Answer:34 cm
Explanation:
Given
mass of meter stick m=80 gm
stick is balanced when support is placed at 51 cm mark
Let us take 5 gm tack is placed at x cm on meter stick so that balancing occurs at x=50 cm mark
balancing torque
Let me give you the procedure like this:
Lets say that F is the fraction of the rope hanging over the table
If its like that then we have to take into account that the <span>friction force keeping on table is given by the following formula:</span>
<span>Ff = u*(1-f)*m*g </span>
and we need to know aso that <span>gravity force pulling off the table Fg is given by this other formula:</span>
<span>Fg = f*m*g </span>
What you need to do is <span>Equate the two and solve for f: </span>
<span>f*m*g = u*(1-f)*m*g </span>
<span>=> f = u*(1-f) = u - uf </span>
<span>=> f + uf = u </span>
=> f = u/(1+u) = fraction of rope
With that you can find the answer
Answer:
The final temperature of the object will be 42.785 °C
Explanation:
When the heat added or removed from a substance causes a change in temperature in it, this heat is called sensible heat.
In other words, sensible heat is the amount of heat that a body absorbs or releases without any changes in its physical state (phase change), so that the temperature varies.
The equation for calculating the heat exchanges in this case is:
Q = c * m * ΔT
where Q is the heat exchanged by a body of mass m, constituted by a substance of specific heat c and where ΔT is the variation in temperature.
In this case:
- Q= 450 J
- c= 2.89
- m= 20 g
- ΔT= Tfinal - Tinitial= Tfinal - 35 °C
Replacing:
450 J= 2.89 *20 g* (Tfinal - 35°C)
Solving for Tfinal:
7.785 °C=Tfinal - 35°C
7.785 °C + 35°C= Tfinal
42.785 °C=Tfinal
<u><em>The final temperature of the object will be 42.785 °C</em></u>
Answer:
D) 42.87 m/s
Explanation:
First, find the time it takes him to land. Given in the y direction:
Δy = 60 m
v₀ = 0 m/s
a = 9.8 m/s²
Find: t
Δy = v₀ t + ½ at²
60 m = (0 m/s) t + ½ (9.8 m/s²) t²
t = 3.5 s
Next, find the speed needed to travel the horizontal distance in that time. Given in the x direction:
Δx = 60 m
a = 0 m/s²
t = 3.5 s
Find: v₀
Δy = v₀ t + ½ at²
150 m = v₀ (3.5 s) + ½ (0 m/s²) (3.5 s)²
v₀ = 42.87 m/s
Answer:
1 greater distances fallen in successive seconds
Explanation:
When a body falls freely it is subjected to the action of the force of gravity, which gives an acceleration of 9.8 m / s2, consequently, we are in an accelerated movement
If we use the kinematic formula we can find the position of the body
Y = Vo t + ½ to t2
Where the initial velocity is zero or constant and the acceleration is the acceleration of gravity
Y = - ½ g t2 = - ½ 9.8 t2 = -4.9 t2
Let's look for the position for successive times
t (s) Y (m)
1 -4.9
2 -19.6
3 -43.2
The sign indicates that the positive sense is up
It can be clearly seen that the distance is greatly increased every second that passes
