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1 year ago

12

Two men are trying to carry a wooden pole. If one of them is weaker than other, how can they carry the pole hence making small l

oad for the weak man?

2 answers:

shepuryov [24]1 year ago

7 0

The stronger man can go more into the middle of the pole and the weak one can take the shorter end

Romashka-Z-Leto [24]1 year ago

6 0

Answer: Answer down below.

Explanation:

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I WILL MARK BRAINLIEST IF CORRECT!!!

julsineya [31]

0.05*0.2=0.05*0.15+0.015*m2',

m2'=1/6 m/s,

m2'=0.17 m/s

N I C E - D A Y!

6 0

2 years ago

A dentist uses a concave mirror (focal length 2.5 cm) to examine some teeth. If the distance from the object to the mirror is 1.

AysviL [449]

Answer:

2.28

Explanation:

From mirror formula,

1/f = 1/u+1/v .......... Equation 1

Where f = focal length of the mirror, v = image distance, u = object distance.

Note: The focal length mirror is positive.

make v the subject of the equation,

v = fu/(u-f)............ Equation 2

Given: f = 2.5 cm, u = 1.4 cm

Substitute into equation 2

v = 2.5(1.4)/(1.4-2.5)

v = 3.5/-1.1

v = -3.2 cm.

Note: v is negative because it is a virtual image.

But,

Magnification = image distance/object distance

M = v/u

Where M = magnification.

Given: v = 3.2 cm, u = 1.4 cm

M = 3.2/1.4

M = 2.28.

Thus the magnification of the tooth = 2.28.

3 0

3 years ago

(10%) Problem 5: The Earth spins on its axis and also orbits around the Sun. For this problem use the following constants. Mass

11Alexandr11 [23.1K]

Incomplete Question.The Complete question is

The Earth spins on its axis and also orbits around the Sun. For this problem use the following constants. Mass of the Earth: 5.97 × 10^24 kg (assume a uniform mass distribution) Radius of the Earth: 6371 km Distance of Earth from Sun: 149,600,000 km

(i)Calculate the rotational kinetic energy of the Earth due to rotation about its axis, in joules.

(ii)What is the rotational kinetic energy of the Earth due to its orbit around the Sun, in joules?

Answer:

(i) KE= 2.56e29 J

(ii) KE= 2.65e33 J

Explanation:

i) Treating the Earth as a solid sphere, its moment of inertia about its axis is

I = (2/5)mr² = (2/5) * 5.97e24kg * (6.371e6m)²

I = 9.69e37 kg·m²

About its axis,

ω = 2π rads/day * 1day/24h * 1h/3600s

ω= 7.27e-5 rad/s,

so its rotational kinetic energy

KE = ½Iω² = ½ * 9.69e37kg·m² * (7.27e-5rad/s)²

KE= 2.56e29 J

(ii) About the sun,

I = mR²

I= 5.97e24kg * (1.496e11m)²

I= 1.336e47 kg·m²

and the angular velocity

ω = 2π rad/yr * 1yr/365.25day * 1day/24h * 1h/3600s

ω= 1.99e-7 rad/s

so

KE = ½ * 1.336e47kg·m² * (1.99e-7rad/s)²

KE= 2.65e33 J

6 0

2 years ago

a projectile is launched at an angle of 30 degrees and lands later at the same level. if it's initial speed is 50 m/s, solve for

Mrrafil [7]

$using \: the \: formula \\ t = \frac{2u \sin( \alpha ) }{g} where \: u = initial \: speed \: \\ \alpha = angle \: of \: projection \\ g = acceleration \: due \: to \: gravity \\ \frac{2 \times 50 \times \sin(30) }{10} \\ \frac{100 \times 0.5}{10} = \frac{50}{10} = 5seconds$

Maximum height
= (Usinα)^2/2g
(50*0.5)^2/20
25^2/20
625/20
=31.25metres
horizontal distance = Range= [U^2 * sin2α]/g
[50^2 * sin60]/10
2500 * 0.8660/10
2165/10=216.5metres

3 0

2 years ago

The pressure on a fluid at rest in a pipe increases by 20 Pa. How does this change in pressure affect the pressure on the fluid

timofeeve [1]

Answer:The change in pressure can affect the pressure on the fluid through the radius and diameter of the pipe.

r^² x Pressure (pa).

Therefore the narrower the other part of the pile, the greater the pressure on the fluid at such part, the wider in other part the lesser the pressure on the fluid at this part.

Explanation:

4 0

3 years ago