Please find attached photograph for your answer. Please do comment whether it is useful or not.
Answer:
The speed of the large cart after collision is 0.301 m/s.
Explanation:
Given that,
Mass of the cart,
Initial speed of the cart,
Mass of the larger cart,
Initial speed of the larger cart,
After the collision,
Final speed of the smaller cart, (as its recolis)
To find,
The speed of the large cart after collision.
Solution,
Let is the speed of the large cart after collision. It can be calculated using conservation of momentum as :
So, the speed of the large cart after collision is 0.301 m/s.
Answer:
W=1055N
Explanation:
In order to solve this problem, we must first do a drawing of the situation so we can visualize theh problem better. (See attached picture)
In this problem, we will ignore the board's weight. As we can see in the free body diagram of the board, there are only three forces acting on the system and we can say the system is in vertical equilibrium, so from this we can say that:
so we can do the sum now:
when solving for the Weight W, we get:
and now we can substitute the given data, so we get:
W=410N+645N
W=1055N
Answer:
(a) 5.43 x 10⁵ J
(b) 3.07 x 10⁵ J
(c) 45 °C
Explanation:
(a)
= Latent heat of fusion of ice to water = 3.33 x 10⁵ J/kg
m = mass of ice = 1.63 kg
= Energy required to melt the ice
Energy required to melt the ice is given as
= m
= (1.63) (3.33 x 10⁵)
= 5.43 x 10⁵ J
(b)
E = Total energy transferred = 8.50 x 10⁵ J
Q = Amount of energy remaining to raise the temperature
Using conservation of energy
E = + Q
8.50 x 10⁵ = 5.43 x 10⁵ + Q
Q = 3.07 x 10⁵ J
(c)
T₀ = initial temperature = 0°C
T = Final temperature
m = mass of water = 1.63 kg
c = specific heat of water = 4186 J/(kg °C)
Q = Amount of energy to raise the temperature of water = 3.07 x 10⁵ J
Using the equation
Q = m c (T - T₀)
3.07 x 10⁵ = (1.63) (4186) (T - 0)
T = 45 °C
1) find speed (8.8 m/s)
2) find acceleration (38.7 m/s^2)
answer is about 38.7 m/s^2