Answer:
to a warm front. Remember to include all data collected on warm fron … ... Remember to include all data collected on warm fronts in this activity to support your answer (examples: interaction of air masses, air pressure, cloud cover, temperature behind/ahead of front, wind direction, precipitation, etc. 1
Explanation:
Answer:
Explanation:
When we shoot the dart upwards the time taken by the dart to go straight up and again come back is given as
here we can say
put t = 4.6 s then we have
Now in order to find the maximum range we can say
so in order to have maximum range we can say
On ground weight of plane will be measured as its actual weight which will be given as
now when plane is in air its weight is measured as combined effect of earth gravitational force and buoyancy force both
so weight in air will be given as
here since net effect due to air is opposite to the direction of weight so in air the plane weight will be measure less than its weight on ground.
so answer will be
A)more than a plane in the air
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)
The solutions and homogeneous mixtures in the given list are:
A. Lead solder, an alloy of tin and lead.
D. Window cleaner, a mixture of ammonia and coloring dissolved in water.
E. Gasoline, a mixture of organic liquids with a fixed composition throughout.
<h3>What is a solution?</h3>
A solution can be defined as a special type of homogeneous mixture that comprises a solute and a solvent.
A homogeneous mixture can be defined as any solid, liquid, or gaseous mixture which has an identical (uniform) composition and properties throughout any given sample of the mixture.
In Science (Physics), all solutions are considered to be a homogeneous mixture because their constituents are uniformly (evenly) distributed.
In conclusion, the solutions and homogeneous mixtures in the given list are:
- Lead solder, an alloy of tin and lead.
- Window cleaner, a mixture of ammonia and coloring dissolved in water.
- Gasoline, a mixture of organic liquids with a fixed composition throughout.
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