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BaLLatris [955]
3 years ago
10

Please please help me please please

Physics
1 answer:
Scorpion4ik [409]3 years ago
3 0

Answer:

<h3>C. </h3>

Explanation:

PO NG SAGOT

<h2>#MARK BRAINLITS PO</h2>

TINUTULUNGAN KO PO KAYO

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an athlete swims the distance from one end of a 50m pool to the other end in a time of 25 s what is The Athlete's average speed
vredina [299]

Answer:

2.5 N

because Average speed is equal to distance divided by time

3 0
3 years ago
A scientist tests the water in a local pond and finds that it has a pH of 7.9. What is true about the water sample?
SSSSS [86.1K]
So the pH level of regular water is about 7.0 so since it is 7.9, it means the water is a little more basic
3 0
3 years ago
The Otto-cycle engine in a Mercedes-Benz SLK230 has a compression ratio of 8.8. What is the ideal efficiency of the engine? Use
vodomira [7]

Answer:

e=58%

Explanation:

Given data

The Otto-cycle engine in a Mercedes-Benz SLK230 has a compression ratio of 8.8.

Solution

We want to calculate the ideal efficiency of the engine when ratio of heat capacity for gas used  γ=1.40. Ideal efficiency (e) of the Otto cycle given by:

e=1-(\frac{1}{r^{Y-1} } )\\

Substitute the given values to find efficiency e

e=1-\frac{1}{8.8^{1.4-1} } \\e=0.58\\

e=58%

8 0
3 years ago
To practice problem-solving strategy 9.1 a strategy for conservation of momentum problems. an 80-kg quarterback jumps straight u
kherson [118]
Refer to the diagram shown below.

By definition momentum =  mass *  velocity.

Before throwing the ball:
The initial momentum is
P₁ = 0.

After throwing the ball:
Let u = the backward velocity of the quarterback.
The momentum is 
P₂ = (0.43 kg)*(15 m/s) + (80 kg)*(- u m/s)

Conservation of momentum requires that
P₂ = P₁
6.45 - 80u = 0
u = 6.45/80 = 0.0806 m/s

Answer: 0.08 m/s backward

6 0
3 years ago
Read 2 more answers
A solid cube of aluminum (density of 2.7 g/cm³) has a volume of 0.9 cm³. how many atoms are contained in the cube?​
Reika [66]

Answer:

0.542*10^{23}\ Aluminum\ Atoms

Explanation:

We\ are\ given:\\Density\ of\ aluminum=2.7\ g/cm^3\\Volume\ of\ aluminum-cube=0.9\ cm^3\\Hence,\\As\ we\ know\ that,\\Density=\frac{Mass}{Volume}\\Mass=Density*Volume\\Hence,\ here,\\Mass\ of\ the\ solid\ iron\ cube=2.7*0.9=2.43\ g\\Now,\\We\ also\ know\ that,\\Gram\ Atomic\ mass\ of\ Aluminum = 26.98 \approx 27\ g\\Hence,\\No.\ of\ particles=\frac{Mass}{GAM}*Avagadro's Constant\\Hence,\ here\\No.\ of\ Aluminum\ atoms=\frac{2.43}{27}*6.022*10^{23} \approx 0.542*10^{23}\ Aluminum\ Atoms

8 0
3 years ago
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