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Brums [2.3K]
2 years ago
15

2. A kg of iron and a kilo of lead, both at 1000 degrees Celsius, are kept on ice. It is discovered after some time that the iro

n has melted more ice than the lead. Why? 2 A vonal containing un or in kont in on oron The tomis motor an hoot​
Physics
1 answer:
Sunny_sXe [5.5K]2 years ago
8 0

The specific heat capacity of iron is greater than that of lead hence the iron will melt the ice more than the lead.

<h3>What is the specific heat capacity?</h3>

The specific heat capacity refers to the amount of heat that is required to raise the temperature of 1Kg of a body by 1 K.

We know that the specific heat capacity of iron is greater than that of lead hence the iron will melt the ice more than the lead.

Learn more about heat capacity:brainly.com/question/13499849

#SPJ1

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Which term refers to a variable that a scientist adjusts during an experiment?
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it’s c

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A T-Bar is similar to a_______
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3 years ago
A box of mass 26 kg is initially at rest on a flat floor. The coefficient of kinetic friction between the box and the floor is 0
Kazeer [188]

Answer:

\Delta K = 52J

Explanation:

The change in kinetic energy will be simply the difference between the final and initial kinetic energies: \Delta K=K_f-K_i

We know that the formula for the kinetic energy for an object is:

K=\frac{mv^2}{2}

where <em>m </em>is the mass of the object and <em>v</em> its velocity.

For our case then we have:

\Delta K = K_f-K_i=\frac{mv_f^2}{2}-\frac{mv_i^2}{2}=\frac{m(v_f^2-v_i^2)}{2}

Which for our values is:

\Delta K = \frac{m(v_f^2-v_i^2)}{2} = \frac{(26Kg)((2m/s)^2-(0m/s)^2)}{2} = 52J

3 0
3 years ago
A hockey stick of mass ms and length L is at rest on the ice (which is assumed to be frictionless). A puck with mass mp hits the
krek1111 [17]

Answer:

L = mp*v₀*(ms*D) / (ms + mp)

Explanation:

Given info

ms = mass of the hockey stick

uis = 0 (initial speed of the hockey stick before the collision)

xis = D (initial position of center of mass of the hockey stick before the collision)

mp = mass of the puck

uip = v₀ (initial speed of the puck before the collision)

xip = 0 (initial position of center of mass of the puck before the collision)

If we apply

Ycm = (ms*xis + mp*xip) / (ms + mp)

⇒  Ycm = (ms*D + mp*0) / (ms + mp)

⇒  Ycm = (ms*D) / (ms + mp)

Now, we can apply the equation

L = m*v*R

where m = mp

v = v₀

R = Ycm

then we have

L = mp*v₀*(ms*D) / (ms + mp)

5 0
3 years ago
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