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3 years ago

9

A student is given a 5.4 g mineral sample. She determines that the volume of her sample is 2 cm3. What is the density of her min

eral

1 answer:

Xelga [282]3 years ago

7 0

Answer:

6.7

Explanation:

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A boy flying a kite is standing 30 ft from a point directly under the kite. if the string to the kite is 50 ft long, what is the

vagabundo [1.1K]

Draw a diagram to illustrate the problem as shown in the figure below.
h = height of the kite above ground.

By definition, the angle of elevation is
$cos \theta = \frac{30}{50} =0.6$
Therefore
$\theta = cos^{-1} 0.6 = 53.1 \, deg.$

Answer: 53° (nearest integer)

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3 years ago

If a certain mass of mercury has a volume of 0.002 m3 at a temperature of 20°C, what will be the volume at 50°C?

Klio2033 [76]

From tables, the density of mercury is
13545 kg/m^3 at 20°C,
13472 kg/m^3 at 50°C.

Because mass = density * volume, the mass of mercury at 20°C is
m = (13545 kg/m^3)*(0.002 m^3) = 27.09 kg

Let V = volume of mercury at 50°C.
Because the mass of mercury does not change, therefore at 50°C,
(13472 kg/m^3)*(V m^3) = 27.09
V = 27.09/13472 = 0.0020108 m^3

Answer: B. 0.002010812 m³

5 0

3 years ago

Particle accelerators, such as the Large Hadron Collider, use magnetic fields to steer charged particles around a ring Consider

coldgirl [10]

Answer:

$\beta= 3.49x10^{-8}T$

Explanation:

The magnetic field can be find using the equation

$m*v^2/r=q*v*\beta$

You can cancel a element of v'

$m*v/r=q*\beta$

$C=36*1m=2\pi*r$

$r=\frac{36}{2\pi } =5.7295m$

Solve to magnetic field

$\beta=\frac{m*v^2}{r*q}$

The charge and mass of the proton are:

$m_p=1.6x10^{27}kg$ , $q_p=1.6x10^{-19}C$

Replacing numeric

$\beta=\frac{1.6x10^{-27}kg*2x10m/s}{1.6x10^{-19}C*5.73m}$

$\beta= 3.49x10^{-8}T$

6 0

3 years ago

A 1 kg ball and a 10 kg ball are dropped from a height of 10 m at the same time. In the absence of air resistance,

RideAnS [48]

Answer:

the two balls will hit the ground at the same time.

Explanation:

The time of dropping, in the following equation, is related to both the distance travel s and the gravitational acceleration g, which are the same for both ball (if we neglect air resistance), no matter what their mass are.

$s = gt^2/2$

$t^2 = 2s/g$

So the time it takes to drop 2 balls are the same. They will hit the ground at the same time.

3 0

3 years ago

Suppose a 20-foot ladder is leaning against a building, reaching to the bottom of a second-floor window 15 feet above the ground

Papessa [141]

Answer:

The answer is β=0,85 rads

Explanation:

As the ladder is leaning against the building, we can imagine there´s a triangle where 20ft is the hypotenuse and 15ft is the maximum vertical distance between the ladder and the ground, it means, the leg opposite to β which is the angle we need

Let β(betha) be the angle between the ladder and the ground

We also know that $sin(betha)=(leg opposite)/(hypotenuse)$

In this case we will need to find β, this way:

$betha=sin^-1((15ft/20ft))$

Then β=48,6°

We also have that 2πrads is equal to 360°, in this way we find how much β is in radians:

$betha=(48,6°)*(2pirads/360°)$

then we find β=0,85rads

7 0

3 years ago