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Mariana [72]
3 years ago
15

Extra-solar planets are probably composed of which gases?

Physics
1 answer:
kifflom [539]3 years ago
5 0
Exoplanets are probably 
<span>made of hydrogen and helium </span>gas<span>. ... </span>
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Answer: the third one
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Lwhich graph is a quadratic graph
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The 4 fourth row of the graph is a quadratic graph
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If you are driving 90 km/hkm/h along a straight road and you look to the side for 2.2 ss , how far do you travel during this ina
ICE Princess25 [194]

Answer: 55m

Explanation:

Given the following :

Driving speed = 90km/hr

Inattentive period (time) = 2.2s

Distance during inattentive period =

(driving speed * time)

Converting driving speed from km/hr to m/s

1000m = 1km

3600s = 1hour

Therefore,

90km/hr = (90 * 1000) / 3600

90km/hr = (90000)/ 3600 = 25m/s

Therefore ;

Distance during inattentive period =

(driving speed * time)

Distance during inattentive period = (25m/s × 2.2s) = 55m

Distance traveled during inattentive period is 55m

6 0
3 years ago
The speed limmit on an interstate highway is posted at 75mi/h. What is the speed in kilometers per hour? In feet per second?
jok3333 [9.3K]

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4 0
3 years ago
Imagine two billiard balls on a pool table. Ball A has a mass of 7 kilograms and ball B has a mass of 2 kilograms. The initial v
wlad13 [49]
1) In a perfectly inelastic collision, the two balls stick together after the collision. In this type of collision, the total kinetic energy of the system is not conserved, while the total momentum is conserved.
If we callv_f the final velocity of the two balls that stick together, the conservation of the total momentum before and after the collision can be written as
m_a v_{Ai} + m_b v_{Bi} = (m_A+m_B)v_f (1)
where
m_A=7 kg is the mass of ball A
m_B=2 kg is the mass of ball B
v_{Ai}=6 m/s is the initial velocity of ball A
v_{Bi}=-12 m/s is the initial velocity of ball B (taken with a negative sign, since it goes in the opposite direction of ball A)

If we solve (1) to find v_f, we find that the final velocity of the balls is
v_f= \frac{m_Av_{Ai}+m_Bv_{Bi}}{m_A+m_B}= \frac{(7\cdot 6)+(2 \cdot (- 12))}{7+2}= \frac{18}{9}=2 m/s
and the positive sign means the two balls are going to the right.


2) I assume here we are talking about an elastic collision. In this case, both total momentum and total kinetic energy are conserved:
m_A v_{Ai}+m_B v_{Bi} = m_A v_{fA} + m_B v_{fB}
\frac{1}{2}m_A v_{Ai}^2+ \frac{1}{2}m_B v_{Bi}^2= \frac{1}{2}m_Av_{fA}^2+ \frac{1}{2}m_B v_{fB}^2
where
v_{fA} is the final velocity of ball A
v_{fB} is the final velocity of ball B

If we solve simultaneously the two equations, we find:
v_{fA}= \frac{v_{Ai}(m_A-m_B)+2m_Bv_{Bi}}{m_A+m_B} = \frac{(6)(7-2)+2(2)(-12)}{7+2}=-2 m/s
v_{fB}= \frac{v_{Bi}(m_B-m_A)+2m_Av_{Ai}}{m_A+m_B} = \frac{(-12)(2-7)+2(7)(6)}{7+2}= \frac{144}{9}=16 m/s
So, after the collision, ball A moves to the left with velocity v=-2 m/s and ball B moves to the right with velocity v=16 m/s.

3) The total momentum before and after the collision is conserved.
In fact, the total momentum before the collision is:
p_i = m_A v_{A} + m_B v_{fB} = (7\cdot 6)+(2 \cdot (-12))=42-24=18 m/s
and the total momentum after the collision is:
p_f = m_A v_{A} + m_B v_{fB} = (7\cdot (-2))+(2 \cdot 16)=-14+32=18 m/s

3 0
3 years ago
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