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Verdich [7]
2 years ago
15

Can anyone help me with questions a and c​

Physics
1 answer:
Ganezh [65]2 years ago
5 0

Answer:

a)

n=sin i/sin r

n= -0.305/-0.428

n=0.713

b)

sin c=1/n

sin c=1/0.713

sin c= 1.403

c=sin⁻¹(1.403)

c= 40.842°

Explanation:

i hope it will be helpful

plzzz mark as brainliest

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With a force of 5 Newton's Amanda pushes the stacks of books to the right. At the same time Jeremih her little brother pushes th
Fynjy0 [20]

Alright, so solving for the net force is a rather simple and easy step. I want to briefly explain how you determine the net force, but before I do, let's examine the problem together. What has been provided to us? What direction is it going? Well, we can look at the information!

Amanda's Force: 5 N

Jeremiah's Force: 10 N

They're both pushing on the books, but in different directions. Left and right. If Amanda is pushing the books to the right 5 newtons, and Jeremiah is pushing the books to the left with 10 newtons, that means the net force is 5. The book is being pushed to the left! We can say that because Jeremiah's force is much larger than his sisters.

We can determine the net force from the following;

Net Force → 10 N - 5 N = 5 N

In conclusion, your answer should be a total of five (5) newtons

4 0
3 years ago
Read 2 more answers
(a) Find the size of the smallest detail observable in human tissue with 20.0-MHz ultrasound. (b) Is its effective penetration d
koban [17]

Answer:

(a) λ=77×10⁻⁶m

(b)Yes its enough to examine entire eye,

(c)λ=16.65μm.

Explanation:

The ultrasonic sound are used to detect problems in human body by using device consisting of transmitter and detector .

The speed of sound in human tissue is 1540m/s.

The wavelength of sound is 20MHz.

(a) let the velocity of sound in human tissue is v = 1540m/s.

                v=fλ

           λ= \frac{v}{f} =\frac{1540}{20MHz} = 77×10⁻⁶m

             λ=77×10⁻⁶m

(b)  we can scan effectively upto the depth of  about 500λ

                   500λ=500× (77×10⁻⁶)= 3.85cm

            Yes its enough to examine entire eye,

(c) In air the  velocity of sound is 331m/s.

              λ=\frac{v}{f}  = \frac{331 m/s}{20Mhz}  = 16.65μm

            λ=16.65μm.

           

5 0
3 years ago
Kindly answer the question about Work and Power. Image is attached below.
natita [175]

Answer:

89600 Watts.

Explanation:

From the question given above, the following data were obtained:

Mass (m) of speedboat = 2800 Kg

Initial velocity (u) = 0 m/s

Final velocity (v) = 16 m/s

Time (t) = 8 s

Power (P) expended by the speedboat =?

Next, we shall determine the acceleration of the speedboat. This can be obtained as follow:

Initial velocity (u) = 0 m/s

Final velocity (v) = 16 m/s

Time (t) = 8 s

Acceleration (a) =?

v = u + at

16 = 0 + (a × 8)

16 = 0 + 8a

16 = 8a

Divide both side by 8

a = 16/8

a = 2 m/s²

The acceleration of the speedboat is 2 m/s²

Next, we shall determine the force exerted by the speedboat. This can be obtained as follow:

Mass (m) of speedboat = 2800 Kg

Acceleration (a) of speedboat = 2 m/s²

Force (F) of speedboat =?

F = ma

F = 2800 × 2

F = 5600 N

Finally, we shall determine the expended by the speedboat at maximum speed. This can be obtained as follow:

Force (F) of speedboat = 5600 N

Maximum velocity (v) = 16 m/s

Power (P) expended by the speedboat =?

P = F × v

P = 5600 × 16

P = 89600 Watts

Thus, the power expended by the speedboat is 89600 Watts.

4 0
3 years ago
Need thy help once again
Thepotemich [5.8K]
B cause I took the quiz
8 0
3 years ago
A mass of 148 g stretches a spring 13 cm. The mass is set in motion from its equlibrium position with a downward velocity of 10
Charra [1.4K]

Answer:

u(t)=1.15 \sin (8.68t)cm

0.3619sec

Explanation:

Given that

Mass,m=148 g

Length,L=13 cm

Velocity,u'(0)=10 cm/s

We have to find the position u of the mass at any time t

We know that

\omega_0=\sqrt{\frac{g}{L}}\\\\=\sqrt{\frac{980}{13}}\\\\=8.68 rad/s

Where g=980 cm/s^2

u(t)=Acos8.68 t+Bsin 8.68t

u(0)=0

Substitute the value

A=0\\u'(t)=-8.68Asin8.68t+8.68 Bcos8.86 t

Substitute u'(0)=10

8.68B=10

B=\frac{10}{8.68}=1.15

Substitute the values

u(t)=1.15 \sin (8.68t)cm

Period =T = 2π/8.68

After half period

π/8.68 it returns to equilibruim

π/8.68 = 0.3619sec

6 0
3 years ago
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