The ball rises for v/g seconds; which equals 14.7/9.8=1.5 seconds . After this time, it’s height will be:
h(t)=g/2(1.5)²+14.7(1.5)
=-4.9 x 2.25 + 22.05
=11.025m
The ball then falls for 49+11.025=60.025m, which takes:
g/2t²=60.025
t²=12.25
t=3.5 secs
Total time: 1.5+3.5=5 seconds
It's a Newton Meter Those two are multiplied to get a joule.
Answer:
a.) The main scale reading is 10.2cm
b.) Division 7 = 0.07
c.) 10.27 cm
d.) 10.31 cm
e.) 10.24 cm
Explanation:
The figure depicts a vernier caliper readings
a.) The main scale reading is 10.2 cm
The reading before the vernier scale
b.) Division 7 = 0.07
the point where the main scale and vernier scale meet
c.) The observed readings is
10.2 + 0.07 = 10.27 cm
d.) If the instrument has a positive zero error of 4 division
correct reading = 10.27 + 0.04 = 10.31cm
e.) If the instrument has a negative zero error of 3 division
correct reading = 10.27 - 0.03 = 10.24cm
<h2>
Answer:</h2>
143μH
<h2>
Explanation:</h2>
The inductance (L) of a coil wire (e.g solenoid) is given by;
L = μ₀N²A / l --------------(i)
Where;
l = the length of the solenoid
A = cross-sectional area of the solenoid
N= number of turns of the solenoid
μ₀ = permeability of free space = 4π x 10⁻⁷ N/A²
<em>From the question;</em>
N = 183 turns
l = 2.09cm = 0.0209m
diameter, d = 9.49mm = 0.00949m
<em>But;</em>
A = π d² / 4 [Take π = 3.142 and substitute d = 0.00949m]
A = 3.142 x 0.00949² / 4
A = 7.1 x 10⁻⁵m²
<em>Substitute these values into equation (i) as follows;</em>
L = 4π x 10⁻⁷ x 183² x 7.1 x 10⁻⁵ / 0.0209 [Take π = 3.142]
L = 4(3.142) x 10⁻⁷ x 183² x 7.1 x 10⁻⁵ / 0.0209
L = 143 x 10⁻⁶ H
L = 143 μH
Therefore the inductance in microhenrys of the Tarik's solenoid is 143
Explanation:
F = MA
200 = 100 * A
A = 200/100
A = 2m/sec^2
<h3><em>hope </em><em>it </em><em>helps </em><em>you </em></h3>
