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lutik1710 [3]
3 years ago
9

12. A car is travelling at 30 m/s when the driver sees a red light in the distance and immediately applies the brakes. The car c

omes to a stop 1.5 s later. How far did the car move from the time the driver applied the brake to when it came to a stop?​
Physics
1 answer:
Triss [41]3 years ago
4 0

Answer:

22.5 m

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 30 m/s

Time (t) = 1.5 s

Final velocity (v) = 0 m/s

Distance (s) =?

The distance to which the car move before stopping from the time the driver applied the brake can be obtained as follow:

s = (u + v)t/2

s = (30 + 0)1.5 / 2

s = (30 × 1.5) / 2

s = 45 / 2

s = 22.5 m

Thus, the car will move to a distance of 22.5 m before stopping from the time the driver applied the brake.

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vichka [17]
To make something electrical you need electricity and the wires will turn a magnet into an electromagnet.
or for a short answer: <span>1) soft iron core 2) coil of insulated wire 3)source of electricity</span>
8 0
3 years ago
In a double slit experiment, the distance between the slits is 0.2 mm and the distance to the screen is 100 cm. What is the phas
Anni [7]

Answer:

The phase difference is  \Delta \phi  = 180^o

Explanation:

From the question we are told that

     The distance between the slits is d = 0.2 \ mm = \frac{0.2}{1000}  = 0.2 *10^{-3} \ m

     The distance to the screen is D = 100 cm = \frac{100}{100} = 1 \ m

      The wavelength is  \lambda = 400nm

 The distance of the wave from the  central maximum is L =  5mm = 5*10^{-3} m

   

Generally the path difference of this  waves is mathematically represented as

              y = d sin \theta

Here \theta is the angle between the the line connecting the mid-point of the slits with  the screen and the line  connecting the mid-point of the slits to the central maximum

  This implies that

              tan \theta  = \frac{L}{D}

     =>     \theta = tan ^{-1} \frac{L}{D}

             \theta = tan ^{-1} [\frac{5*10^{-3}}{1}]

           \theta =0.2865

Substituting values into the formula for path difference

       y = 0.2 *10^{-3} sin(0.2864)  

       y = 9.997*10^{-7} \ m  

The phase difference is mathematically represented as

          \Delta \phi = \frac{2 \pi }{\lambda }  * y

Substituting values        

         \Delta \phi = \frac{2 \pi }{400 *10^{-9} }  \ * 9.997*10^{-7}

         \Delta \phi =5 \pi

Converting to degree

         \Delta \phi =5 \pi radians = 5 (180^o)  =  180^o  

the solution is subtracted by 360° in order to get the actual angle

 

             

4 0
3 years ago
Find the range of a projectile launched at an angle of 30° with an initial velocity of 20m/s.​
Tems11 [23]

Answer:

<em>The range is 35.35 m</em>

Explanation:

<u>Projectile Motion</u>

It's the type of motion that experiences an object projected near the Earth's surface and moves along a curved path exclusively under the action of gravity.

Being vo the initial speed of the object, θ the initial launch angle, and g=9.8m/s^2 the acceleration of gravity, then the maximum horizontal distance traveled by the object (also called Range) is:

\displaystyle d={\frac  {v_o^{2}\sin(2\theta )}{g}}

The projectile was launched at an angle of θ=30° with an initial speed vo=20 m/s. Calculating the range:

\displaystyle d={\frac  {20^{2}\sin(2\cdot 30^\circ )}{9.8}}

\displaystyle d={\frac  {400\sin(60^\circ )}{9.8}}

d=35.35\ m

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7 0
3 years ago
Two uncharged metal spheres, spaced 25.0 cm apart, have a capacitance of 26.0 pF. How much work would it take to move 12.0 nC of
viktelen [127]

Answer:

W=2.76\times 10^{-6}\ J

Explanation:

Given that,

The distance between two spheres, r = 25 cm = 0.25 m

The capacitance, C = 26 pF = 26×10⁻¹² F

Charge, Q = 12 nC = 12 × 10⁻⁹ C

We need to find the work done in moving the charge. We know that, work done is given by :

U=\dfrac{Q^2}{2C}

Put all the values,

U=\dfrac{(12\times 10^{-9})^2}{2\times 26\times 10^{-12}}\\\\U=2.76\times 10^{-6}\ J

So, the work done is 2.76\times 10^{-6}\ J.

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