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2 years ago

If a ball was thrown upward at 46.3 m/s how long would the ball stay in the air

1 answer:

8 0

$V = Vo + a.t

$

The ball is against the vector of gravity. Then, the gravity will be negative.

$0 = 46,3 + (-9.8).t \\ 
t = \frac{46.3}{9.8} \\ 
t \approx 4.72 

$

The ball will stop in the air after approx. 4.72 seconds. And will take the same time to hit the ground.

It will stay approx. 9.44 seconds in the air.

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A basketball with mass of 0.8 kg is moving to the right with velocity 6 m/s and hits a volleyball with mass of 0.6 kg that stays

IceJOKER [234]

Answer:

26.67 m/s

Explanation:

From the law of conservation of linear momentum, the initial sum of momentum equals the final sum.

p=mv where p is momentum, m is the mass of object and v is the speed of the object

Initial momentum

The initial momentum will be that of basketball and volleyball, Since basketball is initially at rest, its initial velocity is zero

$p_i= m_bv_b+m_vv_v=8*6+0.6*0=48 Kg.m/s$

Final momentum

$p_f= m_bv_b+m_vv_v=8*4+0.6*v_v=32+0.6v Kg.m/s\\32+0.6v_v=48\\0.6v=16\\v_v=16/0.6=26.66666667\approx 26.67 m/s$

4 0

2 years ago

In an elastic head-on collision, a 0.60 kg cart moving at 5.0 m/s [W] collides with a 0.80 kg cart moving at 2.0 m/s [E]. The co

labwork [276]

Answer:

The answer is given below

Explanation:

u is the initial velocity, v is the final velocity. Given that:

$m_1=0.6kg,u_1=-5m/s(moving \ west),m_2=0.8kg,u_2=2m/s,k=1200N/m$

The final velocity of cart 1 after collision is given as:

$v_1=(\frac{m_1-m_2}{m_1+m_2})u_1+\frac{2m_2}{m_1+m_2}u_2\\ Substituting:\\v_1=\frac{0.6-0.8}{0.6+0.8} (-5)+\frac{2*0.8}{0.6+0.8}(2)= 5/7+16/7=3\ m/s$

The final velocity of cart 2 after collision is given as:

$v_2=(\frac{m_2-m_1}{m_1+m_2})u_2+\frac{2m_1}{m_1+m_2}u_1\\ Substituting:\\v_1=\frac{0.8-0.6}{0.6+0.8} (2)+\frac{2*0.6}{0.6+0.8}(-5)= 2/7-30/7=-4\ m/s$

b) Using the law of conservation of energy:

$\frac{1}{2}m_1u_1+ \frac{1}{2}m_2u_2=\frac{1}{2}m_1v_1+\frac{1}{2}m_2v_2+\frac{1}{2}kx^2\\x=\sqrt{\frac{m_1u_1+m_2u_2-m_1v_1-m_2v_2}{k}}\\ Substituting\ gives:\\x=\sqrt{\frac{0.6*(-5)^2+0.8*2^2-(0.6*3^2)-(0.8*(-4)^2)}{1200}}=\sqrt{0}=0\ cm$

7 0

2 years ago

Does the surrounding air become warm or cool when vapour phase of H2O condenses? Explain

mel-nik [20]

The surrounding air will become warm when water vapor condenses. The vapors when become water will give away latent heat they have, we know that latent heat is required for the object to change states, so, the latent heat the water vapor had when it became water vapor from water will be given out when it again becomes water.

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2 years ago

A person using an oxygen mask is breathing air that is 33% oxygen. what is the partial pressure of the o2, when the air pressure

yawa3891 [41]

The partial pressure of the O2 is 36.3 kiloPascal when the air pressure in the mask is 110 kiloPascal based on the isotherm relation. This problem can be solved by using the isotherm relation equation which stated as Vx/Vtot = px/ptot, where V represents volume, p represents the pressure, x represents the partial gas, and tot represents the total gas<span>. Calculation: 33/100 = px/110 --> px = 36.3</span>

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2 years ago

In order for the ball to be able to make a complete circle around the peg, there must be sufficient speed at the top of its arc

Vesna [10]

Answer:

Explanation:

Let T be the tension in the swing

At top point $mg-T=\frac{mv^2}{r}$