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otez555 [7]
3 years ago
15

CAN SOMEONE TELL ME THE answer for this ?

Physics
2 answers:
Anna007 [38]3 years ago
5 0

Answer:

C. Light from the sun is reflected by the moon's surface

Explanation:

In some parts the Light reaches, it gets lighted up and the parts where light does not reaches we can't see so we see the phrases of the moon .

Paladinen [302]3 years ago
4 0

Answer:

b. light from earth is reflected by the moon surface

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If he leaves the ramp with a speed of 31.0 m/s and has a speed of 29.5 m/s at the top of his trajectory, determine his maximum h
raketka [301]

Answer:

The maximum height reached is 4.63 m.

Explanation:

Given:

Initial speed of the man (u) = 31.0 m/s

Speed at the top of trajectory (u_x) = 29.5 m/s

Acceleration due to gravity (g) = 9.8 m/s²

When the man reaches the top of the trajectory, the vertical component of velocity becomes zero and hence only horizontal component of velocity acts on him.

Also, since there is no net force acting in the horizontal direction, the acceleration is zero in the horizontal direction from Newton's second law. Thus, the horizontal component of velocity always remains the same.

So, speed at the top of trajectory is nothing but the horizontal component of initial velocity.

Now, initial velocity can be rewritten in terms of its components as:

u^2=u_x^2+u_y^2

Where, u_x\ and\ u_y are the initial horizontal and vertical velocities of the man.

Now, plug in the given values and simplify. This gives,

(31.0)^2=(29.5)^2+u_y^2\\\\961=870.25+u_y^2\\\\u_y^2=961-870.25\\\\u_y^2=90.75\ m^2/s^2--------1

Now, we know that, for a projectile motion, the maximum height is given as:

H=\frac{u_y^2}{2g}

Plug in the value from equation (1) and 9.8 for 'g' to solve for 'H'. This gives,

H=\frac{90.75}{2\times 9.8}\\\\H=4.63\ m

Therefore, the maximum height reached is 4.63 m.

3 0
3 years ago
Which of Newton's motion laws BEST explains WHY a rock falls when it is dropped from a bridge?
erma4kov [3.2K]
It’s A because it stays in motion whenever you drop it
4 0
3 years ago
Read 2 more answers
1. Determine the image distance in each of the following.
miskamm [114]

It is given that for the convex lens,

Case 1.

u=−40cm

f=+15cm

Using lens formula

v

1

−

u

1

=

f

1

v

1

−

40

1

=

15

1

v

1

=

15

1

−

40

1

v=+24.3cm

The image in formed in this case at a distance of 24.3cm in left of lens.

Case 2.

A point source is placed in between the lens and the mirror at a distance of 40 cm from the lens i.e. the source is placed at the focus of mirror, then the rays after reflection becomes parallel for the lens such that

u=∞

f=15cm

Now, using mirror’s formula

v

1

+

u

1

=

f

1

v

1

+

∞

1

=

15

1

v=+15cm

The image is formed at a distance of 15cm in left of mirror

6 0
3 years ago
Use the fact that 1inch=2.54cm and convert 52 inches into the equivalent length in centimeters.
Lelu [443]

Answer:

To convert inches to centimeters, use an easy formula and multiply the length by the conversion ratio.

Since one inch is equal to 2.54 centimeters, this is the inches to cm formula to conver

Explanation:

4 0
2 years ago
Determine the CM of a rod assuming its linear mass density λ (its mass per unit length) varies linearly from λ = λ0 at the left
Dahasolnce [82]

Answer:

x_c= \dfrac{5}{9}L

I=\dfrac {7}{12}\lambda_ 0 L^3

Explanation:

Here mass density of rod is varying so we have to use the concept of integration to find mass and location of center of mass.

At any  distance x from point A mass density

\lambda =\lambda_0+ \dfrac{2\lambda _o-\lambda _o}{L}x

\lambda =\lambda_0+ \dfrac{\lambda _o}{L}x

Lets take element mass at distance x

dm =λ dx

mass moment of inertia

dI=\lambda x^2dx

So total moment of inertia

I=\int_{0}^{L}\lambda x^2dx

By putting the values

I=\int_{0}^{L}\lambda_ ox+ \dfrac{\lambda _o}{L}x^3 dx

By integrating above we can find that

I=\dfrac {7}{12}\lambda_ 0 L^3

Now to find location of center mass

x_c = \dfrac{\int xdm}{dm}

x_c = \dfrac{\int_{0}^{L} \lambda_ 0(1+\dfrac{x}{L})xdx}{\int_{0}^{L} \lambda_0(1+\dfrac{x}{L})}

Now by integrating the above

x_c=\dfrac{\dfrac{L^2}{2}+\dfrac{L^3}{3L}}{L+\dfrac{L^2}{2L}}

x_c= \dfrac{5}{9}L

So mass moment of inertia I=\dfrac {7}{12}\lambda_ 0 L^3 and location of center of mass  x_c= \dfrac{5}{9}L

8 0
3 years ago
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