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gizmo_the_mogwai [7]

3 years ago

10

Physics {deceleration}

1 answer:

seropon [69]3 years ago

7 0

Using kinematic equation, v^2 - u^2 = 2as. 5^2 - 3^2 = 2a x 16. a = 0.5m/s^2. So particle will deaccelerate at 0.5m/s^2. ( v = final velocity, u= initial velocity, a= acceleration, s= displacement.)

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Sugarcane is an example of a plant that can be used to produce which kind of fuel

nexus9112 [7]

<span>Your answer is Ethanol - Sugarcane ethanol is an alcohol-based fuel produced by the fermentation of sugarcane juice and molasses. Because it is a clean, affordable and low-carbon biofuel, sugarcane ethanol has emerged as a leading renewable fuel for the transportation sector.</span>

6 0

3 years ago

Read 2 more answers

Three long parallel wires each carry 2.0-A currents in the same direction. The wires are oriented vertically, and they pass thro

blagie [28]

Answer:

$21.2\times 10^{-6}$ T

Explanation:

$i$ = magnitude of current in each wire = 2.0 A

$a$ = length of the side of the square = 4 cm = 0.04 m

$r$ = length of the diagonal of the square = $\sqrt{2}$ a = $\sqrt{2}$ (0.04) = 0.057 m

$B$ = magnitude of magnetic field by wires at A and C

$B = \left ( \frac{\mu _{o}}{4\pi } \right )\left ( \frac{2i}{a} \right )$

$B = (10^{-7}) \left ( \frac{2(2)}{0.04} \right )$

$B = 10\times 10^{-6}$ T

$B'$ = magnitude of magnetic field by wire at B

$B' = \left ( \frac{\mu _{o}}{4\pi } \right )\left ( \frac{2i}{r} \right )$

$B' = (10^{-7}) \left ( \frac{2(2)}{0.057} \right )$

$B' = 7.02\times 10^{-6}$ T

Net magnitude of the magnetic field at D is given as

$B_{net} = \sqrt{B^{2}+B^{2}} + B'$

$B_{net} = \sqrt{2} B + B'$

$B_{net} = \sqrt{2} (10\times 10^{-6}) + (7.02\times 10^{-6})$

$B_{net} = 21.2\times 10^{-6}$ T

8 0

3 years ago

What if you were a mother for a day boys and girls

Dominik [7]

I’m am a mother no drama !!!!

Boom

6 0

2 years ago

To move a mass you need a force true or false

monitta

Answer: True

Explanation:

A force must be applied to set a stationary object in motion.

4 0

3 years ago

A spring with spring constant of 26 N/m is stretched 0.22 m from its equilibrium position. How much work must be done to stretch

Evgen [1.6K]

Answer:

1.503 J

Explanation:

Work done in stretching a spring = 1/2ke²

W = 1/2ke²........................... Equation 1

Where W = work done, k = spring constant, e = extension.

Given: k = 26 N/m, e = (0.22+0.12), = 0.34 m.

Substitute into equation 1

W = 1/2(26)(0.34²)

W = 13(0.1156)

W = 1.503 J.

Hence the work done to stretch it an additional 0.12 m = 1.503 J

8 0

3 years ago