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3 years ago

What is in the crysphere

Physics

1 answer:

butalik [34]3 years ago

3 0

I think you mean the Cryosphere?

But the answer is D- Earths Ice

This word Cryosphere comes from the greek word "kryos" which means cold

Many people think of the cryosphere as being the north and south poles but snow and ice can be found in a lot of places on the Earth

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A ball rolls over the edge of a platform with only a horizontal velocity. The height of the platform is 1.60m and the horizontal

AysviL [449]

Answer:

v = 46.99 m/s

Explanation:

The velocity of the ball just before it touches the ground, is given by the following formula:

$v=\sqrt{v_x^2+v_y^2}$ (1)

vx: horizontal component of the velocity

vy: vertical component of the velocity

The vertical component vy is calculated by using the following formula:

$v_y^2=v_{oy}^2+2gh$ (2)

vy: final velocity

voy: initial vertilal velocity = 0m/s (because it is a semi parabolic motion)

g: gravitational acceleration = 9.8 m/s^2

h: height = 1.60m

You replace the values of the parameters in the equation (2):

$v_y=2(9.8m/s^2)(1.60m)=31.36\frac{m}{s}$

vx is calculated by using the information about the horizontal range of the ball:

$R=v_o\sqrt{\frac{2h}{g}}$ (3)

R: horizontal range of the ball = 20.0 m

You solve the previous equation for vo, the initial horizontal velocity:

$v_o=R\sqrt{\frac{g}{2h}}=(20.0m)\sqrt{\frac{9.8m/s^2}{2(1.60m)}}\\\\v_o=35\frac{m}{s}$

The horizontal component of the velocity is constant in the complete trajectory, hence, you have that

vx = vo = 35 m/s

Finally, you replace the values of vx and vy in the equation (1):

$v=\sqrt{(35m/s)^2+(31.36m/s)^2}=46.99\frac{m}{s}$

The velocity of the ball just before it touches the ground is 46.99 m/s

5 0

4 years ago

A speed-time graph is shown below:

Juliette [100K]

Answer:

It traveled 4 centimeters.

Explanation:

In a speed versus time graph, the distance travelled is given by the area under the graph.

In this graph we have the following:

- The speed of the object is v = 1 cm/s between time t = 0 s and t = 4 s

- The speed of the object is v = 0 cm/s between time t = 4 s and t = 8 s

Since the speed in the second part is zero, the distance travelled in the second part is zero. So, the only distance travelled by the object is the distance travelled during the first part, which is equal to the area of the first rectangle:

$d=v\Delta t=(1)(4-0)=4 cm$

4 0

3 years ago

A ball is thrown into the air with a vertical velocity of 50 m/s and a horizontal

daser333 [38]

$\mathfrak{\huge{\pink{\underline{\underline{AnSwEr:-}}}}}$

Actually Welcome to the Concept of the Projectile Motion.

Since, here given that, vertical velocity= 50m/s

we know that u*sin(theta) = vertical velocity

so the time taken to reach the maximum height or the time of Ascent is equal to

T = Usin(theta) ÷ g, here g = 9.8 m/s^2

so we get as,

T = 50/9.8

T = 5.10 seconds

thus the time taken to reach max height is 5.10 seconds.

5 0

3 years ago

3. Suppose that you have an electrically charged stick. If you divide the stick in half, each half will have half the original c

FrozenT [24]

Answer:

No, you can't keep on dividing the charge forever.

Explanation:

No, you can't keep on dividing the charge in that manner forever because the total charge of the stick is an integer multiples of individual units known as an elementary charge, <em>which is the electron (e) charge (e = 1.602x10⁻¹⁹C)</em>.

Therefore the limit of the division of the original charge will be the electron charge since it is the smallest charge that can exist freely.

I hope it helps you!

8 0

3 years ago

Jacob's family celebrates with him as he jumps with both feet together. Jacob is likely (2 points)

professor190 [17]

Jacob is three years old

3 0

3 years ago

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