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Juliette [100K]
3 years ago
13

If 1.3 grams of air has a volume of 1000cm ^3, what is the density of air?

Physics
2 answers:
sleet_krkn [62]3 years ago
5 0
D)
Density is equal to mass/volume, so in your case:
density=1.3/1000=0.0013 g/cm^3
Daniel [21]3 years ago
5 0

Answer:

Density, d=0.0013\ g/cm^3

Explanation:

Given that,

Mass of the air, m = 1.3 grams

Volume of the air, V=1000\ cm^3

The density of the air is given by total mass divided by total volume. Mathematically, it is given by :

d=\dfrac{m}{V}

d=\dfrac{1.3\ g}{1000\ cm^3}

d=0.0013\ g/cm^3

Hence, this is the required solution.

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You used the right-hand rule to determine the z component of the angular momentum, but as a check, calculate in terms of positio
KengaRu [80]

Answer:

case x py   L is in the positive z direction

case y px   L the negative z direction

Explanation:

The angular amount is defined by the relation

         L = r x p

the bold are vectors, where r is the position vector and p is the linear amount vector.

The module of this vector can be concentrated by the relation

         L = r p sin θ

the direction of the vector L can be found by the right-hand rule where the thumb points in the direction of the displacement vector, the fingers extended in the direction of the moment p which is the same direction of speed and the palm points in the direction of the angular momentum L

in the case x py

the thumb is in the x direction, the fingers are extended in the direction and the palm is in the positive z direction

In the case y px

the thumb is in the y direction, the fingers are in the x direction, the palm is in the negative z direction

7 0
3 years ago
A ball thrown horizontally at vi = 30.0 m/s travels a horizontal distance of d = 55.0 m before hitting the ground. from what hei
tekilochka [14]
Assume no air resistance, and g = 9.8 m/s².

Let
x =  angle that the initial velocity makes with the horizontal.
u = 30 cos(x), horizontal velocity
v = 30 sin(x), vertical launch velocity

The horizontal distance traveled is 55 m, therefore the time of flight is
t = 55/[30 cos(x)] = 1.8333 sec(x)  s

With regard to the vertical velocity, and the time of flight,obtain
[30 sin(x)]*(1.8333 sec(x)) + (1/2)*(-9.8)*(1.8333 sec(x))² = 0
 55 tan(x) - 16.469 sec²x = 0
55 tan(x) - 16.469[1 + tan²x] = 0
16.469 tan²x - 55 tan(x) + 16.469 = 0
tan²x - 3.3396 tan(x) + 1 = 0

Solve with the quadratic formula.
tan(x) = 0.5[3.3396 +/- √(7.153)] = 3.007 or 0.3326
Therefore
x = 71.6° or x = 18.4°

The time of flight is
t = 1.8333 sec(x) = 5.8096 s or 1.932 s
The initial vertical velocity is
v = 30 sin(x) = 28.467 m/s or 9.468 m/s
The horizontal velocity is
u = 30 cos(x) = 9.467 m/s or 28.469 m/s

If t = 5.8096 s,
  u*t = 9.467*5.8096 = 55 m (Correct)
or
 u*t = 28.469*15.8096 = 165.4 m (Incorrect)

Therefore, reject x = 18.4°. The correct solution is
t = 5.8096 s
x = 71.6°
u = 9.467 m/s
v = 28.467 m/s

The height from which the ball was thrown is
h = 28.467*5.8096 - 0.5*9.8*5.8096² = -110.4 m
The ball was thrown from a height of 110.4 m

Answer: h = 110.4 m

7 0
3 years ago
A certain AM radio wave has a frequency of 1.12 x 100 Hz. Given that radio waves travel at
riadik2000 [5.3K]

Answer: 267 m

Explanation:

2.99x10^8 m/s

———————-

1.12 x 10^6 Hz

3 0
3 years ago
Im back and need help
const2013 [10]

Always here to help. Bring it!!!

7 0
2 years ago
The speed of a boat in still water is 20 mph. it travels from one pier to another with the current in 4 hours and back against t
saw5 [17]
The answer is, "the speed of the current is 5 miles per hour."

To calculate the speed of the current,
let's assume speed of current =  xmph. Time taken to travel from one pier to another with the current = 100/(20+x)h


But the time taken to travel from one pier to another with the current, which is given is = 4 hours. So,  4=100/(20+x) 80+4x = 100

4x = 20

x = 5 Thus, the speed of the current is 5 miles per hour.
3 0
3 years ago
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