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ololo11 [35]
3 years ago
9

Electromagnetic force is present when electromagnetoc fields is ?

Physics
1 answer:
pickupchik [31]3 years ago
8 0
Electromagnetic force is present when electromagnetic field become charged as they lose electrons. :)

hope this helped!!
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A small package is attached to a helium-filled balloon rising at 2 m/s. The package drops from the balloon when it is 14 meters
scZoUnD [109]
I would assume air resistance is negligible and so the acceleration of the package would be approximately 9.81 m/s².

Taking downwards as positive, use v²=u²+2as.
v²=(-2)²+2(9.81)(14)
v=16.7 m/s
4 0
3 years ago
A race car traveling at 44m/s slows at a constant rate to a velocity of 22m/s over 11 seconds , how far does it move during this
krok68 [10]

Answer:

add 44m/s and 22m/s then multiply it by 11

Explanation:

8 0
3 years ago
The storm surge of a hurricane can cause
BabaBlast [244]

Answer:

it can cause huricans and flooding

Explanation:

Storm surge is the rising of the sea level due to the low pressure, high winds, and high waves associated with a hurricane as it makes landfall. The storm surge can cause significant flooding and cost people their lives if they're caught unexpected

7 0
3 years ago
Read 2 more answers
the amount of time for one particle of the medium to make one comlpete vibration cycle is a good description of
-Dominant- [34]
That's the definition of the PERIOD of the vibration.
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7 0
3 years ago
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A Foucault pendulum consists of a brass sphere with a diameter of 31.0 cm suspended from a steel cable 11.0 m long (both measure
kozerog [31]

Answer:

43.7 °C

Explanation:

\alpha_b = Coefficient of linear expansion of brass = 18\times 10^{-6}\ ^{\circ}C

\alpha_s = Coefficient of linear expansion of steel = 11\times 10^{-6}\ ^{\circ}C

L_{0b} = Initial length of brass = 31 cm

L_{0s} = Initial length of steel = 11 m

\Delta L = Total change in length = 3 mm

Total change in length would be

\Delta L=\Delta L_b+\Delta L_s\\\Rightarrow \Delta L=L_{0b}\alpha_b\Delta T+L_{0s}\alpha_b\Delta T\\\Rightarrow \Delta T=\frac{\Delta L}{L_{0b}\alpha_b+L_{0s}\alpha_b}\\\Rightarrow \Delta T=\frac{0.003}{0.31\times 18\times 10^{-6}+11\times 10^{-6}\times 11}\\\Rightarrow \Delta T=23.7\ ^{\circ}C

\Delta T=23.7\\\Rightarrow T_f-T_i=23.7\\\Rightarrow T_f=23.7+T_i\\\Rightarrow T_f=23.7+20\\\Rightarrow T_f=43.7\ ^{\circ}C

The final temperature is 43.7 °C

6 0
4 years ago
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