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madam [21]
3 years ago
9

A 1000-kilogram car traveling with a velocity of 20. meters per second decelerates uniformly at -5.0 meters per second2 until it

comes to rest. What is the magnitude of the impulse applied to the car to bring it to rest?
Physics
1 answer:
Elis [28]3 years ago
3 0

Answer:

-20000 kgm/s

Explanation:

Impulse: This can be defined as the product of the mass of a body and its change in velocity. The S.I unit of impulse is kgm/s.

Mathematically, impulse can be expressed as

I = m(v-u).............. Equation 1.

Where I = impulse applied to the car to bring it to rest, m = mass of the car, u = initial velocity of the car, v = final velocity of the car.

Given: m = 1000 kg, u = 20 m/s, v = 0 m/s ( to rest)

Substitute into equation 1

I = 100(0-20)

I = 1000(-20)

I = -20000 kgm/s

Hence the impulse applied to the car to bring it to rest = -20000 kgm/s

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How many moles of h2 gas would be contained in 4.00 of the gas at 202.6KPa and 127c?
JulsSmile [24]

Answer : The volume of one mole of gas is dependent of temperature and pressure.

Explanation:

At <span><span>0o</span>Cand1atm</span> one mole of an ideal gas would be <span>≈22.4L</span>, which is the same as <span>22.4d<span>m3</span></span>

If you double the pressure, the same amount (mole) of gas would take up only half the space.

If the temperature changes (and pressure stays constant), the gas will expand or contract by about <span>1%/<span>3o</span></span>.

4 0
3 years ago
Red light of wavelength 633 nm from a helium-neon laser passes through a slit 0.360 mm wide. The diffraction pattern is observed
Bumek [7]

Answer:

a) 0.0130 m

b') w' = =6.46*10^{-3] m

Explanation:

given data:

\lambda of light = 633 nm

width of siit a =0.360 mm

distance from screen = 3.75 m

a) the first minima is located at

sin\theta = \frac{\lambda}{a}

              == \frac{633 *10^{-9}}{.360*10^{-3}}

           \theta = 0.100

y_1 = dtan\theta_1 = 3.75*tan(0.100) = 6.54 *10^{-3} m

with of central fringe  = 2y_1 = 2*6.54 *10^{-3} = 0.0130 m

b)

width of the first bright fringe on either side of the central one = w' = y_2 -y_1

calculation for y_2

sin\theta = 2\frac{\lambda}{a}

              = = 2*\frac{633 *10^{-9}}{.360*10^{-3}}

             \theta  = 2*0.100 = 0.200

y_2 = dtan\theta_1 = 3.75*tan(0.200) =0.0130 m

w' = 0.0130  -6.54 *10^{-3}

w' = =6.46*10^{-3] m

6 0
3 years ago
air passing over an airplanes wing travels ,and therefore exerts pressure.than air traveling beneath the wing.
pickupchik [31]
Bernoulli's principle of laminar/lamellar air flow, I think. High flow speed = low pressure, low flow speed = high pressure I think. So, the wings/aerofoils are designed to induce a low pressure on the top side of the wing and a high pressure on the underside of the wing, thus producing an "aerodynamic upthrust" (a static upthrust comes from an object in water via Archimedes) and LIFT. 

Two "particles" of air one going topside and the other underside meet again at the end of their motion across the wing. So, top side has to travel faster than bottom side. So top side has a lower "dynamic pressure" than underside.

And all that for 5 points ????????? (If I'm right, of course ... )
5 0
3 years ago
A circuit has a voltage drop of 54.0 V across a 30.0 o resistor that carries a current of 1.80 A. What is the power used by the
Blizzard [7]

Answer:

P = 97.2 W

Explanation:

Given that,

Voltage drop, V = 54 V

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We need to find the power used by the resistor. The formula used to find the power is given by :

P = VI

Putting all the values,

P = 54 V × 1.8 A

P = 97.2 W

So, the power used by the resistor is 97.2 W.

5 0
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