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2 years ago

What should be done to lift the same load by applying less effort on an inclined plane

Physics

1 answer:

Scorpion4ik [409]2 years ago

5 0

Answer:

Reduce the friction at the surface

Explanation:

If you can reduce the friction between the load and the plane less effort will be required as you are not having to apply effort to overcome friction.

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What is the speed of the roller coaster at the top of the loop if the radius of curvature there is 11.0 m and the downward accel

8090 [49]

<span>When an object moves in a circle, the acceleration points toward the center of the circle. This acceleration is called centripetal acceleration. We can use a simple equation to find centripetal acceleration. a = v^2 / r We can use this same equation to find the speed of the car. v^2 = a * r v = sqrt { a * r } v = sqrt{ (1.50)(9.80 m/s^2)(11.0 m) } v = 12.7 m/s The speed of the roller coaster is 12.7 m/s</span>

3 0

3 years ago

A satellite, orbiting the earth at the equator at an altitude of 400 km, has an antenna that can be modeled as a 1.76-m-long rod

ivann1987 [24]

Answer:

The inducerd emf is 1.08 V

Solution:

As per the question:

Altitude of the satellite, H = 400 km

Length of the antenna, l = 1.76 m

Magnetic field, B = $8.0\times 10^{- 5}\ T$

Now,

When a conducting rod moves in a uniform magnetic field linearly with velocity, v, then the potential difference due to its motion is given by:

$e = - l(vec{v}\times \vec{B})$

Here, velocity v is perpendicular to the rod

Thus

e = lvB (1)

For the orbital velocity of the satellite at an altitude, H:

$v = \sqrt{\frac{Gm_{E}}{R_{E}} + H}$

where

G = Gravitational constant

$m_{e} = 5.972\times 10^{24}\ kg$ = mass of earth

$R_{E} = 6371\ km$ = radius of earth

$v = \sqrt{\frac{6.67\times 10^{- 11}\times 5.972\times 10^{24}}{6371\times 1000 + 400\times 1000} = 7670.018\ m/s$

Using this value value in eqn (1):

$e = 1.76\times 7670.018\times 8.0\times 10^{- 5} = 1.08\ V$

5 0

3 years ago

A uniform disk has a moment of inertia that is (1/2)MR2. A uniform disk of mass 13 kg, thickness 0.3 m, and radius 0.2 m is loca

kicyunya [14]

The angular momentum of an object is equal to the product of its moment of inertia and angular velocity.
L = Iω
I = 1/2 MR²
I = 1/2 x 13 x (0.2)
I = 1.3

ω = 2π/t
ω = 2π/0.3
ω = 20.9

L = 1.3 x 20.9
= 27.2 kgm²/s

6 0

3 years ago

Read 2 more answers

the power rating of an electric lawn mower is 2200 watts. if the lawn mower was used for 30 mins, and 650 Newtons of force was u

Tasya [4]

Answer:

Calculate the work done by a 47 N force pushing a 0.025 kg pencil 0.25 m ... A boy on a bicycle drags a wagon full of newspapers at 0.80 m/s for 30 min ... A power mower does 9.00 x 105 J of work in 0.500 h. ... p: W 2200ch: w will320,000 T/ ... How much electrical energy (in kilowatt hours) would a 60.0 W light bulb use in ..

Explanation:

4 0

3 years ago

Two metal plates 15mm apart have a potential difference of 750v between them. The force on a small charged sphere placed between

Romashka [77]

Answer:

50,000 V/m

Explanation:

The electric field between two charged metal plates is uniform.

The relationship between potential difference and electric field strength for a uniform field is given by the equation

$\Delta V=Ed$