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3 years ago

How gravity is related to the weight of an object

2 answers:

6 0

An object with twice as much mass will exert twice as much gravitational pull on other objects. The gravitational force increases as the size of an objectincreases. On the other hand, the strength of gravity is inversely related to the square of the distance between two objects.

jeka943 years ago

3 0

Gravity is related to a weight of an object because when you drop a 500 page book and an apple at the same time the book will land first.

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HELPPpPpPpPpPp!!!! ASAP !!!

DENIUS [597]

Answer:

50 N

Explanation:

8 0

2 years ago

The vapor pressure of benzene, C6H6, is 40.1 mmHg at 7.6°C. What is its vapor pressure at 60.6°C? The molar heat of vaporization

ANEK [815]

Answer:

The vapor pressure at 60.6°C is 330.89 mmHg

Explanation:

Applying Clausius Clapeyron Equation

$ln(\frac{P_2}{P_1}) = \frac{\delta H}{R}[\frac{1}{T_1}- \frac{1}{T_2}]$

Where;

P₂ is the final vapor pressure of benzene = ?

P₁ is the initial vapor pressure of benzene = 40.1 mmHg

T₂ is the final temperature of benzene = 60.6°C = 333.6 K

T₁ is the initial temperature of benzene = 7.6°C = 280.6 K

ΔH is the molar heat of vaporization of benzene = 31.0 kJ/mol

R is gas rate = 8.314 J/mol.k

$ln(\frac{P_2}{40.1}) = \frac{31,000}{8.314}[\frac{1}{280.6}- \frac{1}{333.6}]\\\\ln(\frac{P_2}{40.1}) = 3728.65 (0.003564 - 0.002998)\\\\ln(\frac{P_2}{40.1}) = 3728.65 (0.000566)\\\\ln(\frac{P_2}{40.1}) = 2.1104\\\\\frac{P_2}{40.1} = e^{2.1104}\\\\\frac{P_2}{40.1} = 8.2515\\\\P_2 = (40.1*8.2515)mmHg = 330.89 mmHg$

Therefore, the vapor pressure at 60.6°C is 330.89 mmHg

6 0

4 years ago

Read 2 more answers

Assume that the electric field E is equal to zero at a given point. Does it mean that the electric potential V must also be equa

lyudmila [28]

Answer:

No, this doesn't mean the electric potential equals zero.

Explanation:

In electrostatics, the electric field $\vec{E}$ is related to the gradient of the electric potential V with :

$\vec{E} (\vec{r}) = - \vec{\nabla} V (\vec{r})$

This means that for constant electric potential the electric field must be zero:

$V(\vec{r}) = k$

$\vec{E} (\vec{r}) = - \vec{\nabla} V (\vec{r}) = - \vec{\nabla} k$

$\vec{E} (\vec{r}) = - (\frac{\partial}{\partial x} , \frac{\partial}{\partial y } , \frac{\partial}{\partial z}) k$

$\vec{E} (\vec{r}) = - (\frac{\partial k}{\partial x} , \frac{\partial k}{\partial y } , \frac{\partial k}{\partial z})$

$\vec{E} (\vec{r}) = - (0,0,0)$

This is not the only case in which we would find an zero electric field, as, any scalar field with gradient zero will give an zero electric field. For example:

$V(\vec{r})= (x+2)^2 (y+4)^3 (z+5)^4$

give an electric field of zero at point (0,0,0)

8 0

3 years ago

How much time will it take to do 33j of work with 11 w of power

Yuri [45]

Time required : 3 s

<h3>Further explanation </h3>

Power is the work done/second.

$\tt P=\dfrac{W}{t}\\\\P=power,j/s,watt\\\\W=work, J\\\\t=times=s$

To do 33 J of work with 11 W of power

P = 11 W

W = 33 J

$\tt t=\dfrac{W}{P}\\\\t=\dfrac{33}{11}\\\\t=\boxed{\bold{3~s}}$

8 0

3 years ago

A 60.0 kg girl stands up on a stationary floating raft and decides to go into shore. She dives off the 180 kg floating raft with

padilas [110]

Momentum, p = m.v
m of the girl = 60.0 kg
m of the boat = 180 kg
v of the girl = 4.0 m/s

A) Momentum of the girl as she is diving:
p = m.v = 60.0 kg * 4.0 m/s = 24.0 N/s

B) momentum of the raft = - momentum of the girl = -24.0 N/s

C) speed of the raft

p = m.v ; v = p/m = 24.0N/s / 180 kg = -0.13 m/s [i.e. in the opposite direction of the girl's velocity]

8 0

3 years ago