You get 493.41 when
you subtract 15.54 from 508.953 using the rules for significant figures.
For addition
and subtraction, look at the decimal portion (i.e., to the right of the decimal
point) of the numbers ONLY. Here is what to do:
1) Count the
number of significant figures in the decimal portion of each number in the
problem. (The digits to the left of the decimal place are not used to determine
the number of decimal places in the final answer.)
2) Add or
subtract in the normal fashion.
3) Round the
answer to the LEAST number of places in the decimal portion of any number in
the problem.
The correct answer between all the choices given is the
third choice or letter C. I am hoping that this answer has satisfied your query
and it will be able to help you in your endeavor, and if you would like, feel
free to ask another question.
The magnitude of the pressure : P = 56,250 N/m
<h3>Further explanation</h3>
Given
Area of 0.004 m²
Force 225 N
Required
the magnitude of the pressure
Solution
Pressure = force exerted per unit of surface area(perpendicular)
Can be formulated :
P = F/A
Input the value :
P = 225 N / 0.004 m
P = 56,250 N/m
The law applied here is Hooke's Law which describes the force exerted by the spring with a given distance. The equation for this is F = kΔx, where F is the force in Newtons, k is the spring constant in N/m while Δx is the displacement in meters.
If you want to find work done by a spring, this can be solved by using differential equations. However, derived equations are already ready for use. The equation is
W = k[{x₂-x₁)² - (x₁-xn)²],
where
xn is the natural length
x₁ is the stretched length
x₂ is also the stretched length when stretched even further than x₁
In this case xn =x₁. So, that means that (x₁-xn) = 0 and (x₂-x₁) = 11 cm or 0.11 m.
Then, substituting the values,
2 J = k (0.11² -0²)
k = 165.29 N/m
Finally, we use the value of k to the Hooke's Law to determine the Force.
F = kΔx = (165.29 N/m)(0.11 m)
F = 18.18 Newtons
Answer:
Explanation:
a ) The motion is one dimensional , so motion is along x - axis , starting from origin ( 0 , 0 )
b ) Initial velocity is 18.5 m /s when boat is situated at origin . When he displaces by 250 m along x axis and his position is ( 250 , 0 ) along x axis , his velocity becomes 36 m /s . Both his velocity and acceleration is along x - axis.
c ) Initial velocity vi = 18.5 m /s
final velocity vf = 36 m/s
Displacement x = 250 m
Acceleration a = ?
Most appropriate formula is given below .
vf² = vi² + 2 a x
2ax = vf² - vi²
x = ( vf² - vi² ) / 2 a
d )
Putting the given values
36² - 18.5² / 2 x 250
= 1296 - 342.25 / 500
= 1.9 m /s².
f ) Time interval t = ?
Required formula
vf = vi + at
t = (vf - vi ) / a
Putting the values
t = (30 - 18.5) / 1.9
= 6.05 second .
Answer:
Explanation:
In the whole process , electric potential energy is converted into kinetic energy .
Kinetic energy = 3.83 MeV
= 3.83 x 1.6 x 10⁻¹⁶ J
= 6.128 x 10⁻¹⁶ J .
Let the closest distance of approach be r .
Electric potential energy = k Q q / r , Q is charge on nucleus , q is charge on alpha particle , r is closest distance .
Electric potential energy = 9 x 10⁹ x 79 x 1.6 x 10⁻¹⁹ x 2 x 1.6 x 10⁻¹⁹ / r
= 3640.32 x 10⁻²⁹ / r
So,
6.128 x 10⁻¹⁶= 3640.32 x 10⁻²⁹ / r
r = 3640.32 x 10⁻²⁹ / 6.128 x 10⁻¹⁶
= 594.05 x 10⁻¹³
= 59.405 x 10⁻¹²
= 59.405 pm .
