Answer:
Yes. Something similar occurs here on Earth.
Explanation:
Gravity tends to pull objects perpendicularly to the ground. In space, the absence of this force means there is no compression on the spine due to gravity trying to pull it down. This means that astronauts undergo an increase in height in space.
Here on Earth, we experience gravity pull on our spine during the day. At night when we sleep, we lie down with our spine parallel to the ground, which means that our spine is no longer under compression from gravity force. The result is that we are a few centimetres taller in the morning when we wake up, than we are before going to bed at night. The increase is not much pronounced here on Earth because there is a repeated cycle of compression and decompression of our spine due to gravity, unlike when compared to that of astronauts that spend long duration in space, all the while without gravity forces on their spine
A cylinder is closed by a piston connected to a spring of constant 2.20 10^3 N/m. With the spring relaxed, the cylinder is filled with 5.00 L of gas at a pressure of 1.00 atm and a temperature of 20.0°C. The piston has a cross sectional area of 0.0100 m^2 and negligible mass. What is the pressure of the gas at 250 °C?
Answer:
1.3515x10^5pa
Explanation:
Plss see attached file
In a polar coordinate system, the velocity vector can be written as . The term theta with dot on top is called _______________________ angular velocity transverse velocity radial velocity angular acceleration
Answer:
I believe it's called rapid growth
Explanation:
that is my answer no matter what
On a certain planet a body is thrown vertically upwards with an initial speed of 40 m / s. If the maximum height was 100 m, the acceleration due to gravity is
a) 15 m / s 2
b) 12.5 m / s 2
c) 8 m / s 2
d) 10 m / s 2
Answer:
C) 8 m/s²
Explanation:
Given:
v₀ = 40 m/s
v = 0 m/s
Δy = 100 m
Find: a
v² = v₀² + 2aΔy
(0 m/s)² = (40 m/s)² + 2a (100 m)
a = -8 m/s²
A record player rotates a record at 45 revolutions per minute. When the record player is switched off, it makes 4.0 complete turns at a constant angular acceleration before coming to rest. What was the magnitude of the angular acceleration (in rads/s2) of the record as it slowed down
Answer:
The angular acceleration is [tex]\alpha = 0.4418 \ rad /s^2[/tex]
Explanation:
From the question we are told that
The angular speed is [tex]w_f = 45 \ rev / minutes = \frac{45 * 2 * \pi }{60 }= 4.713 \ rad/s[/tex]
The angular displacement is [tex]\theta =4 \ rev = 4 * 2 * \pi = 25.14 \ rad[/tex]
From the first equation of motion we can define the movement of the record as
[tex]w_f ^2 = w_o ^2 + 2 * \alpha * \theta[/tex]
Given that the record started from rest [tex]w_o = 0[/tex]
So
[tex]4.713^2 = 2 * \alpha * 25.14[/tex]
[tex]\alpha = 0.4418 \ rad /s^2[/tex]
Experts, ACE, Genius... can anybody calculate for the Reactions at supports A and B please? Will give brainliest! Given: fb = 300 kN/m, fc = 100 kN/m, Dy = 300 kN, spanAB = 6m, span BC = 6m, spanCD = 6m
Answer:
Support at Cy = 1.3 x 10³ k-N
Support at Ay = 200 k-N
Explanation:
given:
fb = 300 k-N/m
fc = 100 k-N/m
D = 300 k-N
L ab = 6 m
L bc = 6 m
L cd = 6 m
To get the reaction A or C.
take summation of moment either A or C.
Support Cy:
∑ M at Ay = 0
(( x1 * F ) + ( D * Lab ) + ( D * L bc + D * L cd )
Cy = -------------------------------------------------------------------
( L ab + L bc )
Cy = 1.3 x 10³ k-N
Support Ay:
Since ∑ F = 0, A + C - F - D = 0
A = F + D - C
Ay = 200 k-N
Answer:
i was going to but its to late
Explanation:
In your own words, discuss how energy conservation applies to a pendulum. Where is the potential energy the most? Where is the potential energy the least? Where is kinetic energy the most? Where is kinetic energy the least?
Answer:
Explanation:
Energy conservation applies to the swinging of pendulum . When the bob is at one extreme , it is at some height from its lowest point . So it has some gravitational potential energy . At that time since it remains at rest its kinetic energy is zero or the least . As it goes down while swinging , its potential energy decreases and kinetic energy increases following conservation of mechanical energy . At the At the lowest point , its potential energy is least and kinetic energy is maximum .
In this way , there is conservation of mechanical energy .
Si se deja caer una piedra desde un helicóptero en reposo, entonces al cabo de 20 s cual será la rapidez y la distancia recorrida por la piedra
Answer:
La piedra alcanza una rapidez de 196.14 metros por segundo y una distancia recorrida de 1961.4 metros en 20 segundos.
Explanation:
Si se excluye los efectos del arrastre por la viscosidad del aire, la piedra experimenta un movimiento de caída libre, es decir, que la piedra es acelerada por la gravedad terrestre. La distancia recorrida y la rapidez final de la piedra pueden obtenerse con la ayuda de las siguientes ecuaciones cinemáticas:
[tex]v = v_{o} + g\cdot t[/tex]
[tex]y - y_{o} = v_{o}\cdot t + \frac{1}{2}\cdot g \cdot t^{2}[/tex]
Donde:
[tex]v[/tex], [tex]v_{o}[/tex] - Rapideces final e inicial de la piedra, medidas en metros por segundo.
[tex]t[/tex] - Tiempo, medido en segundos.
[tex]g[/tex] - Aceleración gravitacional, medida en metros por segundo al cuadrado.
[tex]y[/tex]. [tex]y_{o}[/tex] - Posiciones final e inicial de la piedra, medidos en metros.
Si [tex]v_{o} = 0\,\frac{m}{s}[/tex], [tex]g = -9.807\,\frac{m}{s^{2}}[/tex], [tex]y_{o} = 0\,m[/tex], entonces:
[tex]v = 0\,\frac{m}{s} +\left(-9.807\,\frac{m}{s^{2}} \right) \cdot (20\,s)[/tex]
[tex]v = -196.14\,\frac{m}{s}[/tex]
[tex]y-y_{o} = \left(0\,\frac{m}{s} \right)\cdot (20\,s) + \frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right) \cdot (20\,s)^{2}[/tex]
[tex]y-y_{o} = -1961.4\,m[/tex]
La piedra alcanza una rapidez de 196.14 metros por segundo y una distancia recorrida de 1961.4 metros en 20 segundos.
The 2-Mg truck is traveling at 15 m/s when the brakes on all its wheels are applied, causing it to skid for 10 m before coming to rest. The total mass of the boat and trailer is 1 Mg. Determine the constant horizontal force developed in the coupling C, and the friction force developed between the tires of the truck and the road during this time.
Answer:
constant horizontal force developed in the coupling C = 11.25KN
the friction force developed between the tires of the truck and the road during this time is 33.75KN
Explanation:
See attached file
The friction force between the tires of the truck and the road is 22500 N.
Calculating the friction force:It is given that a 2 Mg truck ( m = 2000 Kg) is initially moving with a speed of u = 15 m/s.
Distance traveled before coming to rest, s = 10m
The final velocity of the truck will be zero, v = 0
When the breaks are applied, only the frictional force is acting on the truck and it is opposite to the motion of the truck.
The frictional force is given by:
f = -ma
the acceleration of the truck = -a
The negative sign indicates that the acceleration is opposite to the motion.
Applying the third equation of motion we get:
v² = u² -2as
0 = 15² - 2×a×10
225 = 20a
a = 11.25 m/s²
So the magnitude of frictional force is:
f = ma = 2000 × 11.25 N
f = 22500 N
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Suppose you are looking into the end of a long cylindrical tube in which there is a uniform magnetic field pointing away from you. If the magnitude of the field is decreasing with time the direction of the induced magnetic field is
Answer:
If the magnitude of the field is decreasing with time the direction of the induced magnetic field is CLOCKWISE
Explanation
This is because If the magnetic field decreases with time, the electric field will be produced in order to oppose the change in line with lenz law. Thus The right hand rule can be applied to find that the direction of electric field is in the clockwise direction.
A 46-ton monolith is transported on a causeway that is 3500 feet long and has a slope of about 3.7. How much force parallel to the incline would be required to hold the monolith on this causeway?
Answer:
2.9tons
Explanation:
Note that On an incline of angle a from horizontal, the parallel and perpendicular components of a downward force F are:
parallel ("tangential"): F_t = F sin a
perpendicular ("normal"): F_n = F cos a
At a=3.7 degrees, sin a is about 0.064 and with F = 46tons:
F sin a ~~ (46 tons)*0.064 ~~ 2.9tons
Also see attached file
The required force parallel to the incline to hold the monolith on this causeway will be "2.9 tons".
Angle and ForceAccording to the question,
Angle, a = 3.7 degrees or,
Sin a = 0.064
Force, F = 46 tons
We know the relation,
Parallel (tangential), [tex]F_t[/tex] = F Sin a
By substituting the values,
= 46 × 0.064
= 2.9 tons
Thus the response above is appropriate answer.
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The magnetic force per meter on a wire is measured to be only 45 %% of its maximum possible value. Calculate the angle between the wire and the magnetic field.
Answer:
27°
Explanation:
The force is proportional to the sine of the angle between the wire and the magnetic field. (See the ref.)
So theta = arcsin(0.45)
=27°
The angle between the wire and the magnetic field is 27°.
Calculation of the angle:Since The magnetic force per meter on a wire is measured to be only 45 %
So here we know that The force should be proportional to the sine of the angle between the wire and the magnetic field
Therefore,
theta = arcsin(0.45)
=27°
Hence, The angle between the wire and the magnetic field is 27°.
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