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2 years ago

HELPPP chchcuhckf hfjf

Physics

1 answer:

MrMuchimi2 years ago

7 0

Oxygen lungs and organs
Put them in this order and it should be right

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A standard 1 kilogram weight is a cylinder 41.5 mm in height and 44.0 mm in diameter. what is the density of the material?

Korvikt [17]

Answer;

=15855.40 kg/m^3

Explanation;

Volume (V) of the cylinder = pi x r^2 x h

V = 3.14 x (44/2 x 10^-3)^2 x 41.5 x 10^-3

V = 6.307 x 10^-5 m^3

By density = m/V

mass = 1 kg

density = 1/(6.307 x 10^-5) = 15855.40 kg/m^3

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3 years ago

PLEASE HELP ME ON THIS ONE ANYONE

andreev551 [17]

Answer:

A & B

Explanation:

A & B Would be the right answer since Morse code cannot be represented through the height of the fire.

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2 years ago

Energy transformation in wound spring of a toy car?<br>give your own answrr

BaLLatris [955]

Answer:

The work done in winding the spring gets stored in the wound up spring in the form of elastic potential energy (i.e potential energy due to change in shape). ... During this process, the potential energy stored in it gets converted to kinetic energy. This turns the wheels of the toy car.

Explanation:

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3 years ago

Read 2 more answers

An igneous intrusion whose boundaries lie parallel to the layering in the surrounding country rock is considered to be:

ikadub [295]

<span>...a concordant intrusion. In geology, "concordant" means the same as "sill" -- or, an intrusion that has gotten in between older layers of rock (or even beds of volcanic lava). An intrusion with boundaries parallel to layering in surrounding rocks suggests this, meaning it is considered to be a concordant intrusion.</span>

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4 years ago

A 975-kg sports car (including driver) crosses the rounded top of a hill at determine (a) the normal force exerted by the road o

iVinArrow [24]

There are missing data in the text of the problem (found them on internet):
- speed of the car at the top of the hill: $v=15 m/s$
- radius of the hill: $r=100 m$

Solution:

(a) The car is moving by circular motion. There are two forces acting on the car: the weight of the car $W=mg$ (downwards) and the normal force N exerted by the road (upwards). The resultant of these two forces is equal to the centripetal force, $m \frac{v^2}{r}$ , so we can write:
$mg-N=m \frac{v^2}{r}$ (1)
By rearranging the equation and substituting the numbers, we find N:
$N=mg-m \frac{v^2}{r}=(975 kg)(9.81 m/s^2)-(975 kg) \frac{(15 m/s)^2}{100 m}=7371 N$

(b) The problem is exactly identical to step (a), but this time we have to use the mass of the driver instead of the mass of the car. Therefore, we find:
$N=mg-m \frac{v^2}{r}=(62 kg)(9.81 m/s^2)-(62 kg) \frac{(15 m/s)^2}{100 m}=469 N$

(c) To find the car speed at which the normal force is zero, we can just require N=0 in eq.(1). and the equation becomes:
$mg=m \frac{v^2}{r}$
from which we find
$v= \sqrt{gr}= \sqrt{(9.81 m/s^2)(100 m)}=31.3 m/s$

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4 years ago