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3 years ago

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in a lab, a block weighing 80 n is attached to a spring scale, and both are pulled to the right on a horizontal surface, as show

n above. friction between the block and the surface is negligible. what is the acceleration of the block when the scale reads 32 n?

1 answer:

shutvik [7]3 years ago

6 0

Answer:

4.0 m/s ^2

Explanation:

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NASA has asked your team of rocket scientists about the feasibility of a new satellite launcher that will save rocket fuel. NASA

kkurt [141]

Answer:

The answer is " $q=0.0945\,C$ ".

Explanation:

Its minimum velocity energy is provided whenever the satellite(charge 4 q) becomes 15 m far below the square center generated by the electrode (charge q).

$U_i=\frac{1}{4\pi\epsilon_0} \times \frac{4\times4q^2}{\sqrt{(15)^2+(5/\sqrt2)^2}}$

It's ultimate energy capacity whenever the satellite is now in the middle of the electric squares:

$U_f=\frac{1}{4\pi\epsilon_0}\ \times \frac{4\times4q^2}{( \frac{5}{\sqrt{2}})}$

Potential energy shifts:

$= U_f -U_i \\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{\sqrt{(15)^2+( \frac{5}{\sqrt{2})^2)}}\right ) \\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ 15 +( \frac{5}{2})}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ (\frac{30+5}{2})}}\right )\\\\$

$=\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ (\frac{35}{2})}}\right )\\\\=\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{17.5}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{ 24.74- 5 }{87.5}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{ 19.74- 5 }{87.5}}\right )\\\\ =\frac{4q^2}{\pi\epsilon_0}\left ( 0.2256 }\right )\\\\= \frac{0.28 \times q^2}{ \epsilon_0}\\\\=q^2\times31.35 \times10^9\,J$

Now that's the energy necessary to lift a satellite of 100 kg to 300 km across the surface of the earth.

$=\frac{GMm}{R}-\frac{GMm}{R+h} \\\\=(6.67\times10^{-11}\times6.0\times10^{24}\times100)\left(\frac{1}{6400\times1000}-\frac{1}{6700\times1000} \right ) \\\\ =(6.67\times10^{-11}\times6.0\times10^{26})\left(\frac{1}{64\times10^{5}}-\frac{1}{67\times10^{5}} \right ) \\\\=(6.67\times6.0\times10^{15})\left(\frac{67 \times 10^{5} - 64 \times 10^{5} }{ 4,228 \times10^{5}} \right ) \\\\$

$=( 40.02\times10^{15})\left(\frac{3 \times 10^{5}}{ 4,228 \times10^{5}} \right ) \\\\ =40.02 \times10^{15} \times 0.0007 \\\\$

$\\\\ =0.02799\times10^{10}\,J \\\\= q^2\times31.35\times10^{9} \\\\ =0.02799\times10^{10} \\\\q=0.0945\,C$

This satellite is transmitted by it system at a height of 300 km and not in orbit, any other mechanism is required to bring the satellite into space.

6 0

3 years ago

how many kilograms off mercury would fill a 5litre container if the density of mercury is 13.6grams per cm3

Sidana [21]

Answer:

68kg

Explanation:

1 cm^3 is the same as 1 mL and there are 5000mL in 5L

Therefore if the density is 13.6g/mL we multiply 13.6 by 5000 to get the amount of grams required = 68000g which is 68kg

7 0

3 years ago

With a mass of 109 kg, Baby Bird is the smallest monoplane ever

OLga [1]

Total resultant velocity=5.11-3.27=1.84m/s

m_1=61.4kg
m_2=109kg
v_1=1.84m/s
v_2=?

$\\ \sf\longmapsto ∆P=P$

$\\ \sf\longmapsto m_1v_1=m_2v_2$

$\\ \sf\longmapsto v_2=\dfrac{m_1v_1}{m_2}$

$\\ \sf\longmapsto v_2=\dfrac{61.4(1.84)}{109}$

$\\ \sf\longmapsto v_2=112.976/109$

$\\ \sf\longmapsto v_2\approx 1.3m/s$

4 0

3 years ago

What are the reactants of cellular respiration? Check all that apply. carbon dioxide energy glucose oxygen water

Rufina [12.5K]

Answer:

Glucose and Oxygen

Explanation:

Cellular respiration is the process whereby cells derives energy by the use of glucose and oxygen.

Organisms that use cellular respiration to produce their energy are known as heterotophs. They derive the glucose from food materials obtained from plant sources. They use the oxygen from the environment to liberate energy from the glucose obtained from feeding on plant materials.

Cellular respiration can be simply expressed as shown below:

GLUCOSE + OXYGEN → CO₂ + H₂O + ATP

The reactants are glucose and oxygen.

The products are CO₂, water and ATP

7 0

3 years ago

Read 2 more answers

The rate (in mg carbon/m3/h) at which photosynthesis takes place for a species of phytoplankton is modeled by the function P = 9

Ivanshal [37]

Answer:

At light intensity I = 3, is P a maximum

Explanation:

Given:

$P=\frac{90I}{I^2+I+9}$

now differentiating the above equation with respect to Intensity 'I' we get

$\frac{dp}{dI}=\frac{(I^2+I+9).\frac{d(90I)}{dI}-90I.\frac{d((I^2+I+9)}{dI}}{(I^2+I+9)^2}$

or

$\frac{dp}{dI}=\frac{(I^2+I+9).90-90I.(2I+1)}{(I^2+I+9)^2}$

or

$\frac{dp}{dI}=\frac{90I^2+90I+810)-(180I^2+90I)}{(I^2+I+9)^2}$

or

$\frac{dp}{dI}=\frac{-90I^2+810)}{(I^2+I+9)^2}$

Now for the maxima $\frac{dP}{dI}=0$

thus,

$0=\frac{-90I^2+810)}{(I^2+I+9)^2}$

or

$-90I^2+810=0$

or

$I^2=\frac{810}{90}$

or

$I^2=9$

or

I = 3

thus, <u>for the value of intensity I = 3, the P is maximum</u>

at I = 3

$P=\frac{90\times3}{3^2+3+9}$

or

$P=\frac{270}{21}$

or

$P=12.85$

5 0

4 years ago