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Irina-Kira [14]
2 years ago
13

Pls help for brainliest:)

Physics
1 answer:
Rasek [7]2 years ago
6 0

Answer:

<h2>C. </h2>

Explanation:

<h3>#CARRY ON LEARNING</h3><h3>#MARK ON LEARNING</h3><h3>#HELPING HAND</h3>
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A car goes East from 2 m/s to 16m/s in 3.5s. What is the Car's acceleration?
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A=v1+v2 / t
2+16/3.5=?
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3 years ago
How can you tell the difference between chemical sedimentary rocks and intrusive<br> igneous rocks?
Nookie1986 [14]

Answer:

Sedimentary rocks are usually formed under water when grains of broken rocks are glued together while igneous rocks form when melted rock (magma or lava) cools and metamorphic are rocks that once were igneous or sedimentary rocks but have been changed by pressure and temperature.

4 0
3 years ago
Read 2 more answers
A wire lying along a y axis form y=0 to y=0.25 m carries a current of 2.0 mA in the negative direction of the axis. The wire ful
balu736 [363]

Answer:

FB = 0.187 N

Explanation:

To find the magnetic force FB in the wire you use the following formula:

|\vec{F_B}|=ILBsin\theta\\\\L=0.25m\\\\|\vec{B}|=\sqrt{(0.3y)^2+(0.4y)^2}=0.5y \ T

the angle between B and L is given by:

\vec{L}\cdot\vec{B}=LBcos\theta\\\\\theta=cos^{-1}(\frac{\vec{L}\cdot\vec{B}}{LB})=cos^{-1}(\frac{0*0.3y+0.25*0.4y}{0.25*0.5y})=36.86\°

Due to B depends on "y" you take into account the contribution of each element dy of the wire to the magnitude of the magnetic force. Thus, you have to integrate the following expression:

|\vec{F_B}|=Isin\theta\int_0^{0.25}B(y)dy=Isin\theta\int_0^{0.25}(0.5y)dy\\\\|\vec{F_B}|=(2.0*10^{-3}A)(sin36.86\°)(0.5T)[\frac{0.25^2}{2}m]=0.187\ N

hence, the magnitude of the magnetic force is 0.187N

4 0
3 years ago
10.A car is travelling at a constant speed of 27m/s. The driver looks away from the road for a 2.0s to tune in a station on the
Korvikt [17]

Explanation:

Distance = speed × time

d = (27 m/s) (2.0 s)

d = 54 m

5 0
3 years ago
A resonant circuit using a 286-nFnF capacitor is to resonate at 18.0 kHzkHz. The air-core inductor is to be a solenoid with clos
lukranit [14]

Answer:

The inductor contains N = 523962.32 loops  

Explanation:

From the question we are told that

     The capacitance of the capacitor is  C =  286nF = 286 * 10^{-9} \  F

      The resonance frequency is  f = 18.0 kHz =  18*10^{3} Hz

       The diameter is  d =  1.1 mm = \frac{1.1 }{1000} = 0.00011 \ m

       The  of the air-core inductor is l = 12 \ m

        The permeability of free space is  \mu_o = 4 \pi *10^{-7} \ T \cdot m/A

 

Generally the inductance of this air-core inductor is mathematically represented as

              L =  \frac{\mu_o * N^2 \pi d^2}{4 l}

This inductance can also be mathematically represented as

               L = \frac{1}{w^2}

Where w is the angular speed mathematically given as

             w = 2 \pi f

So

            L =  \frac{1}{4 \pi ^2 f^2}

Now equating the both formulas for inductance

         \frac{\mu_o * N^2 \pi d^2}{4 l}  =  \frac{1}{4 \pi ^2 f^2}

making N the subject of  the formula

              N = \sqrt{\frac{1}{(2 \pi f)^2} * \frac{4 * l }{\mu_o * \pi d^2 C}  }

              N =  \frac{1}{2 \pi f} * \frac{2}{d} * \sqrt{\frac{l}{\pi * \mu_o * C} }

             

 Substituting value

            N =  \frac{1}{ 3.142  * 18*10^{3} * 0.00011 }  \sqrt{\frac{12}{ 3.142  * 4 \pi *10^{-7}* 286 *10^{-9}} }

              N = 523962.32 loops  

4 0
3 years ago
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