Gravity<span> is measured by the acceleration that it gives to freely falling objects. At Earth's surface the acceleration of </span>gravity<span> is about 9.8 metres (32 feet) per second per second.</span>
Answer:
$900
Explanation:
Step 1: Our output value is 9000.
Step 2: We represent the unknown value with x.
Step 3: From step 1 above,$9000=100\%$
Step 4: Similarly, x=10%
Step 5: This results in a pair of simple equations:
$9000=100
Step 6: By dividing equation 1 by equation 2 and noting that both the RHS (right hand side) of both
equations have the same unit (%); we have
\frac{9000}{x}=\frac{100\%}{10\%}
Step 7: Again, the reciprocal of both sides gives
\frac{x}{9000}=\frac{10}{100}$
\Rightarrow x=900$
Therefore, $10\%$ of $9000$ is $900$
Answer:
"2Ω" is the net resistance in the circuit.
Explanation:
The given resistors are:
R1 = 3Ω
R2 = 6Ω
The net resistance will be:
⇒
On substituting the values, we get
⇒
On taking L.C.M, we get
⇒
⇒
⇒
On applying cross-multiplication, we get
⇒
This question is incomplete, the complete question is;
A 8.45μC particle with a mass of 6.15 x 10⁻⁵ kg moves perpendicular to a 0.493-T magnetic field in a circular path of radius 34.1 m.
How much time will it take for the particle to complete one orbit?
a. 92.7 s
b. 0.0927 s
c. 9.27 s
d. 927 s
Answer:
it will take 92.7 seconds for the particle to complete one orbit.
Option a) 92.7 s is the correct option
Explanation:
Given that;
mass m = 6.15 x 10⁻⁵ kg
q = 8.45μC = 8.45 × 10⁻⁶ C
B = 0.493
we know that
Time period T = 2πr / V
where r = mv/qB
so T = 2πm/qB
we substitute
T = (2 × 3.14 × 6.15 x 10⁻⁵) / ( 8.45 × 10⁻⁶ × 0.493)
T = 0.0003862 / 0.000004165
T = 92.7 sec
Therefore it will take 92.7 seconds for the particle to complete one orbit.
Option a) 92.7 s is the correct option