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vitfil [10]
3 years ago
11

Saeed moves 50 m east then 65 m south to reach the garden. What is the displacement of Saeed?​

Physics
1 answer:
vampirchik [111]3 years ago
4 0

Answer:

82.01m

Explanation:

Using the formula:

R^2=A^2+B^2

R= √ A^2+B^2

Where R= displacement

A=50m

B=65m

R= √ 50^2+65^2

R= √ 2500+4225

R= √ 6725

R=82.01m

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The gravitational potential energy is 252 J

Explanation:

The Gravitational Potential Energy (GPE) of an object is given by

GPE=Wh

where

W is the weight of the object

h is the height of the object above the ground

For the carriage and the baby in this problem, we have

W = 21 N is their weight

h = 21 m is their height above the ground

Substituting,

GPE=(12)(21)=252 J

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3 years ago
Choose only one correct option. Explanation needed.
Mama L [17]

Answer:

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Explanation:

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\displaystyle \rho = \frac{m}{V}

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3 years ago
Relationship between prism and lens ​
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Answer:

In essence, optical lenses bend and focus light, known as refraction. Prism lenses, however, refract light a bit differently. ... Light passing through a prism will bend towards the base, while the image of the object viewed with the prism moves toward the peak.

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The flower of the species Rosa verdus can be either green or red. in the species a single gene with two alleles determines flowe
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Since G is a dominate trait the 2 genotypes would be Gg or GG
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A proton having an initial velvocity of 20.0i Mm/s enters a uniform magnetic field of magnitude 0.300 T with a direction perpend
Sonja [21]

The time interval for which the proton remains in the field is -

Δt = $\frac{\pi R}{40}.

We have a proton entering a uniform magnetic field which is in a direction perpendicular to the proton's velocity.

We have to determine time interval during which the proton is in the field.

<h3>What is the magnitude of force on the charged particle moving in a uniform magnetic field?</h3>

The magnitude of force on the charged particle moving in a uniform magnetic field is given by -

F = qvB sinθ



According to the question, we have -

Entering Velocity (v) = 20 i  m/s

Magnetic field intensity (B) = 0.3 T

Leaving velocity (u) = - 20 j  m/s

Now -

The entering and leaving velocity vectors have 90 degrees difference between them. Therefore, only a quarter of distance of the complete circular path of radius 'R' is traced by the proton. Therefore -

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Since, the radius of circular path is not given, we will assume it R.

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Δt = $\frac{\pi R}{40}

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