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makkiz [27]
3 years ago
14

If an object is acted on by a constant unbalanced force, the object will move with a constant ________

Physics
2 answers:
MissTica3 years ago
7 0
There is always a net force acting on the object, so it will have constant acceleration
AveGali [126]3 years ago
4 0

Answer:

C. Acceleration

Explanation:

I just took the quiz! :)

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An 7.9 kg crate is pulled 5.5 m up a 30∘ incline by a rope angled 19 ∘ above the incline. The tension in the rope is 140 N and t
dedylja [7]

Answer:

Wt= 0.73kJ, Wg=0.21kJ Wn= 0J

(B) Wf= 0.096kJ

Explanation:

See attachment below.

5 0
4 years ago
Two 800 cm^3 containers hold identical amounts of a monatomic gas at 20°C. Container A is rigid. Container B has a 100 cm^2 pis
Vikki [24]

Answer:

1) Final Temperature of the gas in A will be GREATER than the temperature in B

2) Diagram of both processes on a single PV has been uploaded below

3) The Initial  pressures in containers A and B is 3039.87 J/liters

4) the final volume of container B is 923.36 cm³

Explanation:

Given that;

Temperature = 20°C = 293 K

mass of piston = 10 kg

Area = 100cm³

Volume V = 800 cm³ = 0.8 L

ideal gas constant R = 8.3 J/K·mol

1)

Final Temperature of the gas in A will b GREATER than the temperature in B

2)

Diagram of both processes on a single PV has been uploaded below,

3)

Initial  pressures in containers A and B

PV = nRT

P = RT/V

we substitute

P = (8.3 × 293) /  0.8

P = 2431.9 / 0.8

P = 3039.87 J/liters

Therefore, The Initial  pressures in containers A and B is 3039.87 J/liters

4)

Given that;

power = 25 W

time t = 15s

the final volume of container B = ?

we know that;

work done = power × time

work done = 25 × 15 = 375

Also work done = P( V₂ - V₁ )

so we substitute

375 = 3039.87 ( V₂ - 0.8 )

( V₂ - 0.8 ) = 375 / 3039.87

V₂ - 0.8 = 0.12336

V₂ = 0.12336 + 0.8

V₂ = 0.92336 Litres

V₂ = 923.36 cm³

Therefore, the final volume of container B is 923.36 cm³

7 0
3 years ago
How do you do this problem?
kvasek [131]

Explanation:

First, find the velocity of the projectile needed to reach a height h when fired straight up.

Given:

Δy = h

v = 0

a = -g

Find: v₀

v² = v₀² + 2aΔy

(0)² = v₀² + 2(-g)(h)

v₀ = √(2gh)

Now find the height reached if the projectile is launched at a 45° angle.

Given:

v₀ = √(2gh) sin 45° = √(2gh) / √2 = √(gh)

v = 0

a = -g

Find: Δy

v² = v₀² + 2aΔy

(0)² = √(gh)² + 2(-g)Δy

2gΔy = gh

Δy = h/2

5 0
3 years ago
In the circuit diagram, what does the symbol made of two Long lines and two short lines with a positive and a negative sign at e
Svet_ta [14]
Yeah it is a source of electric energy
7 0
3 years ago
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Compare and contrast potential and kinetic energy
inysia [295]
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7 0
3 years ago
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