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2 years ago

Need help ASAP please and thank you

Physics

1 answer:

lapo4ka [179]2 years ago

6 0

Answer:

Explanation:

The fault will eventually build mountians given time.

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I throw a ball upward at an initial speed of 20 m/s. How much time does it take before the ball slows to a speed of 0 m/s

masha68 [24]

20/9.8 = 2.0 seconds. The ball stops after 2 seconds.

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3 years ago

Which statement explains how gravity and inertia work together?

Grace [21]

Answer: Its a

Explanation:

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2 years ago

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In a certain time period a coil of wire is rotated from one orientation to another with respect to a uniform 0.38-T magnetic fie

juin [17]

<h2>The emf produced is 7.2 V</h2>

Explanation:

When coil is placed in the magnetic field , the flux attached with it can be found by the relation . Flux Ф = the dot product of magnetic field and area of coil .

Thus Ф = B A cosθ

here B is magnetic field strength and A is the area of coil .

The angle θ is the angle between coil and field direction .

When coil rotates , the angle varies . By which the flux varies . The emf is produced in coil due to variation of flux . The relation for this is

The emf produced ξ = - $\frac{d\phi}{dt}$ = B A sinθ $\frac{d\theta}{dt}$

Now in the given problem

5 = 0.38 x A x $\frac{d\theta}{dt}$ I

Now if the magnetic field is 0.55 T and all the other terms are same , the emf produced

ξ = 0.55 x A x $\frac{d\theta}{dt}$ Ii

dividing II by I , we have

$\frac{\xi}{5}$ = $\frac{0.55}{0.38}$ = 1.45

or ξ = 7.2 V

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3 years ago

Two 0.40 kg soccer ball collide elastically in a head-on collision. The first ball starts at rest, and the second ball has a spe

yulyashka [42]

Explanation:

Mass of two soccer balls, $m_1=m_2=0.4\ kg$

Initial speed of first ball, $u_1=0$

Initial speed of second ball, $u_2=3.5\ m/s$

After the collision,

Final speed of the second ball, $v_2=0$

(a) The momentum remains conserved. Using the conservation of momentum to find it as :

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

$v_1$ is the final speed of the first ball

$0.4\times 0+0.4\times 3.5=0.4v_1+0.4\times 0$

$0.4\times 3.5=0.4v_1$

$v_1=3.5\ m/s$

(b) Let $E_1$ is the kinetic energy of the first ball before the collision. It is given by :

$E_1=\dfrac{1}{2}mu_1^2$

$E_1=\dfrac{1}{2}\times 0.4\times 0$

It is at rest, so, the kinetic energy of the first ball before the collision is 0.

(c) After the collision, the second ball comes to rest. So, the kinetic energy of the second ball after the collision is 0.

Hence, this is the required solution.

7 0

3 years ago

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By approximately how much does your gravitational potential energy change when you jump

Svet_ta [14]

Potential Energy = height*mass*gravity
the height is factually the height of the center of mass, which will be varied the same as the lowest of your feet. So when you hurdle, the height that you jump can be used to distinguish the alteration in height of your center of mass.In other words, it is inclined on the mass of the human, be contingent on how extraordinary the person can jump. Supposing a man is 80 kg and he can jump such that his center of mass rises 0.3 m in Earth's standard gravitational field. This sums to approximately 240 Joules of increase in GPE.

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2 years ago