Answer:
50 N.
Explanation:
On top of a horizontal surface, the normal force acting on an object is equivalent to the force of gravity acting on the object. That is:
![\displaystyle \begin{aligned} F_N = F_g & = ma \\ & = mg\end{aligned}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cbegin%7Baligned%7D%20F_N%20%3D%20F_g%20%26%20%3D%20ma%20%5C%5C%20%26%20%3D%20mg%5Cend%7Baligned%7D)
The mass of the block is 5 kg and the given force due to gravity is 10 N/kg. Substitute and evaluate:
![\displaystyle F_N = F_g = (5\text{ kg})(10 \text{ N/kg}) = 50 \text{ N}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20F_N%20%3D%20F_g%20%3D%20%285%5Ctext%7B%20kg%7D%29%2810%20%5Ctext%7B%20N%2Fkg%7D%29%20%3D%2050%20%5Ctext%7B%20N%7D)
In conclusion, the normal force acting on the block is 50 N.
Answer:
A,B,D,E,F
Explanation:
I took the test for yall.
Answer:
![\theta=23.7^{\circ}](https://tex.z-dn.net/?f=%5Ctheta%3D23.7%5E%7B%5Ccirc%7D)
Explanation:
The ratio of pressure 2 to 1 us 5.48/1= 5.48 rounded off as 5.5.
Referring to table A.2 of modern compressible flow then ![M_{\beta_1}=2.2](https://tex.z-dn.net/?f=M_%7B%5Cbeta_1%7D%3D2.2)
Also
and making
the subject of the formula then
Making reference to
diagram then
![\theta=23.7^{\circ}](https://tex.z-dn.net/?f=%5Ctheta%3D23.7%5E%7B%5Ccirc%7D)
Answer:
274N 0.41
Explanation:
As he is sliding down in a constant speed then the force that accelerates him (weight) and the force that slows his down (friction) are equal.
then
<em>friction=mass x gravity x sin(21)</em>
Fr=78kg x 9.8m/s2 x sin(21)=274N
<em>friction= coefficient of kinetic friction x normal force of from the slope</em>
Fr= u x 78kg x 9.8m/s2 x cos(21)=274N
Fr= u x 78kg x 9.8m/s2 x cos(21)=274Nu=274/677=0.41