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3 years ago

A constant magnetic field can be used to produce an electric current. True or False?

Physics

1 answer:

Umnica [9.8K]3 years ago

6 0

this question is true

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bonufazy [111]

Переходи на сайте irkmix.top и получай много эмоций из Russia/

7 0

3 years ago

The game was played by swinging the big mallet down hard enough to cause the bell to ring. If it took a 44 newton force to ring

nata0808 [166]

I think C. 33 newtons

3 0

3 years ago

In a football game, a 90 kg receiver leaps straight up in the air to catch the 0.42 kg ball the quarterback threw to him at a vi

irinina [24]

To solve this problem it is necessary to apply the equations related to the conservation of momentum. Mathematically this can be expressed as

$m_1v_1+m_2v_2 = (m_1+m_2)v_f$

Where,

$m_{1,2}$ = Mass of each object

$v_{1,2}$ = Initial velocity of each object

$v_f$ = Final Velocity

Since the receiver's body is static for the initial velocity we have that the equation would become

$m_2v_2 = (m_1+m_2)v_f$

$(0.42)(21) = (90+0.42)v_f$

$v_f = 0.0975m/s$

Therefore the velocity right after catching the ball is 0.0975m/s

8 0

3 years ago

How fast can a 4000 kg truck travel around a 70 m radius turn without skidding if its tires share a 0.6 friction coefficient wit

aleksley [76]

Answer:

Velocity of truck will be 20.287 m /sec

Explanation:

We have given mass of the truck m = 4000 kg

Radius of the turn r = 70 m

Coefficient of friction $\mu =0.6$

Centripetal force is given $F=\frac{mv^2}{r}$

And frictional force is equal to $F_{frictional}=\mu mg$

For body to be move these two forces must be equal

So $\frac{mv^2}{r}=\mu mg$

$v=\sqrt{\mu rg}=\sqrt{0.6\times 70\times 9.8}=20.287m/sec$

7 0

3 years ago

What if m1 is initially moving at 3.4 m/s while m2 is initially at rest? (a) find the maximum spring compression in this case?

Lisa [10]

<span>Ans : Initial E = KE = Â˝mvÂ˛ = Â˝ * 1.2kg * (2.2m/s)Â˛ = 2.9 J max spring compression where both velocities are the same: conserve momentum: 1.2kg * 2.2m/s = (1.2 + 3.2)kg * v â†’ v = 0.6 m/s which means the combined KE = Â˝ * (1.2 + 3.2)kg * (0.6m/s)Â˛ = 0.79 J The remaining energy went into the spring: U = (2.9 - 0.79) J = 2.1 J = Â˝kxÂ˛ = Â˝ * 554N/m * xÂ˛ x = 0.0076 m â† (a)</span>

7 0

3 years ago