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3 years ago

8

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

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