The diagram shows a wave traveling through a medium. Identify point B.

A circular loop of radius 11.7 cm is placed in a uniform magnetic field. (a) If the field is directed perpendicular to the plane

Mama L [17]

Answer:

Magnetic field, B = 0.199 T

Explanation:

It is given that,

Radius of circular loop, r = 11.7 cm = 0.117 m

Magnetic flux through the loop, $\phi=8.6\times 10^{-3}\ T/m^2$

The magnetic flux linked through the loop is :

$\phi=B.A$

$\phi=BA\ cos\theta$

Here, $\theta=0$

$B=\dfrac{\phi}{A}$

or

$B=\dfrac{\phi}{\pi r^2}$

$B=\dfrac{8.6\times 10^{-3}}{\pi (0.117 )^2}$

B = 0.199 T

So, the strength of the magnetic field is 0.199 T. Hence, this is the required solution.

4 0

3 years ago

Consider a turnbuckle that has been tightened until the tension in wire AD is 350 N. Draw the FBD that is required to determine

Mars2501 [29]

Answer:

yes

Explanation:

yes

6 0

3 years ago

Gibbons move through the trees by swinging from successive handholds, as we have seen. To increase their speed, gibbons may brin

Sedbober [7]

By bring their legs close to their bodies, they are decreasing the length of the pendulum which help them move more quickly.

A pendulum is nothing but a body suspended from a fixed point so that it can swing back and forth under the influence of gravity.

Here in this case Gibbons are bringing their legs close to their bodies and reducing the length of the pendulum. Since as the length of the pendulum increases the speed of the movement will be reduced. By bringing their legs close to their bodies they are reducing the length and in turn their speed increase and they move quickly.

To learn more about pendulum, visit here

brainly.com/question/14759840

#SPJ4

5 0

2 years ago

What is wrong about the picture below

Oliga [24]

what even is that ??

4 0

3 years ago

when a 0.622kg basketbll hits the floor its velocit changes from 4.23m/s down to 3.85m/s up. if the averge force was 72.9N how m

scoray [572]

Answer:

Time, t = 3.2 ms

Explanation:

It is given that,

Mass of basketball, m = 0.622 kg

Initial velocity, u = 4.23 m/s

Final velocity, v = 3.85 m/s

Average force acting on the ball, F = 72.9 N

We need to find the time of contact of the ball with the floor. Let t is the time of contact. So,

$F=ma\\\\F=\dfrac{m(v-u)}{t}\\\\t=\dfrac{m(v-u)}{F}\\\\t=\dfrac{0.622\times (3.85-4.23)}{72.9}\\\\t=0.0032\ s\\\\\text{or}\\\\t=3.2\ ms$

So, the ball is in contact with the floor for 3.2 ms.

8 0

4 years ago