Answer:
d. 37 °C
Explanation:
= mass of lump of metal = 250 g
= specific heat of lump of metal = 0.25 cal/g°C
= Initial temperature of lump of metal = 70 °C
= mass of water = 75 g
= specific heat of water = 1 cal/g°C
= Initial temperature of water = 20 °C
= mass of calorimeter = 500 g
= specific heat of calorimeter = 0.10 cal/g°C
= Initial temperature of calorimeter = 20 °C
= Final equilibrium temperature
Using conservation of heat
Heat lost by lump of metal = heat gained by water + heat gained by calorimeter
Given data:
mass of the bullet (m) = 25 g = 0.025 kg,
mass of the gun (M) = 0.9 kg,
speed of the bullet (v) =230 m/s,
speed of the bullet (V) = ?
From the given data it is clear that, the momentum is conserved. According to "<em>law of conservation of momentum" </em>the total momentum before and after the collision is equal.
In this problem the momentum before collision (bullet+gun) is zero.
Therefore, after the gun fires a bullet, the momentum must be zero.
Mathematically,
M × V + m × v = 0
where,
M × V = momentum of the gun
m × v = momentum of the bullet
(0. 9 × V) + (0.025 × 230) = 0
0.9 V = -5.75
V = -5.75/0.9
= -6.39 m/s
<em>The gun recoils with a speed of 6.39 m/s</em>
Answer:8.3m/sec 30 sec,
Explanation:
A student practicing for a track meet, ran 250 m in 30 sec. a. What was her average speed? 250 m = 8.3 m/sec 30 sec.
Answer:
a) R `= 3.5 ohms
b) energy decipated = 560J
Explanation:
V = I . R
R = V / I
R `= 70 / 20
R `= 3.5 ohms
2)energy decipated = 1/2ij²
energy decipated = 1/2 x 2.8 x (20)²
energy decipated = 560J
Answer:
d) The 2 athletes reach the same height, because the athletes run with the same speed.
Explanation:
In the whole process , kinetic energy is converted into potential energy .
1/2 m v² = mgh
v² = 2gh
h = v² / 2g
In this expression we see that height attained does not depend upon mass of the object . At the same time it also makes it clear that it depends upon velocity . As the velocity in both the cases are same , height attained by both of them will be same. Hence option d ) is correct.
