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3 years ago

A physicist wants to study the motion of a car.

Physics

1 answer:

Aleonysh [2.5K]3 years ago

3 0

Answer:

Explanation:

in a

$1 mi = 1.61km$

so,in 50 miles

$50 mi = 50 \times 1.61$

$= 80.5 \: km$

in b

$1 \: mi = 1760 \: yd$

so in 22.3 miles

$22.3 = 22.3 \times 1760$

$= 39248 \: yd$

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A jet is travelling at a speed of 1200 km/h and drops cargo from a height of 2.5 km above the ground Calculate the time it takes

OLEGan [10]

a) Time of flight: 22.6 s

To calculate the time it takes for the cargo to reach the ground, we just consider the vertical motion of the cargo.

The vertical position at time t is given by

$y(t) = h +u_y t - \frac{1}{2}gt^2$

where

h = 2.5 km = 2500 m is the initial height

$u_y = 0$ is the initial vertical velocity of the cargo

g = 9.8 m/s^2 is the acceleration of gravity

The cargo reaches the ground when

$y(t) = 0$

So substituting it into the equation and solving for t, we find the time of flight of the cargo:

$0 = h - \frac{1}{2}gt^2\\t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2(2500)}{9.8}}=22.6 s$

b) 7.5 km

The range travelled by the cargo can be calculated by considering its horizontal motion only. In fact, the horizontal motion is a uniform motion, with constant velocity equal to the initial velocity of the jet:

$v_x = 1200 km/h \cdot \frac{1000 m/km}{3600 s/h}=333.3 m/s$

So the horizontal distance travelled is

$d=v_x t$

And if we substitute the time of flight,

t = 22.6 s

We find the range of the cargo:

$d=(333.3)(22.6)=7533 m = 7.5 km$

7 0

3 years ago

A series circuit has a capacitor of 0.25 × 10−6 F, a resistor of 5 × 103 Ω, and an inductor of 1 H. The initial charge on the ca

svetoff [14.1K]

Answer:

$q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C$

Explanation:

Given that $L=1H, R=5000\Omega, \ C=0.25\times10^{-6}F, \ \ E(t)=12V$ , we use Kirchhoff's 2nd Law to determine the sum of voltage drop as:

$E(t)=\sum{Voltage \ Drop}\\\\L\frac{d^2q}{dt^2}+R\frac{dq}{dt}+\frac{1}{C}q=E(t)\\\\\\\frac{d^2q}{dt^2}+5000\frac{dq}{dt}+\frac{1}{0.25\times10^{-6}}q=12\\\\\frac{d^2q}{dt^2}+5000\frac{dq}{dt}+4000000q=12\\\\m^2+5000m+4000000=0\\\\(m+4000)(m+1000)=0\\\\m=-4000 \ or \ m=-1000\\\\q_c=c_1e^{-4000t}+c_2e^{-1000t}$

#To find the particular solution:

$Q(t)=A,\ Q\prime(t)=0,Q\prime \prime(t)=0\\\\0+0+4000000A=12\\\\A=3\times10^{-6}\\\\Q(t)=3\times10^{-6},\\\\q=q_c+Q(t)\\\\q=c_1e^{-4000t}+c_2e^{-1000t}+3\times10^{-6}\\\\q\prime=-4000c_1e^{-4000t}-1000c_2e^{-1000t}\\q\prime(0)=0\\\\-4000c_1-1000c_2=0\\c_1+c_2+3\times10^{-6}=0\\\\#solving \ simultaneously\\\\c_1=10^{-6},c_2=-4\times10^{-6}\\\\q=10^{-6}e^{-4000t}-4\times10^{-6}e^{-1000t}+3\times10^{-6}\\\\q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C$

Hence the charge at any time, t is $q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C$

6 0

3 years ago

Two electric charges, held a distance, dd, apart experience an electric force of magnitude, FF, between them. If one of the char

lorasvet [3.4K]

Answer:

$F'=2F$

Explanation:

The Coulomb's law states that the magnitude of the electrostatic force between two charges is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them:

$F=\frac{kq_1q_2}{d^2}$

In this case, we have $q_1'=2q_1$ :

$F'=\frac{kq'_1q_2}{d^2}\\F'=\frac{k(2q_1)q_2}{d^2}\\F'=2\frac{kq_1q_2}{d^2}\\F'=2F$

3 0

3 years ago

Questions E6 a& b and E7 a&b?

erica [24]

Explanation:

6a) Work = force × distance

W = Fd

W = (60 N) (10 m)

W = 600 J

6b) Change in energy = work

ΔKE = 600 J

7a) Kinetic energy is half the mass times the square of the velocity.

KE = ½ mv²

KE = ½ (0.4 kg) (25 m/s)²

KE = 125 J

7b) Work = change in energy. When the ball is stopped, it has zero kinetic energy.

W = ΔKE

W = 0 J − 125 J

W = -125 J

8 0

3 years ago

Consider a 400 nm thick film of water (n-1.33) on plastic (-1.9). Visible light is incident. on the film firom air. (25 pts) Ske

Gnoma [55]

Answer:

Explanation:

For constructive interference in thin films by reflected light , the condition is

Path difference = 2μd = odd multiple of λ / 2

Where μ is refractive index of water , d is thickness of water medium and λ

is wavelength of light

λ = 4μd or 4μd / 3 or 4μd / 5

= 4 x 1.33 x 400 nm = 2128 nm

or 2128 /3 = 709.33 nm

or 2128 /5 = 425.6 nm

For destructive interference ( dim reflection )

2μd = multiple of λ = λ or 2λ or 3λ

2 x 1.33 x 400 nm =1064

or 1064 x 2 = 2128 nm

or 1064 x 3 = 3192 nm.

7 0

3 years ago