Answer:
If the temperature of the colder object rises by the same amount as the temperature of the hotter object drops, then <u>the specific heats of both objects will be equal.</u>
Explanation:
If the temperature of the colder object rises by the same amount as the temperature of the hotter object drops when the two<u> objects of same mass</u> are brought into contact, then their specific heat capacity is equal.
<u>We can prove this by the equation of heat for the two bodies:</u>
<em>According to given condition,</em>
<em>when there is no heat loss from the system of two bodies then </em>
- Thermal conductivity is ultimately affects the rate of heat transfer, however the bodies will attain their final temperature based upon their mass and their specific heat capacities.
The temperature of the colder object will rise twice as much as the temperature of the hotter object only in two cases:
- when the specific heat of the colder object is half the specific heat of the hotter object while mass is equal for both.
OR
- the mass of colder object is half the mass of the hotter object while their specific heat is same.
Copper is a reddish orange, soft metal with a bright metallic luster. It is an excellent conductor of heat and electricity. Copper surfaces when exposed to air gradually turns to a dull, brownish color.
Answer:
a) fr = 224.3 N
, b) fr = 224.3 N
, c) v = 198.0 m/s
Explanation:
a) For this exercise let's start by calculating the acceleration in the fall
v² = v₀² - 2 a (y-y₀)
When it jumps the initial vertical speed is zero
a = -v² / 2 (y-y₀)
a = -68 2/2 (1000-2000)
a = 2,312 m / s²
Let's use the second net law to enter the average friction force
fr = m a
fr = 97 2,312
fr = 224.3 N
b) let's look for acceleration
v² = v₀² - 2 a y
a = (v² –v₀²) / 2 (y-y₀)
a = (4² - 68²) / 2 (0-1000)
a = 2,304 m / s²
fr = m a
fr = 97 2,304
fr = 223.5 N
c) the speed of the wallet is searched with kinematics
v² = v₀² - 2 g (y-y₀)
v = √ (0-2 9.8 (0-2000))
v = 198.0 m/s
Answer:
1 joule = 0.737 foot-pound
Joule is the unit of work.
1 J = 1 N·m
In SI units
1 J = 1 kg· m/s²
0.737 foot-pound is the amount of work to raise 0.737 pounds one foot or raising one pound to 0.737 ft.
<h2>
Answer:</h2>
(a) 10N
<h2>
Explanation:</h2>
The sketch of the two cases has been attached to this response.
<em>Case 1: The box is pushed by a horizontal force F making it to move with constant velocity.</em>
In this case, a frictional force 
is opposing the movement of the box. As shown in the diagram, it can be deduced from Newton's law of motion that;
∑F = ma -------------------(i)
Where;
∑F = effective force acting on the object (box)
m = mass of the object
a = acceleration of the object
∑F = F -
m = 50kg
a = 0 [At constant velocity, acceleration is zero]
<em>Substitute these values into equation (i) as follows;</em>
F - = m x a
F - = 50 x 0
F - = 0
F = -------------------(ii)
<em>Case 2: The box is pushed by a horizontal force 1.5F making it to move with a constant velocity of 0.1m/s²</em>
In this case, the same frictional force is opposing the movement of the box.
∑F = 1.5F -
m = 50kg
a = 0.1m/s²
<em>Substitute these values into equation (i) as follows;</em>
1.5F - = m x a
1.5F - = 50 x 0.1
1.5F - = 5 ---------------------(iii)
<em>Substitute </em><em> = F from equation (ii) into equation (iii) as follows;</em>
1.5F - F = 5
0.5F = 5
F = 5 / 0.5
F = 10N
Therefore, the value of F is 10N
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