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fredd [130]
3 years ago
11

A common magnifying glass is an example of.​

Physics
2 answers:
Misha Larkins [42]3 years ago
8 0

Answer:

planoconvex lens

Explanation:

maxonik [38]3 years ago
7 0

Answer:
A convex lens. A convex can magnify objects when kept at a particular position.
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An alternating-current (AC) source supplies a sinusoidally varying voltage that can be described with the function v of t is equ
Marrrta [24]

Answer:

ω, the angular frequency of the source equals 377 rad/s

Explanation:

From the question, V(t) = V cosωt.

Now, ω = the angular frequency of the sinusoidal wave is given by

ω = 2πf where f = the frequency of the source = 60 Hz

So, the angular frequency of the source ,ω = 2π × the frequency of the source.

So, ω = 2πf

ω = 2π × 60 Hz

ω = 120π rad/s

ω = 376.99 rad/s

ω ≅ 377 rad/s

So, ω, the angular frequency of the source equals 377 rad/s

3 0
3 years ago
At a distance of 8 m, the sound intensity of one speaker is 66 dB. If we were to place 3 speakers in a circle of radius 8 m, wha
ludmilkaskok [199]

Answer:

dβ = 70. 77 dβ

Explanation:

The intensity of sound in decibels is

         dβ = 10 log I/I₀

let's look for the intensity of this signal

         I / I₀ = 10 dβ/10

         I / I₀ = 3.981 10⁶

the threshold intensity of sound for humans is I₀ = 1 10⁻¹² W / m²

         I = 3.981 10 ⁶ 1 10⁻¹²

         I = 3,981 10⁻⁶ W / m²

It is indicated that 3 cornets are placed in the circle, for which total intensity is

        I_total - 3 I

        I_total = 3  3,981 10⁻⁶

        I_total = 11,943 10⁻⁶ W / m²

let's reduce to decibels

      dβ = 10 log (11,943 10⁻⁶/1 10⁻¹²)

      dβ = 10  7.077

      dβ = 70. 77 dβ

3 0
3 years ago
n astronaut who weighs 800 N on the surface of the earth lifts off from planet Zuton in a space ship. The free-fall acceleration
ANTONII [103]

Answer: 0.29 kN

Explanation:

We have the following data:

W_{E}=800 N is the weight of the astronaut on Earth

g_{E}=9.8 m/s^{2} is the free fall acceleration due gravity on Earth (directed downwards)

g_{Z}=3 m/s^{2} is the free fall acceleration due gravity on Zuton (directed downwards)

a=0.5 m/s^{2} is the acceleration of the spaceship at litoff (directed upwards)

We have to find the <u>magnitude of the force</u> F the space ship exerts on the astronaut.

Firstly, we have to know weight has a direct relation with the mass and the acceleration due gravity. In the case of Earth is:

W_{E}=mg_{E} (1)

Where m is the mass of the atronaut.

Isolating m:

m=\frac{W_{E}}{g_{E}} (2)

m=\frac{800 N}{9.8 m/s^{2}} (3)

m=81.63 kg (4)

Now that we know the mass of the astronaut, we can find its weight on Zuton:

W_{Z}=mg_{Z} (5)

W_{Z}=(81.63 kg)(3 m/s^{2}) (6)

W_{Z}=244.89 N (7)

Then, we can calculate the force the space ship exerts on the astronaut by the following equation:

F-W_{Z}=m.a (8)

Isolating F:

F=m.a+W_{Z} (9)

F=(81.63 kg)(0.5 m/s^{2})+244.89 N (10)

F=285.7 N \frac{1 kN}{1000 N}=0.285 kN (11)

Finally:

F=0.285 kN \approx 0.29 kN

5 0
4 years ago
A cyclist intends to cycle up a 7.70o hill whose vertical height is 126m. Assuming the mass of bicycle plus person is 75.0kg, ca
Ksivusya [100]

Answer:

92704.5 J

596.44737 N

Explanation:

m = Mass of person + bicycle = 75 kg

g = Acceleration due to gravity = 9.81 m/s²

h = Vertical height = 126 m

\theta = Angle = 7.7°

d = Diameter = 0.388 m

Work done against gravity is given by

P=mgh\\\Rightarrow P=75\times 9.81\times 126\\\Rightarrow P=92704.5\ J

Work done is 92704.5 J

Force required is given by

F=\dfrac{mgrsin\theta}{\pi d}\\\Rightarrow F=\dfrac{75\times 9.81\times sin7.7\times 5.12}{\pi\times 0.388}\\\Rightarrow F=596.44737\ N

The force is 596.44737 N

7 0
3 years ago
Why are experiments often performed in laboratories?
kap26 [50]
In my opinion,I think the answer is b.You can control variables more easily by doing different things for different purposes.
4 0
3 years ago
Read 2 more answers
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