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3 years ago

A boat having a length 3m and breadth 2m is floating on a

Physics

1 answer:

brilliants [131]3 years ago

3 0

Answer:

60 kg

Explanation:

The man's weight is equal to the weight of the water he displaces.

mg = ρVg

m = ρV

m = (1000 kg/m³) (3 m × 2 m × 0.01 m)

m = 60 kg

The man's mass is 60 kg.

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As you lift a book into the air, what kind of energy are you increasing in the book?

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Gravitational potential energy.

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3 years ago

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Suppose you are conducting an experiment concerning gravity's effects on falling objects, and you want to minimize air resistanc

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4 years ago

What would happen to a satellite if it’s orbit speed was reduced by half? Also how about if it doubled?

weeeeeb [17]

For finding the orbital speed of the satellite we can say that the centripetal force for the circular motion of satellite is provided by the gravitational force of earth

so here we can say

$F_g = \frac{mv^2}{r}$

$\frac{GMm}{r^2} = \frac{mv^2}{r}$

now we will have

$v = \sqrt{\frac{GM}{r}}$

now here we will say that orbital speed of the satellite is inversely depends on the orbital radius

<em>So here if orbital speed is half then as per above relation we can say that orbital distance will become four times</em>

<em>Also we can say that if orbital speed is double then orbital distance will become one fourth of initial distance.</em>

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3 years ago

Which of the options is the best definition of a vector?

11111nata11111 [884]

The answer is B, the vector represents magnitude and direction

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3 years ago

German physicist Werner Heisenberg related the uncertainty of an object's position ( Δ x ) to the uncertainty in its velocity (

Ierofanga [76]

Answer:

The uncertainty in the position of the electron is $5.79x10^{-9}m$

Explanation:

The Heisenberg uncertainty principle is defined as:

$\Lambda p\Lambda x$ ≥ $\frac{h}{4 \pi}$ (1)

Where $\Lambda p$ is the uncertainty in momentum, $\Lambda x$ is the uncertainty in position and h is the Planck's constant.

The momentum is defined as:

$p =mv$ (2)

Therefore, equation 2 can be replaced in equation 1

$\Lambda (mv) \Lambda x$ ≥ $\frac{h}{4 \pi}$

Since, the mass of the electron is constant, v will be the one with an associated uncertainty.

$m \Lambda v \Lambda x$ ≥ $\frac{h}{4 \pi}$ (3)

Then, $\Lambda x$ can be isolated from equation 3

$\Lambda x$ ≥ $\frac{h}{m \Lambda v 4 \pi}$ (4)

$\Lambda x = \frac{6.626x10^{-34}J.s}{(9.11x10^{-31} kg)(0.01x10^{6}m/s) 4 \pi}$

But $1J = Kg.m^{2}/s^{2}$

$\Lambda x = \frac{(6.624x10^{-34} Kg.m^{2}/s^{2}.s)}{(9.11x10^{-31} kg)(0.01x10^{6}m/s) (4 \pi)}$

$\Lambda x = 5.79x10^{-9}m$

Hence, the uncertainty in the position of the electron is $5.79x10^{-9}m$

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3 years ago