Describe how sound signals can be transmitted from a radio station and received on a radio at your home. Create a labeled diagra
2 answers:
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Answer:
Resistance = 3.35* Ω
Explanation:
Since resistance R = ρ
whereas
resistivity is given for two ends. At the left end resistivity is whereas x at the left end will be 0 as distance is zero. Thus
At the right end x will be equal to the length of the rod, so
Thus resistance will be R = ρ
where A = π
so,
Answer:
v₀ₓ = 15 m / s, = 5.2 m / s
v = 15.87 m / s , θ = 19.1
Explanation:
This is a projectile launch problem. The horizontal speed that is constant throughout the entire path is worth 15 m / s, instead the vertical speed changes in value due to the acceleration of gravity, let's look for the initial vertical speed
Vy² =² - 2 g y
² =
² + 2 g y
= √ (
² + 2 gy
Let's calculate
= √ (1.25² + 2 9.8 1.3)
= √ (27.04)
= 5.2 m / s
The initial speed can be calculated by the initial speed
v = √ v₀ₓ² + ²
v = RA (15² + 5.2²)
v = 15.87 m / s
We look for the angle with trigonometry
tan θ = voy / vox
θ = tan⁻¹ I'm going / vox
θ = tan⁻¹ 5.2 / 15
θ = 19.1
The answer is
v₀ₓ = 15 m / s
= 5.2 m / s
Answer:
The newton’s second law is F=ma
The Gravitational force is
Explanation:
Given that,
The equivalent formulas for velocity, acceleration, and force using the relationships covered for UCM, Newton’s Laws, and Gravitation.
We know that,
Velocity :
The velocity is equal to the rate of position of the object.
....(I)
Acceleration :
The acceleration is equal to the rate of velocity of the object.
....(II)
Newton’s second Laws
The force is equal to the change in momentum.
In mathematically,
Put the value of p
Put the value from equation (II)
This is newton’s second laws.
Gravitational force :
The force is equal to the product of mass of objects and divided by square of distance.
In mathematically,
Where, m₁₂ = mass of first object
m= mass of second object
r = distance between both objects
Hence, The newton’s second law is F=ma
The Gravitational force is
Answer:
The force applied on the big piston is 1306.67 N
Explanation:
Given;
force applied on small piston, F₁ = 200 N
diameter of the small piston, d₁ = 4.37 cm
radius of the small piston, r₁ = d₁/2 = 2.185 cm
Area of the small piston, A₁ = πr₁² = π(2.185 cm)² = 15 cm²
Area of the big piston, A₂ = 98 cm²
The pressure of the piston is given by;
Where;
F₂ is the force on big piston
Therefore, the force applied on the big piston is 1306.67 N