mass of the second ball: $400\; \rm g = 0.4 \; \rm kg$ .

The two balls stick to each other after the collision. In other words, this collision is a perfectly inelastic collision. Kinetic energy will not be conserved. The velocity of the two balls after the collision can only be found using the conservation of momentum.

Assume that the system of the two balls is isolated. Thus, the sum of the momentum of the two balls will stay the same before and after the collision.

The momentum of an object of mass $m$ and velocity $v$ is: $p = m \cdot v$ .

Momentum of the two balls before collision:

First ball: $p = m \cdot v = 0.1\; \rm kg \times 100\; \rm m \cdot s^{-1} = 10\; \rm kg \cdot m \cdot s^{-1}$ .
Second ball: $p = m \cdot v = 0.4\; \rm kg \times 50\; \rm m \cdot s^{-1} = 20\; \rm kg \cdot m \cdot s^{-1}$ .
Sum: $10\; \rm kg \cdot m \cdot s^{-1} + 20 \; \rm kg \cdot m \cdot s^{-1} = 30 \; \rm kg \cdot m \cdot s^{-1}$ given that the two balls are moving in the same direction.

Based on the assumptions, the sum of the momentum of the two balls after collision should also be $30\; \rm kg \cdot m \cdot s^{-1}$ . The mass of the two balls, combined, is $0.1\; \rm kg + 0.4\; \rm kg = 0.5\; \rm kg$ . Let the velocity of the two balls after the collision $v\; \rm m \cdot s^{-1}$ . (There's only one velocity because the collision had sticked the two balls to each other.)

Momentum after the collision from $p = m \cdot v$ : $(0.5\, v)\; \rm kg \cdot m \cdot s^{-1$ .
Momentum after the collision from the conservation of momentum: $30\; \rm kg \cdot m \cdot s^{-1}$ .

These two values are supposed to describe the same quantity: the sum of the momentum of the two balls after the collision. They should be equal to each other. That gives the equation about $v$ :

$0.5\, v = 30$ .

$v = 60$ .

In other words, the velocity of the two balls right after the collision should be $60\; \rm m \cdot s^{-1}$ .

<h3>Kinetic Energy</h3>

The kinetic energy of an object of mass $m$ and velocity $v$ is $\displaystyle \frac{1}{2}\, m \cdot v^{2}$ .

Kinetic energy before the collision:

First ball: $\displaystyle \frac{1}{2} \, m \cdot v^2 = \frac{1}{2}\times 0.1\; \rm kg \times \left(100\; \rm m \cdot s^{-1}\right)^{2} = 500\; \rm J$ .
Second ball: $\displaystyle \frac{1}{2} \, m \cdot v^2 = \frac{1}{2}\times 0.4\; \rm kg \times \left(50\; \rm m \cdot s^{-1}\right)^{2} = 500\; \rm J$ .
Sum: $500\; \rm J + 500\; \rm J = 1000\; \rm J$ .

The two balls stick to each other after the collision. Therefore, consider them as a single object when calculating the sum of their kinetic energies.

Mass of the two balls, combined: $0.5\; \rm kg$ .
Velocity of the two balls right after the collision: $60\; \rm m\cdot s^{-1}$ .

Sum of the kinetic energies of the two balls right after the collision:

$\displaystyle \frac{1}{2} \, m \cdot v^{2} = \frac{1}{2}\times 0.5\; \rm kg \times \left(60\; \rm m \cdot s^{-1}\right)^2 = 900\; \rm J$ .

Therefore, $1000\; \rm J - 900\; \rm J = 100\; \rm J$ of kinetic energy would be lost during this collision.

7 0

3 years ago

What is the color of lava at its hottest￼? Help asap

What is the color of lava at its hottest? Help asap