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MAXImum [283]
2 years ago
9

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

Explanation:

From the question we are told that:

Speed of Train V=3.0m.s

Angle \theta=12\textdegree

Height of window h_w=0.90m

Width of window w_w=2.0m

The Horizontal distance between B and A from Trigonometric Laws is mathematically given by

 b=\frac{0.9}{tan12}

 b=4.23

Therefore

Distance from A-A

 d_a=2.0+4.23

 d_a=6.23

Therefore

Time Required to travel trough d is mathematically given as

 T=\frac{d_a}{v}

 T=\frac{6.23}{3}

 T=2.08s

5 0
2 years ago
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