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2 years ago

Pls help me this is being timed.

Physics

1 answer:

GrogVix [38]2 years ago

5 0

Answer:

Increase air resistance

Explanation:

Gravity forces the parachute down but air resistance pushing up on the flat surface of the parachute causes it to fall slower to the ground.

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The Andromeda Galaxy is faintly visible to the naked eye in the constellation Andromeda. Suppose instead it were located in the

Kaylis [27]

Answer:

a) We could not see it at all.

Explanation:

The most distant object that can be seen is the andromeda galaxy, which we may have a slight view of. The andromeda galaxy is a large galaxy that along with the previous two is also part of the local group. Spiral-type galaxy that is approximately 250,000 light years in diameter (more than twice the diameter of the Milky Way!) And is about 2.9 million light years away from our galaxy. Because of its distance, we have difficulty visualizing this galaxy, we would have this difficulty even if the andromeda galaxy was in the center of the Milky Way, but maintaining its current distance. That is, even if the andromeda galaxy were located in the same direction in space as the center of the Milky Way (but still at its current distance), we could not see it at all.

8 0

2 years ago

A positively charged particle of mass 7.2 x 10-8 kg is traveling due east with a speed of 88 m/s and enters a 0.6-T uniform magn

Marianna [84]

Answer:

q = 8.57 10⁻⁵ mC

Explanation:

For this exercise let's use Newton's second law

F = ma

where force is magnetic force

F = q v x B

the bold are vectors, if we write the module of this expression we have

F = qv B sin θ

as the particle moves perpendicular to the field, the angle is θ= 90º

F = q vB

the acceleration of the particle is centripetal

a = v² / r

we substitute

qvB = m v² / r

qBr = m v

q = $\frac{m\ v}{B\ r}$

The exercise indicates the time it takes in the route that is carried out with constant speed, therefore we can use

v = d / t

the distance is ¼ of the circle,

d = $\frac{1}{4} \ 2\pi r$

d = $\frac{\pi }{2r}$

we substitute

v = $\frac{\pi r}{2t}$

r = $\frac{2 \ t \ v}{\pi }$

let's calculate

r = $\frac{2 \ 2.2 \ 10^{-3} \ 88}{\pi }$ 2 2.2 10-3 88 /πpi

r = 123.25 m

let's substitute the values

q = $\frac{ 7.2 \ 10^{-8} \ 88}{ 0.6 \ 123.25}$ 7.2 10-8 88 / 0.6 123.25

q = 8.57 10⁻⁸ C

Let's reduce to mC

q = 8.57 10⁻⁸ C (10³ mC / 1C)

q = 8.57 10⁻⁵ mC

4 0

2 years ago

Define , gravitational acceleration

Vitek1552 [10]

The simplest answer would be "acceleration due to gravity."

The exact value of this acceleration changes depending on which planet your on (for example).

3 0

2 years ago

Cheetahs, the fastest of the great cats, can reach 50.0 miles/hour in 2.22 s starting from rest. Assuming that they have constan

elena55 [62]

Answer:

$10.07m/s^2$

Explanation:

Information we have:

velocities:

initial velocity: $v_{i}=0mph$ (starts from rest)

final velocity: $v_{f}=50mph$

time: $t=2.22s$

Since we need the answer in $m/s^2$ , we nees to convert the speed to meters per second:

$v_{f}=\frac{50miles}{1hour}(\frac{1hour}{3,600s} )(\frac{1609.34meters}{1mile} ) \\\\v_{f}=\frac{50*1609.34}{3600} m/s\\\\v_{f}=22.35m/s$

We find the acceleration with the following formula:

$a=\frac{v_{f}-v_{i}}{t}$

substituting the known values:

$a=\frac{22.35m/s-0m/s}{2.22s}\\ \\a=10.07m/s^2$

the acceleration is 10.07 $m/s^2$

8 0

2 years ago

Read 2 more answers

Two rigid tanks of equal size and shape are filled with different gases. The tank on the left contains oxygen, and the tank on t

fredd [130]

Answer:

The number of oxygen molecules in the left container greater than the number of hydrogen molecules in the right container.

Explanation:

Given:

Molar mass of oxygen, $M_O=32$

Molar mass of hydrogen, $M_H=2$

We know ideal gas law as:

$PV=nRT$

where:

P = pressure of the gas

V = volume of the gas

n= no. of moles of the gas molecules

R = universal gs constant

T = temperature of the gas

∵ $n=\frac{m}{M}$

where:

m = mass of gas in grams

M = molecular mass of the gas

∴Eq. (1) can be written as:

$PV=\frac{m}{M}.RT$

$P=\frac{m}{V}.\frac{RT}{M}$

as: $\frac{m}{V}=\rho\ (\rm density)$

So,

$P=\rho.\frac{RT}{M}$

Now, according to given we have T,P,R same for both the gases.

$P_O=P_H$

$\rho_O.\frac{RT}{M_O}=\rho_H.\frac{RT}{M_H}$

$\Rightarrow \frac{\rho_O}{32}=\frac{\rho_H}{2}$

$\rho_O=16\rho_H$

∴The molecules of oxygen are more densely packed than the molecules of hydrogen in the same volume at the same temperature and pressure. So, <em>the number of oxygen molecules in the left container greater than the number of hydrogen molecules in the right container.</em>

5 0

2 years ago