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2 years ago

5

Which is a unique feature of Venus? A. Its surface is well below freezing. B. Its surface is hot enough to melt lead. C. It has

liquid water on the surface. D. It is heavily cratered.

2 answers:

Mkey [24]2 years ago

7 0

The correct answer is B. can you please help me answer my latest question too?

Sophie [7]2 years ago

6 0

It has a surface that is well below freezing.

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A mass weighing 24 pounds, attached to the end of a spring, stretches it 4 inches. Initially, the mass is released from rest fro

svetoff [14.1K]

Answer:

The equation of motion is $x(t)=-$ $\frac{1}{3} cos4\sqrt{6t}$

Explanation:

Lets calculate

The weight attached to the spring is 24 pounds

Acceleration due to gravity is $32ft/s^2$

Assume x , is spring stretched length is ,4 inches

Converting the length inches into feet $x=\frac{4}{12} =\frac{1}{3}feet$

The weight (W=mg) is balanced by restoring force ks at equilibrium position

mg=kx

$W=kx$ ⇒ $k=\frac{W}{x}$

The spring constant , $k=\frac{24}{1/3}$

= 72

If the mass is displaced from its equilibrium position by an amount x, then the differential equation is

$m\frac{d^2x}{dt} +kx=0$

$\frac{3}{4} \frac{d^2x}{dt} +72x=0$

$\frac{d^2x}{dt} +96x=0$

Auxiliary equation is, $m^2+96=0$

$m=\sqrt{-96}$

= $\frac{+}{} i4\sqrt{6}$

Thus , the solution is $x(t)=c_1cos4\sqrt{6t}+c_2sin4\sqrt{6t}$

$x'(t)=-4\sqrt{6c_1} sin4\sqrt{6t}+c_2$ $4\sqrt{6}$ $cos4\sqrt{6t}$

The mass is released from the rest x'(0) = 0

$=-4\sqrt{6c_1} sin4\sqrt{6(0)}+c_2$ $4\sqrt{6}$ $cos4\sqrt{6(0)}$ =0

$c_2$ $4\sqrt{6} =0$

$c_2=0$

Therefore , $x(t)=c_1$ $cos 4\sqrt{6t}$

Since , the mass is released from the rest from 4 inches

$x(0)= -4$ inches

$c_1 cos 4\sqrt{6(0)} =-\frac{4}{12}$ feet

$c_1=-\frac{1}{3}$ feet

Therefore , the equation of motion is $-\frac{1}{3} cos4\sqrt{6t}$

7 0

2 years ago

Determine the amount of work done on an ideal gas as it is heated in an enclosed thermally insulated cylinder topped with a free

sp2606 [1]

Answer:

W = 3/2 n (T₁- T₂)

Explanation:

Let's use the first law of thermodynamics

ΔE = Q + W

in this case the cylinder is insulated, so there is no heat transfer

ΔE = W

internal energy can be related to the change in temperature

ΔE = 3/2 n K ΔT

we substitute

3/2 n (T₂-T₁) = W

as the work is on the gas it is negative

W = 3/2 n (T₁- T₂)

3 0

2 years ago

A 12 n cart is moving on a horizontal surface with a coefficient of kinetic friction of 0.20. what force of friction must be ove

jonny [76]

We must remember that the total net force equation at constant velocity is:

<span>F – Ff = 0</span>

of

F - µN = 0

Using Newton's 2nd Law of Motion:<span>

F = m a

<span>Where,

F = net force acting on the body
m = mass of the body
a = acceleration of the body

Since the cart is moving at a constant velocity, then acceleration is zero, hence the working equation simplifies to

F = net Force = 0

Therefore,

F - µN = 0

where

µ = coefficient of friction = 0.20
N = normal force acting on the cart = 12 N

Therefore,

F - 0.20(12) = 0

<span>F = 2.4 N </span></span></span>

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3 years ago

How to do this question

Anna71 [15]

Answer:

(a) 10 m/s

(b) 22.4 m/s

Explanation:

(a) Draw a free body diagram of the car when it is at the top of the loop. There are two forces: weight force mg pulling down, and normal force N pushing down.

Sum of forces in the centripetal direction (towards the center):

∑F = ma

mg + N = mv²/r

At minimum speed, the normal force is 0.

mg = mv²/r

g = v²/r

v = √(gr)

v = √(10 m/s² × 10.0 m)

v = 10 m/s

(b) Energy is conserved.

Initial kinetic energy + initial potential energy = final kinetic energy

½ mv₀² + mgh = ½ mv²

v₀² + 2gh = v²

(10 m/s)² + 2 (10 m/s²) (20.0 m) = v²

v = 22.4 m/s

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3 years ago

Which of these is NOT an example of a reference direction?

Vikentia [17]

Answer:

A

Explanation:

8 0

3 years ago