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3 years ago

I drop an egg from a certain distance and it takes the egg 3.74 seconds to reach the ground. How high up was the egg?

Physics

1 answer:

Ulleksa [173]3 years ago

5 0

Answer:

Explanation:

<u>Free Fall Motion </u>

When a body is left to move in the air with no friction, the motion is ruled only by the force of gravity. The vertical distance a body travels in the air after a time t is .

$\displaystyle y=\frac{gt^2}{2}$

We know the egg takes 3.74 seconds to reach the ground. The height it was launched from is

$\displaystyle y=\frac{(9.8)(3.474)^2}{2}$

$\displaystyle y=68.54\ m$

The closest correct option is

B. 68.6m

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A box of paper is labeled 24lb paper. This means that 500 sheets (counted number) of paper size 17in x 22in weighs 24 pounds (Th

Tresset [83]

Answer:

The mass of a single paper is approximately 0.047 lb/paper which in SI Units is approximately 21.77 g/paper

Explanation:

The given information on the size and the weight of paper are;

The mass of a box of 500 sheets of paper = 24 lb

The number of sheets in the paper = 500 sheets

The dimensions of the paper = 17 in. × 22 in., which is equivalent to 43.18 cm × 55.88 cm

The mass of a single paper = The mass of the box of paper/(The number of sheets of paper present in the box)

The mass of a single paper = 24 lb/500 = 0.047 lb/paper

Given that 1 lb = 453.6 g, we have;

0.047 lb/paper = 0.047 lb/paper×453.6 g/(lb) = 21.77 g/paper

The mass of a single paper = 0.047 lb/paper = 21.77 g/paper.

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An electron experiences a magnetic force with a magnitude of 4.90×10−15 n when moving at an angle of 60.0 ∘ with respect to a ma

Leya [2.2K]

Missing questions: "find the speed of the electron".

Solution:
the magnetic force experienced by a charged particle in a magnetic field is given by
$F=qvB \sin \theta$
where
q is the particle charge
v its velocity
B the magnitude of the magnetic field
$\theta$ the angle between the directions of v and B.

Re-arranging the formula, we find:
$v= \frac{F}{qB \sin \theta}$
and by substituting the data of the problem (the charge of the electron is $q=1.6 \cdot 10^{-19} C$ ), we find the velocity of the electron:
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