Number 10. But I would like all of them so I can double check. Thank you.

A horizontal wire 1 m long carries 0.4 A and is oriented 30° N of W. The Earth's magnetic field runs due north at this location

alexira [117]

Answer:

Explanation:

Force on a current carrying conductor in a magnetic field is given by the following expression

F = B i L where B is magnetic field perpendicular to wire or current , i is current and L is length of the wire.

Magnetic field of 1.5 x 10⁻⁵ T is making an angle of 60 degree with the wire so the component of field perpendicular to it

B = 1.5 X 10⁻⁵ Cos 30° = 1.3 x 10⁻⁵ T.

Force = BiL

= 1.3 X 10⁻⁵ X .4 X 1

= 5.2 X 10⁻⁶ N.

The direction of force can be found out from Fleming's left hand rule . It will be along downward direction.

5 0

4 years ago

You put a room temperature metal spoon into a bowl of hot soup. What will happen to

Semenov [28]

Answer:

the soup will transfer its thermal energy to the spoon

Hope this helps! Can you please mark me brainliest? (also can you guys answer my recent questions? thanks ^^ )

7 0

3 years ago

A 50kg hiker is standing on the edge of a cliff. Find the hiker’s gravitational potential energy if the cliff is 30m high.

Yuri [45]

Answer:

14700 J

Explanation:

The gravitational potential energy of an object is given by:

$U=mgh$

where

m is the mass

g is the gravitational acceleration

h is the height of the object above the ground

In this problem:

m = 50 kg is the mass of the hiker

$g=9.8 m/s^2$

h = 30 m is the heigth of the cliff

Substituting, we find

$U=(50)(9.8)(30)=14700 J$

3 0

4 years ago

Two moles of an ideal monatomic gas are at a temperature of 345 K. Then, 2250 J of heat is added to the gas, and 870 J of work i

Lapatulllka [165]

Answer: The final temperature is 470K

Explanation: Using the relation;

Q= ΔU +W

Given, n = 2mol

Initial temperature T1= 345K

Heat =Q= 2250J

Workdone=W=-870J(work is done on gas)

T2 =Final temperature =?

ΔU =3/2nR(T2-T1)

ΔU=3/2 × 2 ×8.314 (T2 - 345)

ΔU=24.942(T2-345)

Therefore Q = 24.942(T2-345)+ (-870)

2250=24.942(T2-345)+ (-870)

125.09=(T2-345)

T2 =470K

Therfore the final temperature is 470K

5 0

3 years ago

You have a 192 −Ω−Ω resistor, a 0.409 −H−H inductor, a 4.95 −μF−μF capacitor, and a variable-frequency ac source with an amplitu

Elodia [21]

Answer:

Explanation:

Given that,

The resistance of the resistor is

R = 192 Ω

The inductance of the inductor is

L = 0.409 H

The capacitance of the capacitor is

C = 4.95 μF

a. At what frequency is current max?

The current will be maximum at resonance, and resonance frequency is given as

f = 1/2π√L•C

f = 1/2π√(0.409×4.95×10^-6)

f = 111.86Hz

b. Currency at frequency 400rad/s.

w = 400rad/s

So, we need to find the inductive reactance

XL = wL

XL = 400 ×0.409

XL = 163.6 ohms

We also need the capacitive reacttance

XC = 1/wC

XC = 1/400×4.95×10^-6

XC = 505.05 ohms

Then, the impedance of the circuit is given as

Z = √(R²+(XL-XC)²)

Z = √(192²+(505.05—163.6)²)

Z = √(192²+341.9²)

Z = 392.12 ohms

Then, the current that flows can be calculated using

V= IZ

I = V/Z

I = 3.02/392.12

I = 7.7 × 10^-3 A

I = 7.7 mA

Since Xc>XL, then, the source voltage lags the current

3 0

4 years ago