Answer:
Explanation:
a=v-u/t
a=acceleration
v=final velocity
u=initial velocity
t=tme taken
we need to convert from kph to ms⁻¹
v= 150*1000/60*60= 41.67ms⁻¹
u= 120*1000/60*60= 33.33ms⁻¹
t= 2*60= 120s
a=41.67-33.33/120
a=8.34/120
a=0.0694ms⁻²
Answer:
Explanation:
24 - gauge wire , diameter = .51 mm .
Resistivity of copper ρ = 1.72 x 10⁻⁸ ohm-m
R = ρ l / s
1.72x 10⁻⁸ / [3.14 x( .51/2)² x 10⁻⁶ ]
= 8.42 x 10⁻² ohm
= .084 ohm
B ) Current required through this wire
= 12 / .084 A
= 142.85 A
C )
Let required length be l
resistance = .084 l
2 = 12 / .084 l
l = 12 / (2 x .084)
= 71.42 m
I think the answer is c. but I think it depends on how many zebras you have
If it produces 20J of light energy in a second, then that 20J is the 10% of the supply that becomes useful output.
20 J/s = 10% of Supply
20 J/s = (0.1) x (Supply)
Divide each side by 0.1:
Supply = (20 J/s) / (0.1)
<em>Supply = 200 J/s </em>(200 watts)
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Here's something to think about: What could you do to make the lamp more efficient ? Answer: Use it for a heater !
If you use it for a heater, then the HEAT is the 'useful' part, and the light is the part that you really don't care about. Suddenly ... bada-boom ... the lamp is 90% efficient !
Potential energy of spring equals K times X squared divided by 2 where X is displacement
4 times squared equals 16
choose 1st answer
