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3 years ago

Calculate the vertical distance an object dropped from rest would cover in 12 seconds if it fell freely without air resistance

Physics

1 answer:

Nikolay [14]3 years ago

4 0

Answer:

705.6 m

Explanation:

time, t = 12 sends

initial velocity, u = 0

Acceleration due to gravity, g = - 9.8 m/s^2

As the object is falling freely so the initial velocity is zero.

Let the distance traveled by the body is h in time t = 12 s.

Use second equation of motion

$h = u t - 1/2 gt^{2}$

Ad the distance is downwards so take it as negative

So, $-h = 0 - 1/2\times 9.8 \times 12 \times 12$

h = 705.6 m

Thus, the distance traveled in 12 second is 705.6 m

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You're caught running a red light on Hwy 316. Attempting to impress the skeptical patrol officer with your physics knowledge, yo

timurjin [86]

Answer:

Explanation:

We shall apply here Doppler's effect in optics . The formula is as follows

$\frac{\triangle\lambda }{\lambda } = \frac{v}{c}$

Δλ is change in wavelength , λ is original wavelength , v is velocity and c is velocity of light

Δλ = 685 - 590 = 95 nm

λ = 685

95 / 685 = v / 3 x 10⁸

v = .416 x 10⁸ m / s

= 4.16 x 10⁷ m /s

3 0

3 years ago

A thin spherical spherical shell of radius R which carried a uniform surface charge density σ. Write an expression for the volum

ozzi

Answer:

Explanation:

From the given information:

We know that the thin spherical shell is on a uniform surface which implies that both the inside and outside the charge of the sphere are equal, Then

The volume charge distribution relates to the radial direction at r = R

∴

$\rho (r) \ \alpha \ \delta (r -R)$

$\rho (r) = k \ \delta (r -R) \ \ at \ \ (r = R)$

$\rho (r) = 0\ \ since \ r< R \ \ or \ \ r>R---- (1)$

To find the constant k, we examine the total charge Q which is:

$Q = \int \rho (r) \ dV = \int \sigma \times dA$

$Q = \int \rho (r) \ dV = \sigma \times4 \pi R^2$

∴

$\int ^{2 \pi}_{0} \int ^{\pi}_{0} \int ^{R}_{0} \rho (r) r^2sin \theta \ dr \ d\theta \ d\phi = \sigma \times 4 \pi R^2$

$\int^{2 \pi}_{0} d \phi* \int ^{\pi}_{0} \ sin \theta d \theta * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2$

$(2 \pi)(2) * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2$

Thus;

$k * 4 \pi \int ^{R}_{0} \delta (r -R) * r^2dr = \sigma \times R^2$

$k * \int ^{R}_{0} \delta (r -R) r^2dr = \sigma \times R^2$

$k * R^2= \sigma \times R^2$

$k = R^2 --- (2)$

Hence, from equation (1), if k = $\sigma$

$\mathbf{\rho (r) = \delta* \delta (r -R) \ \ at \ \ (r=R)}$

$\mathbf{\rho (r) =0 \ \ at \ \ rR}$

To verify the units:

$\mathbf{\rho (r) =\sigma \ * \ \delta (r-R)}$

↓ ↓ ↓

c/m³ c/m³ × 1/m

Thus, the units are verified.

The integrated charge Q

$Q = \int \rho (r) \ dV \\ \\ Q = \int ^{2 \ \pi}_{0} \int ^{\pi}_{0} \int ^R_0 \rho (r) \ \ r^2 \ \ sin \theta \ dr \ d\theta \ d \phi \\ \\ Q = \int ^{2 \pi}_{0} \ d \phi \int ^{\pi}_{0} \ sin \theta \int ^R_{0} \rho (r) r^2 \ dr$

$Q = (2 \pi) (2) \int ^R_0 \sigma * \delta (r-R) r^2 \ dr$

$Q = 4 \pi \sigma \int ^R_0 * \delta (r-R) r^2 \ dr$

$Q = 4 \pi \sigma *R^2$ since $( \int ^{xo}_{0} (x -x_o) f(x) \ dx = f(x_o) )$

$\mathbf{Q = 4 \pi R^2 \sigma }$

6 0

3 years ago

Based on what you learn about jovian moons by watching the videos or reading your textbook, what types of evidence for recent or

Valentin [98]

Answer:

b. craters, river valleys feeding into surface lakes of very cold liquids

Explanation:

Jovian moons are the four largest satellites like the moon of the Jupiter ie the Lo, Europa, Ganymede, Callisto and were first seen by the galileo. They are amiugly the largest moons with radii larger than the dwarf planet.

Lo has more than 400 active volcanoes and dotted more than 100 mountains and has an extremely thin atmosphere made up of sulfur dioxide. The Europa has deep oceans of liquid water, and the layer of ice, and are characteristic of the tidal heating.

<u>While the surface of Callisto is heavily cratered and has salty liquid water.</u>

7 0

4 years ago

You have a tungsten sphere (emissivity ε = 0.35) of radius 25 cm at a temperature of 25°C. If the sphere is enclosed in a room w

egoroff_w [7]

Answer:

Explanation:

Stefan's formula for emission of radiation is

E = e σ A ( T⁴ - T₀⁴ )

E is energy radiated , e is emissivity , σ is stefan's constant , T is temperature of object and T₀ is temperature of surrounding. A is area of surface .

E = .35 x 5.67 x 10⁻⁸ ( 298⁴ - 268⁴ ) x 4π x .25²

= 1.9845 x 10⁻⁸ ( 78.86 - 51.58 ) x 10⁸ x .0625

= 3.38 J /s

8 0

3 years ago

The cable holding a 2125 kg elevator has a maximum strength of 21,750 N. What is the maximum upward acceleration the cable can g

vivado [14]

Answer:

10.23m/s^2

Explanation:

GIven data

mass of elevator = 2125 kg

Force= 21,750 N

Required

The maximum acceleration upward

F= ma

a= F/m

a=21,750/2125

a= 10.23m/s^2

Hence the acceleration is 10.23m/s^2

4 0

3 years ago