Answer:
Maximum height, h = 1.74 meters
Explanation:
It is given that,
A potato is shot out of the cylinder. It is a case of projectile motion. The potato makes an angle of 17 degrees above the horizontal.
Initial speed with which the potato is shot out, u = 20 m/s
We have to find the maximum height of the potato. The maximum height of a projectile (h) is given by the following formula as :
Where
= angle between the projectile and the surface
g = acceleration due to gravity
h = 1.74 m
or h = 1.74 meters
Hence, this is the required solution.
The initial force of the throw overcomes gravity quite easily. Then, gravity begins to bring it back down to earth, making a curved path.
Answer:
a) v = 31.67 cm / s
, b) v = -29.36 cm / s
, c) v= 29.36 cm/s, d) x = 3.46 cm
Explanation:
The angular velocity in a simple harmonic movement is
w = √ K / m
w = √ 23.2 / 0.37
w = 7,918 rad / s
a) the expression against the movement is
x = A cos (wt + Ф)
Speed is
v = dx / dt = - A w sin (wt + Ф)
The maximum speed occurs for cos = ± 1
v = A w
v = 4.0 7,918
v = 31.67 cm / s
b) as the object is released from rest
0 = -A w sin (0+ Фi)
sin Ф = 0
Ф = 0
The equation is
x = 4.0 cos (7,918 t)
v = -4.0 7,918 sin (7,918 t)
v = - 31.67 sin (7.918t)
Let's look for the time for a displacement of x = 1.5 cm, remember that the angles must be in radians
7,918 t = cos⁻¹ 1.5 / 4.0
t = 1,186 / 7,918
t = 0.1498 s
We look for speed
v = -31.67 sin (7,918 0.1498)
v = -29.36 cm / s
c) if the object passes the equilibrium equilibrium position again at this point the velocity has the same module, but the opposite sign
v = 29.36 cm / s
d) let's look for the time for the condition v = v_max / 2
31.67 / 2 = 31.67 sin ( 7,918 t)
7.918t = sin⁻¹ 0.5
t = 0.5236 / 7.918
t = 0.06613
With this time let's look for displacement
x = 4.0 cos (7,918 0.06613)
x = 3.46 cm
First question (upper left):
1/Req = 1/12 + 1/24 = 1/8
Req = 8 ohms
Voltage is equal through different resistors, and V1 = V2 = 24 V.
Current varies through parallel resistors: I1 = V1/R1 = 24/12 = 2 A. I2 = 24/24 = 1 A.
Second question (middle left):
V1 = V2 = 6 V (parallel circuits)
I1 = 2 A, I2 = 1 A, IT = 2+1 = 3 A.
R1 = V1/I1 = 6/2 = 3 ohms, R2 = 6/1 = 6 ohms, 1/Req = 1/2 + 1/1, Req = 2/3 ohms
Third question (bottom left):
V1 = V2 = 12 V
IT = 3 A, meaning Req = V/It = 12 V/3 A = 4 ohms
1/Req = 1/R1 + 1/R2, 1/4 = 1/12 + 1/R2, R2 = 6 ohms
I1 = V/R1 = 1 A, I2 = V/R2 = 2 A
Fourth question (top right):
1/Req = 1/20 + 1/20, Req = 10 ohms
IT = 4 A, so VT = IT(Req) = 4*10 = 40 V
Parallel circuits, so V1 = V2 = VT = 40 V
Since the resistors are identical, the current is split evenly between both: I1 = I2 = IT/2 = 2 A.
Fifth question (middle right):
1/Req = 1/5 + 1/20 + 1/4, Req = 2 ohms
IT = VT/Req = 40 V/2 ohms = 20 A
V1 = V2 = V3 = 40 V
The current of 20 A will be divided proportionally according to the resistances of 5, 20, and 4, the factors will be 5/(5+20+4), 20/(5+20+4), and 4/(5+20+4), which are 5/29, 20/29, and 4/29.
I1 = 20(5/29) = 100/29 A
I2 = 20(20/29) = 400/29 A
I3 = 20(4/29) = 80/29 A
Sixth question (bottom right):
V2 = 30V is given, but since these are parallel circuits, V1 = VT = 30 V.
Then I1 = V1/R1 = 30 V/10 ohms = 3 A.
I2 = 30 V/15 ohms = 2 A.
IT = 3 + 2 = 5 A
1/Req = 1/10 + 1/15, Req = 6 ohms
Answer:
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