Answer:
The force applied is 32 N
Explanation:
F = ma
F = 8 × 4
F = 32 N
Answer:
A cyclotron is a type of particle accelerator.
Explanation:
Cyclotron frequency is the frequency of a charged particle moving perpendicular to the direction of a uniform magnetic field B, since that motion is always circular, the cyclotron frequency is given by equality of centripetal force and magnetic Lorentz force.
Answer:
a.) The main scale reading is 10.2cm
b.) Division 7 = 0.07
c.) 10.27 cm
d.) 10.31 cm
e.) 10.24 cm
Explanation:
The figure depicts a vernier caliper readings
a.) The main scale reading is 10.2 cm
The reading before the vernier scale
b.) Division 7 = 0.07
the point where the main scale and vernier scale meet
c.) The observed readings is
10.2 + 0.07 = 10.27 cm
d.) If the instrument has a positive zero error of 4 division
correct reading = 10.27 + 0.04 = 10.31cm
e.) If the instrument has a negative zero error of 3 division
correct reading = 10.27 - 0.03 = 10.24cm
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Answer:
mass of the neutron star =3.45185×10^26 Kg
Explanation:
When the neutron star rotates rapidly, a material on its surface to remain in place, the magnitude of the gravitational acceleration on the central material must be equal to magnitude of the centripetal acc. of the rotating star.
That is
![\frac{GM_{ns}}{R^2}= \omega^2 R](https://tex.z-dn.net/?f=%5Cfrac%7BGM_%7Bns%7D%7D%7BR%5E2%7D%3D%20%5Comega%5E2%20R)
M_ns = mass odf the netron star.
G= gravitational constant = 6.67×10^{-11}
R= radius of the star = 18×10^3 m
ω = 10 rev/sec = 20π rads/sec
therefore,
![M_{ns}= \frac{\omega^2R^3}{G} = \frac{4\pi^2\times(18\times10^3)^3}{6.67\times10^{-11}}](https://tex.z-dn.net/?f=M_%7Bns%7D%3D%20%5Cfrac%7B%5Comega%5E2R%5E3%7D%7BG%7D%20%3D%20%5Cfrac%7B4%5Cpi%5E2%5Ctimes%2818%5Ctimes10%5E3%29%5E3%7D%7B6.67%5Ctimes10%5E%7B-11%7D%7D)
= 3.45185... E26 Kg
= 3.45185×10^26 Kg