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galina1969 [7]
4 years ago
13

Someone please help me to answer this questions...​

Physics
1 answer:
tatyana61 [14]4 years ago
5 0

Answer:

a.25J

Explanation:

m=0.5kg

g=10

h=5

potential energy=m×g×h

=0.5×10×5

=25J

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A wire of resistance 5.9 Ω is connected to a battery whose emf ε is 4.0 V and whose internal resistance is 1.2 Ω. In 2.9 min, ho
dexar [7]

Answer:

a) 390J

b) 322J

c) 68J

Explanation:

We need to calculate the power given by the battery. the power is given by:

P=V*I\\I=\frac{V}{R}\\I=\frac{4}{5.9+1.2}\\I=0.56A\\P=2.24W

Watts is J/s so:

E=P*t\\E=2.24\frac{J}{s}*2.9min*(60\frac{s}{min})=390J

The thermal energy in the wire is given by:

E_w=P_w*t\\P_w=I^2*R_w=0.56^2*5.9=1.85W\\E_w=1.85\frac{J}{s}*2.9min*(60\frac{s}{min})=322J

And the the dissipated thermal energy in the battery will be the remainig energy:

E_b=E-E_w\\E_b=390-322=68J

4 0
4 years ago
In Millikan's oil drop experiment an oil drop is at rest between two large plates separated by 1.3 cm when the potential differe
sweet-ann [11.9K]

Answer:

Mass of the oil drop, m=3.01\times 10^{-15}\ kg

Explanation:

Potential difference between the plates, V = 400 V

Separation between plates, d = 1.3 cm = 0.013 m

If the charge carried by the oil drop is that of six electrons, we need to find the mass of the oil drop. It can be calculated by equation electric force and the gravitational force as :

qE=mg

m=\dfrac{qE}{g}

q=6e, e is the charge on electron

E is the electric field, E=\dfrac{V}{d}=\dfrac{400}{0.013}=30769.23\ V/m

m=\dfrac{6\times 1.6\times 10^{-19}\times 30769.23}{9.8}

m=3.01\times 10^{-15}\ kg

So, the mass of the oil drop is 3.01\times 10^{-15}\ kg. Hence, this is the required solution.

5 0
3 years ago
Two horses on opposite sides of a narrow stream are pulling a barge up the stream. Each hors pulls with a force of 720N. The rop
Sonja [21]
For the answer to the question above, each horse's force forms a right angle triangle with the barge and subtends an angle of 60/2 = 30°. The resultant in the direction of the barge's motion is:
Fx = Fcos(∅)
We can multiply this by 2 to find the resultant of both horses.
Fx = 2Fcos(∅)
Fx = 2 x 720cos(30)
Fx = 1247 N
8 0
3 years ago
POTENTIAL I KINETIC ENERGY
TEA [102]

7kinetic energy is decreasing in B

4 0
2 years ago
a bubble of air of volume 1cm^3 is released by a deep sea diver at a depth where the pressure is 4.0 atmospheres. assuming its t
lys-0071 [83]

Answer:

hope this helps!

Explanation:

Volume of the air bubble, V1=1.0cm3=1.0×10−6m3

Bubble rises to height, d=40m

Temperature at a depth of 40 m, T1=12oC=285K

Temperature at the surface of the lake, T2=35oC=308K

The pressure on the surface of the lake: P2=1atm=1×1.103×105Pa 

The pressure at the depth of 40 m: P1=1atm+dρg

Where,

ρ is the density of water =103kg/m3

g is the acceleration due to gravity =9.8m/s2

∴P1=1.103×105+40×103×9.8=493300Pa

We have T1P1V1=T2P2V2

Where, V2 is the volume of the air bubble when it reaches the surface.

V2=

8 0
3 years ago
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