An astronaut, of total mass 86.0 kg including her suit, stands on a spherical satellite of mass 360 kg, both at rest relative a
1 answer:
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Answer:
<em>J=36221 Kg.m/s</em>
Explanation:
<u>Impulse-Momentum Theorem</u>
These two magnitudes are related in the following way. Suppose an object is moving at a certain speed and changes it to
. The impulse is numerically equivalent to the change of linear momentum. Let's recall the momentum is given by
The initial and final momentums are, respectively
The change of momentum is
It is numerically equal to the Impulse J
We are given
The impulse the car experiences during that time is
J=-36221 Kg.m/s
The magnitude of J is
J=36221 Kg.m/s
This question involves the concepts of th magnetic field and current.
The magnetic field created by the current at the house is "6.75 x 10⁻⁷ T".
<h3>Magnetic Field</h3>The magnetic field created by a current carrying wire can be given by the following formula:
where,
- B = magnetic field = ?
= permeabiliy of free space =4π x 10⁻⁷
- I = current = 152 A
- r = distance = 45 m
B = 6.75 x 10⁻⁷ T
Learn more about magnetic field here:
#SPJ1
Answer:
a) the total electric potential is 2282000 V
b) the total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V
Explanation:
Given the data in the question and as illustrated in the image below;
a) Determine the total electric potential (in V) at the origin.
We know that; electric potential due to multiple charges is equal to sum of electric potentials due to individual charges
so
Electric potential at p in the diagram 1 below is;
Vp = V1 + V2
Vp = kq1/r1 + kq2/r2
we know that; Coulomb constant, k = 9 × 10⁹ C
q1 = 4.60 uC = 4.60 × 10⁻⁶ C
r1 = 1.25 cm = 0.0125 m
q2 = -2.06 uC = -2.06 × 10⁻⁶ C
location x2 = −1.80 cm; so r2 = 1.80 cm = 0.018 m
so we substitute
Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0125 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.018 )
Vp = (3312000) + ( -1030000 )
Vp = 3312000 -1030000
Vp = 2282000 V
Therefore, the total electric potential is 2282000 V
b)
the total electric potential (in V) at the point with coordinates (0, 1.50 cm).
As illustrated in the second image;
r1² = 0.015² + 0.0125²
r1 = √[ 0.015² + 0.0125² ]
r1 = √0.00038125
r1 = 0.0195
Also
r2² = 0.015² + 0.018²
r2 = √[ 0.015² + 0.018² ]
r2 = √0.000549
r2 = 0.0234
Now, Electric Potential at P in the second image below will be;
Vp = V1 + V2
Vp = kq1/r1 + kq2/r2
we substitute
Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0195 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.0234 )
Vp = 2123076.923 + ( -762962.962 )
Vp = 2123076.923 -792307.692
Vp = 1330769.23 V
Therefore, the total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V
