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just olya [345]
3 years ago
12

Why do the passengers in high-altitude jet planes feel the sensation of weight while passengers in an orbiting space vehicle, su

ch as a space shuttle, do not?
Physics
1 answer:
kykrilka [37]3 years ago
7 0

Passengers in an aircraft are subject to the Normal and Gravity Force acting on them at a low 'orbit', so tiny that it can be many times compared to the same surface of the earth when speaking in general terms.

In a high orbit space vehicle or in the same space, said force decreases considerably or simply disappears, generating the sensation of weightlessness.

Remember that the Force of Gravity is given under the principle

F_g = \frac{GMm}{r^2}

Where,

G = Gravitational Universal constant

M = Mass of the planet

m = mass of the object

r = Distance from center of the planet

When the radius grows considerably the gravitational force begins to decrease.

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7. A 400 kg cannon fires an 80 kg student at +15 m/s. At what velocity will
telo118 [61]

Answer:

?

Explanation:

6 0
3 years ago
A ball is thrown straight up with enough speed so that it is in the air for several seconds. Assume the positive direction is up
Andrew [12]

Answer:

a) v= 0 m/s b) v= 6.86 m/s

Explanation:

a) When the ball reaches to its highest point, under the influence of gravity, before starting to fall down, it momentarily comes to an stop (this is needed prior to change direction in any movement), so, applying the definition of acceleration, and replacing the acceleration a by g, we have:

vf = v₀ - g*t (1)

The minus sign means that the acceleration due to gravity is always downward, so if we assume that the positive direction is upwards it must be negative.

At the highest point, vf= 0.

b) Prior to solve this point, we need to know which is the time when the ball reaches to its highest point.

As we know vf=0, we can solve (1) for t, as follows:

th = v₀/g

Now, for a time that is 0.7 s before this time, applying the acceleration definition and solving for v again, we have:

v = v₀ -(g *(th-0.7 s)), but th= v₀/g, so we get:

v= v₀ -g((v₀/g)-0.7 s) = v₀ - v₀ + g*0.7 s

⇒ v=g*0.7 s = 9.8 m/s²*0.7 s

⇒ v = 6.86 m/s

6 0
3 years ago
A narrow beam of light from a laser travels through air (n = 1.00) and strikes the surface of the water (n = 1.33) in a lake at
Natalka [10]

Answer:

A) d = 11.8m

B) d = 4.293 m

Explanation:

A) We are told that the angle of incidence;θ_i = 70°.

Now, if refraction doesn't occur, the angle of the light continues to be 70° in the water relative to the normal. Thus;

tan 70° = d/4.3m

Where d is the distance from point B at which the laser beam would strike the lakebottom.

So,d = 4.3*tan70

d = 11.8m

B) Since the light is moving from air (n1=1.00) to water (n2=1.33), we can use Snell's law to find the angle of refraction(θ_r)

So,

n1*sinθ_i = n2*sinθ_r

Thus; sinθ_r = (n1*sinθ_i)/n2

sinθ_r = (1 * sin70)/1.33

sinθ_r = 0.7065

θ_r = sin^(-1)0.7065

θ_r = 44.95°

Thus; xonsidering refraction, distance from point B at which the laser beam strikes the lake-bottom is calculated from;

d = 4.3 tan44.95

d = 4.293 m

4 0
3 years ago
child mass 25kg moves with speed of 1.5m/s when 7.8 from the central of merry-go round. calculate the centripetal acceleration o
MAVERICK [17]

Answer:

0.2885 m/s²

Explanation:

The formula for centripetal acceleration is given as;

a_c=v^2/r

Given that;

speed = v = 1.5m/s

radius = r = 7.8

a_c=v^2/r\\\\\\a_c=1.5^2/7.8\\\\\\a_c=0.2885m/s^{2}

8 0
3 years ago
A 220g mass is on a frictionless horizontal surface at the end of a spring that has a force constant of 7.0N/m The mass is displ
sladkih [1.3K]

Answer:

Explanation:

Your numbers seem wonky, so I'll just assume that the initial displacement is a distance A (Amplitude) from the equilibrium position. Spring constant = k

Initial potential energy is

PE = ½kA²

As potential energy and kinetic energy are constantly exchanging in SHM,

the position x where half of the original spring potential exists is found where

½kx² = ½(½kA²)

    x² = ½A²

    x = (√0.5)A

    x ≈ 0.707A

just plug in your actual starting position A

With A = 5.2 cm

x = 3.67695... 3.7 cm

5 0
3 years ago
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