The net force = sum of all forces acting on the body
If we take left side as -ve and right side as +ve,
then,
The net force here would be equal to,
10N + (- 3N)
= 7N.
Therefore, a net force of +7N ( + indicates it's moving towards right) is acting on the book of mass 2kg.
<span>3598 seconds
The orbital period of a satellite is
u=GM
p = sqrt((4*pi/u)*a^3)
Where
p = period
u = standard gravitational parameter which is GM (gravitational constant multiplied by planet mass). This is a much better figure to use than GM because we know u to a higher level of precision than we know either G or M. After all, we can calculate it from observations of satellites. To illustrate the difference, we know GM for Mars to within 7 significant figures. However, we only know G to within 4 digits.
a = semi-major axis of orbit.
Since we haven't been given u, but instead have been given the much more inferior value of M, let's calculate u from the gravitational constant and M. So
u = 6.674x10^-11 m^3/(kg s^2) * 6.485x10^23 kg = 4.3281x10^13 m^3/s^2
The semi-major axis of the orbit is the altitude of the satellite plus the radius of the planet. So
150000 m + 3.396x10^6 m = 3.546x10^6 m
Substitute the known values into the equation for the period. So
p = sqrt((4 * pi / u) * a^3)
p = sqrt((4 * 3.14159 / 4.3281x10^13 m^3/s^2) * (3.546x10^6 m)^3)
p = sqrt((12.56636 / 4.3281x10^13 m^3/s^2) * 4.458782x10^19 m^3)
p = sqrt(2.9034357x10^-13 s^2/m^3 * 4.458782x10^19 m^3)
p = sqrt(1.2945785x10^7 s^2)
p = 3598.025212 s
Rounding to 4 significant figures, gives us 3598 seconds.</span>
Mass of the object m = 2.9 kg
Force F1 = 28.449 N
F1 = m1 x a => a = F / m => 28.449 / 2.9 => a = 9.81, which is gravitational acceleration.
In the same lab, a = g = 9.81, second object F2 = 48.7N = m2 x a
m2 = F2 / a => 48.7 / 9.81 => m2 = 4.96 kg
Mass of the second object m2 = 4.96 kg
55 meters. If she started at 10 meters and ran 45 more, 10+45=55.
Answer:
For the complete question provided in explanation, if the elevator moves upward, then the apparent weight will be 1035 N. While for downward motion the apparent weight will be 435 N.
Explanation:
The question is incomplete. The complete question contains a velocity graph provided in the attachment. This is the velocity graph for an elevator having a passenger of 75 kg.
From the slope of graph it is clear that acceleration at t = 1 sec is given as:
Acceleration = a = (4-0)m/s / (1-0)s = 4 m/s^2
Now, there are two cases:
1- Elevator moving up
2- Elevator moving down
For upward motion:
Apparent Weight = m(g + a)
Apparent Weight = (75 kg)(9.8 + 4)m/s^2
<u>Apparent Weight = 1035 N</u>
For downward motion:
Apparent Weight = m(g - a)
Apparent Weight = (75 kg)(9.8 - 4)m/s^2
<u>Apparent Weight = 435 N</u>
