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3 years ago

7

Magnetic field lines are more concentrated at the poles of the magnet True False

1 answer:

yKpoI14uk [10]3 years ago

8 0

Answer:

True! UwU

Explanation:

the magnetic field is strongest near to the poles of the magnet were the lines of flux are more closely spaced.

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How does exhailing remove waste from the body? Explain the systems that make this happen‚ using complete sentences

aliina [53]

Answer:

The lungs get rid of carbon dioxide and water vapor. The liver gets rid of bile, which, in addition to breaking down fats, is partially made up of the breakdown of red blood cells. The kidneys get rid of toxins from the blood. The large intestine gets rid of undigested food

3 0

2 years ago

In a Broadway performance, an 84.5-kg actor swings from a R = 4.30-m-long cable that is horizontal when he starts. At the bottom

Arada [10]

Answer:

1.57772 m

Explanation:

M = Mass of actor = 84.5 kg

m = Mass of costar = 55 kg

v = Velocity of costar

V = Velocity of actor

$h_i$ = Intial height of actor = 4.3 m

g = Acceleration due to gravity = 9.81 m/s²

As the energy of the system is conserved

$\frac{1}{2}MV^2=Mgh_i\\\Rightarrow V=\sqrt{2gh_i}\\\Rightarrow V=\sqrt{2\times 9.81\times 4.3}\\\Rightarrow V=9.18509\ m/s$

As the linear momentum is conserved

$MV=(m+M)v\\\Rightarrow v=\frac{MV}{m+M}\\\Rightarrow V=\frac{84.5\times 9.18509}{84.5+55}\\\Rightarrow v=5.56372\ m/s$

Applying conservation of energy again

$\frac{1}{2}(m+M)v^2=(m+M)gh_f\\\Rightarrow h_f=\frac{v^2}{2g}\\\Rightarrow h_f=\frac{5.56372^2}{2\times 9.81}\\\Rightarrow h_f=1.57772\ m$

The maximum height they reach is 1.57772 m

3 0

2 years ago

A cannon at rest fires a 32.5 kg cannonball forward at 388 m/s. After firing, the cannon recoils at 7.42 m/s. What is the mass o

Bond [772]

Answer:

1700 kg

Explanation:

Let’s use conservation of momentum

32.5 * 388 = 7.42 * mc

mc = 1699.46

mc = 1700 kg

3 0

2 years ago

Sound waves cannot carry energy through. A water B air C a mirror D a vacuum

Ber [7]

I looked up the question and got D- a vacuum

3 0

2 years ago

Suppose a yo-yo has a center shaft that has a 0.230 cm radius and that its string is being pulled.

Fofino [41]

Answer:

Part a)

$\alpha = 782.6 rad/s^2$

Part B)

$\omega = 587 rad/s$

Part c)

$a_t = 24.3 m/s^2$

Explanation:

Part a)

As we know that

$a = R \alpha$

so we will have

$a = 1.80 m/s^2$

$R = 0.230 cm$

$\alpha = \frac{a}{R}$

$\alpha = \frac{1.80}{0.230 \times 10^{-2}}$

$\alpha = 782.6 rad/s^2$

Part B)

Angular speed of the yo-yo

$\omega = \alpha t$

so we have

$\omega = 782.6 \times 0.750$

$\omega = 587 rad/s$

Part c)

Tangential acceleration is given as

$a_t = R \alpha$

$a_t = (3.10 \times 10^{-2})(782.6)$

$a_t = 24.3 m/s^2$

6 0

3 years ago