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4 years ago

How can you tell the difference between and erosion feature and a deposition feature?

1 answer:

5 0

Deposition moves that have already been eroded, and builds up sediment.
erosion is just the moving of sediment from one place to another

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How does a 12-month lunar calendar differ from our 12-month solar calendar?.

horrorfan [7]

Answer:

It has about 11 fewer days. It does not have seasons. Its new year always occurs in February instead of on January 1.

Explanation:

8 0

2 years ago

rickey approaches third base. He dives head first, hitting the ground at 6.75 m/s and reaching the base at 5.91 m/s in 2.5 secon

Gekata [30.6K]

Answer:

15.825 m

Explanation:

t = Time taken = 2.5 s

u = Initial velocity = 6.75 m/s

v = Final velocity = 5.91 m/s

s = Displacement

a = Acceleration

Equation of motion

$v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow a=\frac{5.91-6.75}{2.5}\\\Rightarrow a=-0.336\ m/s^2$

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{5.91^2-6.75^2}{2\times -0.336}\\\Rightarrow s=15.825\ m$

The distance Rickey slides across the ground before touching the base is 15.825 m

4 0

3 years ago

Amoebas have projections called pseudopods. What are they used for

Dvinal [7]

Answer:

Amoebas have projections called pseudopods

Explanation:

Pseudopodia is the locomatary organ of amoeba. It helps them in movement and transportation.

8 0

3 years ago

When the ball of the pendulum moves from (x) to (y) in a duration of 0.02 sec the frequency equals

Mariulka [41]

0.4 sec the frequency equals

5 0

2 years ago

A projectile is thrown with velocity v at an angle θ with horizontal. When the projectile is at a height equal to half of the ma

raketka [301]

Let, the maximum height covered by projectile be $\sf{H_m}$

$\purple{ \longrightarrow \bf{h_m = \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2g} }}$

Projectile is thrown with a velocity = v
Angle of projection = θ

Velocity of projectile at a height half of the maximum height covered be $\sf{v_0}$

$\qquad$ ______________________________

Then –

$\qquad$ $\pink{ \longrightarrow \bf{ \dfrac{h_m}{2} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }}$

$\qquad$ $\longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2g} \times \dfrac{1}{2} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }$

$\qquad$ $\longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{4g} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }$

$\qquad$ $\longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2} = {v_0}^{2} \: {sin}^{2} \theta }$

$\qquad$ $\longrightarrow \sf{ \dfrac{ {v}^{2} }{2} = {v_0}^{2} }$

$\qquad$ $\longrightarrow \bf{v_0 = \sqrt{ \dfrac{ {v}^{2} }{2} } = \dfrac{v}{ \sqrt{2} } }$

Now, the vertical component of velocity of projectile at the height half of $\sf{h_m}$ will be –

$\qquad$ $\longrightarrow \bf{v_{(y)}=v_0 \: sin \theta }$

$\qquad$ $\longrightarrow \bf{v_{(y)} = \dfrac{v}{ \sqrt{2} } \: sin \theta = \dfrac{v \: sin \: \theta}{ \sqrt{2} } }$

Therefore, the vertical component of velocity of projectile at this height will be–

☀️ $\qquad$ $\pink {\bf{ \dfrac{v \: sin \: \theta}{ \sqrt{2} }} }$

6 0

3 years ago

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