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Sauron [17]
3 years ago
10

Which type of stress causes deformation that leads to earthquakes at converging plate boundaries?

Physics
1 answer:
ki77a [65]3 years ago
4 0
The answer is <em>Compressional Stress
</em>

In geology, stress<span> is the force per unit area that is placed on a rock. ... This is called confining </span>stress<span>. Compression squeezes rocks together, </span>causing<span> rocks to fold or fracture (break). Compression is the most common </span>stress<span> at </span>convergent plate boundaries<span>.

I hope this helped!! Have a great day :D</span>
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6. State whether each of the following is an example of conduction, convection, and/or radiation. Explain if you think there is
hodyreva [135]
A).  Convection is heating the soup in the pot.
When you stick the spoon into the hot soup,
conduction heats the spoon all the way up to the end.

b).  Water conducts heat a little bit. 
But convection is much more responsible for the
uniform distribution of temperature in the kiddie pool.

c).  The heat from the metal bench conducts directly
to the buttus epidermis when you sit on it.

d).  You feel the heat on your face ... but not on the back of your
neck ... on account of radiation from the fire and the hot grill.
3 0
3 years ago
Calculate the horizontal force that must be applied to a 1,300 kg vehicle to give it an acceleration of 2.6 m/s² on a level road
ANEK [815]
The answer is 3,380 N
4 0
3 years ago
A generator is designed to produce a maximum emf of 190 V while rotating with an angular speed of 3800 rpm. Each coil of the gen
Zinaida [17]

Answer:

The number of turns of wire needed is 573.8 turns

Explanation:

Given;

maximum emf of the generator, = 190 V

angular speed of the generator, ω = 3800 rev/min =

area of the coil, A = 0.016 m²

magnetic field, B = 0.052 T

The number of turns of the generator is calculated as;

emf = NABω

where;

N is the number of turns

\omega = 3800 \frac{rev}{min} \times \frac{2\pi}{1 \ rev} \times \frac{1 \min}{60 \ s } = 397.99 \ rad/s

N = \frac{emf}{AB\omega } \\\\N = \frac{190}{0.016 \times 0.052\times 397.99} \\\\N = 573.8 \ turns

Therefore, the number of turns of wire needed is 573.8 turns

4 0
3 years ago
What was the direction of the ball’s velocity
Tju [1.3M]

Part of the question is missing. Here it is:

<em>A 72 g autographed baseball slides off of a 1.3 m high table and strikes the floor a horizontal distance of 0.7m away from the table.     The acceleration of gravity is 9.81 m/s2. What was the direction of the ball’s velocity  just before it hit the floor? </em>

Answer:

\theta=-75.7^{\circ}

Explanation:

The motion of the ball is a projectile motion, which consists of two separate motions:

- A horizontal motion at constant velocity

- A vertical motion at constant acceleration (free fall)

We start by analyzing the vertical motion, to find the time of flight of the ball. This can be done by using the suvat equation

s=ut+\frac{1}{2}at^2

where, choosing downward as positive direction:

s =1.3 m is the vertical displacement of the ball

u = 0 is the initial vertical velocity

a=g=9.8 m/s^2 is the acceleration of gravity

t is the time

Solving for t,

t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2(1.3)}{9.8}}=0.52 s

Now we can find the final vertical velocity of the ball, using:

v_y=u+at

And susbtituting t = 0.52 s, we find

v_y = 0 +(9.8)(0.52)=5.1 m/s

It is important to keep in mind that the direction of this velocity is downward, since we chose downward as positive direction.

The horizontal velocity of the ball instead is constant; we know that the ball covers a horizontal distance of

d = 0.7 m

In a time of

t = 0.52 s

So, the horizontal velocity is

v_x = \frac{0.7}{0.52}=1.3 m/s

So now we can find the direction of the ball's velocity using:

\theta=tan^{-1}(\frac{v_y}{v_x})=tan^{-1}(\frac{5.1}{1.3})=75.7^{\circ}

And since the vertical direction is downward, this means that this velocity is below the horizontal, so the answer is

\theta=-75.7^{\circ}

8 0
3 years ago
An object is located 25.0 cm from a convex mirror. The image distance is -50.0 cm. What is the magnification?
Lynna [10]

Answer:

\boxed{\sf Magnification \ (m) = 2}

Given:

Object distance (u) = 25.0 cm

Image distance (v) = -50.0 cm

To Find:

Magnification (m)

Explanation:

\boxed{\bold{\sf Magnification  \: (m) = - \frac{Image  \: distance  \: (v)}{Object  \: distance  \: (u)}}}

Substituting values of Image distance(v) & Object distance (u) in the equation:

\sf \implies m =  -  \frac{( - 50)}{25}

-(-50) = 50:

\sf \implies m =  \frac{50}{25}

\sf \implies m =  \frac{2 \times  \cancel{25}}{ \cancel{25}}

\sf \implies m = 2

4 0
3 years ago
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