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2 years ago

What constant acceleration is required to increase the speed of a car from 20 mi/h to 51 mi/h in 3 seconds

Physics

1 answer:

sdas [7]2 years ago

8 0

Answer:

Explanation:

acceleration is the time rate of change of velocity

a = (51 - 20) / 3 = 10⅓ mi/hr/s

which should probably be converted to standard units.

10⅓ mi/hr/s(5280 ft/mi) / 3600 s/hr) = 15.15555... ≈ 15.2 ft/s²

which is roughly half the acceleration of gravity.

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Are outer planets gaseous

bija089 [108]

Answer:

Yes

Explanation:

The four outer planets are all gas giants made primarily of hydrogen and helium. They have thick gaseous outer layers and liquid interiors.

6 0

2 years ago

Consider the following waves representing electromagnetic radiation: An illustration shows two waves representing electromagneti

Murljashka [212]

Answer:

a) red wave hs a longer wavelength than the green wave

b)f = 1.875 10¹¹ Hz , f_green = 5.45 10¹⁴Hz

c) E = 1.24 10⁻²² J , E_green = 3.6 10⁻¹⁹ J

d) The red wave is in the infrared range, heat waves

The green wave is in the visible wavelength

Explanation:

a) The green wave are on the left in the electromagnetic spectrum so the red wave has a longer wavelength than the green wave

The green wavelength is in the range of 550 10⁻⁹ m

The speed of the wave is

c = λ f

f = c /λ

b) The frequency of the red wave is

f = 3 10⁸ / 1.6 10⁻³

f = 1.875 10¹¹ Hz

For the green wave

f_green = 3 10⁸/550 10⁻⁹

f_green = 5.45 10¹⁴Hz

c) The photon energy is given by the Planck equation

E = h f

E = 6.63 10⁻³⁴ 1.875 10¹¹

E = 1.24 10⁻²² J

For the green wave

E_green = 6.63 10⁻³⁴ 5.45 10¹⁴

E_green = 3.6 10⁻¹⁹ J

d) The speed of electromagnetic waves is constant and has a value of 3 108 m / s

The red wave is in the infrared range, heat waves

The green wave is in the visible wavelength

6 0

3 years ago

If 6.24×10¹⁸ electron pass through a wire in 1s how many pass through it during a time interval of 2hr, 47min and 10s

levacccp [35]

Answer = 6.24x10^18 x ((2 x 3600) + (47 x 60) + 10)

3 0

3 years ago

A force of 9 pounds stretches a spring 1 foot. A mass weighing 6.4 pounds is attached to the spring, and the system is then imme

mart [117]

Answer:

$\frac{d^2x}{dt^2}+\frac{\beta}{m}\frac{dx}{dt}+\frac{k}{m}x=0$

Explanation:

let $m$ be the mass attached, let $k$ be the spring constant and let $\beta$ be the positive damping constant.

-By Newton's second law:

$m\frac{d^2x}{dt^2}=-kx-\beta \frac{dx}{dt}$

where $x(t)$ is the displacement from equilibrium position. The equation can be transformed into:

$\frac{d^2x}{dt^2}+\frac{\beta}{m}\frac{dx}{dt}+\frac{k}{m}x=0$ shich is the equation of motion.

7 0

3 years ago

A spring stretches 1.68cm vertically when a 2.50kg object is suspended from it.Find the distance (in cm) the spring stretches if

Llana [10]

Answer:

2.96 cm

Explanation:

By Hook's law

Force(F) = Spring constant(k) × Extension(d)

F = k × d

Force is the weight of the object, F = W = mg

So we get, mg = kd ⇒ m ∝ d

2.5 ∝ 1.68 --------------(1)

4.4 ∝ d' --------------(2)

From (1) & (2), 4.4/2.5 = d'/1.68

d' = 2.96 cm ⇒ the required extension.

7 0

3 years ago