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3 years ago

A painter lifts a 3 kg can of paint 3 m above the floor. How much work was done?

Physics

1 answer:

Sindrei [870]3 years ago

3 0

Hey there!

P= 3kg=3

W=m x g x d =3kg x 9 8 m/s² x 3m = 88J

Hope this helped you and solve your problem

TobeySnaps.

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Use the equation d = m/v [stack fraction), where d= density, m= mass, and v

sladkih [1.3K]

Answer:

54.

Explanation:

6 (v) x 9 (m) = 54.

54 (m) / 6 (v) = 9 (d)

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3 years ago

How do climate differences affect the movement at the Mariana Trench

vovangra [49]

It pushes the currents to opposite sides

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3 years ago

A light ray incident on a block of glass makes an incident angle of 50.0° with the normal to the surface. The refracted ray in t

Vladimir [108]

Answer:

The index of refraction of the glass is 1.3

Explanation:

Given data:

i = incident angle = 50°

r = refracted angle = 36.1°

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3 years ago

Malcolm and Ravi raced each other. The average of their maximum speeds was 260 km/h If doubled, Malcolm's maximum speed would be

AfilCa [17]

Answer

Given,

Average speed of Malcolm and Ravi = 260 km/h

Let speed of the Malcolm be X and speed of the Ravi Y.

From the given statement

$\dfrac{X+Y}{2}=260$

$X + Y = 520$ ....(i)

$2X - Y = 80$ ....(ii)

Adding both the equations

3 X = 600

X = 200 km/h

Putting value in equation (i)

Y = 520 - 200

Y = 320 Km/h

Speed of Malcolm = 200 Km/h

Speed of Ravi = 320 Km/h

8 0

3 years ago

A felt-covered beanbag is fired into a empty wooden crate that sits on a concrete floor and is open on one side. The beanba

Phantasy [73]

Answer:

31.42383 m/s

Explanation:

g = Acceleration due to gravity = 9.81 m/s²

$\mu$ = Coefficient of kinetic friction = 0.48

s = Displacement = 0.935 m

$m_1$ = Mass of bean bag = 0.354 kg

$m_2$ = Mass of empty crate = 3.77 kg

$v_1$ = Speed of the bean bag

$v_2$ = Speed of the crate

Acceleration

$a=-\frac{f}{m}\\\Rightarrow a=-\frac{\mu mg}{m}\\\Rightarrow a=-\mu g$

$a=--9.81\times 0.48=4.7088\ m/s^2$

From equation of motion

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 4.7088\times 0.935+0^2}\\\Rightarrow v=2.96739\ m/s$

In this system the momentum is conserved

$m_1v_1=(m_1+m_2)v_2\\\Rightarrow v_1=\frac{(m_1+m_2)v_2}{m_1}\\\Rightarrow u=\frac{(0.354+3.77)\times 2.69739}{0.354}\\\Rightarrow u=31.42383\ m/s$

The speed of the bean bag is 31.42383 m/s

8 0

3 years ago