Answer:
Explanation:
In order to answer this question, we simply have to refer to the laws of the equations of gravitational mechanics.
The equation given by Newton tells us that
In the case where we compare a specific place where the Force of Gravity is greater or lesser, we focus on the term assigned to the Planet's Radius.
In the case of 
, we understand that they are constant.
We can easily notice that the more the Radius (Height seen from a viewer on the ground), the lower the force will be.
In other words, the smaller the radius in which the measurement is made with respect to the center of the earth, the greater the gravitational force.
In that order of ideas the smallest radio has South Pole, which is about 6356 km from the center of the Earth on the Equator line
Answer:
A. The time taken for the car to stop is 3.14 secs
B. The initial velocity is 81.64 ft/s
Explanation:
Data obtained from the question include:
Acceleration (a) = 26ft/s2
Distance (s) = 256ft
Final velocity (V) = 0
Time (t) =?
Initial velocity (U) =?
A. Determination of the time taken for the car to stop.
Let us obtain an express for time (t)
Acceleration (a) = Velocity (V)/time(t)
a = V/t
Velocity (V) = distance (s) /time (t)
V = s/t
a = s/t^2
Cross multiply
a x t^2 = s
Divide both side by a
t^2 = s/a
Take the square root of both side
t = √(s/a)
Now we can obtain the time as follow
Acceleration (a) = 26ft/s2
Distance (s) = 256ft
Time (t) =..?
t = √(s/a)
t = √(256/26)
t = 3.14 secs
Therefore, the time taken for the car to stop is 3.14 secs
B. Determination of the initial speed of the car.
V = U + at
Final velocity (V) = 0
Deceleration (a) = –26ft/s2
Time (t) = 3.14 sec
Initial velocity (U) =.?
0 = U – 26x3.14
0 = U – 81.64
Collect like terms
U = 81.64 ft/s
Therefore, the initial velocity is 81.64 ft/s
First of all, we know that one mole is equal to the atomic number of an element.
The atomic number of gold is <span>197.0g Au
And we need to find 5 moles.
5 * 197.0 g Au = </span><span>985.0g
Grams is used to measure mass.
Answer: </span>985.0g
To solve this problem it is necessary to apply the concepts related to wavelength depending on the frequency and speed. Mathematically, the wavelength can be expressed as
Where,
v = Velocity
f = Frequency,
Our values are given as
L = 3.6m
v= 192m/s
f= 320Hz
Replacing we have that
The total number of 'wavelengths' that will be in the string will be subject to the total length over the size of each of these undulations, that is,
Therefore the number of wavelengths of the wave fit on the string is 6.
