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anzhelika [568]
3 years ago
7

the mass of one water drop is 0.0008kg and the gravitational field strength is 10N/kg what is its weight

Physics
1 answer:
djyliett [7]3 years ago
8 0
Weight = (mass) x (gravity)

Weight = (8 x 10⁻⁴ kg) x (10 N/kg) = 0.008 Newton
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Jenny was applying her makeup when she drove into the student parking lot last Friday morning . Unaware that Cheryl was stopped
Akimi4 [234]

Answer: F = 102141N

Explanation: <em><u>Newton's 2nd Law</u></em> states that a force can change the motion of a body. The relation is given by

F = m.a

whose units are:

[F] = N

[m] = kg

[a] = m/s²

Jenny's car, at the moment of the break, had acceleration:

a=\frac{\Delta v}{\Delta t}

a=\frac{11}{0.14}

a = 78.57 m/s²

Then, Force is

F = 1300*78.57

F = 102141 N

<u>Jenny's car experienced a force of </u><u>magnitude 102141N.</u>

6 0
3 years ago
An astronaut lands on a new, recently discovered planet in a different star system. The astronaut measures the acceleration due
Nezavi [6.7K]

Answer:

The radius of the new planet is ~2.04 * 10⁶ m, or 2,041,752 m.

Explanation:

We can use Newton's Law of Universal Gravitation:

  • \displaystyle F_g=G\frac{Mm}{r^2}

Let's look at Newton's 2nd Law:

  • F=ma

We can set these equations equal to each other:

  • \displaystyle G\frac{Mm}{r^2} =ma

The mass of the second mass (astronaut) cancels out. We are left with:

  • \displaystyle G\frac{M}{r^2} =a

We are solving for the radius of the new planet, so we can rearrange the equation:

  • \displaystyle r=\sqrt{\frac{GM}{a} }

Substitute in our known values given in the problem (<u><em>G = 6.67 * 10⁻¹¹ </em></u><em> ; </em><u><em>M = 7.5 * 10²³</em></u><em> ; </em><u><em>a = 12</em></u>).

  • \displaystyle r =\sqrt{\frac{(6.67\times 10^{-11})(7.5 \times 10^{23}}{12} }
  • r=2.04 \times 10^6

The radius of the new planet is ~2.04 * 10⁶ m.

3 0
2 years ago
A photon of wavelength 192 nm strikes an aluminum surface along a line perpendicular to the surface and releases a photoelectron
alex41 [277]

Answer:

KE=3.529\times10^{−27}\ J

Explanation:

Given that

Wavelength λ=192 nm

So energy of photon,E

E=\dfrac{hC}{\lambda }

Now by putting the values

h=6.6\times 10^{-34}\ m^2.kg/s

C=3\times 10^{8}\ m/s

E=\dfrac{6.6\times 10^{-34}\times 3\times 10^{8}}{192\times 10^{-9} }

E=1.03\times 10^{-18} J

We know that

Kinetic energy given as

KE=\dfrac{P^2}{2m}

KE=\dfrac{E^2}{2mC^2}

KE=\dfrac{(1.03\times 10^{-18})^2}{2\times 1.67\times 10^{-27}(3\times 10^8)^2}

KE=3.529\times10^{−27}\ J

5 0
3 years ago
Why electric forces are assential to forming compunds?
lidiya [134]
Ions were once atoms with the same number of electrons and protons. Since they have opposite charges atoms are neutral. When they become ions the lose or gain electrons and become unbalanced.These different charges are attracted to each other via electric forces<span>.</span>
8 0
3 years ago
Read 2 more answers
A ball is dropped from rest from the top of a building of height h. At the same instant, a second ball is projected vertically u
uranmaximum [27]

Answer:

a) t = \sqrt{\frac{h}{2g}}

b) Ball 1 has a greater speed than ball 2 when they are passing.

c) The height is the same for both balls = 3h/4.

Explanation:

a) We can find the time when the two balls meet by equating the distances as follows:

y = y_{0_{1}} + v_{0_{1}}t - \frac{1}{2}gt^{2}  

Where:

y_{0_{1}}: is the initial height = h

v_{0_{1}}: is the initial speed of ball 1 = 0 (it is dropped from rest)

y = h - \frac{1}{2}gt^{2}     (1)

Now, for ball 2 we have:

y = y_{0_{2}} + v_{0_{2}}t - \frac{1}{2}gt^{2}    

Where:

y_{0_{2}}: is the initial height of ball 2 = 0

y = v_{0_{2}}t - \frac{1}{2}gt^{2}    (2)

By equating equation (1) and (2) we have:

h - \frac{1}{2}gt^{2} = v_{0_{2}}t - \frac{1}{2}gt^{2}

t=\frac{h}{v_{0_{2}}}

Where the initial velocity of the ball 2 is:

v_{f_{2}}^{2} = v_{0_{2}}^{2} - 2gh

Since v_{f_{2}}^{2} = 0 at the maximum height (h):

v_{0_{2}} = \sqrt{2gh}

Hence, the time when they pass each other is:

t = \frac{h}{\sqrt{2gh}} = \sqrt{\frac{h}{2g}}

b) When they are passing the speed of each one is:

For ball 1:

v_{f_{1}} = - gt = -g*\sqrt{\frac{h}{2g}} = - 0.71\sqrt{gh}

The minus sign is because ball 1 is going down.

For ball 2:

v_{f_{2}} = v_{0_{2}} - gt = \sqrt{2gh} - g*\sqrt{\frac{h}{2g}} = (\sqrt{1} - \frac{1}{\sqrt{2}})*\sqrt{gh} = 0.41\sqrt{gh}

Therefore, taking the magnitude of ball 1 we can see that it has a greater speed than ball 2 when they are passing.

c) The height of the ball is:

For ball 1:

y_{1} = h - \frac{1}{2}gt^{2} = h - \frac{1}{2}g(\sqrt{\frac{h}{2g}})^{2} = \frac{3}{4}h

For ball 2:

y_{2} = v_{0_{2}}t - \frac{1}{2}gt^{2} = \sqrt{2gh}*\sqrt{\frac{h}{2g}} - \frac{1}{2}g(\sqrt{\frac{h}{2g}})^{2} = \frac{3}{4}h

Then, when they are passing the height is the same for both = 3h/4.

I hope it helps you!                  

7 0
3 years ago
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