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Sedbober [7]
2 years ago
10

Andrew pokes a marble, and the marble rolls down a ramp. The marble moves with speed. Which forces are acting on the marble in t

his situation?
When Andrew pokes the marble, he gives it [A tension/A friction/An applied/] force. As the marble rolls down the ramp, [Tension/Normal/Gravitational}, force pulls it down. The movement of the marble is slowed down by [Gravitational force/Friction/Elastic force] from the ramp
Physics
2 answers:
kvasek [131]2 years ago
5 0

Answer:

When Andrew pokes the marble, he gives it An applied force. As the marble rolls down the ramp, Gravitational force pulls it down. The movement of the marble is slowed down by Friction from the ramp.

Explanation:

At the beginning, Andrew pushes the marble, so is applying a force, which is called applied. force. Then, this force is removed (as Andrew no longer touches the marble), but now the weigth of the marble (the gravitational force) pushes the marble down along the ramp, because the gravitational force acts downward, so it has a donward component along the ramp. At the same time, the surface of the ramp exerts a force of friction against the motion of the marble, and the frictional force is always directed against the motion of the object.

slega [8]2 years ago
4 0
An applied force
gravitational force
friction
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You are flying a hang glider at 14 mph in the northeast direction (45°). The wind is blowing at 4 mph from due north.
Afina-wow [57]

Answer:

<em>a) 17.05 mph</em>

<em>b) 54.7°  northeast direction</em>

<em>c) 10.71 mph</em>

<em>The direction is -22.58° relative to the east.</em>

<em></em>

<em>To head northeast, you must either increase your gliding speed or increase your angle relative to the x-axis greater than 45°.</em>

Explanation:

The question is a little confusing but, I guess the correct question should be;

You are flying a hang glider at 14 mph in the northeast direction (45°). The wind is blowing at 4 mph due north.

a) What is your airspeed?

b) What angle (direction) are you flying?

c) The wind increases to 14 mph from north. Now what is your airspeed and what direction are you flying? If your destination is to the northeast, how would you change your speed or direction so you might make it there?

NB: The difference in the question and my suggestion is highlighted boldly.

Your speed = 14 mph

direction is 45° northeast

Th wind speed = 4 mph

direction is north

We resolve the your speed and the wind speed into the horizontal and vertical components

For vertical the component component

V_{y} = 14(sin 45) + 4 = 9.89 + 4 = 13.89 mph

For the horizontal speed component

V_{x} = 14(cos 45) + 0 = 9.89 + 0 = 9.89 mph

Resultant speed = \sqrt{V^{2} _{y}+V^{2} _{x}  }

==> \sqrt{13.89^{2} +9.89^{2}   } = <em>17.05 mph  This is your airspeed</em>

b) To get your direction, we use

tan ∅ = V_{y} /V_{x}

tan ∅ = 13.89/9.89 = 1.413

∅ = tan^{-1}(1.413) = <em>54.7°  northeast direction</em>

c) If the wind increases to 14 mph from the north, then it means the wind blows due south. As before, only the vertical component is affected .

In this case,

V_{y} = 14(sin 45) - 14 = 9.89 - 14 = -4.11 mph

Resultant speed = \sqrt{V^{2} _{y}+V^{2} _{x}  }

==> \sqrt{4.11^{2} +9.89^{2}   } = <em>10.71 mph  This is your airspeed</em>

Your direction will be,

tan ∅ = V_{y} /V_{x}

tan ∅ = -4.11/9.89 = -0.416

∅ = tan^{-1}(-0.416) =<em> -22.58°  this is the angle you'll travel relative to the east.</em>

<em>To head northeast, you must either increase your gliding speed or increase your angle relative to the x-axis greater than 45°.</em>

5 0
2 years ago
A spherical drop of water carrying a charge of 30 pC has a potential of 500 V at its surface (with V 0 at infinity). (a) What is
Delicious77 [7]

Answer:

A. 5.4 * 10^(-4) m

B. 500V

Explanation:

A. Electric potential, V is given as:

V = kq/r

This means that radius, r is

r = kq/V

r = (9 * 10^9 * 30 * 10^(-12))/500

r = (270 * 10^(-3))/500

r = 5.4 * 10^(-4) m

B. Now the radius is doubled and the charge is doubled,

V = (9 * 10^9 * 2 * 30 * 10^(-12))/(2 * 5.4 * 10^(-4) * 2)

V = 500V

7 0
2 years ago
A beam of monochromatic light (f =5.09 ×1014 Hz) has a wavelength of 589 nanometers in air. What is the wavelength of this light
frosja888 [35]
Lucite has a refractive index of n=1.50. This means that the speed of the light in lucite is decreased according to:
v=\frac{c}{n}
where c=3 \cdot 10^8 m/s is the speed of light in air. Putting the number in the formula, we find that the speed of light in lucite is
v=\frac{3 \cdot 10^8 m/s}{1.50}=2\cdot 10^8 m/s
The frequency of the light is f=5.09 \cdot 10^{14}Hz, so now we can calculate the wavelength in lucite by using the formula:
\lambda=\frac{v}{f}=\frac{2\cdot 10^8 m/s}{5.09 \cdot 10^{14} Hz}=3.93 \cdot 10^{-7} m=393 nm
<span>Therefore, the correct answer is (2) 393 nm.</span>
7 0
2 years ago
Ron fills a beaker with glycerin (n = 1.473) to a depth of 5.0 cm. if he looks straight down through the glycerin surface, he wi
Tju [1.3M]

By law of refraction we know that image position and object positions are related to each other by following relation

\frac{\mu_1}{h_o} = \frac{\mu_2}{h_i}

here we know that

\mu_1 = 1.473

h_o = 5 cm

\mu_2 = 1

now by above formula

\frac{1.473}{5} = \frac{1}{h_i}

h_i = 3.39 cm

so apparent depth of the bottom is seen by the observer as h = 3.39 cm

7 0
2 years ago
A rocket moves upward, starting from rest with an acceleration of +29.4 for 3.98 s. it runs out of fuel at the end of the 3.98 s
topjm [15]
U = 0, initial upward speed
a = 29.4 m/s², acceleration up to 3.98 s
a = -9.8 m/s², acceleration after 3.98s

Let h₁ =  the height at time t, for t ≤ 3.98 s
Let h₂ =  the height at time t > 3.98 s

Motion for  t ≤ 3.98 s:
h₁ = (1/2)*(29.4 m/s²)*(3.98 s)² = 232.854 m
Calculate the upward velocity at t = 3.98 s
v₁ = (29.4 m/s²)*(3.98 s) = 117.012 m/s

Motion for t  > 3.98 s
At maximum height, the upward velocity is zero.
Calculate the extra distance traveled before the velocity is zero.
(117.012 m/s)² + 2*(-9.8 m/s²)*(h₂ m) = 0
h₂ = 698.562 m

The total height is
h₁ + h₂ = 232.854 + 698.562 = 931.416 m

Answer: 931.4 m (nearest tenth)

6 0
2 years ago
Read 2 more answers
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