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Jlenok [28]
3 years ago
7

The student moves his leg 34 cm sideways as shown.

Physics
2 answers:
Nana76 [90]3 years ago
8 0

(i) • there is force applied to an objects

• the object moves

• the object moves in the same direction as the direction of the force

(ii) workdone = force x distance

= 23 x 34

= 782Joules

jenyasd209 [6]3 years ago
5 0

Answer:

Question A1:

Workdone=force x distance

Question A11:

7.82 joules

Explanation:

question A1:

The relationship between work done, force and distance is that workdone can be calculated by multiplying force and distance

Question A11:

Force=23N

Distance=34cm=0.34m

Work done=force x distance

Work done=23 x 0.34

Workdone=7.82J

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From the following statements about mechanical waves, identify those that are true for transverse mechanical waves only, those t
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'In transverse waves, the particles of the medium move perpendicular to the direction of the flow of energy' is true for transverse waves only.

'In longitudinal waves, the particles of the medium move parallel to the direction of the flow of energy' is true for longitudinal waves only.

'Many wave motions in nature are a combination of longitudinal and transverse motion' is true for both longitudinal and transverse waves.

<u>Explanation:</u>

Longitudinal waves are those where the direction of propagation of particles are parallel to the medium' particles. While transverse waves propagate perpendicular to the medium' particles.

As wave motions are assumed to be of standing waves which comprises of particles moving parallel as well as perpendicular to the medium, most of the wave motions are composed of longitudinal and transverse motion.

So the option stating the medium' particle moves perpendicular to the direction of the energy flow is true for transverse waves. Similarly, the option stating the medium' particle moves parallel to the direction of flow of energy is true for longitudinal waves only.

And the option stating that wave motions comprises of combination of longitudinal and transverse motion is true for both of them.

5 0
3 years ago
Where is most of the mass of an atom located? A.in the nucleus B.in the orbits C.in the electrons Dit is split between the nucle
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I think it is A........................................ 
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3 years ago
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1. Identify which of the following will not increase the current induced in a wire loop moving through a magnetic field. a. incr
djyliett [7]

Answer:

Rotating the loop until it is perpendicular to the field  

Explanation:

Current is induced in a conductor when there is a change in magnetic flux.

The strength of the induced current in a wire loop moving through a magnetic field can be increased or decreased by the following methods:

By increasing the strength of the magnetic field there will be increased in the induced current. If the strength of the magnetic field is decreased then there is a decrease in induced current.    

By increasing the speed of the wire there will be increased in the induced current. When the speed of the wire is decreased then there is a decrease in induced current.

By increasing the number of turns of the coil the strength of the induced current can be increased. when there is less number of turns in coils then there is a decrease in induced current.  

Rotating the loop until it is perpendicular to the field will not increase the current induced in a wire loop moving through a magnetic field.

Therefore, the option is (c) is correct.

4 0
3 years ago
A box is held at rest by two ropes that form 30° angles with the vertical. The tension T in either rope is 42 N. What is the wei
iragen [17]
<h3><u>Answer;</u></h3>

= 73 N

<h3><u>Explanation</u>;</h3>

Using the formula

2 T cos(30°) = w

Where; T is the tension on each string, while w is the weight of the box given by mg

Therefore;

W = 2Tcos 30°

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<u>   ≈ 73 N</u>

7 0
3 years ago
In a thunderstorm, electric charge builds up on the water droplets or ice crystals in a cloud. Thus, the charge can be considere
muminat

Answer:

2.1\cdot 10^{21} electrons

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The magnitude of the electric field outside an electrically charged sphere is given by the equation

E=\frac{kQ}{r^2}

where

k is the Coulomb's constant

Q is the charge stored on the sphere

r is the distance (from the centre of the sphere) at which the field is calculated

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r=1.00 km = 1000 m is the radius of the sphere

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Assuming that the cloud is negatively charged, then

Q=-333.3 C

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e=-1.6\cdot 10^{-19}C

The number of excess electrons on the cloud is

N=\frac{Q}{e}=\frac{-333.3}{-1.6\cdot 10^{-19}}=2.1\cdot 10^{21}

5 0
3 years ago
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