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givi [52]
3 years ago
14

SYW A force of 175 N is needed to keep a stationary engine of weight 640 N on wooden skids from

Physics
2 answers:
Katen [24]3 years ago
8 0

Answer:

no

Explanation:

ok

grigory [225]3 years ago
4 0

Answer:

phle follow karo yrr tab hee bat u ga

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Someone help please i need to finish this
Tom [10]

Answer:

4th answer

Explanation:

The gradient of a distance-time graph gives the speed.

gradient = distance / time = speed

Here, the gradient is a constant till 30s. So it has travelled at a constant speed. It means it had not accelarated till 30s. and has stopped moving at 30s.

6 0
2 years ago
A child rides a wagon down a hill. Eventually, the wagon comes to a stop. Which is most responsible for causing the wagon to sto
Dvinal [7]

Answer:

well, the hill isn't constantly going down hill, there's an ending point or goes back up hill making a v/u shape or there's nothing helping the wagon being pushed or pull currently

Explanation:

8 0
3 years ago
for an ideal monoatomic gas, the internal energy U os due to the kinetic energy and U=3/2RT per mole.show that cv=3/2R per mole
sladkih [1.3K]

Answer:

i. Cv =3R/2

ii. Cp = 5R/2

Explanation:

i. Cv = Molar heat capacity at constant volume

Since the internal energy of the ideal monoatomic gas is U = 3/2RT and Cv = dU/dT

Differentiating U with respect to T, we have

= d(3/2RT)/dT

= 3R/2

ii. Cp - Molar heat capacity at constant pressure

Cp = Cv + R

substituting Cv into the equation, we have

Cp = 3R/2 + R

taking L.C.M

Cp = (3R + 2R)/2

Cp = 5R/2

3 0
2 years ago
PLEASE HELP A cup of water and a swimming pool full of water are at 24°C. The _____ of the swimming pool is higher. 1.heat conte
Dimas [21]

Answer:

4. kinetic energy

4 0
3 years ago
A 72.9-kg base runner begins his slide into second base when moving at a speed of 4.02 m/s. The coefficient of friction between
elena-14-01-66 [18.8K]

Answer:

-589.05 J

Explanation:

Using work-kinetic energy theorem, the work done by friction = kinetic energy change of the base runner

So, W = ΔK

W = 1/2m(v₁² - v₀²) where m = mass of base runner = 72.9 kg, v₀ = initial speed of base runner = 4.02 m/s and v₁ = final speed of base runner = 0 m/s(since he stops as he reaches home base)

So, substituting the values of the variables into the equation, we have

W = 1/2m(v₁² - v₀²)

W = 1/2 × 72.9 kg((0 m/s)² - (4.02 m/s)²)

W = 1/2 × 72.9 kg(0 m²/s² - 16.1604 m²/s²)

W = 1/2 × 72.9 kg(-16.1604 m²/s²)

W = 1/2 × (-1178.09316 kgm²/s²)

W = -589.04658 kgm²/s²

W = -589.047 J

W ≅ -589.05 J

4 0
2 years ago
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