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max2010maxim [7]
3 years ago
7

If a bus travels 6 km in 10 minutes, what distance does it travel in 1 second ​

Physics
2 answers:
loris [4]3 years ago
4 0

Answer:

Your answer is 10 m/s...

Masteriza [31]3 years ago
4 0

Answer:

I think it is 1 foot???

Explanation:

sorry if this is not what u are looking for..

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A 1200 kg sports car accelerates from 0 m/s to 30 m/s in 10 s. What is the average power of the engine?
DIA [1.3K]

Answer:

3600N

Explanation:

Given: m = 1200kg, Vo = 0m/s, Vf = 30m/s, Δt = 10s

ΣF = ma

we need to find 'a' first, using the definition of 'a' we get equation:

a = (Vf-Vo)/Δt

a = (30m/s)/10s

a = 3 m/s^2

now substitute into top equation

ΣF = ma

Fengine = (1200kg)(3m/s^2)

Fengine = 3600N

5 0
2 years ago
Read 2 more answers
PLEASE HELP!! DUE SOON!! THANKS BRAINLIEST OFFERED!!
pav-90 [236]

Answer:

real, inverted, and smaller than the object

Explanation:

When the object is placed beyond the center of curvature, the image will formed between the focus and the center of curvature. The size of the image is diminished and its nature is real and inverted.

The whole description is shown in the attached figure. It is clear that the size of the image is smaller than the object.

7 0
4 years ago
In a third class lever, the distance from the effort to the fulcrum is ____________ the distance from the load/resistance to the
padilas [110]
The complete sentence is:
In a third class lever, the distance from the effort to the fulcrum is SMALLER the distance from the load/resistance to the fulcrum.
In fact, in a third class lever, the fulcrum is on one side of the effort and the load/resistance is on the other side, so the effort is located somewhere between the two of them. This means that the distance effort-fulcrum is smaller than the distance load-fulcrum.
4 0
3 years ago
If the plotted points on a speed-time graph do not form a straight line, what do you know about the object's acceleration?
slega [8]
Remember, that while sped is constant, acceleration is not. Acceleration is when velicity changes. So the graph which shows the slop <span>of a velocity vs time describes acceleration.
</span>If we have the straight line on the graph it means that the slope is always the same whereas the <span>non-linear graphs has a variable slope that changes depending on your point in the graph.
</span>To conclude - if your graph is not a straight line it has variable acc at many points.<span>

</span>
6 0
3 years ago
Read 2 more answers
A 150 kg line backer sacks the 120 kg quarterback. With what force is the quarterback sacked if the line backer has an accelerat
Gekata [30.6K]

Answer:

The force required to move the quarterback with linebacker is <u>1215 N</u>

Explanation:

\text { Mass of linebacker } \mathrm{m}_{2}=150 \mathrm{kg}

\text { Mass of quarterback } \mathrm{m}_{2}=120 \mathrm{kg}

\text { Moved at an acceleration }(a)=4.5 \mathrm{m} / \mathrm{s}^{2}

Using Newton's second law, it is established that  F = Ma

Where F is net force acting on the system, a is the acceleration and M is mass of the two object \left(m_{1}+m_{2}\right)

Now consider both \mathrm{m}_{1} \text { and } \mathrm{m}_{2}as a system, so net force acting on the system is \text { Force }=\left(m_{1}+m_{2}\right) a

Substitute the given values in the above formula,

\text { Force }=(150+120) \mathrm{kg} \times 4.5 \mathrm{m} / \mathrm{s}^{2}

\text { Force }=270 \mathrm{kg} \times 4.5 \mathrm{m} / \mathrm{s}^{2}

Force = 1215 N

<u>1215 N </u>is the force required to move the quarterback with linebacker.

5 0
3 years ago
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