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Lyrx [107]
3 years ago
11

A person trying to lose weight should

Physics
2 answers:
erastova [34]3 years ago
7 0

You should take it slow. To much weight loss at once could have affects on you.

I would say 1 or 2 pounds a week.

drek231 [11]3 years ago
5 0
This is a very wierd question but I will answer.It is A

Hope I helped
You might be interested in
What is first and second conditions of equilibrium?
sdas [7]

I think the answer is the first condition sum of forces acting on a body is zero ( ∑ F =0 ) and the second condition sum of torque acting on a body is ( ∑ τ = 0 )

Could you put me as brainliest?

7 0
2 years ago
A diffraction grating is placed 1.00 m from a viewing screen. Light from a hydrogen lamp goes through the grating. A hydrogen sp
Colt1911 [192]

Answer:

λ = 396.7 nm

Explanation:

For this exercise we use the diffraction ratio of a grating

           d sin θ = m λ

in general the networks works in the first order m = 1

we can use trigonometry, remembering that in diffraction experiments the angles are small

           tan θ = y / L

           tan θ = \frac{sin \theta}{cos \theta} = sin θ

           sin θ = y / L

we substitute

          d \  \frac{y}{L} = m λ

with the initial data we look for the distance between the lines

           d = \frac{m \lambda \ L}{y}

           d = 1 656 10⁻⁹ 1.00 / 0.600

            d = 1.09 10⁻⁶ m

for the unknown lamp we look for the wavelength

           λ = d y / L m

           λ = 1.09 10⁻⁶ 0.364 / 1.00 1

           λ = 3.9676 10⁻⁷ m

           λ = 3.967 10⁻⁷ m

         

we reduce nm

           λ = 396.7 nm

8 0
2 years ago
A converging lens of focal length 20 cm is used to form a real image 1.0 m away from the lens. How far from the lens is the obje
Galina-37 [17]

Answer:

0.25 m

Explanation:

We can solve the problem by using the lens equation:

\frac{1}{f}=\frac{1}{p}+\frac{1}{q}

where

f is the focal length

p is the distance of the object from the lens

q is the distance of the image from the lens

In this problem, we have

f = +20 cm=+0.20 m (the focal length is positive for a converging lens)

q = +1.0 m (the image distance is positive for a real image)

Solving the equation for p, we find

\frac{1}{p}=\frac{1}{f}-\frac{1}{q}=\frac{1}{0.20 m}-\frac{1}{1 m}=4 m^{-1}\\p=\frac{1}{4 m^{-1}}=0.25 m

6 0
3 years ago
A car starts from rest with an acceleration of 5 ft/s. What is its velocity after it has gone 600 ft?
Soloha48 [4]

Answer:

First, let’s correct the question. Acceleration is the rate of change in velocity. Its unit therefore is ft/sec/sec. If S is the distance traveled for a given duration, S = Vot + (1/2)at^2 where Vo is the initial velocity, a is the acceleration and t is the time. For Vo = 0, a = 6m/sec/sec and t = 3 sec. The distance traveled is S = 0 + (1/2) x 6 x 3^2 = 27 meters

3 0
2 years ago
A light wave travels through air in equals 1.00 at an angle of 35 degrees what angle does it have when it passes from the air in
lyudmila [28]

Answer: Angle 59 degree

Explanation: Given that the

n1 = 1.0

n2 = 1.5

Øi = 35 degree

From Snell law, which says that

n1/n2 = sinØ1/ sinØ2

Substitute all the parameters into the formula

1/1.5 = sin 35/sinØ2

Cross multiply

Sin Ø2 = 1.5 sin35

SinØ2 = 1.5 × 0.573 = 0.860

Ø2 = sin^-1(0.860)

Ø2 = 59.36 degree

Ø2 = 59 degree ( approximately)

It has angle 59 degree when passing from air to glass

5 0
2 years ago
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