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yuradex [85]
2 years ago
15

On a cold winter day, a penny (mass 2.50 g) and a nickel (mass 5.00 g) are lying on the smooth (frictionless) surface of a froze

n lake. With your finger, you flick the penny toward the nickel with a speed of 2.35 m/s.A) The coins collide elastically; calculate both of their final speeds
Physics
1 answer:
sp2606 [1]2 years ago
6 0

Answer:

0.78333 m/s in the opposite direction

1.566 m/s in the same direction

Explanation:

m_1 = Mass of penny = 0.0025 kg

m_2 = Mass of nickel = 0.005 kg

u_1 = Initial Velocity of penny = 2.35 m/s

u_2 = Initial Velocity of nickel = 0 m/s

v_1 = Final Velocity of penny

v_2 = Final Velocity of nickel

As momentum and Energy is conserved

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

{\tfrac {1}{2}}m_{1}u_{1}^{2}+{\tfrac {1}{2}}m_{2}u_{2}^{2}={\tfrac {1}{2}}m_{1}v_{1}^{2}+{\tfrac {1}{2}}m_{2}v_{2}^{2}

From the two equations we get

v_{1}=\frac{m_1-m_2}{m_1+m_2}u_{1}+\frac{2m_2}{m_1+m_2}u_2\\\Rightarrow v_1=\frac{0.0025-0.005}{0.0025+0.005}\times 2.35+\frac{2\times 0.5}{0.4005+0.5}\times 0\\\Rightarrow v_1=-0.78333\ m/s

The final velocity of the penny is 0.78333 m/s in the opposite direction

v_{2}=\frac{2m_1}{m_1+m_2}u_{1}+\frac{m_2-m_1}{m_1+m_2}u_2\\\Rightarrow v_2=\frac{2\times 0.0025}{0.0025+0.005}\times 2.35+\frac{0.005-0.0025}{0.005+0.0025}\times 0\\\Rightarrow v_2=1.566\ m/s

The final velocity of the nickel is 1.566 m/s in the same direction

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Answer:

a). 53.78 m/s

b) 52.38 m/s

c) -75.58 m

Explanation:

See attachment for calculation

In the c part, The negative distance is telling us that the project went below the lunch point.

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2 years ago
What is the difference between clastic and bioclast?
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The main difference is the source of the sediment that the rock is formed from. Clastic sedimentary rocks are formed mostly from silicate sediment derived by the breakdown of pre-existing rocks. Bioclastic rocks are formed by the accumulation of fragmented organic remains (such as shell-sand) - i.e. the sediment is of biological rather than non-biological origin.
8 0
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A 235 kg crate is pulled across a horizontal surface with a force of 760 N applied at an
AnnyKZ [126]

Answer:

658.16N

Explanation:

Step one:

given data

mass m= 235kg

Force F= 760N

angle= 30 degrees

Required

The horizontal component of the force

Step two:

The horizontal component of the force

Fh= 760cos∅

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Fh=760*0.8660

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3 0
3 years ago
A tin can has a volume of 1100 cm³ and a mass of 80 g. Approximately how many grams of lead shot can it carry without sinking in
Kruka [31]

Answer:

1020g

Explanation:

Volume of can=1100cm^3=1100\times 10^{-6}m^3

1cm^3=10^{-6}m^3

Mass of can=80g=\frac{80}{1000}=0.08kg

1Kg=1000g

Density of lead=11.4g/cm^3=11.4\times 10^{3}=11400kg/m^3

By using 1g/cm^3=10^3kg/m^3

We have to find the mass of lead which shot can it carry without sinking in water.

Before sinking the can  and lead inside it they are floating in the water.

Buoyancy force =F_b=Weight of can+weight of lead

\rho_wV_cg=m_cg+m_lg

Where \rho_w=10^3kg/m^3=Density of water

m_c=Mass of can

m_l=Mass of lead

V_c=Volume of can

Substitute the values then we get

1000\times 1100\times 10^{-6}=0.08+m_l

1.1-0.08=m_l

m_l=1.02 kg=1.02\times 1000=1020g

1 kg=1000g

Hence, 1020 grams of lead shot can it carry without sinking water.

4 0
3 years ago
Suppose a car travels 106 km at a speed of 28 m/s and uses 1.9 gals of gasoline in the process. Only 30% of the gasoline goes in
USPshnik [31]

Answer:

a) The magnitude of the force is 968 N

b) For a constant speed of 30 m/s, the magnitude of the force is 1,037 N

Explanation:

<em>NOTE: The question b) will be changed in other to give a meaningful answer, because it is the same speed as the original (the gallons would be 1.9, as in the original).</em>

Information given:

d = 106 km = 106,000 m

v1 = 28 m/s

G = 1.9 gal

η = 0.3

Eff = 1.2 x 10^8 J/gal

a) We can express the energy used as the work done. This work has the following expression:

W=F\cdot d

Then, we can derive the magnitude of the force as:

F=\frac{W}{d}=\frac{\eta\cdot (G\cdot Eff)}{d}=\frac{0.3*1.9*(1.8*10^8)}{106*10^3} =968\,N

b) We will calculate the force for a speed of 30 m/s.

If the force is proportional to the speed, we have:

F_2=F_1(\frac{v_2}{v_1} )=968(\frac{30}{28} )=968*1.0714=1,037\,N

6 0
2 years ago
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