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Diano4ka-milaya [45]
3 years ago
8

if a metal object was cut on half how much heat will be required to raise both half to a specific temperature

Physics
1 answer:
il63 [147K]3 years ago
3 0

suppose the mass of a metal object be m and its specific heat capacity be s, and, H joules of heat is required to raise its temperature by t degrees Celsius

Than H is given by relation, H = mst

Now if this object is cut in two half's, than mass of half part will also be half <em>i.e. </em>M/2

So heat required to heat the half part will be = (\frac{m}{2}) st =\frac{mst}{2} = H/2  

Hence, the heat required to raise the half object to a specific temperature will also become half.

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A massless, rigid board is placed across two bathroom scales that are separated by a distance of 1.71 m. A person lies on the bo
Nata [24]

Answer:

Let the weight of the person be W and be located at a distance 'a' from the left scale as shown in the figure

Since the body is in equilibrium we can use equations of statics to analyse the problem.

Taking Sum of Moments about A we have

316\times 1.71-W\times a=0\\\\

Taking Sum of Moments about B we have

475\times 1.71-W\times (1.71-a)=0\\\\

Solving the above 2 equations for W and 'a' we get

316\times 1.71=W\times a\\\\475\times 1.71-W\times 1.71=-Wa\\\\\therefore W=\frac{316+475}{1}=791N\\\\\therefore a=\frac{316\times 1.71}{791}=0.683m

8 0
3 years ago
To what temperature must you raise a silver wire (c = 0.0038), originally at 20.0°C, to double its resistance, neglecting any ch
balu736 [363]

Answer:

T=283^{\circ}C

Explanation:

Given a material with temperature coefficient of resistance <em>c</em>, the equation that relates the resistance R_0 at temperature T_0 and the resistance R at temperature T is

\frac{R-R_0}{R_0}=c(T-T_0)

We want to double our resistance, so R=2R_0, thus having:

\frac{2R_0-R_0}{R_0}=\frac{R_0}{R_0}=1=c(T-T_0)

For this T must be:

1=cT-cT_0

T=\frac{1+cT_0}{c}

which for our values means (with T=20^{\circ}C=293^{\circ}K, remember to write temperature in S.I., and that for silver c=0.0038^{\circ}K^{-1}):

T=\frac{1+(0.0038^{\circ}K^{-1})(293^{\circ}K)}{(0.0038^{\circ}K^{-1})}=556^{\circ}K=283^{\circ}C

3 0
2 years ago
Use the equation for motion to answer the question.
Anika [276]
I choose the option D.
The velocity is constant, so it’s acceleration is 0 m/s^2.
X = 2 + 15 x 1 + 0 = 17 m
8 0
2 years ago
A 20-kg block is held at rest on the inclined slope by a peg. A 2-kg pendulum starts at rest in a horizontal position when it is
gregori [183]

Complete Question

The diagram of this question is shown on the first uploaded image

Answer:

The distance the block slides before stopping is d = 0.313 \ m

Explanation:

The free body diagram for the diagram in the question is shown

From the diagram the angle is \theta = 25 ^o

         sin \theta  = \frac{h}{d}

Where h = h_b - h_a

     So      d sin \theta  = h_b - h_a

From the question we are told that

      The mass of the block is  m = 20 \ kg

       The mass of the pendulum is  m_p = 2 \ kg

       The velocity of the pendulum at the bottom of swing is v_p = 15 m/s

        The coefficient of restitution is  e =0.7

         The coefficient of kinetic friction is  \mu _k = 0.5

The velocity of the block after the impact is mathematically represented as

            v_2 f = \frac{m_b - em_p}{m_b + m_p}  * v_2 i + \frac{[1 + e] m_1}{m_1 + m_2 } v_p

Where  v_2 i is the velocity of the block  before collision which is  0

                  = \frac{20 - (0.7 * 2)}{(2 + 20)} * 0 + \frac{(1 + 0.7) * 2 }{2 + 20}   * 15

Substituting value

                   v_2 f = 2.310\  m/s

According to conservation of energy principle

      The energy at point a  =  energy at point b

So    PE_A + KE _A = PE_B + KE_B  +  E_F

Where  

         PE_A is the potential energy at A which is mathematically represented as

          PE_A = m_b gh_a = 0 at the bottom

      KE _A is the kinetic energy at A  which is mathematically represented as

               K_A = \frac{1}{2} m_b * v_2f^2                  

         PE_B is the potential energy at B which is mathematically represented as  

            PE_B = m_b gh

From the diagram h = h_b -h_a

       PE_B = m_b g(h_b - h_a)

KE _B is the kinetic energy at B  which is 0 (at the top )

Where is E_F is the workdone against velocity  which from the diagram is

      \mu_k m_b g cos 25 *d

So

   \frac{1}{2} m_b v_2 f^2  = m_b g h_b + \mu_k m_b g cos \25 * d

Substituting values

   \frac{1}{2}  * 20 * 2.310^2 = 20 * 9.8 * d sin(25)  + 0.5* 20 * 9.8 * cos 25 * d    

So

       d = 0.313 \ m

       

   

6 0
3 years ago
A mirror forms an image because of which behavior of light?
Sveta_85 [38]
The correct answer is refraction.
4 0
3 years ago
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