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3 years ago

Hat is the approximate pressure of air at sea level?

Physics

2 answers:

Tanya [424]3 years ago

5 0

c) 101kPa

Hope I helped! ( Smiles )

Elenna [48]3 years ago

4 0

the answer is c.) 101kPa

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A convex mirror has a focal length of -10.8 cm. An object is placed 32.7 cm from the mirror's surface. Determine the image dista

KonstantinChe [14]

Answer:

-353.16

Explanation:

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3 years ago

When joshua brakes his speeding bicycle to a stop, kinetic energy is transformed to select one:?

photoshop1234 [79]

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3 years ago

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Hooke's law describes a certain light spring of unstretched length 34.8 cm. When one end is attached to the top of a doorframe a

DaniilM [7]

Answer:

a) $k = 1343.6\,\frac{N}{m}$ , b) $l = 0.501\,m\,(50.1\,cm)$

Explanation:

a) The Hooke's law states that spring force is directly proportional to change in length. That is to say:

$F \propto \Delta l$

In this case, the force is equal to the weight of the object:

$F = m\cdot g$

$F = (8.22\,kg)\cdot (9.807\,\frac{m}{s^{2}} )$

$F = 80.614\,N$

The spring constant is:

$k = \frac{F}{\Delta l}$

$k = \frac{80.614\,N}{0.408\,m-0.348\,m}$

$k = 1343.6\,\frac{N}{m}$

b) The length of the spring is:

$F = k\cdot (l-l_{o})$

$l = l_{o} + \frac{F}{k}$

$l=0.348\,m+\frac{205\,N}{1343.6\,\frac{N}{m} }$

$l = 0.501\,m\,(50.1\,cm)$

6 0

3 years ago

The charges of two particles are as follows: Q1=2 x 10 -8 C and Q2 = 3 x 10 -7 C. Find the magnitude of the force between these

Fantom [35]

Answer:

F = 3.86 x 10⁻⁶ N

Explanation:

First, we will find the distance between the two particles:

$r = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2+(z_{2}-z_{1})^2}\\$

where,

r = distance between the particles = ?

(x₁, y₁, z₁) = (2, 5, 1)

(x₂, y₂, z₂) = (3, 2, 3)

Therefore,

$r = \sqrt{(3-2)^2+(2-5)^2+(3-1)^2}\\r = 3.741\ m\\$

Now, we will calculate the magnitude of the force between the charges by using Coulomb's Law:

$F = \frac{kq_{1}q_{2}}{r^2}\\$

where,

F = magnitude of force = ?

k = Coulomb's Constant = 9 x 10⁹ Nm²/C²

q₁ = magnitude of first charge = 2 x 10⁻⁸ C

q₂ = magnitude of second charge = 3 x 10⁻⁷ C

r = distance between the charges = 3.741 m

Therefore,

$F = \frac{(9\ x\ 10^9\ Nm^2/C^2)(2\ x\ 10^{-8}\ C)(3\ x\ 10^{-7}\ C)}{(3.741\ m)^2}\\$

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3 years ago

It is estimated that uranium is relatively common in the earth's crust, occurring in amounts of 4 g / metric ton. A metric ton i

timama [110]

Answer:

The mass of Uranium present in a 1.2mg sample is $4.8 \cdot 10^{-6}\,mg$

Explanation:

The ration between Uranium mass and total sample mass is: $\frac{4g}{1000kg} =\frac{4g}{1000000g}=\frac{1}{250000}$

For a sample of mass 1.2 mg, the amount of uranium is:

$1.2\, mg \cdot \frac{1}{250000}=4.8 \cdot 10^{-6}\,mg$

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3 years ago