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HACTEHA [7]
3 years ago
7

In the Earth's mantle, heat is transferred in large convection currents. Within these currents,

Physics
2 answers:
chubhunter [2.5K]3 years ago
7 0
The answer is B. The hotter rock is less dense so rises. The cooler rock is more dense so sinks.
sertanlavr [38]3 years ago
4 0

B.) Hotter rock rises cooler rock sinks

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What are alkaline earth metals used for?
Tomtit [17]

Answer:

C firework

Explanation:

from Quizlet

6 0
3 years ago
Read 2 more answers
3. A football is kicked with a speed of 35 m/s at an angle of 40°.
jarptica [38.1K]

a) 22.5 m/s

The initial vertical velocity is given by:

u_y = u sin \theta

where

u = 35 m/s is the initial speed

\theta=40^{\circ} is the angle of projection of the ball

Substituting into the equation, we find

u_y = (35)(sin 40)=22.5 m/s

b) 26.8 m/s

The initial horizontal velocity is given by:

u_x = u cos \theta

where

u = 35 m/s is the initial speed

\theta=40^{\circ} is the angle of projection of the ball

Substituting into the equation, we find

u_x = (35)(cos 40)=26.8 m/s

c) 2.30 s

The time it takes for the ball to reach the maximum heigth can be found by considering the vertical motion only. This is a uniformly accelerated motion (free-fall), so we can use the suvat equation

v_y = u_y + at

where

v_y is the vertical velocity at time t

u_y = 22.5 m/s

a=g=-9.8 m/s^2 is the acceleration of gravity (negative because it is downward)

At the maximum height, the vertical velocity becomes zero, v_y =0; substituting, we find the time t at which this happens:

0=u_y + gt\\t=-\frac{u_y}{g}=-\frac{22.5}{-9.8}=2.30 s

d) 25.8 m

The maximum height can also be found by considering the vertical motion only. We can use the following suvat equation:

s=u_y t + \frac{1}{2}gt^2

where

s is the vertical displacement at time t

u_y = 22.5 m/s

g=-9.8 m/s^2

Substituting t = 2.30 s, we find the displacement at maximum height, so the maximum height:

s=(22.5)(2.30)+\frac{1}{2}(-9.8)(2.30)^2=25.8 m

e) 123.3 m

In order to find how far does the ball lands, we have to consider the horizontal motion.

First of all, the time it takes for the ball to go back to the ground is twice the time needed for reaching the maximum height:

t=2(2.30 s)=4.60 s

Then, we consider the horizontal motion. There is no acceleration along this direction, so the horizontal velocity is constant:

v_x = 26.8 m/s

Therefore, the horizontal distance travelled during the whole motion is

d=v_x t = (26.8)(4.60)=123.3 m

So, the ball lands 123.3 m far from the initial point.

4 0
3 years ago
A beam of light strikes a sheet of glass at an angle of 56.6° with the normal in air. You observe that red light makes an angle
Yuri [45]

Answer:

(a). Index of refraction are n_{red} = 1.344 & n_{violet} = 1.406

(b). The velocity of red light in the glass v_{red} = 2.23 ×10^{8} \ \frac{m}{s}

The velocity of violet light in the glass v_{violet} =2.13 ×10^{8} \ \frac{m}{s}

Explanation:

We know that

Law of reflection is

n_1 \sin\theta_{1} = n_2 \sin\theta_{2}

Here

\theta_1 = angle of incidence

\theta_2 = angle of refraction

(a). For red light

1 × \sin 56.6 = n_{red} × \sin 38.4

n_{red} = 1.344

For violet light

1 × \sin 56.6 = n_{violet} × \sin 36.4

n_{violet} = 1.406

(b). Index of refraction is given by

n = \frac{c}{v}

n_{red} = 1.344

v_{red} = \frac{c}{n_{red} }

v_{red} = \frac{3(10^{8} )}{1.344}

v_{red} = 2.23 ×10^{8} \ \frac{m}{s}

This is the velocity of red light in the glass.

The velocity of violet light in the glass is given by

v_{violet} = \frac{3(10^{8} )}{1.406}

v_{violet} =2.13 ×10^{8} \ \frac{m}{s}

This is the velocity of violet light in the glass.

8 0
3 years ago
Question #2: Where does hope come from? Just you? The<br> people around you? Explain.
Tanya [424]

Answer:

it depends ether people can give you hope or you can have hope

Explanation:

8 0
2 years ago
You want to examine the hairy details of your favorite pet caterpillar, using a lens of focal length 8.9 cm 8.9 cm that you just
Zepler [3.9K]

Answer:

The angular magnification is M = 2.808

Explanation:

From the question we are told

           The focal length is  f = 8.9cm

          The near point is n_p = 25.0cm

The angular magnification is mathematically represented as

                          M = \frac{n_p}{f}

Substituting values

                        M = \frac{25}{8.9}

                           = 2.808

4 0
3 years ago
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