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Akimi4 [234]
3 years ago
13

As a science fair project, you want to launch an 800 g model rocket st raight up and hit a horizontally moving target as it pass

es 30 m above the launch point. The rocket engine provides a constant thrust of 15.0 N. The target is approaching at a speed of 15 m/s. At what horizontal dista nce between the target a nd the rocket should you launch
Physics
1 answer:
fredd [130]3 years ago
4 0

Sum the forces in the y (upward) direction

\sum F_y = ma

F_t +F_w = ma

15N - (0.800kg)(9.81m/s^2) = 0.800kg * a

a = 8.94 m/s^2

Applying the kinematic equations of linear motion we have that the displacement as a function of the initial speed, acceleration and time is

s = v_0t + \frac{1}{2} at^2

30 = 0 +\frac{1}{2} (8.94) t^2

t = 2.59 s

Again through the kinematic equation of linear motion that describes velocity as the change of displacement in a given time, we have to

v = \frac{d}{t} \rightarrow d = vt

d = (15m/s)(2.59s)

d = 38.85m

Therefore the horizontal distance between the target and the rocket should be 38.83m

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Which kind of energy does the soccer player transfer to the ball?
djyliett [7]

1. Kinetic

He makes the ball move by kicking it, which increases the kinetic energy

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3 years ago
A common misconception is that an object always moves when a force acts on it. Why is this statement incorrect? Explain the conc
dsp73

Answer:

The statement is incorrect because, a force acting on an object does not necessarily have to produce motion.

People have the misconception that when a force acts on an object it always produces motion

Explanation:

The statement is incorrect because, a force acting on an object does not necessarily have to produce motion. It could be in static equilibrium where the net force is zero and produces not motion. The body could also be in dynamic equilibrium when  no net force acts on it moving at a constant velocity. But here we are concerned with static equilibrium since the body does not move at all.

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3 0
4 years ago
If a 20 g cannonball is shot from a 5 kg cannon with a velocity of 100
balu736 [363]

Strange as it may seem, the statement in the question appears to be <em>TRUE</em>.  

-- Before the shot, neither the cannon nor the ball is moving, so their combined momentum is zero.  

-- Since momentum is conserved, we know immediately that their combined momentum AFTER the shot also has to be zero.

-- (20g is rather puny for a "cannonball" ... about the same weight as four nickels. But we'll take your word for it and just do the Math and the Physics.)

-- Momentum = (mass) x (velocity)

After the shot, the momentum of the cannonball is

(0.02 kg) x (100 m/s ==> that way)

Momentum of the ball = 2 kg-m/s ==> that way.

-- In order for both of them to add up to zero, the momentum of the cannon must be (2 kg-m/s this way <==) .

Momentum of cannon = (5 kg) x (V m/s this way <==)

2 kg-m/s this way <== = (5 kg) x (V m/s this way <==)

Divide each side by (5 kg):

V m/s  =  (2/5) m/s this way <==

Speed of recoil of the cannon = <em>-- 0.4 m/s</em>

3 0
3 years ago
What is the conductor's resistance if its length is 2m and has a cross-sectional area of 0.7m with a resistivity of4Ωm? a. 1.4 Ω
Firlakuza [10]

Answer:

The answer is c. 11.42 Ohm

Explanation:

The conductor's resistance is calculated by the formula in the figure.

So, you have to replace the given values into the formula.

Resistance of a conductor is equal to the product of rho by the lengh of the conductor divided the cross-sectional area of the conductor.

R= 4 ohm.m . (2m/o.7m^{2} )\\R=11.42ohm

3 0
4 years ago
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