Answer:
The gravitational force exerted on the object is 75 N (answer D)
Explanation:
Hi there!
The gravitational force is calculated as follows:
F = m · g
Where:
F = force of gravity.
m = mass of the object.
g = acceleration due to gravity (unknown).
For a falling object moving in a straight line, its height at a given time can be calculated using the following equation:
y = y0 + v0 · t + 1/2 · a · t²
Where:
y = position at time t.
y0 = initial position.
v0 = initial velocity.
t = time.
g = acceleration due to gravity.
Let´s place the origin of the frame of reference at the point where the object is released so that y0 = 0. Let´s also consider the downward direction as negative.
Then, after 2 seconds, the height of the object will be -30 m:
y = y0 + v0 · t + 1/2 · g · t²
-30 m = 0 m + 0 m/s · 2 s + 1/2 · g · (2 s)²
-30 m = 1/2 · g · 4 s²
-30 m = 2 s ² · g
-30 m/2 s² = g
g = -15 m/s²
Then, the magnitude of the gravitational force will be:
F = m · g
F = 5 kg · 15 m/s²
F = 75 N
The gravitational force exerted on the object is 75 N (answer D)
Have a nice day!
Answer: Normal force, N = 141.64 Newton
Explanation:
All the forces acting on the system and described in free body diagram are:
1) gravitational pull in downward direction
2) Normal force in upward direction
3) External force of 40 N acting at an angle of 37° with the horizontal can be resolved in two rectangular components:
i) F Cos 37° along the horizontal plane in forward direction and
ii) F Sin 37° along the vertical plane in downward direction
Applying the Newton's second law, net forces in the vertical plane are:
Net force, f = N - (mg + F Sin 37°)
As there is no acceleration in the vertical plane hence, net force f = 0.
So,
N - (mg + F Sin 37°) = 0
Adding (mg + F Sin 37°) both the sides in above equation, we get
N = mg + F Sin 37°
N = 12 
9.8 + 40 0.601 because (Sin 37° = 0.601)
N = 117.6 + 24.04
N = 141.64 Newton
Answer:
The value is 
Explanation:
From the we are told that
The radius of the sphere is 
The temperature is 
The average temperature of the rest of the universe is 
Generally the change in entropy of the entire universe per second is mathematically represented as
Here 
is the entropy of the rest of the universe which is mathematically represented as
Here Q is the quantity of heat radiated by the star which is mathematically represented as
Here 
is the Stefan-Boltzmann constant with value
=> 
=> 
So
=> 
Here 
is the entropy of the rest of the universe which is mathematically represented as
=> 
=> 
So
=>
Answer:
A thin layer of oil with index of refraction ng = 1.47 is floating above the water. The index of refraction of water is nw = 1.3. The index of refraction of air is na= 1. A light with wavelength λ = 775 nm goes in from the air to oil and water.
Part (a) Express the wavelength of the light in the oil, λ₀, in terms of λ and n⁰ (b) Express the minimum thickness of the film that will result in destructive interference, t min, in terms of λ o
(c) Express tmin in terms of λ and no.
(d) Solve for the numerical value of tmin in nm.
Explanation:
n₀ = 1.47
refraction of water = 1.3
refraction of air = 1
wavelength λ = 775 nm
(a) wavelength of light in water ⇒ λ₀ = λ / n₀
(b) minimum thickness of the film that will result in destructive interference
t(min) = λ₀ / 2
(c) the express t(min)
t = λ /2n₀
(d) the thickness is
t = 775 / 2(1.47)
= 263.61 nm
Answer:
E = 1.25×10¹³ N/m²
Explanation:
Young's modulus is defined as:
E = stress / strain
E = (F / A) / (dL / L)
E = (F L) / (A dL)
Given:
F = 10 kg × 9.8 m/s² = 98 N
L = 1 m
dL = 10⁻⁵ m
A = π/4 (0.001 m)² = 7.85×10⁻⁷ m²
Solve:
E = (98 N × 1 m) / (7.85×10⁻⁷ m² × 10⁻⁵ m)
E = 1.25×10¹³ N/m²
Round as needed.
