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AnnZ [28]
3 years ago
9

Every year, new records in track and field events are recorded. Let's take an historic look back at some exciting races.

Physics
2 answers:
Alexeev081 [22]3 years ago
5 0

Answer:

Average velocity = 10.08 m/s

Explanation:

Given

Length of the track d = 100 m

Duration of the run t = 9.92 s

Solution

Velocity = \frac{Distance}{Time} \\\\Velocity = \frac{100}{9.92} \\\\Velocity = 10.08 m/s

Gala2k [10]3 years ago
4 0
100 meters in 9.92 seconds,
=distance/time
=100m/9.92s 
=10.0806 m/s
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podryga [215]
Basic solutions are hydroxides therefore the answer is A ca(OH)2
3 0
3 years ago
A small balloon is released at a point 150 feet away from an observer, who is on level ground. If the balloon goes straight up a
Elza [17]

Answer:

\dfrac{dz}{dt}=0.65\ ft/s

Explanation:

Given that

x= 150 ft

\dfrac{dy}{dt}= 7\ ft/s

y= 14 ft

From the diagram

z^2=x^2+y^2

When ,x= 150 ft and y= 14 ft

z^2=150^2+14^2

z=\sqrt{150^2+15^2}

z=150.74 ft

z^2=x^2+y^2

By differentiating with respect to time t

2z\dfrac{dz}{dt}= 2x\dfrac{dx}{dt}+2y\dfrac{dy}{dt}

z\dfrac{dz}{dt}= x\dfrac{dx}{dt}+y\dfrac{dy}{dt}

Here x is constant that is why

\dfrac{dx}{dt}=0

z\dfrac{dz}{dt}= y\dfrac{dy}{dt}

Now by putting the values in the above equation we get

150.74\times \dfrac{dz}{dt}=14\times 7

\dfrac{dz}{dt}=\dfrac{14\times 7}{150.74}\ ft/s

\dfrac{dz}{dt}=0.65\ ft/s

Therefore the distance between balloon and observer increasing with 0.65 ft/s.

5 0
3 years ago
Multiple-Concept Example 6 reveiws the principles that play a role in this problem. A nuclear power reactor generates 2.3 x 109
r-ruslan [8.4K]

Answer:

change in mass = 2.41*10^{8}kg

Explanation:

The change in the mass can be computed by using the relation

E=\Delta mc^2\\\Delta m=\frac{E}{c^2}(1)

That is, the energy liberated comes from the mass of the nuclear fuel. The energy generated in one year is

E=Pt=2.3*10^{9}\frac{J}{s}*1 year*\frac{365.25 day}{1 year}*\frac{24h}{1 day}*\frac{3600s}{1h}=7.25*10^{16}J

Hence, by replacing in the equation (1) you have  (c=3*10^{8}m/s)

\Delta m=\frac{7.25*10^{16}J}{3*10^{8}\frac{m}{s}}=2.41*10^{8}kg

HOPE THIS HELPS!!

3 0
3 years ago
Read 2 more answers
Eric has a mass of 70 . He is standing on a scale in an elevator that is accelerating downward at 1.7 . What is the approximate
pychu [463]

Answer:

B)

Explanation:

The value the scale shows is the reaction force to the normal force (they are equal by Newton's 3rd Law) that the scale exerts on Eric.

The forces on Eric are his weight (downward) and this normal force (upward), so we can write the net force over him as (also using Newton's 2nd Law):

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and for our values this is:

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5 0
3 years ago
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Paladinen [302]

Answer:

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The sulfur oxide and water vapor must have been derived from volcanic activities in geologic times past.

3 0
3 years ago
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