Answer:
the friction force in the reverse direction is 200 *0.4=80 N.
the net forward force acting on the box is therefore
Fnet= 100 - 80 N
= 20 N
acceleration = Fnet / mass
=Fnet *g/(weight)
=20 *9.8/200 = 0.98 m/s^2
Explanation:
Because of the build up of pressure. There is so much steam coming from such a compressed point, it’s coming out in force.
Now think of that same spot being closed, it only has one place to go but it can’t leave, so that pressure will build and build and then BOOM, it explodes.
In short, the answer is the pressure being released from a small point, and how that energy is released.
Answer:
B. 2 m/s
B. Acceleration = 4.05 m/s² and Tension = 297.5 N.
Explanation:
A force is applied on a mass m whose acceleration is 4 m/s
Force = mass × acceleration
a = F/m = 4 m/s
4 m/s = F/m
F = 4 m/s (m)
If Force of 2F is applied on a mass of 4m ; it acceleration is as follows:
2F/4 m = F/ 2m
4m/s (m) / 2m = 2 m/s
a = 2 m/s
2.
Given that
mass 
= 30 kg
mass 
= 50 kg
= 0.1
From the question; we can arrive at two cases;
That :
----- equation (1)
---- equation (2)
50 a = 50 g - T
30 a = T - 30 g sin 30 - 4 × 30 g cos 30
By summation
80 a =![[ 50 - 30 * \frac{1}{2} - 0.1 *30* \frac{\sqrt{3}}{2}]](https://tex.z-dn.net/?f=%5B%2050%20%20-%2030%20%2A%20%5Cfrac%7B1%7D%7B2%7D%20-%200.1%20%2A30%2A%20%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B2%7D%5D)
g
80 a = 32. 4 × 10 m/s ² (using g as 10m/s²)
80 a = 324 m/s ²
a = 324/80
a = 4.05 m/s²
From equation , replace a with 4.05
50 × 4.05 = 50 × 10 - T
T = 500 -202.5
T =297.5 N
Answer:
a) Batteries and fuel cells are examples of galvanic cell
b) Ag-cathode and Zn-anode
c) Cell notation: Zn(s)|Zn²⁺(aq) || Ag⁺(aq)|Ag(s)
Explanation:
a) A galvanic cell is an electrochemical cell in which chemical energy is converted to electrical energy. The chemical reaction which drives a galvanic cell is a redox reaction i.e. a reduction-oxidation process.
A typical galvanic cell is composed of two electrodes immersed in a suitable electrolyte and connected via a salt bridge. One of the electrodes serves as a cathode where reduction or gain of electrons takes place. The other half cell functions as an anode where oxidation or loss of electrons occurs. Batteries and fuel cells are examples of galvanic cells.
b) The nature of the electrode that will serve as an anode or cathode depends on the value of the standard reduction potential (E⁰) of that electrode. The electrode with a higher or more positive the value of E⁰ serves as the cathode and the other will function as an anode.
In the given case, the E⁰ values from the standard reduction potential table are:
E⁰(Zn/Zn2+) = -0.763 V
E°(Ag/Ag+)=+0.799 V
Therefore, Ag will be the cathode and Zn will be the anode
c) In the standard cell notation, the anode half cell is written on the left followed by the salt bridge '||' and finally the cathode half cell to the right.
Zn(s)|Zn²⁺(aq) || Ag⁺(aq)|Ag(s)
