Answer: The increase in temperature of the nail after the three blows is 8.0636 Kelvins. The correct option is (d).
Explanation:
Kinetic energy of the hammer ,K.E.=
Half of the kinetic energy of the hammer is transformed into heat in the nail.
Energy transferred to the nail in one blow =
Total energy transferred after 3 blows,Q =
Mass of the nail = 15 g = 0.015 kg
Change in temperature =
Specif heat of the steel = c = 448 J/kg K
The increase in temperature of the nail after the three blows is 8.1 Kelvins.Hence, correct option is (d).
Answer:
No. The protostellar cloud spins faster in the collapsing stage (stage 1) and becomes much slower in the contraction stage (stage 2)
Explanation:
Once the cloud is so dense that the heat which is being produced in its center cannot easily escape, pressure rapidly rises, and catches up with the weight, or whatever external force is causing the cloud to collapse, and the cloud becomes stable, as a protostellar cloud.
The protostellar cloud will become more dense over thousands of years. This stage of decreasing size is known as a contraction, rather than a collapse. In the contraction stage the cloud has become much slower, and because weight and pressure are more or less in balance. In the first stage of formation, the decrease of size is very rapid, and compressive forces completely overwhelm the pressure of the gas, and we say that the cloud is collapsing.
voltage across 2.0μf capacitor is 5.32v
Given:
C1=2.0μf
C2=4.0μf
since two capacitors are in series there equivalent capacitance will be
[tex] \frac{1}{c} = \frac{1}{c1} + \frac{1}{c2} [/tex]
=1.33μf
As the capacitance of a capacitor is equal to the ratio of the stored charge to the potential difference across its plates, giving: C = Q/V, thus V = Q/C as Q is constant across all series connected capacitors, therefore the individual voltage drops across each capacitor is determined by its its capacitance value.
Q=CV
given,V=8v
charge on 2.0μf capacitor is
=5.32v
learn more about series capacitance from here: brainly.com/question/28166078
#SPJ4
The wavelength of the third line in the Lyman series, and identify the type of EM radiation
In this series, the spectral lines are obtained when an electron makes a transition from any high energy level (n=2,3,4,5... ). The wavelength of light emitted in this series lies in the ultraviolet region of the electromagnetic spectrum.
1 / lambda = R(h)* ( 
- 
)
= 109678 ( 
- 
)
= 109678 (8/9)
Lambda = 9 / (109678 * 8 )
= 102.6 * 
m = 102.6 nm
To learn more about Lyman series here
brainly.com/question/5762197
#SPJ4
