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Alina [70]
4 years ago
13

If a box is labeled 50 lbs, what is it's mass?

Physics
1 answer:
aleksandrvk [35]4 years ago
4 0
I would assume it is 50 pounds. that sounds weird?
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The distance between a proton and an electron is cut in half. How does the force of attraction change?
miv72 [106K]

According to Coulomb’s Law, the force will quadruple.

However…

Let’s assume, for the sake of argument, that both proton and electron behave entirely like particles.

The trajectory of the electron will stretch, and its distance from the nucleus will oscilate until is estabilizes again at the original value.

4 0
3 years ago
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In a new lab experiment, two parallel vertical metal rods are separated by L = 1.4 m . A R = 2.0-Ω resistor is connected from th
artcher [175]

Answer:

Explanation:

Let v be the terminal velocity of the bar .

emf induced in the bar of length L

= B L v where B is the value of magnetic field.

current  i in the circuit containing resistance R

i = induced emf / R

BLv / R

Magnetic force in upward direction in the bar

F = BiL

= BL x BLv / R

B²L²v / R

For attainment of uniform velocity

magnetic force = weight

B²L²v / R = mg

so current

i = BLv / R

8 0
3 years ago
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If an object floats, the volume of the displaced water is equal to the volume of the portion of the object that's above the surf
diamong [38]

Answer:

FALSE

Explanation:

When object float in water then it means that weight of the object is counterbalanced by the buoyancy force.

Here as we know that buoyancy force is the weight of water displaced by the object.

So here when object is submerged into the water then the submerged part of the volume of object will displace exactly the same volume of water and that volume of water will exert upwards buoyancy force on the object.

So here this statement is not true as it says that volume of water displaced is volume above the surface of water while it must be like volume of water displaced must be equal to volume of submerged part of the object.

4 0
3 years ago
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The string constrains the rotational and translational motion of the falling cylinder, given that it doesn't slip. what is the r
Sindrei [870]
<span>Answer: First we need to find the acceleration. torque on cylinder τ = T * r where T is the string tension; T = m(g - a) where a is the acceleration of the cylinder. Then τ = m(g - a)r But also τ = Iα. For a solid cylinder, I = ½mr², and if the string doesn't slip, then α = a / r, so τ = ½mr² * a/r = ½mra. Since τ = τ, we have m(g - a)r = ½mra → m, r cancel, leaving g - a = ½a g = 3a/2 a = 2g/3 where g, of course, is gravitational acceleration. We know that v(t) = a*t, so for our cylinder v(t) = 2gt / 3 ◄ linear velocity and ω = v(t) / r = 2gt / 3r ◄ angular velocity</span>
3 0
4 years ago
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Ryan places 0.150 kg of boiling water in a thermos bottle. How many kgs of ice at –12.0 °C must Ryan add to the thermos so that
Greeley [361]

Answer:

The  value  is   m_i =  0.0234 \  kg

Explanation:

Generally from the calorimetry principle we have that

      Heat \ loss\ by\ ice  =  heat\ gained\ by\  water\

So here heat gained water is mathematically represented as i.e

      Q_w  = m_w  *  c_w *  (T_w - T )

substituting  0.150 kg for m_w , 4200 J/kg.°C for  c_w , 100°C for   T_w and  75°C for  T

We have  

       Q_w  = 0.150  *  4200 *  (100 - 75 )

         Q_w  =15750 \  J

The Heat loss by the ice is mathematically represented as

      Q_i  = Q_1 + Q_2 +  Q_3

Here     Q_1 is the energy to move the ice to its melting point which is evaluated as  

        Q_1  =  m_i *  c_i * ( T_o -T_i)

Here  m_i is the mass of  ice

       c_i is the specific heat of ice with value  2.05 * 10^3   J/kg.°C

          T_o temperature of ice at melting point with value 0°C

           T_i is the temperature of ice with value  -12°C

Q_2 is the energy to move the ice from its  its melting point to liquid which is evaluated as  

     Q_2  = m_i  *  L

        Here  L  is the Latent heat of melting of ice with value    334 * 10^3   J/kg

Q_3 is the energy to move the ice from  liquid  to  the equilibrium temperature  which is evaluated as        

       Q_1  =  m_i *  c_w * ( T -T_o)

So  

     Q_i  = m_i [ c_i * ( T_o -T_i) + L  + c_w * ( T -T_o) ]

=>   Q_i  = m_i [ 2.05 * 10^3 * ( 0 -(-12)) + 334 * 10^3  +  4200 * ( 75 - 0) ]

From

 Heat \ loss\ by\ ice  =  heat\ gained\ by\  water\

We have that

  m_i *  673600  =15750

=>     m_i =  \frac{15750}{673600}

=>     m_i =  0.0234 \  kg

8 0
4 years ago
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