Please help, thank you!

If a cup of coffee is at 90°C and a person with a body temperature of 36'C touches it,how will heat flow between them

Arlecino [84]

Answer:

the heat always transfers from high temperature to low temperature body without aid of any external energy to this law the heat transfers from cup of coffee to the person body until both bodies reaches to the equilibrium state

Explanation:

5 0

2 years ago

A silver wire with resistivity 1.59 × 10-8 Ω-m carries a current density of 4.0 A/mm2, What is the magnitude of the electric fie

Nat2105 [25]

Answer:

Electric field, E = 0.064 V/m

Explanation:

It is given that,

Resistivity of silver wire, $\rho=1.59\times 10^{-8}\ \Omega-m$

Current density of the wire, $J=4\ A/mm^2=4\times 10^6\ A/m^2$

We need to find the magnitude of the electric field inside the wire. The relationship between electric field and the current density is given by :

$E=J\times \rho$

$E=4\times 10^6\times 1.59\times 10^{-8}$

E = 0.0636 V/m

or

E = 0.064 V/m

So, the magnitude of electric field inside the wire is 0.064 V/m. Hence, this is the required solution.

6 0

2 years ago

What will happen to this current if a magnet is brought near the cord? A. It will exert a force on the voltage. B. The electric

Andreyy89

Answer:

The correct answer is D)

Explanation:

When an electric magnet is brought near a cord with an electric current, the cord will most likely deflect away from the magnet because electric fields flowing through a wire generates its own magnetic field.

Cheers!

8 0

2 years ago

Help im starving... can u help me with homework

IrinaVladis [17]

Go eat food! Don't starve yourself because of homework! XD

7 0

2 years ago

Read 2 more answers

A glider moves along an air track with constant acceleration a. It is projected from the start of the track (x = 0 m) with an in

Pie

Answer:

vo = 0.175m/s

a = -0.040625 m/s^2

Explanation:

To solve this problem, you will need to use the equations for constant acceleration motion:

$x = \frac{1}{2}at^2 +v_ot+x_o \\v_f^2 - v_o^2 = 2(x-x_o)a$

In the first equation you relate final position with the time elapsed, in the second one, you relate final velocity at any given position. In both equations, you will have both the acceleration a and the initial velocity vo as variables. We can simplify with the information we have:

1. $x = \frac{1}{2}at^2 +v_ot+x_o\\0.1m = \frac{1}{2}a(8s)^2 +v_o(8s)+0m \\0.1 = 32a + 8v_o$

2. $v_f^2 - v_o^2 = 2(x-x_o)a\\(-0.15m/s)^2 - v_o^2 = 2(0.1m-0m)a\\0.0225 - v_o^2 = 0.2a\\a = \frac{0.0225 - v_o^2}{0.2} = 0.1125 - 5v_o^2$

Replacing in the first equation:

$0.1 = 32(0.1125 - 5v_o^2) + 8v_o\\0.1 = 3.6 - 160v_o^2 + 8v_o\\160v_o^2 - 8v_o - 3.5 = 0$

$v_0 = \frac{-(-8) +- \sqrt{(-8)^2 - 4(160)(-3.5)}}{2(160)} \\ v_o = 0.175 m/s | -0.125 m/s$

But as you are told that the ball was projected om the air track, it only makes sense for the velocity to be positive, otherwise it would have started moving outside the air track, so the real solution is 0.175m/s. Then, the acceleration would be:

$a = 0.1125 - 5v_o^2\\a = -0.040625 m/s^2$

3 0

3 years ago