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serious [3.7K]
3 years ago
15

Please help me .... ​

Physics
1 answer:
miss Akunina [59]3 years ago
8 0

Explanation:

Orbital speed= 2pi x radius / time period

=2pi x 1.5x10^11 / 365.25

=2.58x10^9m/day

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EXPERTS/ACE and people that wanna help 4 sure only!
mario62 [17]
That first one you have selected (3,-3) works in both equations so it's correct.
good job.

you can do this guess and test method with multiple choice answers. If it works in both equations it is the solution. Otherwise use substitution or elimination to combine the two into one equation in only one variable. Then you can solve for the one variable first and use it to solve for the other.

3 0
3 years ago
Consider two spaceships, each traveling at 0.50c in a straight line. Ship A is moving directly away from the Sun and ship B is a
attashe74 [19]

Answer:

The velocity of the light will be 1.0c only

Explanation:

The velocity of the light measured in the case given in question will be 1.0c only.

This is due to the fact that the velocity of light is never relative. The velocity of the light is maximum

The velocity of the light cannot be scaled down in no case

Thus, the velocity of the light remains as constant.

Hence, the velocity of the light measured will be 1.0c although the ships have relative velocity.

3 0
2 years ago
1. What are the 2 main categories of nonmetals?
Anettt [7]

the 2 main categories of nonmetals are REACTIVE NONMETALS an NOBLE GAS.

hope it helps

7 0
3 years ago
My buddy and I have just finished a dive to 15 metres/50 feet for 60 minutes. We want to return to the same site and depth and s
marishachu [46]

Answer:

1) Periodically check the no stop or NDL time on their computers

2) The dive computer planning mode can be used if available

3) Make use of a dive planning app

4) Check data from the RDP table or an eRDPML

Explanation:

The no stop times information from the computer gives the no-decompression limit (NDL) time allowable which is the time duration a diver theoretically is able to stay at a given depth without a need for a decompression stop

The dive computer plan mode or a downloadable dive planning app are presently the easiest methods of dive planning

The PADI RDP are dive planners based on several years of experience which provide reliable safety limits of depth and time.

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
3 years ago
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