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PSYCHO15rus [73]
4 years ago
7

Which description includes a scalar quantity? The car is moving at 70 miles/hr. Objects accelerate on Earth at 9.8 m/s2 downward

. The plane is flying west at 850 miles/hr. The box moved 20 meters to the left.
Physics
2 answers:
Shtirlitz [24]4 years ago
5 0

Answer:

Car moving at 70 miles per hour

Explanation:

A scalar is a quantity that can be fully described by magnitude alone.

Examples of scalar quantities include: Distance and speed

Something is not a scalar quantity when it is given direction. When this happens, it becomes a vector quantity.

Thus, from the choices, we can see that the car moving at 70 miles per hour is the scalar quantity as it is the only choice without a direction.

Likurg_2 [28]4 years ago
5 0

Answer:

the car is moving at 70 mph

Explanation:

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Read 2 more answers
A solenoid 25.0 cmcm long and with a cross-sectional area of 0.550 cm^2 contains 460 turns of wire and carries a current of 90.0
ankoles [38]

Answer:

a.  B = 0.20T

b.  u = 17230.6 J/m³

c.  E = 0.236J

d.  L = 5.84*10^-5 H

Explanation:

a. In order to calculate the magnetic field in the solenoid you use the following formula:

B=\frac{\mu_o n i}{L}               (1)

μo: magnetic permeability of vacuum = 4π*10^-7 T/A

n: turns of the solenoid = 460

L: length of the solenoid = 25.0cm = 0.25m

i: current  = 90.0A

You replace the values of the parameters in the equation (1):

B=\frac{(4\pi*10^{-7}T/A)(460)(90.0A)}{0.25m}=0.20T

The magnetic field in the solenoid is 0.20T

b. The magnetic permeability of air is approximately equal to the magnetic permeability of vacuum. To calculate the energy density in the solenoid you use:

u=\frac{B^2}{2\mu_o}=\frac{(0.20T)^2}{2(4\pi*10^{-7}T/A)}=17230.6\frac{J}{m^3}

The energy density is 17230.6 J/m³

c. The total energy contained in the solenoid is:

E=uV           (2)

V is the volume of the solenoid and is calculated by assuming the solenoid as a perfect cylinder:

V=AL

A: cross-sectional area of the solenoid = 0.550 cm^2 = 5.5*10^-5m^2

V=(5.5*10^{-5}m^2)(0.25m)=1.375*10^{-5}m^3

Then, the energy contained in the solenoid is:

E=(17230.6J/m^3)(1.375*10^{-5}m^3)=0.236J

The energy contained is 0.236J

d. The inductance of the solenoid is calculated as follow:

L=\frac{\mu_o N^2 A}{L}=\frac{(4\pi*10^{-7}T/A)(460)^2(5.5*10^{-5}m^2)}{0.25m}\\\\L=5.84*10^{-5}H

The inductance of the solenoid is 5.84*10^-5 H

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4 years ago
1) the strength of an electromagnet can be increased by
miss Akunina [59]
Using coils of fewer turns on the electromagnet
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