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3 years ago

The height of a typical playground slide is about 6 ft and it rises at an angle of 30 ∘ above the horizontal.

Physics

1 answer:

garri49 [273]3 years ago

6 0

Answer:

What is the coefficient of kinetic friction = 0.432

Explanation:

The detailed steps and derivation with appropriate substitution is as shown in the attached file.

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A transparent oil with index of refraction 1.28 spills on the surface of water (index of refraction 1.33), producing a maximum o

Ad libitum [116K]

Answer:

The thickness of the oil slick is $1.95\times10^{-7}\ m$

Explanation:

Given that,

Index of refraction = 1.28

Wave length = 500 nm

Order m = 1

We need to calculate the thickness of oil slick

Using formula of thickness

$2nt= m\lambda$

Where, n = Index of refraction

t = thickness

$\lambda$ = wavelength

Put the value into the formula

$2\times1.28 \times t=1\times\times500\times10^{-9}$

$t = \dfrac{1\times\times500\times10^{-9}}{2\times1.28 }$

$t=1.95\times10^{-7}\ m$

Hence, The thickness of the oil slick is $1.95\times10^{-7}\ m$

4 0

2 years ago

A car moving with an initial speed of 25 m/s slows down to a speed of 5 m/s in 10 seconds Calculate a) the acceleration of the c

stealth61 [152]

Answer :

(a) The acceleration of the car is, $-2m/s^2$

(b) The distance covered by the car is, 150 m

Explanation :

By the 1st equation of motion,

$v=u+at$ ...........(1)

where,

v = final velocity = 5 m/s

u = initial velocity = 25 m/s

t = time = 10 s

a = acceleration of the car = ?

Now put all the given values in the above equation 1, we get:

$5m/s=25m/s+a\times (10s)$

$a=-2m/s^2$

The acceleration of the car is, $-2m/s^2$

By the 2nd equation of motion,

$s=ut+\frac{1}{2}at^2$ ...........(2)

where,

s = distance covered by the car = ?

u = initial velocity = 25 m/s

t = time = 10 s

a = acceleration of the car = $-2m/s^2$

Now put all the given values in the above equation 2, we get:

$s=(25m/s)\times (10s)+\frac{1}{2}\times (-2m/s^2)\times (10s)^2$

By solving the term, we get:

$s=150m$

The distance covered by the car is, 150 m

8 0

3 years ago

Rahul has 2 bulbs connected across two cells in a simple circuits shown. How can he make the bulbs glow dimmer?

Ray Of Light [21]

Answer:

in the parallel connection the light bulbs shine less than in the series connection

Explanation:

In a series circuit the current through the whole circuit is the same, therefore the power (brightness) of each bulb is

P = i² R

where R is the resistance of each bulb and i the current of the circuit.

If we connect the light bulbs and the cells in parallel, the current in the circuit is the sum of the east that passes through each light bulb,

i = i₁ + i₂

if the two light bulbs are the same

i = 2 i₁

i₁ = i / 2

so the power of each bulb is is

P = i₁² R

P = R i² / 4

P = ¼ P_initial

Therefore we see that in the parallel connection the light bulbs shine less than in the series connection

3 0

3 years ago

3. Maverick and Goose are flying a training mission in their F-14. They are

Elanso [62]

Answer:

A. The bomb will take <em>17.5 seconds </em>to hit the ground

B. The bomb will land <em>12040 meters </em>on the ground ahead from where they released it

Explanation:

Maverick and Goose are flying at an initial height of $y_0=1500m$ , and their speed is v=688 m/s

When they release the bomb, it will initially have the same height and speed as the plane. Then it will describe a free fall horizontal movement

The equation for the height y with respect to ground in a horizontal movement (no friction) is

$y=y_0 - \frac{gt^2}{2}$ [1]

With g equal to the acceleration of gravity of our planet and t the time measured with respect to the moment the bomb was released

The height will be zero when the bomb lands on ground, so if we set y=0 we can find the flight time

The range (horizontal displacement) of the bomb x is

$x = v.t$ [2]

Since the bomb won't have any friction, its horizontal component of the speed won't change. We need to find t from the equation [1] and replace it in equation [2]:

Setting y=0 and isolating t we get

$t=\sqrt{\frac{2y_0}{g}}$

Since we have $y_0=1500m$

$t=\sqrt{\frac{2(1500)}{9.8}}$

$t=17.5 sec$

Replacing in [2]

$x = 688\ m/sec \ (17.5sec)$

$x = 12040\ m$

A. The bomb will take 17.5 seconds to hit the ground

B. The bomb will land 12040 meters on the ground ahead from where they released it

6 0

2 years ago

A common flashlight bulb is rated at 0.32 A and 4.3 V (the values of the current and voltage under operating conditions). If the

sleet_krkn [62]

Answer:

1176.01 °C

Explanation:

Using Ohm's law,

V = IR................. Equation 1

Where V = Voltage, I = current, R = Resistance when the bulb is on

make R the subject of the equation

R = V/I.................. Equation 2

R = 4.3/0.32

R = 13.4375 Ω

Using

R = R'(1+αΔθ)............................. Equation 3

Where R' = Resistance of the bulb at 20°, α = Temperature coefficient of resistivity, Δθ = change in temperature

make Δθ the subject of the equation

Δθ = (R-R')/αR'.................. Equation 4

Given: R = 13.4375 Ω, R' = 1.6 Ω, α = 6.4×10⁻³ K⁻¹

Substitute into equation 4

Δθ = (13.4375-1.6)/(1.6×0.0064)

Δθ = 11.8375/0.01024

Δθ = 1156.01 °C

But,

Δθ = T₂-T₁

T₂ = T₁+Δθ

Where T₂ and T₁ = Final and initial temperature respectively.

T₂ = 20+1156.01

T₂ = 1176.01 °C

5 0

2 years ago

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