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PolarNik [594]
3 years ago
10

An 877 kg drag race car accelerates from rest to 116 km/h in 0.951 s. What change in momentum does the force produce? Answer in

units of kg. m/s.​
Physics
1 answer:
Eduardwww [97]3 years ago
7 0

Answer:

The change in momentum is 28265.71  kg-m/s.

Explanation:

Given that,

Mass of a car, m = 877 kg

Initial velocity of the car, u = 0 (at rest)

Final velocity of the car, v = 116 km/h = 32.23 m/s

Time, t = 0.951 s

We need to find the change in momentum produced by the force. It can be calculated as the difference of final momentum and the initial momentum.

\Delta P=m(v-u)\\\\=877\times (32.23-0)\\\\=28265.71\ kg-m/s

So, the change in momentum is 28265.71  kg-m/s.

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Air pollution caused by the reaction of nitrogen compounds and other pollutants in the presence of sunlight is
ehidna [41]

Nitrogen oxides play a critical role in photochemical smog. They give the smog its yellowish-brown hue. Indoor residential appliances like gas stoves and gas or wood heaters can be significant emitters of nitrogen oxides in poorly ventilated environments.

  • Nitrogen dioxide (NO₂), ozone (O₃), peroxyacetyl nitrate (PAN), and chemical compounds with the -CHO group are the main harmful elements of photochemical smog (aldehydes). If present in high enough amounts, PAN and aldehydes can harm plants and irritate the eyes.
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Therefore , on conclusion i.e. two gases with molecules consisting of nitrogen and oxygen atoms are nitric oxide (NO) and nitrogen dioxide (NO₂). These nitrogen oxides play a part in the development of smog and acid rain, adding to the issue of air pollution.

To know more about photochemical smog

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7 0
1 year ago
A charge of 25 nC is uniformly distributed along a straight rod of length 3.0 m that is bent into a circular arc with a radius o
Greeley [361]

Answer:

E = 31.329 N/C.

Explanation:

The differential electric field dE at the center of curvature of the arc is

dE = k\dfrac{dQ}{r^2}cos(\theta ) <em>(we have a cosine because vertical components cancel, leaving only horizontal cosine components of E. )</em>

where r is the radius of curvature.

Now

dQ = \lambda rd\theta,

where \lambda is the charge per unit length, and it has the value

\lambda = \dfrac{25*10^{-9}C}{3.0m} = 8.3*10^{-9}C/m.

Thus, the electric field at the center of the curvature of the arc is:

E = \int_{\theta_1}^{\theta_2} k\dfrac{\lambda rd\theta  }{r^2} cos(\theta)

E = \dfrac{\lambda k}{r} \int_{\theta_1}^{\theta_2}cos(\theta) d\theta.

Now, we find \theta_1 and \theta_2. To do this we ask ourselves what fraction is the arc length  3.0 of the circumference of the circle:

fraction = \dfrac{3.0m}{2\pi (2.3m)}  = 0.2076

and this is  

0.2076*2\pi =1.304 radians.

Therefore,

E = \dfrac{\lambda k}{r} \int_{\theta_1}^{\theta_2} cos(\theta)d\theta= \dfrac{\lambda k}{r} \int_{0}^{1.304}cos(\theta) d\theta.

evaluating the integral, and putting in the numerical values  we get:

E = \dfrac{8.3*10^{-9} *9*10^9}{2.3} *(sin(1.304)-sin(0))\\

\boxed{ E = 31.329N/C.}

4 0
3 years ago
A football is place kicked with a velocity having a vertical component of 12 m/s and a horizontal component of 6 m/s. Find the r
SSSSS [86.1K]

The velocity is given by:

V = √(Vx²+Vy²)

V = velocity, Vx = horizontal velocity, Vy = vertical velocity

Given values:

Vx = 6m/s, Vy = 12m/s

Plug in and solve for V:

V = √(6²+12²)

V = 13.42m/s

Now find the direction:

θ = tan⁻¹(Vy/Vx)

θ = angle of velocity off horizontal, Vy = vertical velocity, Vx = horizontal velocity

Given values:

Vx = 6m/s, Vy = 12m/s

Plug in and solve for θ:

θ = tan⁻¹(12/6)

θ = 63.4°

The resultant velocity is 13.42m/s at an angle of 63.4° off the horizontal.

6 0
3 years ago
A car travelling at 10 m/s accelerates at 3 m/s^2 for 6 seconds. What is the final velocity of the car?
tankabanditka [31]

Answer:28m/s

Explanation:I got one point so yeah hope this help

7 0
3 years ago
At what frequency should a 200-turn, flat coil of cross sectional area of 300 cm2 be rotated in a uniform 30-mT magnetic field t
stellarik [79]

Answer:

The frequency of the coil is 7.07 Hz

Explanation:

Given;

number of turn of the coil, N = 200 turn

area of the coil, A = 300 cm² = 0.03 m²

magnitude of magnetic field, B = 30 mT = 0.03 T

maximum value of induced emf, E = 8 V

The maximum induced emf in the coil is given by;

E = NBAω

E = NBA(2πf)

f = \frac{E_{max}}{2\pi*NBA}

where;

f is the frequency of the coil

f = \frac{E_{max}}{2\pi*NBA}\\\\f = \frac{8}{2\pi(200)(0.03)(0.03)} \\\\f = 7.07 \ Hz

Therefore, the frequency of the coil is 7.07 Hz

5 0
3 years ago
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