Actually Welcome to the concept of Efficiency.
Here we can see that, the Input work is given as 2.2 x 10^7 J and the efficiency is given as 22%
The efficiency is => 22% => 22/100.
so we get as,
E = W(output) /W(input)
hence, W(output) = E x W(input)
so we get as,
W(output) = (22/100) x 2.2 x 10^7
=> W(output) = 0.22 x 2.2 x 10^7 => 0.484 x 10^7
hence, W(output) = 4.84 x 10^6 J
The useful work done on the mass is 4.84 x 10^6 J
Answer:
(A) Velocity will be 1.88 m/sec
(b) Force will be 187.45 N
Explanation:
We have given work done = 4780 j
Distance d = 25.5 m
(A) Mass of the truck m = 
We know that kinetic energy is given by
So 
(B) We know that work done is given by
W = Fd
So 
Answer:
The (s) indicates that the state of matter for NaHCO3 is solid.
Explanation:
When a chemical reaction is written, the state of matter for each components of the reactants and products are mentioned in brackets along with their names or formulas.
For example, NaHCO3 has (s) mentioned in the brackets. The s shows that the state of matter for NaHCO3. (l) represents liquid format. (g) represents that the state of matter is gas.
Answer:
65.87 s
Explanation:
For the first time,
Applying
v² = u²+2as.............. Equation 1
Where v = final velocity, u = initial velocity, a = acceleration, s = distance
From the question,
Given: u = 0 m/s (from rest), a = 1.99 m/s², s = 60 m
Substitute these values into equation 1
v² = 0²+2(1.99)(60)
v² = 238.8
v = √238.8
v = 15.45 m/s
Therefore, time taken for the first 60 m is
t = (v-u)/a............ Equation 2
t = (15.45-0)/1.99
t = 7.77 s
For the final 40 meter,
t = (v-u)/a
Given: v = 0 m/s(decelerates), u = 15.45 m/s, a = -0.266 m/s²
Substitute into the equation above
t = (0-15.45)/-0.266
t = 58.1 seconds
Hence total time taken to cover the distance
T = 7.77+58.1
T = 65.87 s
4.6 j more. To get this take 7 and multiply it by 3.5 to get 24.5 take the x which is what you’re looking for and multiply it by the 2.1 to get 2.1x. Take 24.5 and divide it by 2.1 x and get 11.6. Subtract 11.6 by 7 and get 4.6
