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natulia [17]
3 years ago
15

A box can slide with negligible friction on a horizontal table top. a string of negligible mass runs horizontally over a dowel w

ith negligible friction, and then connects to an identical box which is falling vertically. m m m~g how does the tension in the string between the boxes compare to the net force on the boxes? 1. it is less, about 50%.
Physics
1 answer:
Lina20 [59]3 years ago
3 0
<span>it is less, about 50%. Since we're ignoring friction, let's do the math and see what happens. I'll use the variables: M = mass of a single box. g = gravitational acceleration. The only force available is from the box that's suspended, so we'll have (1) Mg Newtons of force available. Now since the 2 boxes are connected together, that force will have to accelerate 2M. So the acceleration of the group is gM/2M = g/2 So both boxes are accelerating at half the local gravitational acceleration. This makes sense since the gravitational acceleration is only affecting half of the mass being accelerated. Now, to see the tension in the string, you need to ask the question, "How much force is required to accelerate the mass sliding horizontally at g/2?" The answer to that question is obviously Mg/2 newtons. So the net force being supplied is Mg newtons (See equation (1) above), and the force being transmitted via the string is Mg/2 Newtons, it's pretty obvious that the tension in the string is 50% of the total net force.</span>
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A 3.53-g lead bullet traveling at 428 m/s strikes a target, converting its kinetic energy into thermal energy. Its initial tempe
Taya2010 [7]

Complete question:

A 3.53-g lead bullet traveling at 428 m/s strikes a target, converting its kinetic energy into thermal energy. Its initial temperature is 40.0°C. The specific heat is 128 J/(kg · °C), latent heat of fusion is 24.5 kJ/kg, and the melting point of lead is 327°C.

(a) Find the available kinetic energy of the bullet. J

(b) Find the heat required to melt the bullet. J

Answer:

Part (a) the available kinetic energy of the bullet is 323.32 J

Part (b) the heat required to melt the bullet is 216.17 J

Explanation:

Given;

mass of the bullet = 3.53 g = 0.00353 kg

velocity of the bullet = 428 m/s

initial temperature of the bullet = 40.0°C

final temperature of the bullet =  327°C

specific heat capacity, c= 128 J/(kg · °C)

latent heat of fusion, Hf  = 24.5 kJ/kg

Part (a) the available kinetic energy of the bullet. J

KE = ¹/₂ × mv²

KE = ¹/₂ × 0.00353 × 428²

     = 323.32 J

Part (b) the heat required to melt the bullet. J

This is the thermal energy required to increase the temperature of the bullet and the heat energy required to melt the bullet.

Quantity of heat required to raise the temperature of the bullet:

Q = mcΔT

   = 0.00353 × 128 × (327-40)

   = 0.00353 × 128 × 287

   = 129.68 J

Quantity of heat required to melt the bullet:

Q = mH_f

Q = 0.00353 × 24500

   = 86.49 J

TOTAL energy required to melt the bullet = 129.68 J + 86.49 J

                                                                      = 216.17 J

3 0
3 years ago
A baseball player was hit by a baseball in the left eye and complains of double vision. What should you do? A. Restrict spinal m
Alex
I think the smartest answer would be C sorry if i’m writing though
4 0
3 years ago
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A 12.0kg microwave oven is pushed 14.0m up the sloping surface of a loading ramp inclined at an angle 37 degrees above the horiz
Semenov [28]

Answer:

Answered

Explanation:

a) What is the work done on the oven by the force F?

W = F * x

W = 120 N * (14.0 cos(37))

<<<< (x component)

W = 1341.71

b) F_f=\mu_k N

F_f=0.25\times12\times9.8

= 29.4 N

W_f= F_f\times x

W_f= 29.0\times 14 cos(37)

W_f= 328.72 J = 329 J

c) increase in the internal energy

U_2 = mgh

= 12*9.81*14sin(37)

= 991 J

d) the increase in oven's kinetic energy

U_1 + K_1 + W_other = U_2 + K_2

0 + 0 + (W_F - W_f ) = U_2 + K_2

1341.71 J - 329 J - 991 J = K_2

K_2 = 21.71 J

e) F - F_f = ma

(120N - 29.4N ) / 12.0kg = a

a = 7.55m/s^2

vf^2 = v0^2 + 2ax

vf^2 = 2(7.55m/s)(14.0m)  

V_f = 14.5396m/s

K = 1/2(mv^2)

K = 1/2(12.0kg)(14.5396m/s)

K = 87.238J

4 0
3 years ago
A sample of helium has a volume of 12.7 m3. The temperature is raised to 323 K at which time the gas occupies 32.5 m3? Assume pr
jasenka [17]

Answer: The original temperature was

T_{1}=126.51K

Explanation:

Let's put the information in mathematical form:

V_{1}=12.7m^{3}

T_{1}=?

V_{2}=32.5m^{3}

T_{2}=323K

P_{1}=P_{2}=3atm

If we consider the helium as an ideal gas, we can use the Ideal Gas Law:

PV=nRT

were <em>R</em> is the gas constant. And <em>n</em> is the number of moles (which we don't know yet)

From this, taking R=0.08205746\frac{atm.l}{mol.K},  we have:

n=\frac{P_{2}V_{2}}{RT_{2}}

⇒n=3.67mol

Now:

T_{1}=\frac{P_{1}V_{1}}{nR}

⇒T_{1}=126.51K

7 0
3 years ago
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A student drops two metallic objects into a 120-g steel container holding 150 g of water at 25°C. One object is a 206-g cube of
spayn [35]

Answer:

Mass of the aluminium chunk = 278.51 g

Explanation:

For an isolated system as given the energy lost and gains in the system will be zero therefore sum of all transfer of energy will be zero,as the temperature will also remain same

A specific heat formula is given as                  

Energy Change = Mass of liquid x Specific Heat Capacity x Change in temperature

                                       Q =  m×c×ΔT

                        Heat gain by aluminium + heat lost by copper  = 0    (1)

For Aluminium:

      Q = m\times0.897\frac{J}{g.k}\times(25-5)

      Q = m x 17.94 joule

For Copper:

Q= 206g\times0.385\frac{J}{g.k} \times(88-25)

       Q= 4996.53 Joule

from eq 1

     m x 17.94 = 4996.53

     mass of aluminium = \frac{4996.53}{17.94} g

    Mass of the aluminium chunk = 278.51 g

                         

3 0
3 years ago
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