To place the poles of a 1. 5 v battery to achieve the same electric field is 1.5×10−2 m
The potential difference is related to the electric field by:
∆V=Ed
where,
∆V is the potential difference
E is the electric field
d is the distance
what is potential difference?
The difference in potential between two points that represents the work involved or the energy released in the transfer of a unit quantity of electricity from one point to the other.
We want to know the distance the detectors have to be placed in order to achieve an electric field of
E=1v/cm=100v/cm
when connected to a battery with potential difference
∆v=1.5v
Solving the equation,we find
learn more about potential difference from here: brainly.com/question/28166044
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Based on Newton's principle, whenever objects A and B interact with each other, they exert forces upon each other.
When a horse pulls on a cart, t<span>he horse exerts a force only to the cart. But that force applies only to the cart, not to the horse.
The cart in turn exerts a force on the horse. But that force applies only to the horse, not the cart also.
</span>
There are two forces resulting from this interaction - a force on the horse and a force on the cart. T<span>he net force on the cart remains as it was --- a positive force in the direction of the horse's movement. Therefore, the cart begins to accelerate and move.</span><span>
</span>
Answer:
Explanation:
a) Energy stored in spring = 1/2 k x² = .5 x k 0.1²
500 = 5 x 10⁻³ k ,
k = (500/5) x 10³ = 10⁵ N/m
b )
k = 4.5 x 10¹ = 45 N/m
Stored energy = 1/2 k x² = .5 x 45 x 8² x 10⁻⁴ =1440 x 10⁻⁴ J
This energy gets dissipated by friction .
work done by friction = μ mg d
d is the distance traveled under friction
so 1440 x 10⁻⁴ = μ x 3 x 9.8 x 2
μ = 245 x 10⁻⁴ or 0.00245 which appears to be very small. .
Answer:
A) 1568.60 Hz
B) 1437.15 Hz
Explanation:
This change is frequency happens due to doppler effect
The Doppler effect is the change in frequency of a wave in relation to an observer who is moving relative to the wave source
where
C = the propagation speed of waves in the medium;
Vr= is the speed of the receiver relative to the medium,(added to C, if the receiver is moving towards the source, subtracted if the receiver is moving away from the source;
Vs= the speed of the source relative to the medium, added to C, if the source is moving away from the receiver, subtracted if the source is moving towards the receiver.
A) Here the Source is moving towards the receiver(C-Vs)
and the receiver is standing still (Vr=0) therefore the observed frequency should get higher
B)Here the Source is moving away the receiver(C+Vs)
and the receiver is still not moving (Vr=0) therefore the observed frequency should be lesser
Answer:
The answer to your question is: D.
Explanation:
Distance refers to the amount of space between two points, it is a scalar quantity.
Displacement refers to the space between two points, measure from the minimum path linking them, it is a vector quantity.
I'm not agree with these answers, because the total distance is approximately 500km.
A) The distance traveled is 300 km. This answer is not correct.
B) Distance is 300 km and displacement is 0 km. This answer is not correct because the displacement is also 500 km.
C) Distance is 300 km/hour and displacement is 300 km.
300 km/h is a measure of speed not of distance, this option is wrong.
D) Both distance traveled and displacement are 300 km. I think this is the correct answer because distance and displacement measure the same. but I think both measure 500 km.
