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sattari [20]
3 years ago
15

BRAINLIEST WILL BE MARKED FOR THE FIRST ANSWER

Physics
1 answer:
slamgirl [31]3 years ago
4 0

Answer:

The weight of the body in the new planet is 100 newtons.

Explanation:

From Newton's Law of Gravitation we find that gravitational force is directly proportional to mass of the planet and inversely proportional to the square of its radius. From this fact we can build the following relationship:

\frac{F_{1}\cdot R_{1}^{2}}{M_{1}} = \frac{F_{2}\cdot R_{2}^{2}}{M_{2}} (1)

Where:

F_{1}, F_{2} - Gravitational force, measured in newtons.

M_{1}, M_{2} - Mass of planet, measured in kilograms.

R_{1}, R_{2} - Radius of the planet, measured in meters.

If we know that F_{1} = 800\,N, \frac{M_{2}}{M_{1}} = \frac{1}{2} and \frac{R_{2}}{R_{1}} = 2, then the expected gravitational force in the new planet is:

F_{2} = F_{1}\cdot \left(\frac{M_{2}}{M_{1}} \right)\cdot \left(\frac{R_{1}} {R_{2}}\right)^{2}

F_{2} = (800\,N)\cdot \left(\frac{1}{2} \right)\cdot \left(\frac{1}{2}\right)^{2}

F_{2} = 100\,N

The weight of the body in the new planet is 100 newtons.

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Find the net force and acceleration. 15 points. Will give brainliest!
gladu [14]

Answer:

\boxed{F_{net} = 28.7 \ N}

\boxed{a = 2.1 \ m/s^2}

Explanation:

<u><em>Finding the net force:</em></u>

<u><em>Firstly , we'll find force of Friction:</em></u>

F_{k} = (micro)_{k}mg

Where (micro)_{k} is the coefficient of friction and m = 13.6 kg

F_{k} = (0.16)(13.6)(9.8)\\

F_{k} = 21.32 \ N

<u><em>Now, Finding the net force:</em></u>

F_{net} = F - F_{k}\\F_{net} = 50 - 21.32\\

F_{net} = 28.7 \ N

<u><em>Finding Acceleration:</em></u>

a = \frac{F_{net}}{m}

a = \frac{28.7}{13.6}

a = 2.1 \ m/s^2

8 0
4 years ago
Imagine whirling a ball on a string over you head. Suppose the knot holding the ball comes loose and the ball is instantly relea
zaharov [31]

B is the answer to the question

8 0
3 years ago
Consider the following reaction proceeding at 298.15 K: Cu(s)+2Ag+(aq,0.15 M)⟶Cu2+(aq, 1.14 M)+2Ag(s) If the standard reduction
lutik1710 [3]

Answer : The cell potential for this cell 0.434 V

Solution :

The balanced cell reaction will be,  

Cu(s)+2Ag^{+}(aq)\rightarrow Cu^{2+}(aq)+2Ag(s)

Here copper (Cu) undergoes oxidation by loss of electrons, thus act as anode. silver (Ag) undergoes reduction by gain of electrons and thus act as cathode.

First we have to calculate the standard electrode potential of the cell.

E^o_{[Cu^{2+}/Cu]}=0.34V

E^o_{[Ag^{+}/Ag]}=0.80V

E^o=E^o_{[Ag^{+}/Ag]}-E^o_{[Cu^{2+}/Cu]}

E^o=0.80V-(0.34V)=0.46V

Now we have to calculate the concentration of cell potential for this cell.

Using Nernest equation :

E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Cu^{2+}][Ag]^2}{[Cu][Ag^+]^2}

where,

n = number of electrons in oxidation-reduction reaction = 2

E_{cell} = ?

Now put all the given values in the above equation, we get:

E_{cell}=0.46-\frac{0.0592}{2}\log \frac{(1.14)\times (1)^2}{(1)\times (0.15)}

E_{cell}=0.434V

Therefore, the cell potential for this cell 0.434 V

8 0
4 years ago
When is the pressure of the man on the ground more – lying position or standing
Wittaler [7]

Hi,

<u>The man on the ground in standing position has more pressure</u>. This is because when he stands, only his legs are in contact with the ground. While lying, his body is more in contact with the ground, therefore, he exerts less pressure.

To the point, a man standing position on the ground had more pressure.

More is the area of contact, less is the pressure efforted.

Thank you...

7 0
3 years ago
A small mailbag is released from a helicopter that is descending steadily at 3 m/s.
mario62 [17]

<u>Answer:</u>

a) Speed of mailbag after 3 seconds = 32.4 m/s

b) Package is 44.1 meter below helicopter

c) If the helicopter was rising steadily at 3.00 m/s

       Speed of mailbag after 3 seconds = 26.4 m/s

       Package is 44.1 meter below helicopter

<u>Explanation:</u>

a)  We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

   Initial velocity = 3 m/s, acceleration = 9.8 m/s^2 and time = 3 seconds.

   v = 3+9.8*3 = 32.4 m/s

  Speed of mailbag after 3 seconds = 32.4 m/s

b) We have equation of motion , s= ut+\frac{1}{2} at^2, s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

 Velocity of helicopter = 3 m/s, time taken = 3 seconds, acceleration = 0 m/s^2.

    s= 3*3+\frac{1}{2} *0*3^2\\ \\ s=9m

    Distance traveled by helicopter = 9 meter.

 Velocity of package = 3 m/s, time taken = 3 seconds, acceleration = 9.8 m/s^2.

  s= 3*3+\frac{1}{2} *9.8*3^2\\ \\ s= 53.1m

  Distance traveled by package  = 53.1 meter.

So package is (53.1-9)meter below helicopter = 44.1 m

c) Initial velocity = -3 m/s, acceleration = 9.8 m/s^2 and time = 3 seconds.

  v = -3+9.8*3 = 26.4 m/s

  Speed of mailbag after 3 seconds = 26.4 m/s

 Velocity of helicopter = -3 m/s, time taken = 3 seconds, acceleration = 0 m/s^2.

    s= -3*3+\frac{1}{2} *0*3^2\\ \\ s=-9m

    Distance traveled by helicopter = 9 meter.

 Velocity of package = -3 m/s, time taken = 3 seconds, acceleration = 9.8 m/s^2.

  s= -3*3+\frac{1}{2} *9.8*3^2\\ \\ s= 35.1m

  Distance traveled by package  = 35.1 meter.

So package is (35.1+9)meter below helicopter = 44.1 m

4 0
3 years ago
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