Answer:
<em>F= 99 N</em>
Explanation:
<u>Dynamics and Kinematics</u>
This problem is a combination of dynamics and kinematics because we need formulas and concepts of both branches of physics to solve it.
We need to calculate the force required to launch horizontally a pumpkin of m=1 Kg a distance of d=10 m away from a height of h=5 m.
Since the force is:
F = m.a
We need to calculate the acceleration required to move the pumpkin from rest (vo=0) to the launching speed (vf) in a time t=0.1 seconds.
The acceleration can be calculated by using the kinematic equation:
![\displaystyle a=\frac{v_f-v_o}{t}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20a%3D%5Cfrac%7Bv_f-v_o%7D%7Bt%7D)
The final launching speed vf can be calculated knowing the height and maximum horizontal distance reached by the pumpkin.
When an object is thrown horizontally with a speed vf from a height h, the range or maximum horizontal distance traveled by the object can be calculated as follows:
![\displaystyle d=v_f\cdot\sqrt{\frac {2h}{g}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20d%3Dv_f%5Ccdot%5Csqrt%7B%5Cfrac%20%20%7B2h%7D%7Bg%7D%7D)
Solving for vf:
![\displaystyle v_f=d\cdot\sqrt{\frac {g}{2h}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20v_f%3Dd%5Ccdot%5Csqrt%7B%5Cfrac%20%20%7Bg%7D%7B2h%7D%7D)
Substituting:
![\displaystyle v_f=10\cdot\sqrt{\frac {9.8}{2\cdot 5}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20v_f%3D10%5Ccdot%5Csqrt%7B%5Cfrac%20%20%7B9.8%7D%7B2%5Ccdot%205%7D%7D)
Calculating:
![v_f=9.9\ m/s](https://tex.z-dn.net/?f=v_f%3D9.9%5C%20m%2Fs)
Now we calculate the acceleration:
![\displaystyle a=\frac{9.9-0}{0.1}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20a%3D%5Cfrac%7B9.9-0%7D%7B0.1%7D)
![a= 99\ m/s^2](https://tex.z-dn.net/?f=a%3D%2099%5C%20m%2Fs%5E2)
Thus, the force required is:
![F=1\ Kg\cdot 99\ m/s^2](https://tex.z-dn.net/?f=F%3D1%5C%20Kg%5Ccdot%2099%5C%20m%2Fs%5E2)
F= 99 N
Explanation:
In di electric medium
The force is
So the force on air is decreased k times
Option C is correct
Answer:
The speed of the block when it reaches the bottom is 17 m/s.
(b) is correct option.
Explanation:
Given that,
Height = 15 m
Mass = 2 kg
We need to calculate the speed of the block when it reaches the bottom
Using conservation of energy
![mgh=\dfrac{1}{2}mv^2](https://tex.z-dn.net/?f=mgh%3D%5Cdfrac%7B1%7D%7B2%7Dmv%5E2)
![v^2=2gh](https://tex.z-dn.net/?f=v%5E2%3D2gh)
![v=\sqrt{2gh}](https://tex.z-dn.net/?f=v%3D%5Csqrt%7B2gh%7D)
Where, g = acceleration due to gravity
h = height
Put the value into the formula
![v=\sqrt{2\times9.8\times15}](https://tex.z-dn.net/?f=v%3D%5Csqrt%7B2%5Ctimes9.8%5Ctimes15%7D)
![v=17\ m/s](https://tex.z-dn.net/?f=v%3D17%5C%20m%2Fs)
Hence, The speed of the block when it reaches the bottom is 17 m/s.
Because many objects in space don't radiate any optical (visible) radiation at all.
And other objects, like stars, radiate a lot of invisible radiation in addition to the
visible light from them. So the ability to detect and measure invisible radiation
makes it possible to learn a lot more about objects in space than we could if
we could only use their visible light.
Answer:
μk = 0.26885
Explanation:
Conceptual analysis
We apply Newton's second law:
∑Fx = m*a (Formula 1)
∑F : algebraic sum of the forces in Newton (N)
m : mass in kilograms (kg)
a : acceleration in meters over second square (m/s²)
Data:
a= -0.9 m/s²,
g = 9.81 m/s² : acceleration due to gravity
W= 75 N : Block weight
W= m*g
m = W/g = 75/9.8= 7.65 kg : Block mass
Friction force : Ff
Ff= μk*N
μk: coefficient of kinetic friction
N : Normal force (N)
Problem development
We apply the formula (1)
∑Fy = m*ay , ay=0
N-W-25 = 0
N = 75
+25
N= 100N
∑Fx = m*ax
20-Ff= m*ax
20-μk*100
= 7.65*(-0.90 )
20+7.65*(0.90) = μk*100
μk = ( 20+7.65*(0.90)) / (100)
μk = 0.26885