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3 years ago

Someone help me?

Physics

1 answer:

kap26 [50]3 years ago

3 0

Answer:

-27.3 m/s

Explanation:

Given:

y₀ = 38 m

y = 0 m

v₀ = 0 m/s

a = -9.8 m/s²

Find: v

v² = v₀² + 2a (y − y₀)

v² = (0 m/s)² + 2 (-9.8 m/s²) (0 m − 38 m)

v = -27.3 m/s

Or, you can solve with energy.

PE = KE

mgh = ½ mv²

v² = 2gh

v = -27.3 m/s

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What are four elements on the periodic table

Nataly_w [17]

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If we only stick to the 92 natural ones, then there are

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3 years ago

Velocity (m/s)

alekssr [168]

Explanation:

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(a) The segment A shows acceleration as velocity increases with the increase in time.

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(b) The segment C shows the object is slowing down as the time increases in segment C, the velocity decreases and afterwards it comes to rest.

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(c) The velocity is segment B is 40m/s. And in the diagram there is no change in velocity.

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2 years ago

How many moles of gas must be forced into a 4.6 l tire to give it a gauge pressure of 31.2 psi at 26 âc? the gauge pressure is r

Bess [88]

Use the Ideal Gas Law to the air in the tire :

( P ) ( V ) = ( n ) ( R ) ( T )
n = ( P ) ( V ) / ( R ) ( T )
P = P gauge + P baro = 31.2 psig + 14.8 psia = 46 psia
P = ( 46 psia ) ( 1 atm / 14.696 psia ) = 3.13 atm
n = ( P ) ( V ) / ( R ) ( T )
n = ( 3.13 atm ) ( 4.6 L ) / ( 0.08206 atm - L / mol - K ) ( 26.0 + 273.2 K )
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7 0

3 years ago

A device for acclimating military pilots to the high accelerations they must experience consists of a horizontal beam that rotat

Natali [406]

centripetal acceleration is given by formula

$a_c = \omega^2*R$

given that

$a_c = 34.1 m/s^2$

$R = 5.91 m$

now we have

$\omega^2 R = 34.1$

$\omega^2 * 5.91 = 34.1$

$\omega^2 = 5.77$

$\omega = 2.4 rad/s$

so the ratationa frequency is given by

$\omega = 2 \pi f$

$2.4 = 2 \pi f$

$f = \frac{2.4}{2\pi}$

$f = 0.38 Hz$

7 0

3 years ago

At a particular instant the magnitude of the momentum of a planet is 2.60 × 10^29 kg·m/s, and the force exerted on it by the sta

aleksley [76]

Answer:

F=(-4.8*10^22,0,0) N

Explanation:

Given :

We are given the magnitude of the momentum of the planet and let us call this momentum (p_now) and it is given by p_now = 2.60 × 10^29 kg·m/s. Also, we are given the force exerted on the planet F = 8.5 × 10^22 N. and the angle between the planet and the star is Ф = 138°

Solution :

We are asked to find the parallel component of the force F The momentum here is not constant, where the planet moving along a curving path with varying speed where the rate change in momentum and the force may be varying in magnitude and direction. We divide the force here into two parts: a parallel force F to the momentum and a perpendicular force F' to the momentum.

The parallel force exerted to the momentum will speed or reduce the velocity of the planet and does not change its moving line. Let us apply the direction cosines, we could obtain the parallel force as next

F=|F|cosФp (1)

Where the parallel force F is in the opposite direction of p as the angle between them is larger than 90°. Now we can plug our values for 0 and I F I into equation (1) to get the parallel force to the planet

F=|F|cosФp

=-4.8*10^22 N*p

As this force is in one direction, we could get its vector as next

F=(-4.8*10^22,0,0) N

F=(0,-4.8*10^22,0) N

F=(0,0-4.8*10^22) N

The cosine of 138°, the angle between F and p is, is a negative number, so F is opposite to p. The magnitude of the planet's momentum will decrease.

8 0

3 years ago