Answer:
The frictional force needed to overcome the cart is 4.83N
Explanation:
The frictional force can be obtained using the following formula:
where 
is the coefficient of friction = 0.02
R = Normal reaction of the load = 
= 
= 
Now that we have the necessary parameters that we can place into the equation, we can now go ahead and make our substitutions, to get the value of F.
F = 4.83 N
Hence, the frictional force needed to overcome the cart is 4.83N
The answer is gravity. I hope this helps.
Answer:
215955.06 m/s^2
Explanation:
length of barrel, s = 0.89 m
initial velocity of the bullet, u = 0 m/s
Final velocity of the bullet, v = 620 m/s
Let a be the acceleration of the bullet in the barrel
Use third equation of motion, we get
a = 215955.06 m/s^2
Thus, the acceleration of the bullet inside the barrel is 215955.06 m/s^2.
Answer:
1 Ampere.
Explanation:
From the question given above, the following data were obtained:
Resistor 1 (R₁) = 20 ohm
Resistor (R₂) = 20 ohm
Voltage (V) = 10 V
Current (I) =?
Next, we shall determine the equivalent resistance in the circuit. This can be obtained as follow:
Resistor 1 (R₁) = 20 ohm
Resistor (R₂) = 20 ohm
Equivalent Resistance (R) =?
Since the resistors are in parallel connection, the equivalent resistance can be obtained as follow:
R = (R₁ × R₂) / (R₁ + R₂)
R = (20 × 20) / (20 + 20)
R = 400 / 40
R = 10 ohm
Finally, we shall determine the total current in the circuit. This can be obtained as illustrated below:
Voltage (V) = 10 V
Equivalent Resistance (R) = 10 ohm
Current (I) =?
V = IR
10 = I × 10
Divide both side by 10
I = 10 / 10
I = 1 Ampere
Therefore, the total current in the circuit is 1 Ampere.
Answer:
H = 1/2 g t^2 where t is time to fall a height H
H = 1/8 g T^2 where T is total time in air (2 t = T)
R = V T cos θ horizontal range
3/4 g T^2 = V T cos θ 6 H = R given in problem
cos θ = 3 g T / (4 V) (I)
Now t = V sin θ / g time for projectile to fall from max height
T = 2 V sin θ / g
T / V = 2 sin θ / g
cos θ = 3 g / 4 (T / V) from (I)
cos θ = 3 g / 4 * 2 sin V / g = 6 / 4 sin θ
tan θ = 2/3
θ = 33.7 deg
As a check- let V = 100 m/s
Vx = 100 cos 33.7 = 83,2
Vy = 100 sin 33,7 = 55.5
T = 2 * 55.5 / 9.8 = 11.3 sec
H = 1/2 * 9.8 * (11.3 / 2)^2 = 156
R = 83.2 * 11.3 = 932
R / H = 932 / 156 = 5.97 6 within rounding
