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satela [25.4K]
3 years ago
13

Smaller the surface area greater the gravitational force and vice–versa yes or no? justify the statement. ​

Physics
1 answer:
Sholpan [36]3 years ago
6 0

Answer:

It is true that Smaller the surface area greater the gravitational force and vice versa

Explanation:

The gravitational force depends on the distance between the two objects

F=GMm/r2

where G = gravitational constant

M = mass of one object

m = mass of second object

r is the distance between the two objects

Gravitational force is inversely proportional to the square of the distance between two objects. Hence, it can be said the Gravitational force is inversely proportional to the area and hence as the area increases gravitational force will decrease or vice versa

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Which organelle acts like the transportation or circulatory system of the cell?
REY [17]
The endoplasmic rectiulum... hope this helps!
7 0
3 years ago
Place /our live /music / important / has / in / an​
oksian1 [2.3K]

Answer:

Music has an important place in our lives.

3 0
4 years ago
A parcel of land is in the shape of an isosceles triangle. The base has a length of 425 ft.; the other sides, which are of equal
kogti [31]

Answer:

The answer to your question is 636.6 ft    

Explanation:

Data

base = 425 ft

angle = 39°

See the picture below

1.- Divide the triangle to get two right triangles.

    Now the superior angle will measure 19.5° and the opposite side will measure 212.5 ft

2.- Use the trigonometric function sine to find the hypotenuse

     sin 19.5 = 212.5/hyp

solve for hyp

    hyp = 212.5 / sin 19.5

Result

    hyp = 212.5/ 0.333

    hyp = 636.6 ft    

4 0
3 years ago
A 6.99-g bullet is moving horizontally with a velocity of +341 m/s, where the sign + indicates that it is moving to the right (s
Ratling [72]

Answer:

a). 1.218 m/s

b). R=2.8^{-3}

Explanation:

m_{bullet}=6.99g*\frac{1kg}{1000g}=6.99x10^{-3}kg

v_{bullet}=341\frac{m}{s}

Momentum of the motion the first part of the motion have a momentum that is:

P_{1}=m_{bullet}*v_{bullet}

P_{1}=6.99x10^{-3}kg*341\frac{m}{s} \\P_{1}=2.3529

The final momentum is the motion before the action so:

a).

P_{2}=m_{b1}*v_{fbullet}+(m_{b2}+m_{bullet})*v_{f}}

P_{2}=1.202 kg*0.554\frac{m}{s}+(1.523kg+6.99x10^{-3}kg)*v_{f}

P_{1}=P_{2}

2.529=0.665+(1.5299)*v_{f}\\v_{f}=\frac{1.864}{1.5299}\\v_{f}=1.218 \frac{m}{s}

b).

kinetic energy

K=\frac{1}{2}*m*(v)^{2}

Kinetic energy after

Ka=\frac{1}{2}*1.202*(0.554)^{2}+\frac{1}{2}*1.523*(1.218)^{2}\\Ka=1.142 J

Kinetic energy before

Kb=\frac{1}{2}*mb*(vf)^{2}\\Kb=\frac{1}{2}*6.99x10^{-3}kg*(341)^{2}\\Kb=406.4J

Ratio =\frac{Ka}{Kb}

R=\frac{1.14}{406.4}\\R=2.8x10^{-3}

3 0
4 years ago
A delivery truck travels 2.8 km North, 1.0 km East, and 1.6 km South. The final displacement from the origin is ___km to the ___
34kurt

Answer:

The final displacement from the origin is <u>1.6</u> km to the <u>NE</u>

Explanation:

The directions in which the delivery truck travels are;

1) 2.8 km North = 2.8·\hat j, in vector form

2) 1.0 km East = 1.0·\hat i, in vector form

3) 1.6 km South = -1.6·\hat j, in vector form

Therefore, to find the final displacement, Δx, of the delivery truck, we add the individual displacements as follows;

Final displacement, Δd = 2.8·\hat j + 1.0·\hat i +(-1.6·\hat j) = 1.2·\hat j + 1.0·\hat i

Final displacement, = 1.0·\hat i + 1.2·\hat j

Where;

Δx = The displacement in the x-direction = 1.0·\hat i

Δy = The displacement in the y-direction = 1.2·\hat j

The magnitude of the resultant displacement vector is given as follows

\left | d \right | = √((Δx)² + (Δy)²) = √(1² + 1.2²) ≈ 1.6 (To the nearest tenth)

The magnitude of the resultant displacement vector ≈ 1.6 km

The direction of the resultant vector is positive for both the east and north direction, therefore, the direction of the resultant vector = NE

Therefore, the resultant displacement of the delivery truck is approximately 1.6 km, NE from the origin.

3 0
3 years ago
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