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2 years ago

How much would you weigh on an imaginary planet that has no gravitational force?

Physics

1 answer:

GarryVolchara [31]2 years ago

5 0

0kg

If the gravitational pull is zero and I multiply by mass I get a zero

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Given the diagram below determine the net force acting on the object

crimeas [40]

A) 0

because all of the forces cancel out, so it is not moving with balanced forces.

7 0

2 years ago

Read 2 more answers

Suppose I have a vector that is 7 units long and that makes an angle of +30 degrees from the positive x-axis. I want to add to t

Vinil7 [7]

Answer:

sum of these two vectors is 6.06i+3.5j-3.5i+6.06j = 2.56i+9.56j

Explanation:

We have given first vector which has length of 7 units and makes an angle of 30° with positive x-axis

So x component of the vector $=7cos30^{\circ}=7\times 0.866=6.06$

y component of the vector $=7sin30^{\circ}=7\times 0.5=3.5$

So vector will be 6.06i+3.5j

Now other vector of length of 7 units and makes an angle of 120° with positive x-axis

So x component of vector $=7cos120^{\circ}=7\times -0.5=-3.5i$

y component of the vector $=7sin120^{\circ}=7\times 0.866=6.06j$

Now sum of these two vectors is 6.06i+3.5j-3.5i+6.06j = 2.56i+9.56j

5 0

3 years ago

Can anyone help ????

user100 [1]

Explanation:

1. serie

2.Red

3.16 ohms

I=V/R

6=V/16

V=16×6

V=96 volts

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3 years ago

The figure shows a graph of electric potential versus position along the x-axis. A proton is originally at point A, moving along

Lostsunrise [7]

Can we see the figure?

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2 years ago

Two stationary positive point charges, charge 1 of magnitude 3.90 nC and charge 2 of magnitude 1.80 nC, are separated by a dista

soldi70 [24.7K]

Answer:

$v = 7793150 m/s$

Explanation:

First, we are going to calculate the electrical potential in the point middle between the two charges

Remember that the electrical potential can be calculated as:

$v = \frac{kQ}{r}$

Where $k = 8.9874 x 10^{9} \frac{Nm^{2} }{C^{2} }$

and it is satisfy the superposition principle, thus

$v = \frac{8.9874x10^{9}(3.90x10^{-9} ) }{0.23} + \frac{8.9874x10^{9}(1.80x10^{-9} ) }{0.23}$

$v = 222.73v$

The electrical potential at 10 cm from charge 1 is:

$v = \frac{8.9874x10^{9}(3.90x10^{-9} ) }{0.1} + \frac{8.9874x10^{9}(1.80x10^{-9} ) }{0.36}$

$v = 395.44 v$

Since the work - energy theorem, we have:

$q\Delta v = \frac{mv^{2} }{2}$

where q is the electron's charge and m is the electron's mass

Therefore:

$v = \sqrt{\frac{2q\Delta v}{m} }$

$v = 7793150 m/s$

6 0

2 years ago

How much would you weigh on an imaginary planet that has no gravitational force?​

How much would you weigh on an imaginary planet that has no gravitational force?