Answer:
speed of puck acc. to the radar gun = 138 km/h
speed of player = 15 km/h
since the player is in motion when he shoots, the speed of the puck will be the sum of the speed of the player and the speed at which he shot. so,
speed of puck = speed of player + speed of puck acc. to player
138 = 15 + speed of puck acc. to player
speed of puck acc. to player = 138 -15
speed of puck acc. to player = 123 km/h
Brainly this answer if you think it deserves it
Answer:
a) 2.87 m/s
b) 3.23 m/s
Explanation:
The avergare velocity can be found dividing the length traveled d by the total time t.
a)
For the first part we easily know the total traveled length which is:
d = 50.2 m + 50.2 m = 100.4 m
The time can be found dividing the distance by the velocity:
t1 = 50.2 m / 2.21 m/s = 22.7149 s
t2 = 50.2 m / 4.11 m/s = 12.2141 s
t = t1 +t2 = 34.9290 s
Therefore, the average velocity is:
v = d/t =2.87 m/s
b)
Here we can easily know the total time:
t = 1 min + 1.16 min = 129.6 s
Now the distance wil be found multiplying each velocity by the time it has travelled:
d1 = 2.21 m/s * 60 s = 132.6 m
d2 = 4.11 m/s *(1.16 * 60 s) = 286.056 m
d = 418.656 m
Therefore, the average velocity is:
v = d/t =3.23 m/s
Answer:
Capacitance of cylindrical capacitor does not depends on the amount of charge on the conductors
Explanation:
Consider a cylindrical capacitor of length L, inner radius R₁ and outer radius R₂, permitivity ε₀ constant then capacitance of cylindrical capacitor is given by:
From this equation it is clear that capacitance of cylindrical capacitor is independent of the amount of charge on the conductors where as directly proportional permitivity constant and length of cylinder where as inversely proportional to natural log of ratio of R₂ and R₁
Answer: 13.94 tons/s
Explanation:
On adding heat energy to a substance, the temperature would be changed by a particular amount. This relationship between heat energy and temperature is often different for each material. The specific heat, is a value that describes how they relate.
Heat energy = mass flow rate * specific heat * Δ T
Q = MC (ΔΦ)
Heat energy, Q= 3.5*10^8J
Mass flow rate, M= ?
Specific heat, C= 4184j/KgC
Change in temperature, ΔΦ= 6°C
M = Q/CΔΦ
M = (3.5*10^8)/4184*6
M = 13942kg/s
M = 13.94 tons/s
Work of the force = 10 N
Time required for the work = 50 sec
Height = 7 m
We are given with the value of work and time in the question.
Substitute the values in the formula of power and then you'll get the power required.
We know that,
w = Work
p = Power
t = Time
By the formula,
Given that,
Work (w) = 7 m = 70 Joules
Time (t) = 50 sec
Substituting their values,
p = 70/50
p = 1.4 watts
Therefore, the power required is 1.4 watts.
Hope it helps!
