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Helen [10]
3 years ago
13

If you neglect air resistance, the horizontal component of a projectile A. Is always changing B. Remains constant throughout the

flight C. Increases, goes to zero, then increases D. Depends on who threw it
Physics
1 answer:
Shtirlitz [24]3 years ago
8 0

Answer:

B. Remains constant throughout the flight

Explanation:

If we completely neglect air resistance then the projectile will not have any horizontal resistances to deal with, therefore the horizontal component will remain constant throughout the flight. This would continue to be the case until it meets some form of resistance. Which assuming that everything else is normal would be the case since the force of gravity will push the projectile down (vertical component) until it hits the ground which in that instance would act as an opposing force to the horizontal component as well.

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Two objects with masses of 2.75 kg and 4.45 kg are connected by a light string that passes over a light frictionless pulley to f
Llana [10]

Answer:

Explanation:

F = ma

4.45g - 2.75g = (4.45 + 2.75)a

a = 9.81(4.45 - 2.75) / (4.45 + 2.75) = 2.31625 ≈ 2.32 m/s²

a)

T = 2.75(9.81 + 2.32) = 33.3 N

or

T = 4.45(9.81 - 2.32) = 33.3 N

b) 2.32 m/s² upward for 2.75 kg mass

    2.32 m/s² downward for 4.45 kg mass

c) y = ½at² = ½(2.31625/3)1² = 1.158125 ≈ 1.16 m

4 0
3 years ago
Please help me find out this answer
Elanso [62]
The direction of work.........

4 0
3 years ago
A golfer hits her tee-shot due north towards the fairway. Her shot has an initial velocity of 60 m/s. A 15 m/s wind is blowing i
vovangra [49]

Answer:

c=71.4m/s

\theta=8.54\textdegree

Explanation:

From the question we are told that

Initial velocity of 60 m/s

Wind speed V_w= 15 m/s \angle  45 \textdegree

Generally Resolving vector mathematically

  sin(45\textdegree)15=10.6\\cos(45\textdegree)15=10.6

Generally the equation Pythagoras theorem is given mathematically by

c^2=a^2+b^2

c^2=10.6^2 +(10.6+60)^2

c=\sqrt{10.6^2 +(10.6+60)^2}

Therefore Resultant velocity (m/s)

c=71.4m/s

b)Resultant direction

Generally the equation for solving Resultant direction

\theta=tan^-1(\frac{y}{x})

Therefore

\theta=tan^-1(\frac{10.6}{70.6})

\theta=8.54\textdegree

7 0
2 years ago
Chandra is doing an experiment to find out which type of dog food her dog prefers. The following are the steps of her experiment
ELEN [110]

Answer:

A the type of dog food because it was different and every other thing was the same.

8 0
3 years ago
A balloon filled with helium gas has an average density of Q,-0.41 kg/m'. The density of the air is Qa-1.23 kg/m3. The volume of
Citrus2011 [14]

Answer:

a) (Qa*g*Vb)-(Qh*Vb*g)=(Qh*Vb*a)\\where \\g=gravity [m/s^2]\\a=acceleration [m/s^2]

b) a = 19.61[m/s^2]

Explanation:

The total mass of the balloon is:

massball=densityheli*volumeheli\\\\massball=0.41 [kg/m^3]*0.048[m^3]\\massball=0.01968[kg]\\\\

The buoyancy force acting on the balloon is:

Fb=densityair*gravity*volumeball\\Fb=1.23[kg/m^3]*9.81[m/s^2]*0.048[m^3]\\Fb=0.579[N]

Now we need to make a free body diagram where we can see the forces that are acting over the balloon and determinate the acceleration.

In the attached image we can see the free body diagram and the equation deducted by Newton's second law

6 0
3 years ago
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