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3 years ago

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What is the common formula for work? Assume that W is the work, F is a constant force, 656-06-01-00-00_files/i0060000.jpgv is th

e change in velocity, and d is the displacement.

1 answer:

klio [65]3 years ago

6 0

The common formula for work is:
Work = force x displacement
W = F x d

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A ball of clay is moving at a speed of 12 m/s collides and sticks to a stationary ball of clay. If each ball has a mass of 13 kg

LUCKY_DIMON [66]

Answer:

p_{f} = 6 m / s

Explanation:

We can solve this exercise using conservation of momentum. For this we define a system formed by the two balls, so that the forces during the collision have been intense and the moment is preserved

Initial instant. Before the crash

p₀ = m v +0

Final moment. Right after the crash

$p_{f}$ = (m + m) v_{f}

how the moment is preserved

p₀ = p_{f}

m v = 2 m v_{f}

v_{f} = v / 2

we calculate

v_{f} = 12/2

p_{f} = 6 m / s

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3 years ago

What temperature resistivity of tungsten is 4 times resistivity of silver?

mihalych1998 [28]

Answer:

Slightly above 20°c

Explanation:

The resistivity of silver is 1.59 x 10-8Ωm{ as discovered by experiment from scientist}

The resistivity of tungsten at 20°c is 5.6x10-8Ωm

We see the value is just an approximation of 4 times not exactly.

6 0

3 years ago

The diagram shows waves of sound travelling through the air. By which factor is the sound intensity decreased at 3 meters?

SIZIF [17.4K]

The diagram is missing; however, we know that the intensity of a sound wave is inversely proportional to the square of the distance from the source:
$I(r)= \frac{1}{r^2}$
where I is the intensity and r is the distance from the source.

We can assume for instance that the initial distance from the source is r=1 m, so that we put
$I= \frac{1}{r^2}= \frac{1}{(1)^2}=1$
The intensity at r=3 m will be
$I= \frac{1}{r^2}= \frac{1}{(3)^2}= \frac{1}{9}$
Therefore, the sound intensity has decreased by a factor $1/9$ .

8 0

3 years ago

Read 2 more answers

Lasers are classified according to the eye-damage danger they pose. Class 2 lasers, including many laser pointers, produce visib

Alexus [3.1K]

Answer:

<em>a) 318.2 W/m^2</em>

<em>b) 2.5 x 10^-4 J</em>

<em>c) 1.55 x 10^-8 v/m</em>

<em></em>

Explanation:

Power of laser P = 1 mW = 1 x 10^-3 W

exposure time t = 250 ms = 250 x 10^-3 s

If beam diameter = 2 mm = 2 x 10^-3 m

then

cross-sectional area of beam A = $\pi d^{2} /4$ = (3.142 x $(2*10^{-3} )^{2}$ )/4

A = 3.142 x 10^-6 m^2

a) Intensity I = P/A

where P is the power of the laser

A is the cros-sectional area of the beam

I = ( 1 x 10^-3)/(3.142 x 10^-6) = <em>318.2 W/m^2</em>

<em></em>

b) Total energy delivered E = Pt

where P is the power of the beam

t is the exposure time

E = 1 x 10^-3 x 250 x 10^-3 = <em>2.5 x 10^-4 J</em>

<em></em>

c) The peak electric field is given as

E = $\sqrt{2I/ce_{0} }$

where I is the intensity of the beam

E is the electric field

c is the speed of light = 3 x 10^8 m/s

$e_{0}$ = 8.85 x 10^9 m kg s^-2 A^-2

E = $\sqrt{2*318.2/3*10^8*8.85*10^9}$ = <em>1.55 x 10^-8 v/m</em>

6 0

3 years ago

A police car is located 50 feet to the side of a straight road. A red car is driving along the road in the direction of the poli

NNADVOKAT [17]

Answer:74.33 feet per sec

Explanation:

Given

Police car is 50 feet side of road

Red car is 140 feet up the road

Distance between them is decreasing at the rate of 70 feet per sec

From figure

$x^2+y^2=z^2$

$z=\sqrt{140^2+50^2}$

z=148.66 feet

as the red car is moving

therefore its velocity magnitude is given by

$\frac{\mathrm{d} x}{\mathrm{d} t}$

$2x\times \frac{\mathrm{d} x}{\mathrm{d} t}+0=2z\frac{\mathrm{d} z}{\mathrm{d} t}$

$x\frac{\mathrm{d} x}{\mathrm{d} t}=z\frac{\mathrm{d} z}{\mathrm{d} t}$

$140\times \frac{\mathrm{d} x}{\mathrm{d} t}=148.66\times 70$

$\frac{\mathrm{d} x}{\mathrm{d} t}=74.33 feet\ per\ sec$

6 0

3 years ago