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Anettt [7]
3 years ago
11

Which has higher resistence a 50W lamp bulb or a 25W lamp​ which greater resistance

Physics
1 answer:
Goryan [66]3 years ago
3 0
25Wwwwwwwww has a greater resists
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What is the value of work done on an object when a 70–newton force moves it 9.0 meters in the same direction as the force?
slava [35]

Answer:

630 J

Explanation:

W = Fdcosθ

W = 70(9)cos0

W = 630 J

5 0
3 years ago
Please help, thank you,
Elena-2011 [213]

same, stapler, gravity, motion, acceleration

6 0
3 years ago
Which name is given to the type of friction that objects falling through air experience?
Pani-rosa [81]

Some call it "air resistance", and others just call it "drag".

4 0
3 years ago
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light of a wavelength 600 nm shines on a soap bubble film. For what soap film thickness will destructive interference occur
VashaNatasha [74]

Answer:

The minimum thickness of the soap bubble for destructive interference to occur is 225.56 nm.

Explanation:

Given;

wavelength of light, λ = 600 nm

The minimum thickness of the soap bubble for destructive interference to occur is given by;

t = \frac{\lambda/n}{2}\\\\t = \frac{\lambda}{2n}

where;

n is refractive index of soap film = 1.33

t = \frac{\lambda}{2n} \\\\t = \frac{600*10^{-9}}{2(1.33)}\\\\t = 2.2556 *10^{-7} \ m\\\\t =  225.56 *10^{-9} \ m\\\\t = 225.56 \ nm

Therefore, the minimum thickness of the soap bubble for destructive interference to occur is 225.56 nm.

4 0
3 years ago
A man with a mass of 65.0 kg skis down a frictionless hill that is 5.00 m high. At the bottom of the hill the terrain levels out
anzhelika [568]

Answer:

The horizontal distance is 4.823 m

Solution:

As per the question:

Mass of man, m = 65.0 kg

Height of the hill, H = 5.00 m

Mass of the backpack, m' = 20.0 kg

Height of ledge, h = 2 m

Now,

To calculate the horizontal distance from the edge of the ledge:

Making use of the principle of conservation of energy both at the top and bottom of the hill (frictionless), the total mechanical energy will remain conserved.

Now,

KE_{initial} + PE_{initial} = KE_{final} + PE_{final}

where

KE = Kinetic energy

PE = Potential energy

Initially, the man starts, form rest thus the velocity at start will be zero and hence the initial Kinetic energy will also be zero.

Also, the initial potential energy will be converted into the kinetic energy thus the final potential energy will be zero.

Therefore,

0 + mgH = \frac{1}{2}mv^{2} + 0

2gH = v^{2}

v = \sqrt{2\times 9.8\times 5} = 9.89\ m/s

where

v = velocity at the hill's bottom

Now,

Making use of the principle of conservation of momentum in order to calculate the velocity after the inclusion, v' of the backpack:

mv = (m + m')v'

65.0\times 9.89 = (65.0 + 20.0)v'

v' = 7.56\ m/s

Now, time taken for the fall:

h = \frac{1}{2}gt^{2}

t = \sqrt{\frac{2h}{g}}

t = \sqrt{\frac{2\times 2}{9.8} = 0.638\ s

Now, the horizontal distance is given by:

x = v't = 7.56\times 0.638 = 4.823\ m

7 0
3 years ago
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