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nadezda [96]
3 years ago
5

A 0.50-T magnetic field is directed perpendicular to the plane of a circular loop of radius 0.25 m. What is the magnitude of the

magnetic flux through the loop
Physics
1 answer:
Cloud [144]3 years ago
6 0

Answer:

The magnitude of the magnetic flux through the loop is 0.0982 T.m²

Explanation:

Given;

magnitude of magnetic field, B = 0.5 T

radius of the loop, r = 0.25 m

Area of the loop is given by;

A = πr²

A = 3.142 x (0.25)²

A = 0.1964 m²

The magnitude of the magnetic flux through the loop is given by;

Ф = BA

Where;

B is the magnitude of the magnetic field

A is area of the field

Ф = 0.5 x 0.1964

Ф = 0.0982 T.m²

Therefore, the magnitude of the magnetic flux through the loop is 0.0982 T.m²

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Which term best describes those processes that move weathered rock materials and soils downslope?
mote1985 [20]

Answer:

mass wasting

Explanation:

5 0
2 years ago
Four astronauts are in a spherical space station. (a) If, as is typical, each of them breathes about 500 cm
mario62 [17]

Answer:

A) if each astronaut breathes about 500 cm³, the total volume of air breathed in a year is 14716.8m³.

B) The Diameter of this spherical space station should be 30.4m

Explanation:

The breathing frequency (according to Rochester encyclopedia) is about 12-16 breath per minute. if we take the mean value (14 breath per minute), we can estimate the total breaths of a person along a year:

f_b=14\frac{br}{min} \cdot \frac{60min}{1hr} \frac{24hr}{1day}\frac{365day}{1year}=29433600\frac{br}{year}

If we multiply this for the number of people in the station and the volume each breath needs, we obtain the volume breathed in a year.

The volume of a sphere is:

V_{sph}=\frac{4\pi}{3}r^3

So the diameter is:

D=2r=2\sqrt[3]{\frac{3V_{sph}}{4\pi}} =30.4m

6 0
3 years ago
A cabbie is trying to stop when he notices a fare is whistling them over. The
liberstina [14]
  • K.E=18750J
  • Mass=m=2100kg
  • Velocity=v

\boxed{\sf K.E=\dfrac{1}{2}mv^2}

\\ \sf\longmapsto 18750=\dfrac{1}{2}2100v^2

\\ \sf\longmapsto 18750=1050v^2

\\ \sf\longmapsto v^2=\dfrac{18750}{1050}

\\ \sf\longmapsto v^2=17.85m^2

\\ \sf\longmapsto v=\sqrt{17.85}

\\ \sf\longmapsto v=4.1m/s

7 0
3 years ago
Two identical copper blocks are connected by a weightless, unstretchable cord through a frictionless pulley at the top of a thin
likoan [24]

Answer:

13.6 N

Explanation:

Since one side of the wedge is vertical and the wedge makes and angle of 33 with the horizontal, the angle between the weight of the copper block on the incline and the incline is thus 90 - 33 = 57.

Let M be the mass of the block that hangs, m be the mass of the block on the incline and T be the tension in the weightless unstretchable cord.

We assume the motion is downwards in the direction of the hanging block, M.

We now write equations of motion for each block.

So

Mg - T = Ma    (1) and T - mgcos57 - F = ma where F is the frictional force on the block on the incline and a is their acceleration.

Now, since both blocks do not move, a = 0.

So, Mg - T = M(0) = 0     and T - mgcos57 - F = m(0) = 0

Mg - T = 0    (3) and T - mgcos57 - F = 0 (4)

From (3), T = Mg

Substituting T into (4), we have

T - mgcos57 - F = 0

Mg - mgcos57 - F = 0

So, Mg - mgcos57 = F  

F = Mg - mgcos57

F = (M - mcos57)g

Since g = acceleration due to gravity = 9.8 m/s², and M = 2.94 kg and m = 2.85 kg.

We find F, thus

F = (2.94 kg - 2.85 kgcos57)9.8 m/s²

F = (2.94 kg - 2.85 kg × 0.5446)9.8 m/s²

F = (2.94 kg - 1.552 kg)9.8 m/s²

F = (1.388 kg)9.8 m/s²

F = 13.6024 kgm/s²

F ≅ 13.6 N

6 0
3 years ago
Students who embrace an innate mindset might view their poor performance in math as inevitable or beyond their control
PilotLPTM [1.2K]

Answer:

True

Explanation:

4 0
3 years ago
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