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frozen [14]
4 years ago
13

Two objects exert a gravitational force on 8 N on one another. What would that force be if the mass of BOTH objects were doubled

?
Physics
1 answer:
seropon [69]4 years ago
4 0
<span>Based on Newton's law of universal gravitation, the equation for the gravitational force exerted by an object on another object is given by:
F = Gm1m2/(r^2)
where G is the universal gravitational constant, F is the gravitational force exerted, m1 is the mass of the first object, m2 is the mass of the second object, and r is the separation distance between the two objects.
If the mass of both objects were doubled, then we would have: m1' * m2' = (2m1) * (2m2) = 4m1m2. Assuming r stays constant (G is a constant so that won't change anyway), then this means that the new force will be 4 times greater, ie 8N * 4 = 32N of gravitational force. </span>
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Why are we integrating the differential x-component of the e field of this charged ring?
iragen [17]
Please give us the picture and tell us what quantity you're asked to find.
6 0
3 years ago
Changes in behavior of managers, coworkers, and subordinates can be a warning sign before a layoff. True or false
Yakvenalex [24]

Answer:

The correct answer is - true.

Explanation:

Changes in the higher authorities or managers or coworkers can be a warning sign before a layoff, however, a company or organization needs to give an official notice before a particular time period before the layoff.

The layoff is a temporary suspension or permanent termination from the employment of an employee due to reasons related to business such as reducing manpower or staff and less work in the organization. These reasons can lead to a change in the behavior of the managers, coworkers, and subordinates.

3 0
3 years ago
A fighter plane flying at constant speed 420 m/s and constant altitude 3300 m makes a turn of curvature radius 11000 m. On the g
Arada [10]

Answer:

"Apparent weight during the "plan's turn" is  519.4 N

Explanation:

The "plane’s altitude" is not so important, but the fact that it is constant tells us that the plane moves in a "horizontal plane" and its "normal acceleration" is \mathrm{a}_{\mathrm{n}}=\frac{v^{2}}{R}

Given that,

v = 420 m/s

R = 11000 m

Substitute the values in the above equation,

a_{n}=\frac{420^{2}}{11000}

a_{n}=\frac{176400}{11000}

a_{n}=16.03 \mathrm{m} / \mathrm{s}^{2}

It has a horizontal direction. Furthermore, constant speed implies zero tangential acceleration, hence vector a = vector a N. The "apparent weight" of the pilot adds his "true weight" "m" "vector" "g" and the "inertial force""-m" vector a due to plane’s acceleration, vectorW_{\mathrm{app}}=m(\text { vector } g \text { -vector a })

In magnitude,

| \text { vector } g-\text { vector } a |=\sqrt{\left(g^{2}+a^{2}\right)}

| \text { vector } \mathrm{g}-\text { vector } \mathrm{a} |=\sqrt{\left(9.8^{2}+16.03^{2}\right)}

| \text { vector } \mathrm{g}-\text { vector } \mathrm{a} |=\sqrt{(96.04+256.96)}

| \text { vector } \mathrm{g}-\text { vector } \mathrm{a} |=\sqrt{353}

| \text { vector } \mathrm{g}-\text { vector } \mathrm{a} |=18.78 \mathrm{m} / \mathrm{s}^{2}

Because vector “a” is horizontal while vector g is vertical. Consequently, the pilot’s apparent weight is vector

\mathrm{W}_{\mathrm{app}}=(18.78 \mathrm{m} / \mathrm{s}^ 2)(53 \mathrm{kg})=995.77 \mathrm{N}

Which is quite heavier than his/her true weigh of 519.4 N

7 0
4 years ago
Read 2 more answers
Red light of wavelength 633 nm from a helium-neon laser passes through a slit 0.400mm wide. The diffraction pattern is observed
kogti [31]

Answer:

a)y_{first}=5.3mm

b)y_{second}=10.6-5.3 =5.3 mm  

Explanation:

a)

The width of the central bright in this diffraction pattern is given by:

y=\frac{m\lambda D}{a} when m is a natural number.

here:

  • m is 1 (to find the central bright fringe)                
  • D is the distance from the slit to the screen
  • a is the slit wide
  • λ is the wavelength

So we have:

y_{first}=\frac{633*10^{9}*3.35}{0.0004}

y_{first}=5.3mm

b)

Now, if we do m=2 we can find the distance to the second minima.

y_{2}=\frac{2*633*10^{9}*3.35}{0.0004}

y_{2}=10.6 mm

Now we need to subtract these distance, to get the width of the first bright fringe :

y_{second}=10.6-5.3 =5.3 mm    

I hope it heps you!

     

4 0
3 years ago
How do contact and non-contact forces affect the design of different modes of transportation?
padilas [110]

Contact and non contact forces are big factors to consider in designing of different modes of transportation because this factors are resistances for the mode of transportation. These contact and non- contact forces should be minimized in order for the energy requirement to be also minimum. But not the extent of risking the safety, for example a non contact force is the wieght, the should be optimum, safety of the design should not be compromised just to reduce weight, just like by removing essential parts, support just to remove weight is not good. 

7 0
4 years ago
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