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melomori [17]
3 years ago
11

Give an example to explain that motion is relative in nature

Physics
1 answer:
Maurinko [17]3 years ago
3 0
Consider a long train moving at speed v. Now consider a passenger throwing a ball inside this train, towards the back of the train, with same velocity v (but in the opposite direction of the train movement). 

- A passenger inside the train will see the ball moving with speed v
- For an observer outside the train, however, the ball will appear as still. In fact, for him the ball will have a speed v (given by the movement of the train) -v (velocity of the ball but moving in the opposite direction), so the net velocity will be v+(-v)=0. 
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A baseball m=.34kg is spun vertically on a massless string of length l=.52m. the string can only support a tension of tmax=9.9n
larisa86 [58]
<span>4.5 m/s This is an exercise in centripetal force. The formula is F = mv^2/r where m = mass v = velocity r = radius Now to add a little extra twist to the fun, we're swinging in a vertical plane so gravity comes into effect. At the bottom of the swing, the force experienced is the F above plus the acceleration due to gravity, and at the top of the swing, the force experienced is the F above minus the acceleration due to gravity. I will assume you're capable of changing the velocity of the ball quickly so you don't break the string at the bottom of the loop. Let's determine the force we get from gravity. 0.34 kg * 9.8 m/s^2 = 3.332 kg m/s^2 = 3.332 N Since we're getting some help from gravity, the force that will break the string is 9.9 N + 3.332 N = 13.232 N Plug known values into formula. F = mv^2/r 13.232 kg m/s^2 = 0.34 kg V^2 / 0.52 m 6.88064 kg m^2/s^2 = 0.34 kg V^2 20.23717647 m^2/s^2 = V^2 4.498574938 m/s = V Rounding to 2 significant figures gives 4.5 m/s The actual obtainable velocity is likely to be much lower. You may handle 13.232 N at the top of the swing where gravity is helping to keep you from breaking the string, but at the bottom of the swing, you can only handle 6.568 N where gravity is working against you, making the string easier to break.</span>
7 0
3 years ago
Read 2 more answers
I need a quick way to answer a question on graphs in physics grade12
DIA [1.3K]
For you to answer a question on graphs, you have to first, identify the variables and coefficients given in the problem. Then, assess the Problem what is required given the <span>variables and coefficients. Lastly, develop a solution that would answer the required variables in the problem.</span>
6 0
2 years ago
As a train accelerates uniformly, it passes successive 5-kilometer markers while traveling at velocities 10 m/s and 25 m/s. what
gregori [183]
<span>122.0 km/hr. First let’s make sure all of our units are in the base meter form: i.e. convert 5km to 5000m. (We will convert back to km later). The first thing to do is look at the equation relating velocity, acceleration, and distance: Vf^2 = Vi^2 + 2*a*d, where Vf is final velocity, Vi is initial velocity, a is acceleration, and d is distance. 25^2 = 10^2 + 2*a*5000 =?> 625 = 100 +10000a => a= 0.0525m/s^2. Now that we have acceleration, we can use the same equation again with different numbers.: Vf^2 = Vi^2 + 2*a*d = 25^2 + 2*0. 0525m*5000 = 625 + 525 =1150 => Vf^2 = 1150 => 33.9m/s. Convert to km/hour: 33.9m/s * 1km/1000m *60s/1min * 60min/ 1 hr = 122.0 km/hr.</span>
8 0
3 years ago
Describe in your own words what is meant by the statement that the Sun is in hydrostatic equilibrium.
Jet001 [13]

Answer and Explanation:

Hydrostatic equilibrium is the condition in which force is balance that is upward force and downward force the downward force is due to gravitational force and the upward force is due to the pressure. The Sun is said to be in hydrostatic equilibrium means the force acting on it is balance means upward force which is due to pressure is same as the force exerted by gravitation.

8 0
3 years ago
Two particles of equal mass m are at the vertices of the base of an equilateral triangle. The triangle’s center of mass is midwa
Helga [31]

Answer:Twice of given mass

Explanation:

Given

Two Particles of Equal mass placed at the base of an equilateral Triangle

let mass of two equal masses be m and third mass be m'

Taking one of the masses at origin

Therefore co-ordinates of first mass be (0,0)

Co-ordinates of other equal mass is (a,0)

if a is the length of triangle

co-ordinates of final mass (\frac{a}{2},\frac{\sqrt{3}a}{2})

Given its center of mass is at midway between base and third vertex therefore

x_{cm},y_{cm}=\frac{a}{2},\frac{\sqrt{3}a}{4}

y_{cm}=\frac{m_1y_1+m_2y_2+m_3y_3}{m_1+m_2+m_3}

\frac{\sqrt{3}a}{4}=\frac{m\cdot 0+m\cdot 0+m'\cdot \frac{\sqrt{3}a}{2}}{m+m+m'}

2m+m'=4\times (\frac{m'}{2})

2m+m'=2m'

m'=2m

8 0
3 years ago
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