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Arisa [49]
4 years ago
9

A system releases 680 kj of heat and does 150 kj of work on the surroundings.what is the change in internal energy of the system

?
Physics
1 answer:
egoroff_w [7]4 years ago
4 0
I need a answer choice
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A 30-g car rolls from a hill 12 cm high and is traveling at 154 cm/s as it travels along a 275 cm horizontal track. What is the
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Momentum of the car is 46.2 kgm/s
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A body from the top of a 180 m high mountain is abandoned. Disregarding air resistance and adopting g = 10m / s2, determine:
elena55 [62]

Answer:

1) 6 seconds

2) 60 m/s

Explanation:

Given:

Δy = 180 m

v₀ = 0 m/s

a = 10 m/s²

1) Find t.

Δy = v₀ t + ½ at²

180 m = (0 m/s) t + ½ (10 m/s²) t²

t = 6 s

2) Find v.

v² = v₀² + 2aΔy

v² = (0 m/s)² + 2 (10 m/s²) (180 m)

v = 60 m/s

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3 years ago
Wave A has an amplitude of 2 and wave B has an amplitude of 2 as shown below. What will happen when the crest of wave A meets th
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Since the two waves have equal amplitudes, if the crest of one wave
meets the trough of the other one, they'll add to produce a level of zero
at that location.
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3 years ago
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The Astronomer Carl Sagan is famous. Why?
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Mainly because he was Johnny Carson's advisor and consultant
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a vector has an x-component of 19.5m and a y-component of 28.4m. Find the magnitude and direction of the vector
Delicious77 [7]

Answer:

magnitude=34.45 m

direction=55.52\°

Explanation:

Assuming the initial point P1 of this vector is at the origin:

P1=(X1,Y1)=(0,0)

And knowing the other point is P2=(X2,Y2)=(19.5,28.4)

We can find the magnitude and direction of this vector, taking into account a vector has a initial and a final point, with an x-component and a y-component.

For the magnitude we will use the formula to calculate the distance d between two points:

d=\sqrt{{(Y2-Y1)}^{2} +{(X2-X1)}^{2}} (1)

d=\sqrt{{(28.4 m - 0 m)}^{2} +{(19.5 m - 0m)}^{2}} (2)

d=\sqrt{1186.81 m^{2}} (3)

d=34.45 m (4) This is the magnitude of the vector

For the direction, which is the measure of the angle the vector makes with a horizontal line, we will use the following formula:

tan \theta=\frac{Y2-Y1}{X2-X1}  (5)

tan \theta=\frac{24.8 m - 0m}{19.5 m - 0m}  (6)

tan \theta=\frac{24.8}{19.5}  (7)

Finding \theta:

\theta= tan^{-1}(\frac{24.8}{19.5})  (8)

\theta= 55.52\°  (9) This is the direction of the vector

3 0
3 years ago
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