Answer:
option B. Limestone
<em>h</em><em>o</em><em>p</em><em>e</em><em>f</em><em>u</em><em>l</em><em>l</em><em>y</em><em>,</em>
<em>Z</em><em>a</em><em>r</em><em>a</em><em>♡</em>
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:
the impulse experienced by this object is 4 Ns
Explanation:
Given;
mass of the object, m = 3 kg
velocity of the object, v = 4 m/s
resistive force, F = 20 N
duration of impact, t = 0.2 s
The impulse experienced by this object is calculated as follows;
J = F x t
J = 20 x 0.2
J = 4 Ns
Therefore, the impulse experienced by this object is 4 Ns
The weight of an object is given by:
where
F is the weight
m is the mass of the object
is the gravitational acceleration
The piece of ham in this problem has a mass of
m=630 g = 0.63 kg
therefore its weight is
The phase angle ф for the source voltage relative to the current the and Frequency is 192.14 H.
Phase angle b/w VL and inductor is 90°
Angular frequency, w = 2 pi f
VL = I XL
Frequency, f = VL/2 pi L I = 49/(2 pi x 4.25 x 0.00955) = 192.14 Hz.
The formula = 2/T yields the angular frequency. In radians per second, the angular frequency is expressed. The frequency, f = 1/T. the reciprocal period. The number of full oscillations per unit of time is given by the motion's frequency, f = 1/T = /2. Angular frequency and velocity are related to displacement and revolution velocity, respectively. full response Angular frequency is a scalar measure of rotation rate, also known as angular speed, radial frequency, circular frequency, and orbital frequency.
Learn more about Phase angle here:
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