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Vinvika [58]
3 years ago
8

A diver with a mass of 80.0 kg jumps from a dock into a 130.0 kg boat at rest on the west side of the dock. if the velocity of t

he diver in the air is 4.10 m/s to the west, what is the final velocity of the diver after landing in the boat?
Physics
1 answer:
Bogdan [553]3 years ago
8 0
Applying conservation of momentum;

m_{1}  v_{1} + m_{2}  v_{2} = m_{f}  v_{f}

Where m1 = 80.0 kg; v1 =4.10 m/s; m2 = 130.0 kg; v2 = 0; mf = (80+130) kg; vf = ??

Therefore,
80*4.1 + 130*0 = 210*vf
vf = (80*4.1)/210 = 1.56 m/s
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Two moles of helium are initially at a temperature of 21.0 ∘Cand occupy a volume of 3.30×10−2 m3 . The helium first expands at c
Anettt [7]

Answer:

(B) The total internal energy of the helium is 4888.6 Joules

(C) The total work done by the helium is 2959.25 Joules

(D) The final volume of the helium is 0.066 cubic meter

Explanation:

(B) ∆U = P(V2 - V1)

From ideal gas equation, PV = nRT

T1 = 21°C = 294K, V1 = 0.033m^3, n = 2moles, V2 = 2× 0.033=0.066m^3

P = nRT ÷ V = (2×8.314×294) ÷ 0.033 = 148140.4 Pascal

∆U = 148140.4(0.066 - 0.033) = 4888.6 Joules

(C) P2 = P1(V1÷V2)^1.4 =148140.4(0.033÷0.066)^1.4= 148140.4×0.379=56134.7 Pascal

Assuming a closed system

(C) Wc = (P1V1 - P2V2) ÷ 0.4 = (148140.4×0.033 - 56134.7×0.066) ÷ 0.4 = (4888.6 - 3704.9) ÷ 0.4 = 1183.7 ÷ 0.4 = 2959.25 Joules

(C) Final volume = 2×initial volume = 2×0.033= 0.066 cubic meter

6 0
3 years ago
10. Mario is making chocolate. In part of the chocolate-making process, he changed the phase from a liquid to a solid. What happ
Naily [24]

Answer:

B

Explanation:

in a liquid the particles are widespread and move around each other but in a solid they move in place and are tightly packed

6 0
3 years ago
Can someone please help​
coldgirl [10]

Answer:

Making a quick cut left to intercept a pass

Explanation:

It takes more energe to do than running

7 0
3 years ago
PLEASE HURRYYY
babymother [125]

your answer is make up artist

4 0
3 years ago
A 1-kg rock is suspended from the tip of a horizontal meterstick at the 0-cm mark so that the meterstick barely balances like a
tigry1 [53]

Explanation:

Given that,

Mass if the rock, m = 1 kg

It is  suspended from the tip of a horizontal meter stick at the 0-cm mark so that the meter stick barely balances like a seesaw when its fulcrum is at the 12.5-cm mark.

We need to find the mass of the meter stick. The force acting by the stone is

F = 1 × 9.8 = 9.8 N

Let W be the weight of the meter stick. If the net torque is zero on the stick then the stick does not move and it remains in equilibrium condition. So, taking torque about the pivot.

9.8\times 12.5=W\times (50-12.5)\\\\W=\dfrac{9.8\times 12.5}{37.5}

W = 3.266 N

The mass of the meters stick is :

m=\dfrac{W}{g}\\\\m=\dfrac{3.266}{9.81}\\\\m=0.333\ kg

So, the mass of the meter stick is 0.333 kg.

5 0
3 years ago
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