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Viktor [21]
2 years ago
10

Fossil fuels store energy from the sun as question 5

Physics
1 answer:
qwelly [4]2 years ago
3 0

Answer:

Fossil fuels store energy from the sun as

Explanation:

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O((>ω< ))oo((>ω< ))oo((>ω< ))oo((>ω< ))oo((>ω< ))o
Paladinen [302]

Answer:

(•_•)

Explanation:

7 0
2 years ago
The conductors that carry the current to electrical devices and ? equipment are the heart of all electrical systems. There are a
MariettaO [177]

Answer:

utilization / effects

Explanation:

Utilization equipment are those equipment that makes use of electric energy for the purpose of chemical, electronic, lighting, heating, electro-mechanical or other alike purposes. Hence utilization best suits the first question mark in the question. Secondly, there are associated effects when current flows through a conductor, not responses.

5 0
3 years ago
Regular exercise is positively related to wellness
Harman [31]

Answer:

yes  ( true)

Explanation:

positive  effects on all  the body systems.

7 0
2 years ago
In a jump spike, a volleyball player slams the ball from overhead and toward the opposite floor. Controlling the angle of the sp
MAVERICK [17]

Answer:

The ball would have landed 3.31m farther if the downward angle were 6.0° instead.

Explanation:

In order to solve this problem we must first start by doing a drawing that will represent the situation. (See picture attached).

We can see in the picture that the least the angle the farther the ball will go. So we need to find the A and B position to determine how farther the second shot would go. Let's start with point A.

So, first we need to determine the components of the velocity of the ball, like this:

V_{Ax}=V_{A}cos\theta

V_{Ax}=(21m/s)cos(-14^{o})

V_{Ax}=20.38 m/s

V_{Ay}=V_{A}sin\theta

V_{Ay}=(21m/s)sin(-14^{o})

V_{Ay}=-5.08 m/s

we pick the positive one, so it takes 0.317s for the ball to hit on point A.

so now we can find the distance from the net to point A with this time. We can find it like this:

x_{A}=V_{Ax}t

x_{A}=(20.38m/s)(0.317s)

x_{A}=6.46m

Once we found the distance between the net and point A, we can similarly find the distance between the net and point B:

V_{Bx}=20.88 m/s

V_{By}=-2.195 m/s

y_{Bf}=y_{B0}+V_{0}t-\frac{1}{2}at^{2}

0=2.1m+(-2.195m/s)t-\frac{1}{2}(-9.8m/s^{2})t^{2}

-4.9t^{2}-2.195t+2.1=0

t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}

t=\frac{-(-2.195)\pm\sqrt{(-2.195)^{2}-4(-4.9)(2.1)}}{2(-4.9)}

t= -0.9159s    or   t=0.468s

we pick the positive one, so it takes 0.468s for the ball to hit on point B.

so now we can find the distance from the net to point B with this time. We can find it like this:

x_{B}=V_{Bx}t

x_{B}=(20.88m/s)(0.468s)

x_{B}=9.77m

So once we got the two distances we can now find the difference between them:

x_{B}-x_{A}=9.77m-6.46m=3.31m

so the ball would have landed 3.31m farther if the downward angle were 6.0° instead.

7 0
2 years ago
Which is the correct answer ?
hodyreva [135]
D ............................
7 0
3 years ago
Read 2 more answers
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