Answer:
24.7 ohms
Explanation:
Given in the question,
electric current = 3 A
potential difference = 74 V
Using ohm's law
<h3>V = IR</h3>
R = V/I
<em>here,</em>
<em>R = resistance of wire</em>
<em>V = potential difference</em>
<em>I = electric current</em>
<em />
plug value in the formula
R = 74/3
R = 24.7 ohms
Therefore, the resistance of the wire is, 24.7 ohms
Explanation:
W.D. = F * s = 50N * 2m = 100J.
Answer:
a) 12.8212 N
b) 12.642 N
Explanation:
Mass of bucket = m = 0.54 kg
Rate of filling with sand = 56.0 g/ sec = 0.056 kg/s
Speed of sand = 3.2 m/s
g= 9.8 m/sec2
<u>Condition (a);</u>
Mass of sand = Ms = 0.75 kg
So total mass becomes = bucket mass + sand mass = 0.54 +0.75=1.29 kg
== > total weight = 1.29 × 9.8 = 12.642 N
Now impact of sand = rate of filling × velocity = 0.056 × 3.2 = 0.1792 kg. m /sec2=0.1792 N
Scale reading is sum of impact of sand and weight force ;
i-e
scale reading = 12.642 N+0.1792 N = 12.8212 N
<u>Codition (b);</u>
bucket mass + sand mass = 0.54 +0.75=1.29 kg
==>weight = mg = 1.29 × 9.8 = 12.642 N (readily calculated above as well)
Answer:
x_{cm} = 4.644 10⁶ m
Explanation:
The center of mass is given by the equation
= 1 / 
∑ 
Where M_{total} is the total masses of the system, 
is the distance between the particles and 
is the masses of each body
Let's apply this equation to our problem
M = Me + m
M = 5.98 10²⁴ + 7.36 10²²
M = 605.36 10²² kg
Let's locate a reference system located in the center of the Earth
Let's calculate
x_{cm} = 1 / 605.36 10²² [Me 0 + 7.36 10²² 3.82 10⁸]
x_{cm} = 4.644 10⁶ m
Answer:
There are two significant figures in 2.200 x 10^7
