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Darina [25.2K]
3 years ago
8

Show by dimension that the equation of motion is correct V² = U²+2as​

Physics
1 answer:
borishaifa [10]3 years ago
5 0

Answer:

i think its 56as

Explanation:

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A 2.50-m segment of wire carries 1000 A current and feels a 4.00-N repulsive force from a parallel wire 5.00 cm away. What is th
Stolb23 [73]

Answer:

The current is  I_b  =  400 \ A

Explanation:

From the question we are told that

    The  length of the segment is  l  =  2.50  \  m

     The current is  I_a  =  1000 \ A

     The force felt is  F  =  4.0 \  N

        The distance of the second wire is  d =  5.0 \ cm  = 0.05 \  m

Generally the current on the second wire is mathematically represented as

        I_b  =  \frac{2 \pi * r * F }{ l *  \mu_o  *  I_a }

Here  \mu_o is the permeability of free space with value  \mu_o =  4 \pi * 10^{-7} \ N/A^2

=>      I_b  =  \frac{2 * 3.142  *  0.05 *  4 }{ 2.50  *  4\pi *10^{-7}  * 1000 }

=>      I_b  =  400 \ A

4 0
2 years ago
Gravity on Earth is 9.8 m/s^2, and gravity on Jupiter is 23.1 m/s^2. So, if the mass of a rock is 70 kilograms, it's weight on E
Levart [38]
~686newtons on earth and
~1617 newtons on jupiter
the formula is weight = gravitational acceleration * mass of the object
3 0
3 years ago
Read 2 more answers
You toss a tennis ball straight upward. At the moment it leaves your hand it is at a height of 1.5 m above the ground, and it is
ollegr [7]

My answer was incorrect, please disregard.

4 0
2 years ago
A vertical straight wire 35.0 cmcm in length carries a current. You do not know either the magnitude of the current or whether t
12345 [234]

Answer:

1.714\ \text{A}

Explanation:

F = Magnetic force = 0.018 N

B = Magnetic field = 0.03 T

L = Length of wire = 35 cm

\theta  = Angle between current and magnetic field = 90^{\circ}

Magnetic force is given by

F=IBL\sin\theta\\\Rightarrow I=\dfrac{F}{BL\sin\theta}\\\Rightarrow I=\dfrac{0.018}{0.03\times 35\times 10^{-2}\times \sin90^{\circ}}\\\Rightarrow I=1.714\ \text{A}

The magnitude of the current is 1.714\ \text{A}.

8 0
2 years ago
A whistle of frequency 564 Hz moves in a circle of radius 71.2 cm at an angular speed of 17.1 rad/s. What are (a) the lowest and
trapecia [35]

Answer:

a) f'=544.66 \textup{Hz}

b) f'=584.75 \textup{Hz}

Explanation:

Given:

Frequency of the whistle, f = 564 Hz

Radius of the circle, r = 71.2 cm = 0.712 m

Angular speed, ω = 17.1 rad/s

speed of source, v_s = rω = 0.712 × 17.1 = 12.1752 m/s

speed of sound, v = 343 m/s

Now, applying the Doppler's effect formula, we have

f'=f\frac{v\pm v_d}{v\pm v_s}

where,

v_d = relative speed of the detector with respect to medium = 0

a) for lowest frequency, we have the formula as:

f'=f\frac{v}{v+v_s}

on substituting the values, we get

f'=564\times\frac{343}{343+12.1752}

or

f'=544.66 \textup{Hz}

b) for maximum frequency, we have the formula as:

f'=f\frac{v}{v-v_s}

on substituting the values, we get

f'=564\times\frac{343}{343-12.1752}

or

f'=584.75 \textup{Hz}

3 0
3 years ago
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