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3 years ago

Which statements accurately compare the tectonic activity of the planets? Check all that apply.

Physics

2 answers:

Serhud [2]3 years ago

8 0

Answer:The terrestrial planets experience quakes; the gas giants do not.

Terrestrial planets experience (or experienced) tectonic activity; gas giants do (or did) not.

Activity in the molten interior of the terrestrial planets results (or resulted) in tectonic activity.

Explanation: 1,4, & 6

Andreyy893 years ago

4 0

Answer: 1, 4, and 6

Explanation:

took the assignment

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A large stone block is to be part of a harbour wall. The block is supported beneath the surface of the sea by a cable from a cra

stich3 [128]

Answer:

The top of the block has an area of 0.25m².

i) a Calculate the pressure on the top face of the block.

ii) The atmospheric pressure is 1.0 × 10⁵ Pa.

Calculate the pressure on the top face of the block due to the depth h of water.

iii) The density of sea water is 1020kg/m³.

Calculate the depth h.

4 0

2 years ago

Read 2 more answers

At t = 0 the switch S is closed with the capacitor uncharged. If C = 50 F,  = 20 V, and R = 4.0 k, what is the charge on the

gulaghasi [49]

Hello!

Recall the equation for the current in an RC circuit as a capacitor is charging:
$i(t) = \frac{\epsilon}{R}e^{-\frac{t}{RC}}$

i(t) = Current as a function of time (2.0 mA)

ε = Emf of voltage source (20V)
R = Resistance of resistor (4kΩ)

C = Capacitance of capacitor (50 μF)

We can solve for the time at which the current in the circuit decreases to 2.0 mA by plugging in the given values and solving for 't'.

$0.002 = \frac{20}{4000} e^{-\frac{t}{(4000)(0.00005)}}\\\\0.002 = 0.005 e^{-\frac{t}{(4000)(0.00005)}$

Solve:
$.4 = e^{-\frac{t}{0.2}}\\\\ln(.4) = -\frac{t}{0.2}\\\\t = -0.2(ln(.4)) = 0.1833 s$

Now, we can solve for the charge on the capacitor using the equation for charge of a charging capacitor:
$q(t) = C\epsilon(1 - e^{\frac{-t}{RC}})$

Plug in the given values and the time we solved for.

$q(t) = (0.00005)(20)(1 - e^{\frac{-0.1833}{(4000)(0.00005)}}) \\\\q(t) = 0.0006 = \boxed{600 \mu C}$

5 0

2 years ago

A student is conducting an experiment to determine how far a ball will roll down a ramp based on the angle of incline. What are

icang [17]

Controls or controlled variables are variables that should remain unchanged in the experiment. These variables can have a direct effect on the dependent variable. Take note that controlled variables are NOT what is being tested but can alter results. Three possible controls for this experiment would mainly be features of the ball and the ramp.

Examples and possible answers to your question would be:

1. Mass of the ball
2. Size of the ball
3. The density of the ball
4. Type of ball
5. Length of the ramp
6. Type of ramp
7. The thickness of the ramp

Other controls that you can consider also is the environment. Wind can also affect the ball that is being rolled.

6 0

3 years ago

7.4 m = _________________________ mm 7.4x 10-3 74 7.4 x 103 7.4 x 104

jeyben [28]

7.4 m is equal to 7400 mm

This would make 7.4 * 10^3 the correct answer

7 0

3 years ago

Refer to the first diagram. What is the weight of the person hanging on the end of the seesaw in Newtons?

irina1246 [14]

Due to equilibrium of moments:

1) The weight of the person hanging on the left is 250 N

2) The 400 N person is 3 m from the fulcrum

3) The weight of the board is 200 N

Explanation:

To solve the problem, we use the principle of equilibrium of moments.

In fact, for the seesaw to be in equilibrium, the total clockwise moment must be equal to the total anticlockwise moment.

The moment of a force is defined as:

$M=Fd$

where

F is the magnitude of the force

d is the perpendicular distance of the force from the fulcrum

In the first diagram:

- The clockwise moment is due to the person on the right is

$M_c = W_2 d_2$

where $W_2 = 500 N$ is the weight of the person and $d_2 = 2 m$ is its distance from the fulcrum

- The anticlockwise moment due to the person hanging on the left is

$M_a = W_1 d_1$

where $W_1$ is his weight and $d_1 = 4 m$ is the distance from the fulcrum

Since the seesaw is in equilibrium,

$M_c = M_a$

So we can find the weight of the person on the left:

$W_1 d_1 = W_2 d_2\\W_1 = \frac{W_2 d_2}{d_1}=\frac{(500)(2)}{4}=250 N$

Again, for the seesaw to be in equilibrium, the total clockwise moment must be equal to the total anticlockwise moment.

- The clockwise moment due to the person on the right is

$M_c = W_2 d_2$

where $W_2 = 400 N$ is the weight of the person and $d_2$ is its distance from the fulcrum

- The anticlockwise moment due to the person on the left is

$M_a = W_1 d_1$

where $W_1 = 300 N$ is his weight and $d_1 = 4 m$ is the distance from the fulcrum.

Since the seesaw is in equilibrium,

$M_c = M_a$

So we can find the distance of the person on the right:

$W_1 d_1 = W_2 d_2\\d_2 = \frac{W_1 d_1}{W_2}=\frac{(300)(4)}{400}=3 m$

As before, for the seesaw to be in equilibrium, the total clockwise moment must be equal to the total anticlockwise moment.

- The clockwise moment around the fulcrum this time is due to the weight of the seesaw:

$M_c = W_2 d_2$

where $W_2$ is the weight of the seesaw and $d_2 = 3 m$ is the distance of its centre of mass from the fulcrum

- The anticlockwise moment due to the person on the left is

$M_a = W_1 d_1$

where $W_1 = 600 N$ is his weight and $d_1 = 1 m$ is the distance from the fulcrum

Since the seesaw is in equilibrium,

$M_c = M_a$

So we can find the weight of the seesaw:

$W_1 d_1 = W_2 d_2\\W_2 =\frac{W_1 d_1}{d_2}= \frac{(600)(1)}{3}=200 N$

#LearnwithBrainly

8 0

3 years ago