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2 years ago

The Biot-Savart Law describes the relationship between which of the following? Select all that apply.

Physics

1 answer:

faltersainse [42]2 years ago

8 0

The correct answer is magnetic field, electric field, and charges.

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Water is moving at a velocity of 2.00 m/s through a hose with an Internal diameter of 1.60 cm. What is the volume flow rate at t

Leya [2.2K]

Answer:

15 m/s

Explanation:

3 0

2 years ago

It is known that birds can detect the earth's magnetic field, but the mechanism of how they do this is not known. It has been su

Lubov Fominskaja [6]

Answer:

A) 0.50 mV

Explanation:

In this problem, we can think the wings of the bird as a metal rod moving across a magnetic field. So, and emf will be induced into the wings of the bird, according to the formula:

$\epsilon = BvL sin \theta$

where

$B=5\cdot 10^{-5} T$ is the strength of the magnetic field

v = 13 m/s is the speed of the bird

L = 1.2 m is the wingspan of the bird

$\theta=40^{\circ}$ is the angle between the direction of motion and the direction of the magnetic field

Substituting numbers into the formula, we find

$\epsilon = (5.0\cdot 10^{-5} T)(13 m/s)(1.2 m) sin 40^{\circ}=0.00050 V = 0.50 mV$

8 0

3 years ago

Sheila uses a 45N force on her bowling ball across a 15m lane. What work did she do on the bowling ball? Show your work.

Maurinko [17]

Answer:

675J

Explanation:

Given parameters:

Force = 45N

Distance = 15m

Unknown:

Work done by Sheila = ?

Solution:

Work done by a body is the amount of force applied to make a body move through a distance;

Work done = Force x distance

Now;

Work done = 45 x 15 = 675J

6 0

2 years ago

A 70.0 kg ice hockey goalie, originally at rest, has a 0.110 kg hockey puck slapped at him at a velocity of 31.5 m/s. Suppose th

NISA [10]

Answer

given,

mass of the goalie(m₁) = 70 kg

mass of the puck (m₂)= 0.11 kg

velocity of the puck = 31.5 m/s

elastic collision

$v_1=\dfrac{m_2-m_1}{m_1+m_2}v_1+\dfrac{2m_2}{m_1+m_2}v_2$

$v_{pf}=\dfrac{0.11-70}{0.11+70}31.5+\dfrac{2m_2}{m_1+m_2}\times (0)$

$v_{pf}=-31.4\ m/s$

$v'_2 = \dfrac{2m_1v_1}{m_1+m_2}-\dfrac{(m_2-m_1)v_2}{m_2+m_1}$

$v_{gf} = \dfrac{2\times 0.11\times 31.5}{0.11+70}-\dfrac{(0.11-70)\times 0}{m_1+m_2}$

$v_{gf} = \dfrac{2\times 0.11\times 31.5}{0.11+70}$

$v_{gf} = 0.0988\ m/s$

4 0

3 years ago

50 POINTS!

zysi [14]

The object is moving with constant or uniform acceleration and in average speed

The object is de accelerating

The object deaccelerated and came to rest so fast.

The object moves slowly first then accelerated.

The object accelerated at first so fast then move with constant acceleration then again accelerated .

7 0

2 years ago