Answer:
Explanation:
The formula for potential energy is:
where <em>m </em>is the mass, <em>g</em> is the gravitational acceleration, and <em>h</em> is the height.
The mass of the book is 0.4 kilograms. The gravitational acceleration on Earth is 9.8 m/s². The height of the book is 2 meters.
Substitute the values into the formula.
Multiply the first two numbers.
- 0.4 kg*9.8 m/s²= 3.92 kg*m/s²
- If we convert the units now, the problem will be much easier later on.
- 1 kg*m/s² is equal to 1 Newton. So, our answer of 3.92 kg*m/s² is equal to 3.92 N
Multiply.
- 3.92 N* 2 m=7.84 N*m
- 1 Newton meter is equal to 1 Joule (this is why we converted the units).
- Our answer is equal to<u> 7.84 Joules.</u>
Answer:
0.572
Explanation:
First examine the force of friction at the slipping point where Ff = µsFN = µsmg.
the mass of the car is unknown,
The only force on the car that is not completely in the vertical direction is friction, so let us consider the sums of forces in the tangential and centerward directions.
First the tangential direction
∑Ft =Fft =mat
And then in the centerward direction ∑Fc =Ffc =mac =mv²t/r
Going back to our constant acceleration equations we see that v²t = v²ti +2at∆x = 2at πr/2
So going backwards and plugging in Ffc =m2atπr/ 2r =πmat
Ff = √(F2ft +F2fc)= matp √(1+π²)
µs = Ff /mg = at /g √(1+π²)=
1.70m/s/2 9.80 m/s² x√(1+π²)= 0.572
Answer: Look where the points are.
Explanation:
Answer:
a. Decreases
b. Increases
c. Remains the same
d. Increases
Explanation:
a. Capacitance is given by c= Ak€/d
where A is conductivity plate with Area
K is a constant
€ is dielectric with permittivity.
d is the distance
b. Potential difference is given by
V = Ed, since, the electric field remains the
same, the potential diterence also increases with increase in distance.
Since the capacitance depends upon the distance, and all the other factors are kept constant, the capacitance decreases.
c. Electric field remains the same because charge on the
plate remains the same.
d. since electric field remains the same and capacitance decreases, the energy increases.
E= 1/2c * Q^2
Answer:
(C) 40m/s
Explanation:
Given;
spring constant of the catapult, k = 10,000 N/m
compression of the spring, x = 0.5 m
mass of the launched object, m = 1.56 kg
Apply the principle of conservation of energy;
Elastic potential energy of the catapult = kinetic energy of the target launched.
¹/₂kx² = ¹/₂mv²
where;
v is the target's velocity as it leaves the catapult
kx² = mv²
v² = kx² / m
v² = (10000 x 0.5²) / (1.56)
v² = 1602.56
v = √1602.56
v = 40.03 m/s
v ≅ 40 m/s
Therefore, the target's velocity as it leaves the spring is 40 m/s
