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A volcano erupts and launches a chunk of hot magma horizontally with a speed of 252 m/s. The magma travels a horizontal distance

ArbitrLikvidat [17]

Answer:

The value is $v_y = -48.61 \ m/s$

Explanation:

From the question we are told that

The horizontal speed is $u_x = 252 \ m/s$

The horizontal distance is $d = 1250 \ m$

Generally the time taken by the hot magma in air before landing is mathematically represented as

$t = \frac{d}{u_x}$

=> $t = \frac{ 1250 }{252}$

=> $t = 4.96 \ s$

Generally the initial vertical velocity of the magma when it was lunched is

$u_y = 0 \ m/ s$

Then the final velocity of the magma when it hits the ground is mathematically represented s

$- v_y = u_y + gt$

Here the negative sign mean that the direction of the velocity is towards the negative y -axis

So

$- v_y = 48.61 \ m/s$

=> $v_y = -48.61 \ m/s$

7 0

2 years ago

An object is most likely to sink in water if

LiRa [457]

Answer:

High density D answers to your questions

8 0

2 years ago

Read 2 more answers

At a location near the equator, the earth’s magnetic field is horizontal and points north. An electron is moving vertically upwa

Elan Coil [88]

Answer:

2. west

Explanation:

Given an electron is moving vertically upward from ground.

Now Fleming right hand rule state that: make L shape with thumb and index finger then point middle finger perpendicular to index and thumb.Then index finger points in the direction of moving charge , middle finger points in the direction of the magnetic field and thumb points in the direction of the magnetic force.

According to Fleming right hand rule the direction of the magnetic that acts on the electron is west.

5 0

3 years ago

What feature means the Earth wire will take current away from the case reducing the chance of accidental electrocution?

Fudgin [204]

The answer should be A

6 0

3 years ago

Starting from rest, a boulder rolls down a hill with constant acceleration and travels 3.00m during the first second.

Alexus [3.1K]

Answer:

a) 9.00 m b) 6.00 m/s c) 12.00 m/s

Explanation:

a) If the acceleration is constant, and we know that the displacement during the first second was 3.00 m, as the boulder (assumed that we can treat it as a point mass) started from rest, we can say the following:

Δx = $\frac{1}{2}*a*t^{2}$ = 3.00 m

As t = 1 s, replacing in the expression above, and solving for a, we have:

$a = \frac{2*3.00m}{1s2}$ = 6.00 m/s²

In order to know how far it travels during the second second, we need to know the value of the speed after the first second, as it is the initial velocity when the second second begins:

vf = v₀ + a*t ⇒ vf = 0 + 6 m/s²*1s = 6.00 m/s

The total displacement, during the second second, will be as follows:

Δx = v₀*t + $\frac{1}{2}*a*t^{2}$ = 6.00m/s*1s + $\frac{1}{2}*6.00 m/s2*1s^{2} = 9.00 m$

⇒ Δx = 9.00 m

b) At the end of the first second, the final speed can be obtained as follows:

vf = v₀ + a*t ⇒ vf = 0 + 6 m/s²*1s = 6.00 m/s

c) At the end of the second second, the final speed can be obtained as follows:

vf = v₀ + a*t ⇒ vf = 0 + 6 m/s²*2s = 12.00 m/s

3 0

2 years ago

I need help thank you​

I need help thank you