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Veseljchak [2.6K]
3 years ago
13

A man holding a rock sits on a sled that is sliding across a frozen lake (negligible friction) with a speed of 0.480 m/s. The to

tal mass of the sled, man, and rock is 96.0 kg. The mass of the rock is 0.310 kg and the man can throw it with a speed of 14.5 m/s. Both speeds are relative to the ground. Determine the speed of the sled if the man throws the rock forward (i.e. in the direction the sled is moving).
Also determine the speed of the sled if he throws the rock directly backwards.
Physics
1 answer:
SpyIntel [72]3 years ago
3 0
This is a problem of conservation of momentum

Momentum before throwing the rock: m*V = 96.0 kg * 0.480 m/s = 46.08 N*s

A) man throws the rock forward

=>

rock:
m1 = 0.310 kg
V1 = 14.5 m/s, in the same direction of the sled with the man

sled and man:
m2 = 96 kg - 0.310 kg = 95.69 kg
v2 = ?

Conservation of momentum:
momentum before throw = momentum after throw

46.08N*s = 0.310kg*14.5m/s + 95.69kg*v2

=> v2 = [46.08 N*s - 0.310*14.5N*s ] / 95.69 kg = 0.434 m/s

B) man throws the rock backward

this changes the sign of the velocity, v2 = -14.5 m/s

 46.08N*s = - 0.310kg*14.5m/s + 95.69kg*v2

v2 = [46.08 N*s + 0.310*14.5 N*s] / 95.69 k = 0.529 m/s


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EleoNora [17]

Distance is indeed a scalar amount that also refers to "<em><u>how the ground an object has encased</u></em>", and the Displacement is a vector thing that leads "<em><u>to the extent to which an object is located</u></em>", and the further calculation can be defined as follows:

Given:

distance= 70 miles

displacement = 20 miles

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Please find the graph in the attached file.

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7 0
2 years ago
A 1.0 kg mass is attached to the end of a vertical ideal spring with a force constant of 400 N/m. The mass is set in simple harm
Marina86 [1]

Answer:

v(0)=2m/s

Explanation:

The instantaneous velocity of a point mass that executes a simple harmonic movement is given by:

v(t)=\omega  *A*cos(\omega t + \phi)

Where:

\omega=Angular\hspace{3}frequency\\A=Amplitude\\\phi=Initial\hspace{3}phase

Express the amplitude in meters:

10cm*\frac{1m}{100cm} =0.1m

The angular frequency can be found using the next equation:

\omega=\sqrt{\frac{k}{m} }

Using the data provided:

\omega=\sqrt{\frac{400}{1} } =20

At the equilibrium position:

\phi=0

v(0)=20*(0.1)cos(20*0+0)=2*cos(0)=2*1=2m/s

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The boom of a fire truck raises a fireman (and his equipment – total weight 280 lbf) 60 ft into the air to fight a building fire
Katen [24]

Answer:

a)Work done by fireman=   2.15  Btu

b) Time t= 0.86 sec

Explanation:

Given that

Weight = 280 lbf

We know that 1 lbf = 4.44 N

so 280 lbf = 1245.5 N

Weight =1245.5 N

Height h = 60 ft

We know that

1 ft = 0.3048 m

So 60 ft = 18.28 m

 h =18.28 m

Power = 3.5 hp

We know that

1 hp =0.74 KW

So 3.5 hp = 2.61 KW

Power = 2.61 KJ/s

So the work done by fireman = Weight x h

Now by putting the values

Work done by fireman= 1245.5 x 18.28 J

Work done by fireman=   2267.74 J

Work done by fireman=   2.26774  KJ

We know that 1 Btu= 1.05 KJ

So   2.266 KJ = 2.15 Btu

Work done by fireman=   2.15  Btu

We know that ,rate of work is called power.

Power x time =  work

2.61 x t = 2.26

So t= 0.86 sec

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Answer:

The total work done by the two tugboats on the supertanker is 3.44 *10^9 J

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The force by the tugboats acting on the supertanker is constant and the displacement of the supertanker is along a straight line.

The angle between the 2 forces and displacement is ∅ = 15°.

First we have to calculate the work done by the individual force and then we can calculate the total work.

The work done on a particle by a constant force F during a straight line displacement s is given by following formula:

W = F*s

W = F*s*cos∅

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The total work done can be calculated as followed:

Wtotal = Ftugboat1 s * cos ∅1 + Ftugboat2 s* cos ∅2

Wtotal = 2Fs*cos∅

Wtotal = 2*2.2*10^6 N * 0.81 *10³ m s *cos15°

Wtotal = 3.44*10^9 Nm = <u>3.44 *10^9 J</u>

<u />

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