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sergey [27]
3 years ago
5

What are tectonic plate boundaries?

Physics
2 answers:
Rasek [7]3 years ago
8 0

Answer:

Areas where earth crust experiences a high amount of stress

Explanation:

FCAT review question

Helga [31]3 years ago
4 0

Answer:

c

Explanation:

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A student gives a 5.0 kg box a brief push causing the box to move with an initial speed of 8.0 m/s along a rough surface. The bo
WITCHER [35]

Answer:

The time taken to stop the box equals 1.33 seconds.

Explanation:

Since frictional force always acts opposite to the motion of the box we can find the acceleration that the force produces using newton's second law of motion as shown below:

F=mass\times acceleration\\\\\therefore acceleration=\frac{Force}{mass}

Given mass of box = 5.0 kg

Frictional force = 30 N

thus

acceleration=\frac{30}{5}=6m/s^{2}

Now to find the time that the box requires to stop can be calculated by first equation of kinematics

The box will stop when it's final velocity becomes zero

v=u+at\\\\0=8-6\times t\\\\\therefore t=\frac{8}{6}=4/3seconds

Here acceleration is taken as negative since it opposes the motion of the box since frictional force always opposes motion.

5 0
3 years ago
When a river flows into an ocean, it slows down and deposits materials in its alluvial fan/delta
nikklg [1K]
When a river flows into an ocean, it slows down and deposits materials in its delta
8 0
3 years ago
A 645 g block is released from rest at height h0 above a vertical spring with spring constant k = 530 N/m and negligible mass. T
Nina [5.8K]

Answer:

a)5.88J

b)-5.88J

c)0.78m

d)0.24m

Explanation:

a) W by the block on spring is given by

W= \frac{1}{2}kx² = \frac{1}{2}(530)(0.149)² =  5.88 J

b)  Workdone by the spring = - Workdone by the block = -5.88J

c) Taking x = 0 at the contact point we have U top = U bottom

So, mgh_o = \frac{1}{2}kx² - mgx

And, h_o= ( \frac{1}{2}kx² - mgx )/(mg) = [\frac{1}{2} (530)(0.149^2)-(0.645)(9.8)(0.149)]/(0.645x9.8)    

   h_o=   0.78m            

d) Now, if the initial initial height of block is 3h_o

h_o = 3 x 0.78 = 2.34m

then, \frac{1}{2}kx² - mgx - mgh_o =0

 

\frac{1}{2}(530)x²  - [(0.645)(9.8)x] - [(0.645)(9.8)(2.34) = 0

265x² - 6.321x - 14.8 = 0  

a=265

b=-6.321

c=-14.8

By using quadratic eq. formula, we'll have the roots

x= 0.24 or x=-0.225

Considering only positive root:

x= 0.24m (maximum  compression of the spring)

4 0
3 years ago
A force is a vector quantity. What does that mean?
Inga [223]

Answer: Acceleration due to gravity

Explanation: Force, F = mg, is a vector quantity because the acceleration due to gravity, g, is a vector quantity. Explanation: F = mg Where m is the mass (in kilograms) of the object in question and g is the acceleration due to gravity. Mass is a scalar quantity; mass has no dependence on direction whatsoever.

5 0
2 years ago
In a fast-pitch softball game the pitcher is impressive to watch, as she delivers a pitch by rapidly whirling her arm around so
svetlana [45]

Answer:

(a). The magnitude of the total acceleration of the ball is 239.97 m/s².

(b). The angle of the total acceleration relative to the radial direction is 11.0°

Explanation:

Given that,

Radius of the circle = 0.681 m

Angular acceleration = 67.7 rad/s²

Angular speed =18.6 rad/s

We need to calculate the centripetal acceleration of the ball

Using formula of centripetal acceleration

a_{c}=\omega^2\times r

Put the value into the formula

a_{c}=(18.6)^2\times0.681

a_{c}=235.5\ m/s^2

We need to calculate the tangential acceleration of the ball

Using formula of tangential acceleration

a_{t}=r\alpha

Put the value into the formula

a_{t}=0.681\times67.7

a_{t}=46.104\ m/s^2

(a). We need to calculate the magnitude of the total acceleration of the ball

Using formula of total acceleration

a=\sqrt{a_{c}^2+a_{t}^2}

Put the value into the formula

a=\sqrt{(235.5)^2+(46.104)^2}

a=239.97\ m/s^2

(b). We need to calculate the angle of the total acceleration relative to the radial direction

Using formula of the direction

\theat=\tan^{-1}(\dfrac{a_{t}}{a_{c}})

Put the value into the formula

\theta=\tan^{-1}(\dfrac{46.104}{235.5})

\theta=11.0^{\circ}

Hence, (a). The magnitude of the total acceleration of the ball is 239.97 m/s².

(b). The angle of the total acceleration relative to the radial direction is 11.0°

5 0
3 years ago
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