What day of the year is solar time the same as sidereal time?

What is the electric potential v due to the nucleus of hydrogen at a distance of 5.292×10−11 m ?

viktelen [127]

Answer: 27.21 V

Explanation:

The electric potential $V_{E}$ due to a point charge is expressed as:

$V_{E}=k\frac{q}{r}$

Where:

$k=9(10)^{9}\frac{Nm^{2}}{C^{2}}$ is the electric constant

$q=1.6(10)^{-19}C$ is the electric charge of the hydrogen nucleus, which is positive

$r=5.292(10)^{-11}m$ is the distance

Rewritting the equation with the known values:

$V_{E}=9(10)^{9}\frac{Nm^{2}}{C^{2}}\frac{1.6(10)^{-19}C}{5.292(10)^{-11}m}$

Finally:

$V_{E}=27.21 V$

5 0

4 years ago

A 50 g mass is freely hanging from a horizontal meter stick at a distance of 99 cm from the pivot. Calculate the weight force W

Neko [114]

Answer:

W = 0.49 N

τ = 0.4851 Nm

Force

Explanation:

The weight force can be found as:

W = mg

W = (0.05 kg)(9.8 m/s²)

W = 0.49 N

The torque about the pivot can be found as:

τ = W*d

where,

τ = torque

d = distance between weight and pivot = 99 cm = 0.99 m

Therefore,

τ = (0.49 N)(0.99 m)

τ = 0.4851 Nm

The pivot exerts a FORCE on the meter stick because the pivot applies force normally over the stick and has a zero distance from stick.

6 0

3 years ago

Joe is hiking through the woods when he decides to stop and take in the view. He is particularly interested in three objects: a

Novosadov [1.4K]

Answer:

A) correct answer is C, B) correct answer is b and C) The correct answer is b

Explanation:

In the exercises of geometric optics, the equation of the constructor tells us the location of the image.

1 / f = 1 / p + 1 / q

where f is the focal length of the cornea-crystalline system, p and q are the distances to the object and the image.

In this case, the distance to the image on the retina is constant, about 3 cm. Therefore depending on the distance to the object) p = the focal length must change

1 / q = 1 / f 1 / p

let's apply this expression to our case

A) indicates that the tree is at a medium distance

so that the image is formed on the retina THE SAME AS

correct answer is C

B) The squirrel is at a smaller distance (p ') than the tree (p), therefore if we substitute in the equation above we find that q must decrease. Consequently the image is in front of the retina

The mountain is very far, suppose in infinity, so the image is BEHIND THE RETINA

therefore the correct answer is b

C) The squirrel is very close so the curvature of the lens INCREASES, resulting in a DECREASE in the focal length

The correct answer is b

6 0

3 years ago

Suppose that sunlight is incident upon both a pair of reading glasses and a pair of sunglasses. Which pair would you expect to b

Ainat [17]

Answer: the pair of sunglasses

Explanation:

A good pair of sunglasses are composed of abosorbent lenses that filter the sunlight that affects the eyes retina, especially ultraviolet (UV). So, these sunglasses are used to reduce the amount of light or radiant energy transmitted.

On the other hand, normal reading glasses (in which the lens glass has not been treated to filter ultraviolet sunlight) will let UV rays pass through.

Therefore, if both glasses are exposed to sunlight, the sunglasses are expected to be warmer by absorbing that radiant energy and preventing it from reaching the eyes.

4 0

3 years ago

A projectile is launched at an angle of 36.7 degrees above the horizontal with an initial speed of 175 m/s and lands at the same

Softa [21]

Answer:

a) The maximum height reached by the projectile is 558 m.

b) The projectile was 21.3 s in the air.

Explanation:

The position and velocity of the projectile at any time "t" is given by the following vectors:

r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)

v = (v0 · cos α, v0 · sin α + g · t)

Where:

r = position vector at time "t"

x0 = initial horizontal position

v0 = initial velocity

t = time

α = launching angle

y0 = initial vertical position

g = acceleration due to gravity (-9.80 m/s² considering the upward direction as positive).

v = velocity vector at time t

a) Notice in the figure that at maximum height the velocity vector is horizontal. That means that the y-component of the velocity (vy) at that time is 0. Using this, we can find the time at which the projectile is at maximum height:

vy = v0 · sin α + g · t

0 = 175 m/s · sin 36.7° - 9.80 m/s² · t

- 175 m/s · sin 36.7° / - 9.80 m/s² = t

t = 10.7 s

Now, we have to find the magnitude of the y-component of the vector position at that time to obtain the maximum height (In the figure, the vector position at t = 10.7 s is r1 and its y-component is r1y).

Notice in the figure that the frame of reference is located at the launching point, so that y0 = 0.

y = y0 + v0 · t · sin α + 1/2 · g · t²

y = 175 m/s · 10.7 s · sin 36.7° - 1/2 · 9.8 m/s² · (10.7 s)²

y = 558 m

The maximum height reached by the projectile is 558 m

b) Since the motion of the projectile is parabolic and the acceleration is the same during all the trajectory, the time of flight will be twice the time it takes the projectile to reach the maximum height. Then, the time of flight of the projectile will be (2 · 10.7 s) 21.4 s. However, let´s calculate it using the equation for the position of the projectile.

We know that at final time the y-component of the vector position (r final in the figure) is 0 (because the vector is horizontal, see figure). Then:

y = y0 + v0 · t · sin α + 1/2 · g · t²

0 = 175 m/s · t · sin 36.7° - 1/2 · 9.8 m/s² · t²

0 = t (175 m/s · sin 36.7 - 1/2 · 9.8 m/s² · t)

0 = 175 m/s · sin 36.7 - 1/2 · 9.8 m/s² · t

- 175 m/s · sin 36.7 / -(1/2 · 9.8 m/s²) = t

t = 21.3 s

The projectile was 21.3 s in the air.

7 0

3 years ago