1 question, brainliest for whoever gets it right.

. A magnetic field has a magnitude of 0.078 T and is uniform over a circular surface whose radius is 0.10 m. The field is orient

Dmitriy789 [7]

Answer:

The magnetic flux through surface is $2.22 \times 10^{-3}$ Wb

Explanation:

Given :

Magnitude of magnetic field $B = 0.078$ T

Radius of circle $r = 0.10$ m

Angle between field and surface normal $\theta =$ 25°

From the formula of flux,

$\phi = B.A$

$\phi = BA\cos \theta$

Where $\theta =$ angle between magnetic field line and surface normal, $A =$ area of circular surface.

$A = \pi r^{2}$

$A = 3.14 \times (0.10) ^{2}$

$A = 0.0314$ $m^{2}$

Magnetic flux is given by,

$\phi = 0.078 \times 0.0314 \times \cos 25$

$\phi = 2.22 \times 10^{-3}$ Wb

Therefore, the magnetic flux through surface is $2.22 \times 10^{-3}$ Wb

6 0

4 years ago

A penguin slides at a constant velocity of 3.57 m/s down an icy incline. The incline slopes above the horizontal at an angle of

igomit [66]

Answer:t=0.81 s

Explanation:

Given

Penguin slides down with constant velocity of 3.57 m/s

as the Penguin Slides with constant velocity therefore $F_{net}$ is zero on Penguin

$F_{net}=mg\sin \theta -f_r$

$f_r=$ friction Force

$f_r=\mu mg$

$\mu =$ coefficient of Kinetic friction

$mg\sin \theta =\mu mg$

$\tan \theta =\mu$

$\mu =\tan 9.85=0.45$

after reaching on floor final velocity of penguin will be zero after time t

thus

$v=u+at$

here $a=\mu g$

$a=0.45\times 9.8=4.41$ (deceleration)

$0=3.57-4.41\times t$

$t=\frac{3.57}{4.41}=0.809$

$t=0.81 s$

8 0

3 years ago

A rifle fires a 2.01 10-2-kg pellet straight upward, because the pellet rests on a compressed spring that is released when the t

Zina [86]

Answer:

The value of spring constant is 266.01 $\frac{N}{m}$

Explanation:

Given:

Mass of pellet $m = 2.01 \times 10^{-2}$ kg

Height difference of pellet rise $h_{f} - h_{o} = 6.03$ m

Spring compression $x = 9.45 \times 10^{-2}$ m

From energy conservation law,

Spring potential energy is stored into potential energy,

$mg(h_{f} -h_{o}) = \frac{1}{2} kx^{2}$

Where $k =$ spring constant, $g = 9.8 \frac{m}{s^{2} }$

$k = \frac{2mg(h_{f} -h_{o} )}{x^{2} }$

$k = \frac{2 \times 9.8 \times 6.03\times 2.01 \times 10^{-2} }{(9.45\times 10^{-2} )^{2} }$

$k = 266.01$ $\frac{N}{m}$

Therefore, the value of spring constant is 266.01 $\frac{N}{m}$

6 0

3 years ago

Convert 90 oF to Celcius, (write your answer up to 1st decimal point)

Mice21 [21]

Answer:

The answer is 32.2

Explanation:

Hope this helps

5 0

3 years ago

Read 2 more answers

A police car is traveling north on a straight road at a constant 16.0 m/s. An SUV traveling north at 30.0 m/s passes the police

Nastasia [14]

Answer:

It will take 15.55s for the police car to pass the SUV

Explanation:

We first have to establish that both the police car and the SUV will travel the same distance in the same amount of time. The police car is moving at constant velocity and the SUV is experiencing a deceleration. Thus we will use two distance fromulas (for constant and accelerated motions) with the same variable for t and x:

1. $x=x_{0}+vt$

2. $x=x_{0}+v_{0}t+\frac{at^{2}}{2}$

Since both cars will travel the same distance x, we can equal both formulas and solve for t:

$vt = v_{0}t+\frac{at^2}{2}\\\\ 16\frac{m}{s}t =30\frac{m}{s}t-\frac{1.8\frac{m}{s^{2}} t^{2}}{2}$

We simplify the fraction present and rearrange for our formula so that it equals 0:

$0.9\frac{m}{s^{2}} t^{2}-14\frac{m}{s}t=0 \\\\ t(0.9\frac{m}{s^{2}}t-14\frac{m}{s})=0$

In the very last step we factored a common factor t. There is two possible solutions to the equation at $t=0$ and:

$0.9\frac{m}{s^{2}}t-14\frac{m}{s}=0 \\\\ 0.9\frac{m}{s^{2}}t =14\frac{m}{s} \\\\ t =\frac{14\frac{m}{s}}{0.9\frac{m}{s^{2}}}=15.56s$

What this means is that during the displacement of the police car and SUV, there will be two moments in time where they will be next to each other; at $t=0 s$ (when the SUV passed the police car) and $t=15.56s$ (when the police car catches up to the SUV)

8 0

3 years ago