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kari74 [83]
3 years ago
15

Beating a carpet with a carpet beater.

Physics
1 answer:
prisoha [69]3 years ago
8 0
I would say a short person with muscles considering they are closer to the ground, but they may not be able to build up as much force in such a short time compared to the tall person.
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A 7.0-kilogram cart, A, and a 3.0-kilogram cart, B, are initially held together at rest on a horizontal, frictionless surface. W
7nadin3 [17]
For this problem, we use the conservation of momentum as a solution. Since momentum is mass times velocity, then,

m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'
where
v₁ and v₂ are initial velocities of cart A and B, respectively
v₁' and v₂' are final velocities of cart A and B, respectively
m₁ and m₂ are masses of cart A and B, respectively

(7 kg)(0 m/s) + (3 kg)(0 m/s) = (7 kg)(v₁') + (3 kg)(6 m/s)
Solving for v₁',
v₁' = -2.57 m/s

<em>Therefore, the speed of cart A is at 2.57 m/s at the direction opposite of cart B.</em>
7 0
3 years ago
Which information about the earth would not be represented when a globe is used as a model?
egoroff_w [7]
A globe sitting on the desk can't demonstrate the speed of axial rotation
or the speed of orbital revolution.
5 0
3 years ago
Two rigid rods are oriented parallel to each other and to the ground. The rods carry the same current in the same direction. The
Greeley [361]

Answer:

I = 215.76 A  

Explanation:

The direction of magnetic field produced by conductor 1 on the location of conductor 2 is towards left. Based on Right Hand Rule -1 and taking figure 21.3 as reference, the direction of force Fm due to magnetic field produced at C_2 is shown above. The force Fm balances the weight of conductor 2.  

Fm = u_o*I^2*L/2*π*d

where I is the current in each rod, d = 0.0082 m is the distance 27rId  

between each, L = 0.85 m is the length of each rod.

Fm = 4π*10^-7*I^2*1.1/2*π*0.0083

The mass of each rod is m = 0.0276 kg  

F_m = mg

4π*10^-7*I^2*1.1/2*π*0.0083=0.0276*9.8

I = 215.76 A  

note:

mathematical calculation maybe wrong or having little bit error but the method is perfectly fine

5 0
3 years ago
Consider the following arrangement with a frictionless/massless pulley. Determine the force F required to move block A if the co
Delicious77 [7]

Answer: The free - body diagrams for blocks A and B. frictionless surface by a constant horizontal force F = 100 N. Find the tension in the cord between the 5 kg and 10 kg blocks. The string that attaches it to the block of mass M2 passes over a frictionless pulley of negligible mass. The coefficient of kinetic friction Hk between M.

Explanation: Hope this helped :)

6 0
2 years ago
Nitrogen at 100 kPa and 25oC in a rigid vessel is heated until its pressure is 300 kPa. Calculate (a) the work done and (b) the
nignag [31]

Answer:

A. The work done during the process is W = 0

B. The value of heat transfer during the process Q = 442.83 \frac{KJ}{kg}

Explanation:

Given Data

Initial pressure P_{1} = 100 k pa

Initial temperature T_{1} = 25 degree Celsius = 298 Kelvin

Final pressure P_{2} = 300 k pa

Vessel is rigid so change in volume of the gas is zero. so that initial volume is equal to final volume.

⇒ V_{1} = V_{2} ------------- (1)

Since volume of the gas is constant so pressure of the gas is directly proportional to the temperature of the gas.

⇒ P ∝ T

⇒ \frac{P_{2} }{P_{1}} = \frac{T_{2} }{T_{1}}

⇒ Put all the values in the above formula we get the final temperature

⇒ T_{2} = \frac{300}{100} × 298

⇒ T_{2} = 894 Kelvin

(A). Work done during the process is given by W = P × (V_{2} -V _{1})

From equation (1), V_{1} = V_{2} so work done W = P × 0 = 0

⇒ W = 0

Therefore the work done during the process is zero.

Heat transfer during the process is given by the formula Q = m C_{v} ( T_{2} -T_{1} )

Where m = mass of the gas = 1 kg

C_{v} = specific heat at constant volume of nitrogen = 0.743 \frac{KJ}{kg k}

Thus the heat transfer Q = 1 × 0.743 × ( 894- 298 )

⇒ Q = 442.83 \frac{KJ}{kg}

Therefore the value of heat transfer during the process Q = 442.83 \frac{KJ}{kg}

6 0
3 years ago
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