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4 years ago

7

A normal walking speed is around 2.0 m/s . how much time t does it take the box to reach this speed if it has the acceleration 5

.54 m/s2 that you calculated above?

1 answer:

creativ13 [48]4 years ago

6 0

Given:

u(initial velocity)=0

a=5.54m/s^2

v(final velocity)=2 m/s

v=u +at

Where v is the final velocity.

u is the initial velocity

a is the acceleration.

t is the time

2=0+5.54t

t=2/5.54

t=0.36 sec

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At what altitude h above the north pole is the weight of an object reduced to 67% of its earth-surface value? Assume a spherical

lara31 [8.8K]

Answer:

The answer to the question is

At an altitude of 1413 km or 1.222·R above the north pole the weight of an object reduced to 67% of its earth-surface value

Explanation:

We make use of the gravitational formula as follows

F = G $\frac{m_{1} m_{2} }{R^{2} }$ where

m₁ = mass of the object

m₂ = mass of the earth

d = distance between the two objects and

G = gravitational constant

if at the altitude the weight is reduced to 67 % of its weight on earth then

with all other variables remaining constant, we have

67% F = G $\frac{m_{1} m_{2} }{R_{2} ^{2} }$ =0.67× G× $\frac{m_{1} m_{2} }{R_{1} ^{2} }$

cancelleing like ternss from both sides we have

1/R₂² =0.67/R₁² or r₁²/R₂² =0.67 and R₁/R₂ = 0.8185

or R₂ = R₁/0.8185 or 1.222·R = (6.37×10⁶ m)/0.8185 = 7.783×10⁶ m

Hence at an altitude = 1.413×10⁶ m the weight of the object will reduce to 67% its earth-surface value

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3 years ago

Which describes the best approach when conducting a scientific experiment

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When conducting and experiment you want to have a notebook and something to write down notes with so you can keep everything organized and proper, and to not miss anything in the experiment. Also you want to have everything in order of the way it should be in.

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3 years ago

Help!!!! A loop of area 0.100 m^2 is oriented at 15.5 degree angle to a 0.500 T magnetic field. It rotates until it is at a 45.0

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4 years ago

Consider the equation . The dimensions of the variables v, x, and t are , , and , respectively. The numerical factor 3 is dimens

Vinvika [58]

Answer:

Part of the question is missing but here is the equation for the function;

Consider the equation v = (1/3)zxt2. The dimensions of the variables v, x, and t are [L/T], [L], and [T] respectively.

Answer = The dimension for z = 1/T3 i.e 1/ T - raised to power 3

Explanation:

What is applied is the principle of dimensional homogenuity

From the equation V = (1/3)zxt2.

V has a dimension of [L/T]

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t has a dimension of [T]
from the equation, make z the subject of the relation
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3 years ago

Two coils close to each other have a mutual inductance of 32 mH. If the current in one coil decays according to I=I0e−αt, where

fiasKO [112]

The emf induced in the second coil is given by:

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The current in the first coil is given by:

i = i₀ $e^{-at}$

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i = 5.0e^(-2.0×10³t)

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di/dt = -1.0×10⁴e^(-2.0×10³t)

Calculate a general formula for V. Givens:

M = 32×10⁻³H, di/dt = -1.0×10⁴e^(-2.0×10³t)

Plug in and solve for V:

V = -32×10⁻³(-1.0×10⁴e^(-2.0×10³t))

V = 320e^(-2.0×10³t)

We want to find the induced emf right after the current starts to decay. Plug in t = 0s:

V = 320e^(-2.0×10³(0))

V = 320e^0

V = 320 volts

We want to find the induced emf at t = 1.0×10⁻³s:

V = 320e^(-2.0×10³(1.0×10⁻³))

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3 years ago