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mylen [45]
2 years ago
6

Anne has a sample of a substance. Its volume is 20 cm3, and its mass is 100 grams. What is the sample’s density? The sample’

s density is g/cm3.
Physics
1 answer:
Tanzania [10]2 years ago
6 0

Answer:

<h2>5 g/cm³</h2>

Explanation:

The density of a substance can be found by using the formula

d =  \frac{m}{v}  \\

m is the mass

v is the volume

From the question

m = 100 g

v = 20 cm³

We have

d =  \frac{100}{20}  = 5 \\

We have the final answer as

<h3>5 g/cm³</h3>

Hope this helps you

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A student took a calibrated 250.0 gram mass, weighed it on a laboratory balance, and found it read 266.5 g. What was the student
Kruka [31]

Answer:

B. 6.6%

Explanation:

The percentage error of a measurement can be calculated using the formula;

Percent error = (experimental value - accepted value / accepted value) × 100

In this question, the calibrated 250.0 gram mass is the accepted value while the weighed mass of 266.5 g is the experimental or measured value.

Hence, the percentage error can be calculated thus;

Percent error = (266.5-250.0/250.0) × 100

Percent error = 16.5/250 × 100

Percent error = 0.066 × 100

Percent error = 6.6%

7 0
3 years ago
In order to attain orbit around earth, the ATLAS-V rocket must accelerate up to a
IrinaK [193]
A) average acceleration = final velocity - initial velocity / time

= 7700 - 0 / 11

= 700ms^-2


B) force = mass x acceleration

= (3.05 x 105) x 700

= 320.25 x 700

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3 years ago
Very far from earth (at R- oo), a spacecraft has run out of fuel and its kinetic energy is zero. If only the gravitational force
Margaret [11]

Answer:

Speed of the spacecraft right before the collision: \displaystyle \sqrt{\frac{2\, G\cdot M_\text{e}}{R\text{e}}}.

Assumption: the earth is exactly spherical with a uniform density.

Explanation:

This question could be solved using the conservation of energy.

The mechanical energy of this spacecraft is the sum of:

  • the kinetic energy of this spacecraft, and
  • the (gravitational) potential energy of this spacecraft.

Let m denote the mass of this spacecraft. At a distance of R from the center of the earth (with mass M_\text{e}), the gravitational potential energy (\mathrm{GPE}) of this spacecraft would be:

\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R}.

Initially, R (the denominator of this fraction) is infinitely large. Therefore, the initial value of \mathrm{GPE} will be infinitely close to zero.

On the other hand, the question states that the initial kinetic energy (\rm KE) of this spacecraft is also zero. Therefore, the initial mechanical energy of this spacecraft would be zero.

Right before the collision, the spacecraft would be very close to the surface of the earth. The distance R between the spacecraft and the center of the earth would be approximately equal to R_\text{e}, the radius of the earth.

The \mathrm{GPE} of the spacecraft at that moment would be:

\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}.

Subtract this value from zero to find the loss in the \rm GPE of this spacecraft:

\begin{aligned}\text{GPE change} &= \text{Initial GPE} - \text{Final GPE} \\ &= 0 - \left(-\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\right) = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \end{aligned}

Assume that gravitational pull is the only force on the spacecraft. The size of the loss in the \rm GPE of this spacecraft would be equal to the size of the gain in its \rm KE.

Therefore, right before collision, the \rm KE of this spacecraft would be:

\begin{aligned}& \text{Initial KE} + \text{KE change} \\ &= \text{Initial KE} + (-\text{GPE change}) \\ &= 0 + \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \\ &= \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\end{aligned}.

On the other hand, let v denote the speed of this spacecraft. The following equation that relates v\! and m to \rm KE:

\displaystyle \text{KE} = \frac{1}{2}\, m \cdot v^2.

Rearrange this equation to find an equation for v:

\displaystyle v = \sqrt{\frac{2\, \text{KE}}{m}}.

It is already found that right before the collision, \displaystyle \text{KE} = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}. Make use of this equation to find v at that moment:

\begin{aligned}v &= \sqrt{\frac{2\, \text{KE}}{m}} \\ &= \sqrt{\frac{2\, G\cdot M_\text{e} \cdot m}{R_\text{e}\cdot m}} = \sqrt{\frac{2\, G\cdot M_\text{e}}{R_\text{e}}}\end{aligned}.

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Answer:

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