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myrzilka [38]
3 years ago
7

Consider a person sliding down a water slide at constant velocity. what are the forces acting on the person as they slide? are t

he forces balanced or unbalanced? explain.

Physics
2 answers:
Andru [333]3 years ago
4 0

Explanation :

When a person is sliding with constant velocity, the forces that are acting on him are :

(1) Weight, W = mg where g is acceleration due to gravity.

(2) Friction forces acting opposite to the direction of motion.

(3) Normal force.

The forces acting on the person is balanced because the person is sliding with constant velocity.

From figure it is clear that, friction force is balanced by w\ sin\theta and normal is balanced by w\ cos\theta.

Umnica [9.8K]3 years ago
4 0

Answer:

It is D.

Explanation:

Gravity and friction are acting on the person. The forces are balanced. There is no net force because the person is moving at a constant velocity. In this example, there is no net force when a mass moves at constant velocity. Although friction is acting on the person, there is no change in velocity and friction is not a net force in this case. Friction is only a net force if it changes the velocity of a mass. If friction slowed the person down, then there is a net force.

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Artyom0805 [142]

Answer:

Sound intensity is the amount of energy carried by sound versus loudness is a subjective measurement of the audible sound.

Sound intensity is measured in watt per square meter where loudness is measured in sones (sone is a subjective measurement and not an SI unit)

6 0
3 years ago
NASA is designing a Mars-lander that will enter the Martian atmosphere at high speed. To land safely it must slow to a constant
Viktor [21]

Answer:

a) maximum mass of the Mars lander to ensure it can land safely is 200 kg

b) area of the parachute required is 480 m² which is larger than 400 m²

c) area of the parachute should be 12.68 m²

Explanation:

Given the data in the question;

V = 20 m/s

A = 200 m²

drag co-efficient CD = 1.855

g = 3.71 m/s²

density of the atmospheric pressure β = 0.01 kg/m³

a. Calculate the maximum mass of the Mars lander to ensure it can land safely?

Drag force FD = 1/2 × CD × β × A × V²

we substitute

FD = 1/2 × 1.855 × 0.01 kg/m × 200 m² × ( 20 m/s )²

FD = 742 N

we know that;

FD = Fg

Fg = gravity force

Fg = mg

so

FD = mg

m = FD/g

we substitute

m = 742 N / 3.71 m/s²

m = 200 kg

Therefore, the maximum mass of the Mars lander to ensure it can land safely is 200 kg

b. The mission designers consider a larger lander with a mass of 480 kg. Show that the parachute required would be larger than 400 m²;

Given that;

M = 480 kg

Show that the parachute required would be larger than 400 m²

we know that;

FD = Fg = Mg = 480 kg × 3.71 m/s²

FD = 1780.8 N

Now, FD = 1/2 × CD × β × A × V², we solve for A

A = FD / 0.5 × CD × β × V²

we substitute

A = 1780.8  / 0.5 × 1.855 × 0.1 × (20)²

A = 1780.8 / 3.71

A = 480 m²

Therefore, area of the parachute required 480 m² which is larger than 400 m²

c. To test the lander before launching it to Mars, it is tested on Earth where g = 9.8 m/s^2 and the atmospheric density is 1.0 kg m-3. How big should the parachute be for the terminal speed to be 20 m/s, if the mass of the lander is 480 kg?

Given that;

g = 9.8 m/s²,

β" = 1 kg/m³

v" = 20 m/s

M" = 480 kg

we know that;

FD = Fg = M"g

FD = 480 kg × 9.8 m/s² = 4704 N

from the expression; FD = 1/2 × CD × β × A × V²

A = FD / 0.5 × CD × β" × V"²

we substitute

A = 4704 / 0.5 × 1.855 × 1 × (20)²

A = 4704 / 371

A = 12.68 m²

Therefore area of the parachute should be 12.68 m²

3 0
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What scenario would most likely lead to a higher level of physical fitness among people who live in a city?
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Two radio antennas are 120 m apart on a north-south line, and they radiate in phase at a frequency of 3.4 MHz. All radio measure
chubhunter [2.5K]

Answer:

the smallest angle from the antennas is <em>47.3°</em>

Explanation:

We first need to write the expression for the relation between the wavelength (λ) and the frequency (f) of the wave, and then solve for the wavelength.

Therefore, the relation is:

λ = c /f

where

  • c is the speed of light constant
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  • f is the frequency

Thus,

λ = (3 × 10⁸ m/s) / (3.4 MHz)

  = (3 × 10⁸ m/s) / (3.4 MHz)(10⁶ Hz/1 MHz)

  = 88.235 m

Therefore, the smallest angle measured (from the north of east) from the antennas for the constructive interference of the two-radio wave can be calculated as

θ = sin⁻¹(λ / d)

where

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Thus,

θ = sin⁻¹(88.235 / 120)

<em>θ = 47.3 °</em>

<em></em>

Therefore, the smallest angle from the antennas, measured north of east, at which constructive interference of two radio waves occurs is <em>47.3 °</em>.

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We know that the electric field is obtained as the ratio of the voltage to the distance that separates the plates.

Thus;

E = V/d

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V = voltage

d = distance of separation

E = 24 * 10^3 V/1.52 * 10^-2 m

E = 1.58 * 10^6 V/m

Learn more about electric field :brainly.com/question/8971780

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2 years ago
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