All categories

3 years ago

What is the difference between 3.15 m and 2.006 m with the correct number of significant figures

Physics

2 answers:

snow_lady [41]3 years ago

5 0

The differences is 1.144

tekilochka [14]3 years ago

4 0

Answer:

S = 1.144

Explanation:

As we know that when we add or subtract two or more numbers then the final result must have same number of precision as that of numbers given in the problem

here we have

S = 3.15 - 2.006

so we have to give our result with 3 digits after decimal

so we will have

$S = 1.144$

since it must have more precise digits so correct answer after subtraction must be

S = 1.144

You might be interested in

What is the number of wavelengths that pass a given point each second called?

denpristay [2]

Since waves are moving, we define frequency as the number of waves that pass a given point in a specified unit of time. The unit commonly used is Hertz which is the number of wave cycles that pass a point in one second.

4 0

3 years ago

Darius' boat sails into the harbor with a speed of 80m/s. After 20 seconds, Darius' boat has come to a stop at the dock. What is

Novosadov [1.4K]

Answer:

The boat's acceleration is 4 m/s²

Explanation:

This question seeks to test the knowledge of acceleration and how to calculate acceleration when the speed is provided. Hence, the formula for acceleration would be used here.

Acceleration (m/s²) = speed (in m/s) ÷ time (in seconds)

Acceleration = 80 m/s ÷ 20 secs

Acceleration = 4 m/s² or 4 ms⁻²

The boat's acceleration is 4 m/s²

4 0

2 years ago

Fossils found in the La Brea tar pits indicate a California climate that was A) similar to today's climate. B) similar to the pr

NikAS [45]

Answer:

i believe it is D but not 100% sure

Explanation:

3 0

3 years ago

Read 2 more answers

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:

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

6 0

3 years ago

A 42 Watt light bulb is connected to a 12 volt source of potential

abruzzese [7]

Answer:

The natural medium emanating from the Sun and other very hot sources (now recognised as electromagnetic radiation with a wavelength of 400-750 nm), within which vision is possible.

Explanation:

just the way it is

8 0

3 years ago