We commonly know refer to something 'digital' has to something electronic that can be visibly seen such as a watch, clock, camera, screen, etc. It really refers to stored energy or electricity that's not natural. But the word 'digital' in science refers to the depiction of data<span> or </span>information<span> in </span>figures<span> (such as in a </span>table<span>) in contrast to as a </span>chart<span>, </span>graph<span>, </span>drawing<span>, or other pictorial </span>form.<span>
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Answer:
Velocity: +ve, Acceleration: -ve
Explanation:
Here I've considered downward direction as positive direction.
Answer:
a) Em₀ = 42.96 104 J
, b)
= -2.49 105 J
, c) vf = 3.75 m / s
Explanation:
The mechanical energy of a body is the sum of its kinetic energy plus the potential energies it has
Em = K + U
a) Let's look for the initial mechanical energy
Em₀ = K + U
Em₀ = ½ m v2 + mg and
Em₀ = ½ 50.0 (1.20 102) 2 + 50 9.8 142
Em₀ = 36 104 + 6.96 104
Em₀ = 42.96 104 J
b) The work of the friction force is equal to the change in the mechanical energy of the body
= Em₂ -Em₀
Em₂ = K + U
Em₂ = ½ m v₂² + m g y₂
Em₂ = ½ 50 85 2 + 50 9.8 427
Em₂ = 180.625 + 2.09 105
Em₂ = 1,806 105 J
= Em₂ -Em₀
= 1,806 105 - 4,296 105
= -2.49 105 J
The negative sign indicates that the work that force and displacement have opposite directions
c) In this case the work of the friction going up is already calculated in part b and the work of the friction going down would be 1.5 that job
We have that the work of friction is equal to the change of mechanical energy
= ΔEm
= Emf - Emo
-1.5 2.49 10⁵ = ½ m vf² - 42.96 10⁴
½ m vf² = -1.5 2.49 10⁵ + 4.296 10⁵
½ 50.0 vf² = 0.561
vf = √ 0.561 25
vf = 3.75 m / s
Answer:
1.5 m
Explanation:
a = v/t
a = 2/1,5 = 4/3
x = 1/2 a t² + vt + x
x = 1/2 × 4/3 × 9/4 = 3/2 =1.5 m
Answer:
The current pass through the coil is 6.25 A
Explanation:
Given that,
Diameter = 25 cm
Magnetic field = 1.0 mT
Number of turns = 100
We need to calculate the current
Using the formula of magnetic field


Where, N = number of turns
r = radius
I = current
Put the value into the formula


Hence, The current passes through the coil is 6.25 A