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Assoli18 [71]
3 years ago
5

A 0.59 kg bullfrog is sitting at rest on a level log. how large is the normal force of the log on the bullfrog?

Physics
1 answer:
ladessa [460]3 years ago
7 0
<span>The bullfrog is sitting at rest on the log. The force of gravity pulls down on the bullfrog. We can find the weight of the bullfrog due to the force of gravity. weight = mg = (0.59 kg) x (9.80 m/s^2) weight = 5.782 N The bullfrog is pressing down on the log with a force of 5.782 newtons. Newton's third law tells us that the log must be pushing up on the bullfrog with a force of the same magnitude. Therefore, the normal force of the log on the bullfrog is 5.782 N</span>
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if a current of 0.3 ampere flows through a conductor of resistance of 11.7 ohm's,the voltage across the ends of the conductor is
Gnesinka [82]

Here,

Resistance = 11.7

Ampere = 0.3

Voltage = ?

Now,

R = V/A

11.7 = V/0.3

11.7*0.3 =V

V = 3.51

I hope it's help you...

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7 0
3 years ago
A cat is dropped out of a window and falls to the ground in 12 seconds. It lands on its feet.
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Answer:

The impact was gravity

Explanation:

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3 years ago
A certain pair of slits are separated by a distance d. Monochromatic coherent light falls on this pair of slits and the interfer
DanielleElmas [232]

Answer:

The new separation distance between adjacent bright fringes will be <u>4 mm</u>

Explanation:

Since, the distance between adjacent bright fringes is given by the formula:

Δx₁ = λL/d = 2 mm  -------- eqn (1)

where,

Δx = Distance between adjacent bright fringes

λ = wavelength of light = constant for both cases

L = Distance between the slits and the screen

d = slit separation

Now, for the second case:

Slit Separation = d/2

Therefore,

Δx₂ = λL/(d/2)

Δx₂ = 2(λL/d)

using eqn (1), we get:

Δx₂ = 2 Δx₁

Δx₂ = 2(2 mm)

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5 0
3 years ago
You are trying to decide between two new stereo amplifiers. One is rated at 130 W per channel and the other is rated at 200 W pe
NARA [144]

Answer:

The sound level will be 1.870 dB louder.

Explanation:

Given that,

Power = 130 W

Power = 200 W

We need to calculate the sound level

Using formula of sound level

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For one amplifier,

I_{1}=10\log(\dfrac{130}{I_{0}})...(I)

For other amplifier,

I_{2}=10\log(\dfrac{200}{I_{0}})...(II)

For difference in dB levels

I_{2}-I_{1}=10\log(\dfrac{200}{I_{0}})-10\log(\dfrac{130}{I_{0}})

I_{2}-I_{1}=10\log(\dfrac{200}{I_{0}}\times\dfrac{I_{0}}{130})

I_{2}-I_{1}=10\log(\dfrac{200}{130})

I_{2}-I_{1}=1.870\ dB

Hence, The sound level will be 1.870 dB louder.

7 0
3 years ago
Is the time in Tennessee the same as florida
geniusboy [140]
I would say the same thing as the first answer
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