Answer:
2.9 A
Explanation:
L = 16 cm = 0.16 m
B = 0.19 T
m = 9 g = 0.009 kg
Let the minimum current be i.
Magnetic force is balanced by the gravitational force
B x i x L = m x g
0.19 x i x 0.16 = 0.009 x 9.8
i = 2.9 A
Answer:
f(t) = 28,7 [N]
Explanation: IMPORTANT NOTE: IN PROBLEM STATEMENT CHARGES ARE IN C (COULOMBS) AND IN THE DIAGRAM IN μC. WE ASSUME CHARGES ARE IN μC.
The net force on +q₂ is the sum of the force of +q₁ on +q₂ ( is a repulsion force since charges of equal sign repel each other ) and the force of -q₃ on +q₂ ( is an attraction force, opposite sign charges attract each other)
The two forces have the same direction to the right of charge q₂, we have to add them
Then
f(t) = f₁₂ + f₃₂
f₁₂ = K * ( q₁*q₂ ) / (0,1)²
q₁ = + 8 μC then q₁ = 8*10⁻⁶ C
q₂ = + 3,5 μC then q₂ = 3,5 *10⁻⁶ C
K = 9*10⁹ [ N*m² /C²]
f₁₂ = 9*10⁹ * 8*3,5*10⁻¹²/ 1*10⁻² [ N*m² /C²]* C*C/m²
f₁₂ = 252*10⁻¹ [N]
f₁₂ = 25,2 [N]
f₃₂ = 9*10⁹*3,5*10⁻⁶*2,5*10⁻⁶ /(0,15)²
f₃₂ = 78,75*10⁻³/ 2,25*10⁻²
f₃₂ = 35 *10⁻¹
f₃₂ = 3,5 [N]
f(t) = 28,7 [N]
Hi there!
A.
Since the can was launched from ground level, we know that its trajectory forms a symmetrical, parabolic shape. In other words, the time taken for the can to reach the top is the same as the time it takes to fall down.
Thus, the time to its highest point:
Now, we can determine the velocity at which the can was launched at using the following equation:
In this instance, we are going to look at the VERTICAL component of the velocity, since at the top of the trajectory, the vertical velocity = 0 m/s.
Therefore:
***vsinθ is the vertical component of the velocity.
Solve for 'v':
Now, recall that:
Plug in the expression for velocity:
B.
We can use the same process as above, where T' = 2T and Th = T.
C.
The work done in part B is 4 times greater than the work done in part A.
An interesting problem, and thanks to the precise heading you put for the question.
We will assume zero air resistance.
We further assume that the angle with vertical is t=53.13 degrees, corresponding to sin(t)=0.8, and therefore cos(t)=0.6.
Given:
angle with vertical, t = 53.13 degrees
sin(t)=0.8; cos(t)=0.6;
air-borne time, T = 20 seconds
initial height, y0 = 800 m
Assume g = -9.81 m/s^2
initial velocity, v m/s (to be determined)
Solution:
(i) Determine initial velocity, v.
initial vertical velocity, vy = vsin(t)=0.8v
Using kinematics equation,
S(T)=800+(vy)T+(1/2)aT^2 ....(1)
Where S is height measured from ground.
substitute values in (1): S(20)=800+(0.8v)T+(-9.81)T^2 =>
v=((1/2)9.81(20^2)-800)/(0.8(20))=72.625 m/s for T=20 s
(ii) maximum height attained by the bomb
Differentiate (1) with respect to T, and equate to zero to find maximum
dS/dt=(vy)+aT=0 =>
Tmax=-(vy)/a = -0.8*72.625/(-9.81)= 5.9225 s
Maximum height,
Smax
=S(5.9225)
=800+(0.8*122.625)*(5.9225)+(1/2)(-9.81)(5.9225^2)
= 972.0494 m
(iii) Horizontal distance travelled by the bomb while air-borne
Horizontal velocity = vx = vcos(t) = 0.6v = 43.575 m/s
Horizontal distace travelled, Sx = (vx)T = 43.575*20 = 871.5 m
(iv) Velocity of the bomb when it strikes ground
vertical velocity with respect to time
V(T) =vy+aT...................(2)
Substitute values, vy=58.1 m/s, a=-9.81 m/s^2
V(T) = 58.130 + (-9.81)T =>
V(20)=58.130-(9.81)(20) = -138.1 m/s (vertical velocity at strike)
vx = 43.575 m/s (horizontal at strike)
resultant velocity = sqrt(43.575^2+(-138.1)^2) = 144.812 m/s (magnitude)
in direction theta = atan(43.575,138.1)
= 17.5 degrees with the vertical, downward and forward. (direction)
Answer:
240 kPa
Explanation:
The ideal gas law states:
where
p is the gas pressure
V is the gas volume
n is the number of moles
R is the gas constant
T is the absolute temperature of the gas
For a fixed amount of gas, n and R are constant, so we can rewrite the equation as
For the gas in the problem, which undergoes a transformation, this can be rewritten as
where we have:
is the initial pressure
is the initial volume
is the initial temperature
is the final pressure
is the final volume
is the final temperature
Solving the formula for p2, we find the final pressure of the gas:
