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frozen [14]
2 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]2 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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An object begins at position x = 0 and moves one-dimensionally along the x-axis witļi a velocity v
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Answer:

The answer is "between 20 s and 30 s".

Explanation:

Calculating the value of positive displacement:

\ (x_{+ve}) = \frac{1}{2} \times 15 \times  20 \\\\

          = \frac{1}{2} \times 300 \\\\=  150 \\\\

Calculating the value of negative displacement upon the time t:

(x_{-ve}) = \frac{1}{2} \times 5 \times 20- 20(t-20) \\\\

          = \frac{1}{2} \times 100- 20t+ 400 \\\\= 50- 20t+ 400 \\\\

\to X= X_{+ve} + X_{-ve} \\\\

\to  150 - 50 -20t+400 =0\\\\\to 100 -20t+400 =0 \\\\\to 500 -20t =0\\\\\to 20t =500 \\\\\to t=\frac{500}{20}\\\\\to t=\frac{50}{2}\\\\\to t= 25

That's why its value lie in "between 20 s and 30 s".

6 0
2 years ago
For a particular reaction, the change in enthalpy is 51kJmole and the activation energy is 109kJmole. Which of the following cou
Ronch [10]

Answer

given,

change in enthalpy = 51 kJ/mole

change in activation energy = 109 kJ/mole

when a reaction is catalysed change in enthalpy between the product and the reactant does not change it remain constant.

where as activation energy of the product and the reactant decreases.

example:

ΔH = 51 kJ/mole

E_a= 83 kJ/mole

here activation energy decrease whereas change in enthalpy remains same.

5 0
1 year ago
Use Newton's laws of motion to explain why it is important that baseballs and softballs each have a small acceptable range of ma
tiny-mole [99]

Explanation:

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5 0
1 year ago
When were Earth’s landmasses first recognizable as the continents we know today? 10 million years ago 135 million years ago 180
Bess [88]

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b

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i took the test

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1 year ago
Read 2 more answers
An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The el
Tresset [83]

Complete Question

An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The electric field in the wire changes with time as E(t)=0.0004t2−0.0001t+0.0004 newtons per coulomb, where time is measured in seconds.

I = 1.2 A at time 5 secs.

Find the charge Q passing through a cross-section of the conductor between time 0 seconds and time 5 seconds.

Answer:

The charge is  Q =2.094 C

Explanation:

From the question we are told that

    The diameter of the wire is  d =  0.205cm = 0.00205 \ m

     The radius of  the wire is  r =  \frac{0.00205}{2} = 0.001025  \ m

     The resistivity of aluminum is 2.75*10^{-8} \ ohm-meters.

       The electric field change is mathematically defied as

         E (t) =  0.0004t^2 - 0.0001 +0.0004

     

Generally the charge is  mathematically represented as

       Q = \int\limits^{t}_{0} {\frac{A}{\rho} E(t) } \, dt

Where A is the area which is mathematically represented as

       A =  \pi r^2 =  (3.142 * (0.001025^2)) = 3.30*10^{-6} \ m^2

 So

       \frac{A}{\rho} =  \frac{3.3 *10^{-6}}{2.75 *10^{-8}} =  120.03 \ m / \Omega

Therefore

      Q = 120 \int\limits^{t}_{0} { E(t) } \, dt

substituting values

      Q = 120 \int\limits^{t}_{0} { [ 0.0004t^2 - 0.0001t +0.0004] } \, dt

     Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] }  \left | t} \atop {0}} \right.

From the question we are told that t =  5 sec

           Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] }  \left | 5} \atop {0}} \right.

          Q = 120 [ \frac{0.0004(5)^3 }{3} - \frac{0.0001 (5)^2}{2} +0.0004(5)] }

         Q =2.094 C

     

5 0
2 years ago
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