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3 years ago

How much work in joules must be done to stop a 920-kg car traveling at 90 km/h?

Physics

1 answer:

tankabanditka [31]3 years ago

3 0

Answer:

Work done, W = −287500 Joules

Explanation:

It is given that,

Mass of the car, m = 920 kg

The car is travelling at a speed of, u = 90 km/h = 25 m/s

We need to find the amount of work must be done to stop this car. The final velocity of the car, v = 0

Work done is also defined as the change in kinetic energy of an object i.e.

$W=\Delta K$

$W=\dfrac{1}{2}m(v^2-u^2)$

$W=\dfrac{1}{2}\times 920\ kg(0-(25\ m/s)^2)$

W = −287500 Joules

Negative sign shows the work done is done is opposite direction.

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1)Light of wavelength 588.0 nm is incident on a narrow slit. The diffraction pattern is viewed on a screen 55.5 cm from the slit

Talja [164]

Answer:

These are Diffraction Grating Questions.

Q1. To determine the width of the slit in micrometers (μm), we will need to use the expression for distance along the screen from the center maximum to the nth minimum on one side:

Given as

y = nDλ/w Eqn 1

where

w = width of slit

D = distance to screen

λ = wavelength of light

n = order number

Making x the subject of the formula gives,

w = nDλ/y

Given

y = 0.0149 m

D = 0.555 m

λ = 588 x 10-9 m

and n = 3

w = 6.6x10⁻⁵m

Hence, the width of the slit w, in micrometers (μm) = 66μm

Q2. To determine the linear distance Δx, between the ninth order maximum and the fifth order maximum on the screen

i.e we have to find the difference between distance along the screen (y₉-y₅) = Δx

Recall Eqn 1, y = nDλ/w

given, D = 27cm = 0.27m

λ = 632 x 10-9 m

w = 0.1mm = 1.0x10⁻⁴m

For the 9th order, n = 9,

y₉ = 9 x 0.27 x 632 x 10-9/ 1.0x10⁻⁴m = 0.015m

Similarly, for n = 5,

y₅ = 5x 0.27 x 632 x 10-9/ 1.0x10⁻⁴m = 0.0085m

Recall, Δx = (y₉-y₅) = 0.015 - 0.0085 = 0.0065m

Hence, the linear distance Δx between the ninth order maximum and the fifth order maximum on the screen = 6.5mm

8 0

3 years ago

Hola buen día por favor alguien me puede ayudar por favor1. Dos cargas puntuales q1=+4μC y q2=+ 6μC están separadas por 10 cm. U

luda_lava [24]

Answer:

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Explanation:

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6 0

3 years ago

A man on the 14 th floor of a building sees a bucket (dropped by a window washer) passing his window and notes that it hits the

natta225 [31]

Answer:

The bucket was the dropped from 56 th floor.

Explanation:

Given that,

Height of floor = 4.9 m

Height of 14 floor = 68.6 m

Time taken = 1 sec

We need to calculate the speed of the bucket

Using equation of motion

$s=ut+\dfrac{1}{2}gt^2$

Put the value into the formula

$68.6=v\times1+\dfrac{1}{2}\times9.8\times(1)^2$

$v=68.6-\dfrac{1}{2}\times9.8\times(1)^2$

$v=63.7\ m/s$

We need to calculate the time

Using equation of motion

$v=u+gt$

$t=\dfrac{v}{g}$

Put the value into the formula

$t=\dfrac{63.7}{9.8}$

$t=6.5\ sec$

We need to calculate the distance

Using equation of motion

$s=ut+\dfrac{1}{2}gt^2$

$s=0+\dfrac{1}{2}gt^2$

Put the value into the formula

$s=\dfrac{1}{2}\times9.8\times(6.5)^2$

$s=207.025\ m$

We need to calculate the number of floor

$n=\dfrac{s}{h_{f}}$

Put the value into the formula

$n=\dfrac{207.025}{4.9}$

$n=42.25\approx42$

The bucket was the dropped from

$f=14+42= 56$

Hence, The bucket was the dropped from 56 th floor.

8 0

3 years ago

If an object free-falls for t seconds from rest to d distance, how far will the object fall from rest in twice the elapsed time?

balu736 [363]

Answer:

When the time of fall is doubled, the height of fall will be quadrupled

Explanation:

Given;

height of fall, h = d m

time of fall, t = t s

initial velocity of the object, u = 0 m/d

The height of fall of the object is calculated from the kinematic equation below;

$h = ut + \frac{1}{2} gt^2\\\\h = 0 + \frac{1}{2} gt^2\\\\h = \frac{1}{2} gt^2\\\\2h = gt^2\\\\g = \frac{2h}{t^2}$

where;

g is acceleration due to gravity, which is constant

if the time of fall is doubled, the height of fall is calculated as;

$\frac{2h_1}{t_1^2} = \frac{2h_2}{t_2^2} \\\\\frac{h_1}{t_1^2} = \frac{h_2}{t_2^2}\\\\(Note: h_1 = d, \ and \ t_1 = t)\\\\h_2 = \frac{h_1t_2^2}{t_1^2} \\\\h_2 = \frac{(d)(2t_1)^2}{t_1^2} \\\\h_2 = \frac{(d)(2t)^2}{t^2}\\\\h_2 = \frac{(d)\times 4t^2}{t^2}\\\\h_2 = 4d$

Therefore, when the time of fall is doubled, the height of fall will be quadrupled

6 0

3 years ago

Why do objects move at constant speed

kati45 [8]

Most objects move at a constant speed because of friction and acceleration. The constant speed keeps them in place, and keeps a balance.

7 0

3 years ago