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Pani-rosa [81]
3 years ago
5

If 125 j of heat energy is applied to a block of silver weighing 29.3 g, by how many degrees will the tem- perature of the silve

r increase?
Physics
1 answer:
Aloiza [94]3 years ago
5 0
In order to solve the problem, we should know the heat capacity of silver. Which is equal to 0.240 J/g C  

The formula: H = mCpT
Where h is the energy applied
M is the mass
Cp is the specific heat
T is the change in temperature  

125 J = ( 29.3 g ) ( 0.240 J/g C)
T Solve for T  
<span>T = 17.78 degree C increase </span>
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Consider two spaceships, each traveling at 0.50c in a straight line. Ship A is moving directly away from the Sun and ship B is a
attashe74 [19]

Answer:

The velocity of the light will be 1.0c only

Explanation:

The velocity of the light measured in the case given in question will be 1.0c only.

This is due to the fact that the velocity of light is never relative. The velocity of the light is maximum

The velocity of the light cannot be scaled down in no case

Thus, the velocity of the light remains as constant.

Hence, the velocity of the light measured will be 1.0c although the ships have relative velocity.

3 0
3 years ago
What is the tensile strength of hair ?
Olegator [25]
Human hair

Yield strength (MPa)
140-160

Ultimate tensile strength (MPa)
200-250

Hope that helps.
6 0
3 years ago
Read 2 more answers
An 8 kg mass moving at 8 m/s collides with a 6 kg mass
steposvetlana [31]

Answer:

10 m/s

Explanation:

Momentum before collision = momentum after collision

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

(8 kg)(8 m/s) + (6 kg)(6 m/s) = (8 kg)(5 m/s) + (6 kg) v

64 kg m/s + 36 kg m/s = 40 kg m/s + (6 kg) v

60 kg m/s = (6 kg) v

v = 10 m/s

3 0
3 years ago
Using this formula Vj = V; + at, If a vehicle starts from rest
Furkat [3]

Answer:

<em>The final speed of the vehicle is 36 m/s</em>

Explanation:

<u>Uniform Acceleration</u>

When an object changes its velocity at the same rate, the acceleration is constant.

The relation between the initial and final speeds is:

v_f=v_o+a.t

Where:

vf  = Final speed

vo = Initial speed

a   = Constant acceleration

t   = Elapsed time

The vehicle starts from rest (vo=0) and accelerates at a=4.5 m/s2 for t=8 seconds. The final speed is:

v_f=0+4.5*8

v_f=36\ m/s

The final speed of the vehicle is 36 m/s

5 0
3 years ago
A solid sphere of radius R is placed at a height of 30 cm on a15 degree slope. It is released and rolls, without slipping, to th
photoshop1234 [79]

Answer:

The height is  h_c = 42.857

A circular hoop of different diameter cannot be released from a height 30cm and match the sphere speed because from the conservation relation the speed of the hoop is independent of the radius (Hence also the diameter )

Explanation:

   From the question we are told that

           The height is h_s = 30 \ cm

            The angle of the slope is \theta = 15^o

According to the law of conservation of energy

     The potential energy of the sphere at the top of the slope = Rotational kinetic energy + the linear kinetic energy

                          mgh = \frac{1}{2} I w^2 + \frac{1}{2}mv^2

Where I is the moment of inertia which is mathematically represented as this for  a sphere

                    I = \frac{2}{5} mr^2

  The angular velocity w is mathematically represented as

                         w = \frac{v}{r}

So the equation for conservation of energy becomes

               mgh_s = \frac{1}{2} [\frac{2}{5} mr^2 ][\frac{v}{r} ]^2 + \frac{1}{2}mv^2

              mgh_s = \frac{1}{2} mv^2 [\frac{2}{5} +1 ]

             mgh_s = \frac{1}{2} mv^2 [\frac{7}{5} ]

            gh_s =[\frac{7}{10} ] v^2

              v^2 = \frac{10gh_s}{7}

Considering a circular hoop

   The moment of inertial is different for circle and it is mathematically represented as

             I = mr^2

Substituting this into the conservation equation above

              mgh_c = \frac{1}{2} (mr^2)[\frac{v}{r} ] ^2 + \frac{1}{2} mv^2

Where h_c is the height where the circular hoop would be released to equal the speed of the sphere at the bottom

                 mgh_c  = mv^2

                     gh_c = v^2

                     h_c = \frac{v^2}{g}

Recall that   v^2 = \frac{10gh_s}{7}

                    h_c= \frac{\frac{10gh_s}{7} }{g}

                      = \frac{10h_s}{7}

            Substituting values

                   h_c = \frac{10(30)}{7}

                       h_c = 42.86 \ cm    

       

     

                         

8 0
3 years ago
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