Which option tells the forces that influence the movement of earths plates

Please help I have no idea

DochEvi [55]

Answer:

b

Explanation:

bbbbbbbbbbbbbbvgh c tyvftj xf

3 0

3 years ago

Which of the following is a heterogeneous mixture? A. salt B. dye in water C. sugar water D. a garden salad

dedylja [7]

D garden salad : )

A heterogenous mixture can be easily taken apart visually/physically

7 0

3 years ago

I hope you are able to read this question?? Help ASAP this question is on the quiz tommorow

Masteriza [31]

You would be correct.

Because you have only JUST released the arrow, and how close he is to the target, it would have the same amount of energy when it strikes the target. Yes, the kinetic energy would be destroyed when you hit the target but not right away. And yes, the potential energy would also be destroyed once you release the arrow, but it goes straight back once it stops moving, aka when it hits the target, although it has only just stopped moving.

Hope this helps!

8 0

3 years ago

A voltage of 12 cos(I000t+45) Vis applied to a circuit in which a resistor of 4 .n, aninductor of L H, and a capacitor of 100 μF

vlabodo [156]

Answer:

0.01 H

Explanation:

V = 12 cos (1000t + 45)

C = 100 micro farad

Let the inductance be L .

When the current and the voltage are in the same phase so it is the condition of resonance.

So capacitive reactance = inductive reactance

Xc = XL

1/ωC = ωL

L = 1 / ω²C

By comparisonV = Vo Cos (ωt + Ф)

ω = 1000 rad/s

L = 1 / (1000 x 1000 x 100 x 10^-6)

L = 1 / 100

L = 0.01H

thus, the inductance of the inductor is 0.01 H.

8 0

3 years ago

A pin-supported, vertically-oriented 1-m long thin rod is struck by a pellet at m down from the pin at the top. The mass of the

Natasha_Volkova [10]

Answer:

the angular velocity of the rod immediately after being struck by the pellet, provided that the pellet gets lodged in the rod is = 0.5036 k` rad/s

Explanation:

Using the conservation of momentum of approach.

From the question; the pellet is hitting at a distance of 0.4 m down from the point of rotation of the rod.

So, the angular momentum of the system just before the collision occurs with respect to the axis of the rotation is expressed by the formula:

$L_i ^ { ^ \to } = mp ( r_y } ^ { ^ \to } * v_{pi} ^ { ^ \to } )$ ----- equation (1)

The position vector can now be :

$x ^ { ^ \to }$ = $- 0.4 \ j \ m$

Also, given that :

$v_{p,i} ^ { ^ \to } = (280 \ i - 350 \ j) \ m/s$

Replacing the value into above equation (1); we have:

$L_i ^ { ^ \to } =0.012 ((- \ 0.4 \ j) *(280 \ i - 350 \ j ))$

$L_i ^ { ^ \to } =0.012 * 112 \ k$ (by using cross product )

$L_i ^ { ^ \to } = 1.344 k` \ \ kg m^2 s^{-1}$

However; the moment of inertia of the rod about the axis of rotation is :

$I_{rod} = \frac{1}{3}m_rl^2 \\ \\ I_{rod} = \frac{1}{3}*8*1^2 \\ \\ I_{rod} = \frac{8}{3} \ \ kg \ m^2$

Also, the moment of inertia of the pellet about the axis of rotation is:

$I_{pellet} = m_pr_y^2 \\ \\ I_{pellet} = 0.012 *0.4^2 \\ \\ I_{pellet} = 1.92*10^{-3} kg . m^2$

So, the moment of inertia of the rod +pellet system is:

$I = I_{rod}+I_{pellet}$

$I =( \frac{8}{3}+1.92 *10^{-3} )kg. m^2$

$I = 2.6686 \ kg. m^2$

The final angular momentum is :

$L_f ^ {^ \to} = I \omega { ^ {^ \to} } = 2.6686 \ \omega ^ {^ \to}$

The angular velocity of the rod $\omega$ is determined by equating the angular momentum just before the collision with the final angular momentum (i.e after the collision).

So;

$L_f ^ {^ \to} = L_i ^ { ^ \to}$

$2.6686 \omega ^ {^ \to} = 1.344 \ k ^ {^ \to}$

$\omega ^ {^ \to} = \frac{1.344 \ k`}{2.6686}$

= 0.5036 k` rad/s

Hence; the angular velocity of the rod immediately after being struck by the pellet, provided that the pellet gets lodged in the rod is = 0.5036 k` rad/s

7 0

3 years ago