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Alex17521 [72]
3 years ago
12

What type of reaction is being shown in this energy diagram?

Physics
1 answer:
Ratling [72]3 years ago
8 0

Answer:

O exothermic, because energy is released into the surrounding

Explanation:

From the diagram the energy of the reactant is higher than the energy of the product, thereby making it exothermic. If you study diagram well, exothermic reaction means that the reactions releases energy into the surroundings.

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An object is placed in front of a convex lens of a length 10cm. What is the nature of the image formed if the object distance is
Lady_Fox [76]

{\mathfrak{\underline{\purple{\:\:\: Given:-\:\:\:}}}} \\ \\

\:\:\:\:\bullet\:\:\:\sf{Focal\:length=10\:cm}

\:\:\:\:\bullet\:\:\:\sf{Object \ distance = -15\:cm}

\\

{\mathfrak{\underline{\purple{\:\:\:To \:Find:-\:\:\:}}}} \\ \\

\:\:\:\:\bullet\:\:\:\sf{Nature \: of \:the\:image}

\\

{\mathfrak{\underline{\purple{\:\:\: Solution:-\:\:\:}}}} \\ \\

<h3>☯ <u>By using formula of Lens</u> </h3>

\\

\dashrightarrow\:\: {\boxed{\sf{\dfrac{1}{u} + \dfrac{1}{v} = \dfrac{1}{f}}}}

\\

\dashrightarrow\:\: \sf{\dfrac{1}{v}-\dfrac{1}{-15}=\dfrac{1}{10}}

\\

\dashrightarrow\:\: \sf{\dfrac{1}{v}+\dfrac{1}{15}=\dfrac{1}{10}}

\\

\dashrightarrow\:\: \sf{\dfrac{1}{v} = \dfrac{1}{10} - \dfrac{1}{15}}

\\

\dashrightarrow\:\: \sf{\dfrac{1}{v} = \dfrac{1}{30}}

\\

\dashrightarrow\:\: \sf{ v = 30 \ cm}

\\

<h3>☯ <u>Now, Finding the magnification </u></h3>

\\

\dashrightarrow\:\: \sf{ m = \dfrac{-30}{-15}}

\\

\dashrightarrow\:\: \sf{m = -2}

\\

<h3>☯ <u>Hence</u>,\\</h3>

\:\:\:\:\star\:\:\:\sf{Image \ distance = 30 \ cm}

\:\:\:\:\star\:\:\:\sf{Nature = Real \ \& \ inverted}

3 0
3 years ago
Write<br>any<br>to increase<br>magnitude of current in dynamo​
Svetach [21]

Answer:

Explanation:

It can be increased by: increasing the rate of rotation. Increasing the strength of the magnetic field. Increasing the number of turns on the coil.

Hope this helps

plz mark it as brainliest!!!!!!

4 0
3 years ago
A square coil ℓ = 2cm on a side with 30 turns rotates in a uniform magnetic field, B~ = B0zˆ = 0.1Tˆz, such that the normal of t
kow [346]

Answer:

a) 1.2*10^{-3}cos(1.25t)

b) 0.49mV

Explanation:

a) The coil rotates periodically with period T. Hence, we can write the variation of the magnetic flux with a sinusoidal function, and with max flux NAB. Thus, we have that:

\Phi_B(t)=NABcos(\omega t)\\\\\omega=\frac{2\pi}{T}=1.25\frac{rad}{s}\\\\A=l^2=(0.02m)^2=4*10^{-4}m^2\\\\B=0.1T\\\\\Phi_B(t)=1.2*10^{-3}cos(1.25 t) W

where we have used the values given by the information of the problem for N B and A.

b)

the emf is given by:

emf=-\frac{d\Phi_B}{dt}=-NBA\omega sin(\omega t)\\\\emf(t=12.5s)=-(30)(0.1T)(4*10^{-4})(1.25\frac{rad}{s})sin(1.25*12.5)=1.49*10^{-4}V=0.49mV

hope this helps!!

5 0
3 years ago
Which simple machine is NOT correctly matched with an appropriate task for its use?
Tcecarenko [31]
The simple machine that is not correctly matched with  its appropriate task is the inclined plane because there is no such big ramp that is as high as 1 storey building, the appropriate task would be Lifting a heavy box and moving it across a room. and for the pulley : <span>Moving a heavy box up to the second floor of a building.</span>
6 0
3 years ago
Calculate the orbital speed (in m/s) of a satellite that circles the Earth with a time period of 12.00 hours. The mass of the Ea
Hoochie [10]

Answer:

v = 3869 m/s

Explanation:

As we know that the orbital speed of the satellite is given as

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

also we know that

time period of the revolution is given as

T = \frac{2\pi r}{v}

now from above equation we know that

T = \frac{2\pi (\frac{GM}{v^2})}{v}

T = \frac{2\pi GM}{v^3}

so we will have

v = (\frac{2\pi GM}{T})^{1/3}

now plug in all data in this equation

v = (\frac{2\pi (6.67 \times 10^{-11})(5.97 \times 10^{24})}{12 \times 3600})^{1/3}

v = 3869 m/s

3 0
3 years ago
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