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2 years ago

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Astronaut Neil during his lunar mission recorded his weight. He noticed a big difference in his weight compared to his weight ta

ken when he was on earth. If his weight on earth was 90kgs, what would be his weight on moon and what could be the reason for this difference?
a. His weight will be 5kg; this is because moon’s gravity is 1/10th as compared to that of the earth.
b. His weight will be 45kg; this is because moon’s gravity is 1/7th as compared to that of the earth.
c. His weight will be 5kg; this is because moon’s gravity is 1/8th as compared to that of the earth.
d. His weight will be 15kg; this is because moon’s gravity is 1/6th as compared to that of the earth.

1 answer:

Aleks [24]2 years ago

6 0

Answer:

no entiendo

Explanation:

esta en ingles

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In the most common isotope of Hydrogen the nucleus is made out of a single proton. When this Hydrogen atom is neutral, a single

FinnZ [79.3K]

Answer:

The ratio of electric force to the gravitational force is $2.27\times 10^{39}$

Explanation:

It is given that,

Distance between electron and proton, $r=4.53\ A=4.53\times 10^{-10}\ m$

Electric force is given by :

$F_e=k\dfrac{q_1q_2}{r^2}$

Gravitational force is given by :

$F_g=G\dfrac{m_1m_2}{r^2}$

Where

$m_1$ is mass of electron, $m_1=9.1\times 10^{-31}\ kg$

$m_2$ is mass of proton, $m_2=1.67\times 10^{-27}\ kg$

$q_1$ is charge on electron, $q_1=-1.6\times 10^{-19}\ kg$

$q_2$ is charge on proton, $q_2=1.6\times 10^{-19}\ kg$

$\dfrac{F_e}{F_g}=\dfrac{kq_1q_2}{Gm_1m_2}$

$\dfrac{F_e}{F_g}=\dfrac{9\times 10^9\times (1.6\times 10^{-19})^2}{6.67\times 10^{-11}\times 9.1\times 10^{-31}\times 1.67\times 10^{-27}}$

$\dfrac{F_e}{F_g}=2.27\times 10^{39}$

So, the ratio of electric force to the gravitational force is $2.27\times 10^{39}$ . Hence, this is the required solution.

3 0

3 years ago

A factory emits pollutants at a rate of 25 g/s. The factory is located between two mountain ranges resulting in an effective val

Lostsunrise [7]

Answer:

$1.25\ \mu\text{g/m}^3$

Explanation:

v = Velocity of the breeze = 4 m/s

w = Width of the valley = 5000 m

h = Height of the valley = 1000 m

Volumetric flow rate is given by

$\dot{V}=vwh\\\Rightarrow \dot{V}=4\times 5000\times 1000\\\Rightarrow \dot{V}=2\times10^{7}\ \text{m}^3/\text{s}$

$\dot{m}$ = Mass flow rate of pollutant = 25 g/s = $25\times 10^6\ \mu\text{g/s}$

Concentration is given by

$C=\dfrac{\dot{m}}{\dot{V}}\\\Rightarrow C=\dfrac{25\times 10^6}{2\times 10^7}\\\Rightarrow C=1.25\ \mu\text{g/m}^3$

The steady state concentration of pollutants in the valley, is $1.25\ \mu\text{g/m}^3$ .

6 0

3 years ago

An astronaut drops a rock from the top of a crater on the moon. When the rock is halfway down to the bottom of the crater, its s

Alexxx [7]

Answer: vf1/vf2= 1/ sqrt(2)

Explanation :on the moon no drag force so we have only the force of gravity. aceleration is g(moon)= 1.62m/s2.the rest is basic kinematics

if the rock travels H to the bottom we can calculate velocity:

vo=0m/s (drops the rock) , yo=0

vf*vf= vo*vo+2g(y-yo)

when the rock is halfway y = H/2 so:

vf1*vf1=2*g*H/2 so vf1 = sqrt(gH)

when the rock reach the bottom y=H so:

vf2*vf2=2*g*H so vf2 = sqrt(2gH)

so vf1/vf2= 1/ sqrt(2)

good luck from colombia

8 0

3 years ago

Read 2 more answers

Dr. Kirwan is preparing a slide show that he will present to the executive board at tonight's committee meeting. He places a 3.5

lisov135 [29]

Answer:

A) d_o = 20.7 cm

B) h_i = 1.014 m

Explanation:

A) To solve this, we will use the lens equation formula;

1/f = 1/d_o + 1/d_i

Where;

f is focal Length = 20 cm = 0.2

d_o is object distance

d_i is image distance = 6m

1/0.2 = 1/d_o + 1/6

1/d_o = 1/0.2 - 1/6

1/d_o = 4.8333

d_o = 1/4.8333

d_o = 0.207 m

d_o = 20.7 cm

B) to solve this, we will use the magnification equation;

M = h_i/h_o = d_i/d_o

Where;

h_o = 3.5 cm = 0.035 m

d_i = 6 m

d_o = 20.7 cm = 0.207 m

Thus;

h_i = (6/0.207) × 0.035

h_i = 1.014 m

8 0

2 years ago

A proton traveling at 17.6° with respect to the direction of a magnetic field of strength 3.28 mT experiences a magnetic force o

umka2103 [35]

Answer:

a) The proton's speed is 5.75x10⁵ m/s.

b) The kinetic energy of the proton is 1723 eV.

Explanation:

a) The proton's speed can be calculated with the Lorentz force equation:

$F = qv \times B = qvBsin(\theta)$ (1)

Where:

F: is the force = 9.14x10⁻¹⁷ N

q: is the charge of the particle (proton) = 1.602x10⁻¹⁹ C

v: is the proton's speed =?

B: is the magnetic field = 3.28 mT

θ: is the angle between the proton's speed and the magnetic field = 17.6°

By solving equation (1) for v we have:

$v = \frac{F}{qBsin(\theta)} = \frac{9.14 \cdot 10^{-17} N}{1.602\cdot 10^{-19} C*3.28 \cdot 10^{-3} T*sin(17.6)} = 5.75 \cdot 10^{5} m/s$

Hence, the proton's speed is 5.75x10⁵ m/s.

b) Its kinetic energy (K) is given by:

$K = \frac{1}{2}mv^{2}$

Where:

m: is the mass of the proton = 1.67x10⁻²⁷ kg

$K = \frac{1}{2}mv^{2} = \frac{1}{2}1.67 \cdot 10^{-27} kg*(5.75 \cdot 10^{5} m/s)^{2} = 2.76 \cdot 10^{-16} J*\frac{1 eV}{1.602 \cdot 10^{-19} J} = 1723 eV$

Therefore, the kinetic energy of the proton is 1723 eV.

I hope it helps you!

3 0

3 years ago