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2 years ago

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A car is travelling to the right with a speed of 42\,\dfrac{\text m}{\text s}42 s m 42, space, start fraction, m, divided by,

s, end fraction when the driver slams on the brakes. The car skids for 4.0\,\text s4.0s4, point, 0, space, s with constant acceleration before it comes to a stop. How many meters did the car skid before coming to a stop? Answer using a coordinate system where rightward is positive. Round the answer to two significant digits.

2 answers:

Effectus [21]2 years ago

7 0

Answer:

d = 84 m

Explanation:

As we know that when an object moves with uniform acceleration or deceleration then we can use equation of kinematics to find the distance moved by the object

here we know that

initial speed $v_i = 42 m/s$

final speed $v_f = 0$

time taken by the car to stop

$t = 4s$

now the distance moved by the car before it stop is given as

$d = \frac{v_f + v_i}{2} \times t$

now we have

$d = \frac{42 + 0}{2} \times 4$

$d = 84 m$

wolverine [178]2 years ago

5 0

Answer:

84

Explanation:

khan academy

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The air in a pipe resonates at 150 Hz and 750 Hz, one of these resonances being the fundamental. If the pipe is open at both end

Xelga [282]

Answer:

Explanation:

Two frequencies with magnitude 150 Hz and 750 Hz are given

For Pipe open at both sides

fundamental frequency is 150 Hz as it is smaller

frequency of pipe is given by

$f=\frac{nv}{2L}$

where L=length of Pipe

v=velocity of sound

$f=150\ Hz$ for n=1

and f=750 is for n=5

thus there are three resonance frequencies for n=2,3 and 4

For Pipe closed at one end

frequency is given by

$f=\frac{(2n+1)}{4L}\cdot v$

for n=0

$f_1=\frac{v}{4L}$

$f_1=150\ Hz$

for n=2

$f_2=\frac{5v}{4L}$

Thus there is one additional resonance corresponding to n=1 , between $f_1$ and $f_2$

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2 years ago

The frequency of the applied RF signal used to excite spins is directly proportional to the magnitude of the static magnetic fie

Anni [7]

Answer:

The inverse frequency is $\dfrac{3}{80}\ s$

Explanation:

Given that,

Magnetic field = 20 T

Proportionality constant = 5 Hz/T

Change in magnetic field = 3 T

We know that,

$B=\dfrac{K}{\dfrac{1}{\omega}}$

We need to calculate the inverse frequency

Using formula of frequency

$\Delta(\dfrac{1}{\omega})=\dfrac{\Delta B}{k\times(\dfrac{1}{\omega^2})}$

$\Delta(\dfrac{1}{\omega})=\dfrac{k\times\Delta B}{B^2}$

Put the value into the formula

$\Delta(\dfrac{1}{\omega})=\dfrac{3\times5}{(20)^2}$

$\Delta(\dfrac{1}{\omega})=\dfrac{3}{80}\ s$

Hence, The inverse frequency is $\dfrac{3}{80}\ s$

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2 years ago

Variations in the resistivity of blood can give valuable clues about changes in various properties of the blood. Suppose a medic

Elena-2011 [213]

Answer:

Answer:

1.1 x 10^9 ohm metre

Explanation:

diameter = 1.5 mm

length, l = 5 cm

Potential difference, V = 9 V

current, i = 230 micro Ampere = 230 x 10^-6 A

radius, r = diameter / 2 = 1.5 / 2 = 0.75 x 10^-3 m

Let the resistivity is ρ.

Area of crossection

A = πr² = 3.14 x 0.75 x 0.75 x 10^-6 = 1.766 x 10^-6 m^2

Use Ohm's law to find the value of resistance

V = i x R

9 = 230 x 10^-6 x R

R = 39130.4 ohm

Use the formula for the resistance

$R=\rho \frac{l}{A}$

$\rho =\frac{RA}{l}$

$\rho =\frac{39130.4\times 0.05}{1.766\times 10^{-6}}$

ρ = 1.1 x 10^9 ohm metre

Explanation:

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2 years ago

The pail is rotated at a constant rate so it has the minimum speed at all points along its circular path. The water has mass m.

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Answer:

Explanation:

The pail is rotated at a constant rate in vertical circular path so it has the minimum speed at all points along its circular path . That means at top position the velocity is almost zero. In that case the centripetal force at top position will be provided by its weight or

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v = √ rg

At the bottom position its velocity will be increased due to loss of potential energy

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V =√ 5 gr

If R be the reaction force at the bottom by bottom of pail

R - mg = mV² / r

R = mg +mV² / r

= mg + m x 5gr / r

R = 6mg

This is the magnitude of the force exerted by the water on the bottom of the pail .

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2 years ago

The mass of the Sun is 2multiply1030 kg, and the mass of the Earth is 6multiply1024 kg. The distance from the Sun to the Earth i

spayn [35]

Answer:

<em>a) 3.56 x 10^22 N</em>

<em>b) 3.56 x 10^22 N</em>

<em></em>

Explanation:

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mass of the Earth m = 6 x 10^24 kg

Distance between the sun and the Earth R = 1.5 x 10^11 m

From Newton's law,

F = $\frac{GMm}{R^2}$

where F is the gravitational force between the sun and the Earth

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m is the mass of the Earth

M is the mass of the sun

R is the distance between the sun and the Earth.

Substituting values, we have

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<em></em>

A) The force exerted by the sun on the Earth is equal to the force exerted by the Earth on the Sun also, and the force is equal to <em>3.56 x 10^22 N</em>

b) The force exerted by the Earth on the Sun = <em>3.56 x 10^22 N</em>

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