Answer:
(a). The spring compressed is 
.
(b). The acceleration is 1.5 g.
Explanation:
Given that,
Acceleration = a
mass = m
spring constant = k
(a). We need to calculate the spring compressed
Using balance equation
....(I)
The spring compressed is .
(b). If the compression is 2.5 times larger than it is when the mass sits in a still elevator,
The compression is given by
Here, acceleration is zero
So, 
We need to calculate the acceleration
Put the value of x in equation (I)
Hence, (a). The spring compressed is .
(b). The acceleration is 1.5 g.
No, there isn't. Please consult your doctor if this is the case with yours or someone you know.
Answer:
14869817.395 m
Explanation:
=22 microarcsecond
λ = Wavelength = 1.3 mm
Converting to radians we get
From Rayleigh Criterion
Diameter of the effective primary objective is 14869817.395 m
It is not possible to build one telescope with a diameter of 14869817.395 m. But, we need this type of telescope. So, astronomers use an array of radio telescopes to achieve a virtual diameter in order to observe objects that are the size of supermassive black hole's event horizon.
Answer:
the <em>ratio F1/F2 = 1/2</em>
the <em>ratio a1/a2 = 1</em>
Explanation:
The force that both satellites experience is:
F1 = G M_e m1 / r² and
F2 = G M_e m2 / r²
where
- m1 is the mass of satellite 1
- m2 is the mass of satellite 2
- r is the orbital radius
- M_e is the mass of Earth
Therefore,
F1/F2 = [G M_e m1 / r²] / [G M_e m2 / r²]
F1/F2 = [G M_e m1 / r²] × [r² / G M_e m2]
F1/F2 = m1/m2
F1/F2 = 1000/2000
<em>F1/F2 = 1/2</em>
The other force that the two satellites experience is the centripetal force. Therefore,
F1c = m1 v² / r and
F2c = m2 v² / r
where
- m1 is the mass of satellite 1
- m2 is the mass of satellite 2
- v is the orbital velocity
- r is the orbital velocity
Thus,
a1 = v² / r ⇒ v² = r a1 and
a2 = v² / r ⇒ v² = r a2
Therefore,
F1c = m1 a1 r / r = m1 a1
F2c = m2 a2 r / r = m2 a2
In order for the satellites to stay in orbit, the gravitational force must equal the centripetal force. Thus,
F1 = F1c
G M_e m1 / r² = m1 a1
a1 = G M_e / r²
also
a2 = G M_e / r²
Thus,
a1/a2 = [G M_e / r²] / [G M_e / r²]
<em>a1/a2 = 1</em>
Responder:
20πrads ^ -1; 24πrads ^ -1; 0,1 seg; 10 Hz
Explicación:
Dado lo siguiente:
Radio (r) del círculo = 120 cm
600 revoluciones por minuto en radianes por segundo
(600 / min) * (2π rad / 1 rev) * (1min / 60seg)
(1200πrad / 60sec) = 20π rad ^ -1
Velocidad angular (w) = 20πrads ^ -1
Velocidad lineal = radio (r) * velocidad angular (w)
Velocidad lineal = (120/100) * 20πrad
Velocidad lineal = 1.2 * 20πrads ^ -1 = 24πrads ^ -1
C.) Período (T):
T = 2π / w = 2π / 20π = 0.1 seg
D.) Frecuencia (f):
f = 1 / T = 1 / 0.1
1 / 0,1 = 10 Hz
