When you run outside you transform what energy into what energy

One jet is flying east at 880 km/h,and another is traveling north at 880 km/h.Do they have the same velocity?the same speed?Expl

shusha [124]

Velocity is the speed and direction combined.
So, the two jets both are going at the same speed.
But they are going in different directions, so their velocities are different.

6 0

3 years ago

Problem #1- In the first scene in which Miss Clark appears, she is supposed to be wearing a BLACK dress.

katrin [286]

Answer:

Blue Lighting

Explanation:

In order to make red look black, you must use blue light. The blue would be absorbed and there would be no red light to reflect.

6 0

3 years ago

Read 2 more answers

A car accelerates from rest to a velocity of 5 meters/second in 4 seconds. What is its average acceleration over this period of

disa [49]

The average acceleration is

$\bar a=\dfrac{5\,\frac{\mathrm m}{\mathrm s}-0\,\frac{\mathrm m}{\mathrm s}}{4\,\mathrm s}=1.25\,\dfrac{\mathrm m}{\mathrm s^2}$

5 0

3 years ago

Read 2 more answers

Match the change of state of matter to the term:

Sphinxa [80]

Sup Milk,

Sublimation = Energy is absorbed and a solid turns to a gas.

Condensation = Energy is released and a gas changes to a liquid.

Evaporation = Energy is absorbed into a liquid to turn it into a gas.

6 0

3 years ago

Read 2 more answers

Two long, parallel wires carry unequal currents in the same direction. The ratio of the currents is 3 to 1. The magnitude of the

astraxan [27]

Answer:

3A is the larger of the two currents.

Explanation:

Let the currents in the two wires be I₁ and I₂

given:

Magnitude of the electric field, B = 4.0μT = 4.0×10⁻⁶T

Distance, R = 10cm = 0.1m

Ratio of the current = I₁ : I₂ = 3 : 1

Now, the magnitude of a magnetic field at a distance 'R' due to the current 'I' is given as

$B = \frac{\mu_oI}{2\pi R}$

Where $\mu_o$ is the magnitude constant = 4π×10⁻⁷ H/m

Thus, the magnitude of a magnetic field due to I₁ will be

$B_1 = \frac{\mu_oI_1}{2\pi R}$

$B_2 = \frac{\mu_oI_2}{2\pi R}$

given,

B = B₁ - B₂ (since both the currents are in the same direction and parallel)

substituting the values of B, B₁ and B₂

we get

4.0×10⁻⁶T = $\frac{\mu_oI_1}{2\pi R}$ - $\frac{\mu_oI_2}{2\pi R}$

or

4.0×10⁻⁶T = $\frac{\mu_o}{2\pi R}\times (I_1-I_2 )$

also

$\frac{I_1}{I_2} = \frac{3}{1}$

⇒ $I_1 = 3\times I_2$

substituting the values in the above equation we get

4.0×10⁻⁶T = $\frac{4\pi\times 10^{-7}}{2\pi 0.1}\times (3 I_2-I_2)$

⇒ $I_2 = 1A$

also

$I_1 = 3\times I_2$

⇒ $I_1 = 3\times 1A$

⇒ $I_1 = 3A$

Hence, the larger of the two currents is 3A

3 0

3 years ago