ANSWER QUICK 30 POINTS
1 answer:
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Force is the influence that is capable of bringing about motion
The correct options are;
1. C. The pull of the engine, friction force
2. A. Weight, push of the tracks against the wheel
3. D. The astronaut weighs the most on Earth and least on the Moon
4. Please see attached drawing for example
5. Please see attached drawing for example
6. The data points will fall along a line this that as the force increases, the acceleration increases
Reason:
1. The horizontal forces are the forces acting along the horizontal x-axis, therefore;
A train engine pulling a boxcar what horizontal forces are;
C. The pull of the engine, friction force
2. The vertical forces are the forces acting an the direction of the y-axis of a graph
The vertical forces present when a train pulls a boxcar are;
A. Weight, push of the tracks against the wheel
3. The weight of an astronaut, W = mass × Gravity
- Therefore, the weight increases with increasing acceleration due to gravity
Therefore, the least weight is on the moon, where acceleration due to gravity is 1.6 m/s², and most on Earth where the acceleration due to gravity is approximately 9.81 m/s²
- The correct option is D. The astronaut weighs the most on Earth and least on the Moon
4. The magnitude of the downward force is the sum of the vertical forces acting downwards
5. The net force is given by the vector sum of the forces
6. According to Newton's second law, Force, F = Mass, m × Acceleration, a
- Therefore, where the students conduct the experiments rightly, we have that the data points will fall along a line this that as the force increases, the acceleration increases
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Answer:
Explanation:
First of all we shall calculate electric field near charge 2Q .
electric field due to charge Q = K x Q / (5 x 10⁻² )²
E₁ = KQ / 25 x 10⁻⁴ = KQ x 10⁴ / 25 . It is acting along positive x axis
E₁ = KQ x 10⁴ i / 25
Similarly electric field due to charge 3Q near 2Q
= 3KQ x 10⁴ i / 25 . It is acting along y-axis
E₂ = 3KQ x 10⁴ j / 25
Similarly electric field due to charge 4Q near 2Q
= 4KQ x 10⁴ j / (25 x 2 )
= 2 KQ x 10⁴ / 25 . It is acting acting along north east direction
unit vector in north east direction = ( i + j )/ √2
So E₃ can be represented by
E₃ = 2 KQ x 10⁴ ( i + j ) / 25 x √2
Total field = KQ x 10⁴ i / 25 + 3KQ x 10⁴ j / 25 + 2 KQ x 10⁴ ( i + j ) / ( 25 x √2 )
= KQ x 10⁴ [ i + 3 j + √2 i + √2 j ) / 25
= 400 KQ ( 2.414 i + 4.414 j ) N / C
Force on 2Q = Field x charge = 400 KQ ( 2.414 i + 4.414 j ) x 2Q N
= 800 KQ² ( 2.414 i + 4.414 j ) N
= 800 x 9 x 10⁹ x ( 2.5 x 10⁻⁶ )² x 2.414 x ( i + 2 j ) N
= 108.63 ( i + 2 j ) N .
Magnitude of this force
= 108.63 x √5
= 243 N approx .
