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Gre4nikov [31]
3 years ago
15

As long as the net force on a falling object is not zero, the object will...

Physics
1 answer:
Lady_Fox [76]3 years ago
7 0
The object accelerates downwards due to the force of gravity
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When reaching a boundary between two media (1 and 2), an incident ray is partially reflected and partially refracted. The ray is
lukranit [14]

Answer:

The angle of incidence when the reflected ray is perpendicular to the incident ray = 45°

Explanation:

According to Snell's Law,

n₁ sin θ₁ = n₂ sin θ₂

When the angle between the incident ray and reflected ray is 90°, the angle of incidence is θ₁ and the angle of reflection, θ₂ = 90° - θ₁ and the index of refraction in the Snell's Law for both media would be the same, n₁ = n₂ = n

n sin θ₁ = n sin (90° - θ₁)

Note that from trigonometric relations,

Sin (90° - θ₁) = cos θ₁

n sin θ₁ = n cos θ₁

(sin θ₁)/(cos θ₁) = 1

tan θ₁ = 1

θ₁ = arctan 1 = 45°

Hope this Helps!!!

7 0
3 years ago
Read 2 more answers
How to do this question about lens? Please help.
hichkok12 [17]

Its six cm mate because you are multiply the numbers together then that total with the number of sides there are

4 0
3 years ago
What are (a) the charge and (b) the charge density on the surface of a conducting sphere of radius 0.20 m whose potential is 240
Lady bird [3.3K]

Answer:

(a) charge q=5.33 nC

(b) charge density σ=10.62 nC/m²

Explanation:

Given data

radius r=0.20 m

potential V=240 V

coulombs constant k=9×10⁹Nm²/C²

To find

(a) charge q

(b) charge density σ

Solution

For (a) charge q

As

V_{potential}=kq/r\\ q=rV_{potential}/k\\q=\frac{(0.20)(240)}{9*10^{9} }\\ q=5.333*10^{-9}C\\or\\ q=5.33nC

For (b) charge density

As charge density σ is given as:

σ=q/(4πR²)

σ=(5.333×10⁻⁹) / (4π×(0.20)²)

σ=10.62 nC/m²

3 0
3 years ago
Suppose that you make a series RC circuit with a capacitor and a known resistor that has a 5% tolerance: R= 5.20 ± 0.26kΩ. You p
Masja [62]

Answer:

correct answer is C

Explanation:

The time constant of an RC circuit is

           τ = RC

so to find the capacitance

          C = τ/ R

          C = 2.150 / 5.20 10³

          C = 4.13 10⁻⁴ F

to find the error we use the worst case

         ΔC = | |\frac{dC}{d \tau }| \ \Delta  \tau + | \frac{dC}{dR} | \ \Delta R

the absolute value guarantees that we find the worst case, we evaluate the derivatives

          ΔC = 1 /R Δτ + τ/R²  ΔR

the absolute values ​​of the errors are

          Δτ = 0.002 s

          ΔR = 0.3 kΩ

we substitute

           ΔC = 0.002 /5.20 10³ + 2.150/(5.20 10³)²   0.3 10³

           ΔC = 3.8 10⁻⁷ + 1.74 10⁻⁵

           ΔC = 1.77 10⁻⁵ F

the uncertainty or error must be expressed with a significant figure

            ΔC = 2 10⁻⁵ F

the percentage error is

            Er% =\frac{\Delta C}{C} \ 100

            Er% = \frac{2 \ 10^{-5} }{ 4.13 \ 10^{-4} } \ 100

            Er% = 4.8%

the correct answer is C

3 0
3 years ago
A 2 N and an 8 N force pull on an object to the right and a 4 N force pulls to the left a 0.5 kg object. What is the acceleratio
lara31 [8.8K]

Answer:

Acceleration is 12m/s^2

Explanation:

We have a resultant force of 10N to the right and a resultant of 4N to the left, since the tow forces are acting in opposite directions, we subtract the two forces to find the net force. The net force would be 6N to the right.

We also know that F=ma, where F=force, m=mass, and a=acceleration

we can rearrange the equation like this,

a=F/m

now we can plug in the known variables

a=6N/0.5kg

a=12m/s^2

8 0
2 years ago
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