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Dafna1 [17]
3 years ago
5

An object located near the surface of Earth has a weight of a 245 N

Physics
1 answer:
marusya05 [52]3 years ago
4 0

Answer:

The mass of the object is 24.5 kg and weight of the object on Mars is 91.14 N.

Explanation:

Weight of the object on the surface of Earth, W = 245 N

On the surface of Earth, acceleration due to gravity, g = 10 m/s²

Weight of an object is given by :

W = mg

m is mass

m=\dfrac{W}{g}\\\\m=\dfrac{245\ N}{10\ m/s^2}\\\\=24.5\ kg

So, the mass of the object is 24.5 kg

Acceleration due to gravity on Mars, g' = 3.72 m/s²

Weight of the object on Mars,

W' =mg'

W' = 24.5 kg × 3.72 m/s²

= 91.14 N

So, the weight of the object on Mars is 91.14 N.

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3 years ago
A ball of mass m, attached to the end of a horizontal cord, is rotated in a circle of radius r on a frictionless horizontal surf
kirza4 [7]

Answer:v=\sqrt{\frac{FL}{m}}

Explanation:

Given

Ball of mass m

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Here Tension will Provide Centripetal Force

T=Centripetal Force

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3 years ago
Can A positively charged body attract another positively charged body​
andriy [413]

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3 0
3 years ago
A stationary boat in the ocean is experiencing waves from a storm. The waves move at 59 km/h and have a wavelength of 145 m . Th
krek1111 [17]

Answer:

The time elapses until the boat is first at the trough of a wave is 4.46 seconds.

Explanation:

Speed of the wave, v = 59 km/h = 16.38 m/s

Wavelength of the wave, \lambda=145\ m

If f is the frequency of the wave. The frequency of a wave is given by :

v=f\lambda\\\\f=\dfrac{v}{\lambda}\\\\f=\dfrac{16.38\ m/s}{145\ m}\\\\f=0.112\ Hz

The time period of the wave is given by :

T=\dfrac{1}{f}\\\\T=\dfrac{1}{0.112\ Hz}\\\\T=8.92\ s

We need to find the time elapses until the boat is first at the trough of a wave. So, the time will be half of the time period of the wave.

T=\dfrac{8.92}{2}\\\\T=4.46\ s

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5 0
3 years ago
A trombone can produce pitches ranging from 85 Hz to 660 Hz approximately. When the trombone is producing a 562 Hz tone, what is
tester [92]

To solve this problem we will apply the concept of wavelength, which warns that this is equivalent to the relationship between the speed of the air (in this case in through the air) and the frequency of that wave. The air is in standard conditions so we have the relation,

Frequency = f = 562Hz

Speed of sound in air = v = 331m/s

The definition of wavelength is,

\lambda = \frac{v}{f}

Here,

v = Velocity

f = Frequency

Replacing,

\lambda = \frac{331m/s}{562Hz}

\lambda = 0.589m

Therefore the wavelength of that tone in air at standard conditions is 0.589m

3 0
3 years ago
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