Answer:
3.5m/s^2
Explanation:
From Newton's second Law of Motion
F = ma
Where F is the applied force, m is the mass of the object and a is the acceleration.
F = 350 N
Mass = 100kg
350N = 100×a
a = 350/100
a = 3.5m/s^2
The acceleration of the object will be 3.5m/s^2
The power of the engine is 320 W.
<u>Explanation:</u>
Power may be defined as the rate of doing work (or) work done per unit time. One unit of energy is used to do the one unit of work.
Power = Work done / Time taken
Given, Force = 80 N, height = 5 m , final velocity = 4 m/s
To calculate the power, we must know the time taken.
To find the time, use the distance and speed formula which is given by
Time = Distance / speed
Here distance = 5 m and speed = 4 m/s
Time = 5 / 4 = 1.25 s.
Now, Power = work done / time
= (F * d) / t = (80 * 5) / 1.25
Power = 320 W.
The standard unit of power is watt (W) which is joule per second.
Adaptations can help organisms reproduce and continue living in different conditions
Answer: D <u>(chemical</u> -> <u>heat</u> -> <u>mechanical</u>)
In automobile engines the petrol/diesel fuel enter in to the engine cylinder, due to spark at the end of the compression, fuel burnt increase the temperature and pressure, develops heat <em>(chemical energy -> heat energy). </em><em>This heat energy acts on a piston develops the work on the crankshaft </em><em>( Heat energy -> Mechanical energy)</em><em>. </em>
Answer:
A) 1.4167 × 10^(-11) F
B) r_a = 0.031 m
C) E = 3.181 × 10⁴ N/C
Explanation:
We are given;
Charge;Q = 3.40 nC = 3.4 × 10^(-9) C
Potential difference;V = 240 V
Inner radius of outer sphere;r_b = 4.1 cm = 0.041 m
A) The formula for capacitance is given by;
C = Q/V
C = (3.4 × 10^(-9))/240
C = 1.4167 × 10^(-11) F
B) To find the radius of the inner sphere,we will make use of the formula for capacitance of spherical coordinates.
C = (4πε_o)/(1/r_a - 1/r_b)
Rearranging, we have;
(1/r_a - 1/r_b) = (4πε_o)/C
ε_o is a constant with a value of 8.85 × 10^(−12) C²/N.m
Plugging in the relevant values, we have;
(1/r_a - 1/0.041) = (4π × 8.85 × 10^(−12) )/(1.4167 × 10^(-11))
(1/r_a) - 24.3902 = 7.8501
1/r_a = 7.8501 + 24.3902
1/r_a = 32.2403
r_a = 1/32.2403
r_a = 0.031 m
C) Formula for Electric field just outside the surface of the inner sphere is given by;
E = kQ/r_a²
Where k is a constant value of 8.99 × 10^(9) Nm²/C²
Thus;
E = (8.99 × 10^(9) × 3.4 × 10^(-9))/0.031²
E = 3.181 × 10⁴ N/C
