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2 years ago

ID like to say thx everyone who is on brainly u guys helped me get through my quiz so thx

Physics

2 answers:

VARVARA [1.3K]2 years ago

8 0

Answer:

I want to know how your experience has been on brainly. I hope it is good. I really like spending time helping others here. If you have any other questions please ask me :)

Explanation:

Gnesinka [82]2 years ago

5 0

Answer:

u welcome

Explanation:

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At constant volume, the heat of combustion of a particular compound is − 3550.0 kJ / mol. When 1.075 g of this compound ( molar

swat32

Answer:

$C=1,25\cdot 10^{5} kJ/^{\circ}C$

Explanation:

First of all let's define the specific molar heat capacity.

$C = \frac{-Q}{n\cdot \Delta T}$ (1)

Where:

Q is the released heat by the system

n is the number of moles

ΔT is the difference of temperature of the system

Now, we can find n with the molar mass (M) the mass of the compound (m).

$n=\frac{m}{M}=6.95\cdot 10^{-3} moles$

Using (1) we have:

$C=\frac{-3550}{6.95\cdot 10^{-3} 4.073}$

$C=1,25\cdot 10^{5} kJ/^{\circ}C$

I hope it helps!

6 0

3 years ago

-Starts to suck on ur neck giving you hickeys n runs my hand down your body biting my lip against your neck

Hoochie [10]

Answer:

I would shout fore help if I was being raped or try to make him or her stop

6 0

2 years ago

A physics major is cooking breakfast when he notices that the frictional force between the steel spatula and the Teflon frying p

Semenov [28]

Answer:

$f_n=3.75N$

Explanation:

From the question we are told that:

Frictional force $F=0.150N$

Coefficient of kinetic friction $\mu=0.04$

Generally the equation for Normal for is mathematically given by

$f_n=\frac{F}{\mu}$

Therefore

$f_n=\frac{0.150}{0.04}$

$f_n=3.75N$

5 0

2 years ago

How does your power output in climbing the stairs compare to the power output of a 100-watt light bulb? if your power could have

cricket20 [7]

1) Assuming an adult person has an average mass of m=80 kg, and assuming it takes about 30 seconds to climb 5 meters of stairs, the energy used by the person is
$E=mgh=(80 kg)(9.81 m/s^2)(5 m)=3924 J$
So the power output is
$P= \frac{E}{t}= \frac{3924 J}{30 s} \sim 130 W$

And since the estimate we made is very rough, we can say that the power output of the person is comparable to the power output of the light bulb of 100 W.

2) Based on the results we found in the previous part of the exercise, since the power output of the person is comparable to the power output of 1 light bulb of 100 W, we can say that the person could have kept burning only one 100-W light bulb during the climb.

6 0

2 years ago

Read 2 more answers

You are on a train that is traveling at 3.0 m/s along a level straight track. Very near and parallel to the track is a wall that

loris [4]

Questions Diagram is attached below

Answer:

$T=2.08s$