If the earth's mass were half its actual value but its radius stayed the same, the escape velocity of the earth would be
.
<h3>What is an escape velocity?</h3>
The ratio of the object's travel distance over a specific period of time is known as its velocity. As a vector quantity, the velocity requires both the magnitude and the direction. the slowest possible speed at which a body can break out of the gravitational pull of a certain planet or another object.
The formula to calculate the escape velocity of earth is given below:-

Given that earth's mass was half its actual value but its radius stayed the same. The escape velocity will be calculated as below:-

.
Therefore, If the earth's mass were half its actual value but its radius stayed the same, the escape velocity of the earth would be
.
To know more about escape velocity follow
brainly.com/question/14042253
#SPJ4
Work = force * distance.
We must produce twice as much energy as we are lifting the weight twice as high.
But we are not increasing the force so we must increase the length of the ramp ( distance ) instead.
The new length will be twice as great as the previous length.
So 8 metres is required.
25 kg * 8 m = work = 100 kg * 2 m
Answer:
2.45 J
Explanation:
The following data were obtained from the question:
Mass (m) = 0.5 kg
Height (h) = 1 m
Kinetic energy (KE) =?
Next, we shall determine the velocity of the rock after it has fallen half way. This can be obtained as follow:
Initial velocity (u) = 0 m/s
Acceleration due to gravity (g) = 9.8 m/s²
Height (h) = 1/2 = 0.5 m
Final velocity (v) =?
v² = u² + 2gh
v² = 0² + (2 × 9.8 × 0.5)
v² = 9.8
Take the square root of both side
v = √9.8
v = 3.13 m/s
Finally, we shall determine the kinetic energy of the rock after it has fallen half way. This can be obtained as follow:
Mass (m) = 0.5 kg
Velocity (v) = 3.13 m/s
Kinetic energy (KE) =?
KE = ½mv²
KE = ½ × 0.5 × 3.13²
KE = 0.25 × 9.8
KE = 2.45 J
Therefore, the kinetic energy of the rock after it has fallen half way is 2.45 J
Answer:
![B_T=2.0*10^-5[-\hat{i}+\hat{j}]T](https://tex.z-dn.net/?f=B_T%3D2.0%2A10%5E-5%5B-%5Chat%7Bi%7D%2B%5Chat%7Bj%7D%5DT)
Explanation:
To find the magnitude of the magnetic field, you use the following formula for the calculation of the magnetic field generated by a current in a wire:

μo: magnetic permeability of vacuum = 4π*10^-7 T/A
I: current = 6.0 A
r: distance to the wire in which magnetic field is measured
In this case, you have four wires at corners of a square of length 9.0cm = 0.09m
You calculate the magnetic field in one corner. Then, you have to sum the contribution of all magnetic field generated by the other three wires, in the other corners. Furthermore, you have to take into account the direction of such magnetic fields. The direction of the magnetic field is given by the right-hand side rule.
If you assume that the magnetic field is measured in the up-right corner of the square, the wire to the left generates a magnetic field (in the corner in which you measure B) with direction upward (+ j), the wire down (down-right) generates a magnetic field with direction to the left (- i) and the third wire generates a magnetic field with a direction that is 45° over the horizontal in the left direction (you can notice that in the image attached below). The total magnetic field will be:
![B_T=B_1+B_2+B_3\\\\B_{T}=\frac{\mu_o I_1}{2\pi r_1}\hat{j}-\frac{\mu_o I_2}{2\pi r_2}\hat{i}+\frac{\mu_o I_3}{2\pi r_3}[-cos45\hat{i}+sin45\hat{j}]](https://tex.z-dn.net/?f=B_T%3DB_1%2BB_2%2BB_3%5C%5C%5C%5CB_%7BT%7D%3D%5Cfrac%7B%5Cmu_o%20I_1%7D%7B2%5Cpi%20r_1%7D%5Chat%7Bj%7D-%5Cfrac%7B%5Cmu_o%20I_2%7D%7B2%5Cpi%20r_2%7D%5Chat%7Bi%7D%2B%5Cfrac%7B%5Cmu_o%20I_3%7D%7B2%5Cpi%20r_3%7D%5B-cos45%5Chat%7Bi%7D%2Bsin45%5Chat%7Bj%7D%5D)
I1 = I2 = I3 = 6.0A
r1 = 0.09m
r2 = 0.09m

Then you have:
![B_T=\frac{\mu_o I}{2\pi}[(-\frac{1}{r_2}-\frac{cos45}{r_3})\hat{i}+(\frac{1}{r_1}+\frac{sin45}{r_3})\hat{j}}]\\\\B_T=\frac{(4\pi*10^{-7}T/A)(6.0A)}{2\pi}[(-\frac{1}{0.09m}-\frac{cos45}{0.127m})\hat{i}+(\frac{1}{0.09m}+\frac{sin45}{0.127m})]\\\\B_T=\frac{(4\pi*10^{-7}T/A)(6.0A)}{2\pi}[-16.67\hat{i}+16.67\hat{j}]\\\\B_T=2.0*10^-5[-\hat{i}+\hat{j}]T](https://tex.z-dn.net/?f=B_T%3D%5Cfrac%7B%5Cmu_o%20I%7D%7B2%5Cpi%7D%5B%28-%5Cfrac%7B1%7D%7Br_2%7D-%5Cfrac%7Bcos45%7D%7Br_3%7D%29%5Chat%7Bi%7D%2B%28%5Cfrac%7B1%7D%7Br_1%7D%2B%5Cfrac%7Bsin45%7D%7Br_3%7D%29%5Chat%7Bj%7D%7D%5D%5C%5C%5C%5CB_T%3D%5Cfrac%7B%284%5Cpi%2A10%5E%7B-7%7DT%2FA%29%286.0A%29%7D%7B2%5Cpi%7D%5B%28-%5Cfrac%7B1%7D%7B0.09m%7D-%5Cfrac%7Bcos45%7D%7B0.127m%7D%29%5Chat%7Bi%7D%2B%28%5Cfrac%7B1%7D%7B0.09m%7D%2B%5Cfrac%7Bsin45%7D%7B0.127m%7D%29%5D%5C%5C%5C%5CB_T%3D%5Cfrac%7B%284%5Cpi%2A10%5E%7B-7%7DT%2FA%29%286.0A%29%7D%7B2%5Cpi%7D%5B-16.67%5Chat%7Bi%7D%2B16.67%5Chat%7Bj%7D%5D%5C%5C%5C%5CB_T%3D2.0%2A10%5E-5%5B-%5Chat%7Bi%7D%2B%5Chat%7Bj%7D%5DT)