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zzz [600]
3 years ago
9

A uniform ladder is 10 m long and weighs 400 n. it rests with its upper end against a frictionless vertical wall. its lower end

rests on the ground and is prevented from slipping by a peg driven into the ground. the ladder makes a 30◦ angle with the horizontal. the magnitude of the force exerted on the peg by the ladder is:

Physics
1 answer:
valkas [14]3 years ago
6 0
Check the attached file for the answer.

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HELPPPPP!!!!!
leva [86]

Answer:

Option C

Maximum potential energy is at point R.

Explanation:

Potential energy is a product of mass, acceleration due to gravity and height ie

PE=mgh where PE is the potential energy, m is mass of an object, g is acceleration due to gravity whose value is normally taken as 9.81 and h is height. Since at point R we have the maximum height, the potential energy will be highest at this point.

3 0
3 years ago
A proton in a cyclotron is moving with a speed of 2.97×107 m/s in a circle of radius 0.568 m. 1.67 × 10−27 kg is the mass of the
vivado [14]

Answer:

B = 0.546 T,  F = 2.59 10⁻¹² N

Explanation:

The magnetic force is

            F = q v x B

We can calculate the magnitude of the force and find the direction by the right hand rule

          F = q v B sin θ

Let's use Newton's second law

         F = m a

Acceleration is centripetal

         a = v² / r

We substitute

       q v B sin θ = m v² / r

The angle between the field and the radius of the circle is 90º so sin 90 = 1

        q B = m v / r

        B = m v / q r

Let's calculate ’

       B = 1.67 10⁻²⁷ 2.97 10⁷ / (1.60 10⁻¹⁹ 0.568)

        B = 0.546 T

The foce is

         F = q v B

         F = 1.60 10⁻¹⁹ 2.97 10⁷ 0.546

         F = 2.59 10⁻¹² N

3 0
3 years ago
What is the density of a substance that has a mass of 20 g and a volume of 10mL
jenyasd209 [6]

Answer:

Density =mass/volume 20/10=2

4 0
3 years ago
You stand on a frictionless platform that is rotating with an angular speed of 5.1 rev/s. Your arms are outstretched, and you ho
timurjin [86]

Answer:

\omega'=19.419\ rev.s^{-1}

Explanation:

Given:

angular speed of rotation of friction-less platform, \omega=5.1\ rev.s^{-1}

moment of inertia with extended weight, I=9.9\ kg.m^2

moment of inertia with contracted weight, I'=2.6\ kg.m^2

<u>Now we use the law of conservation of angular momentum:</u>

I.\omega=I'.\omega'

9.9\times 5.1=2.6\times \omega'

\omega'=19.419\ rev.s^{-1}

The angular speed becomes faster as the mass is contracted radially near to the axis of rotation.

5 0
3 years ago
When electrons absorb energy they are able to
MaRussiya [10]
B. Move from their ground states up to excited states

:)
7 0
3 years ago
Read 2 more answers
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