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sineoko [7]
3 years ago
7

EMERGENCY!! Doing my Physics homework now and I need help on this one please - all information provided, I'd be very VERY gratef

ul for help!! 15 points!!

Physics
1 answer:
IrinaK [193]3 years ago
6 0

(a) The vertical motion is accelerated by gravity. The horizontal component is constant (neglecting air resistance, if this is not a projectile motion, the horizontal component would also be accelerated)

(b) Vertical:

v_y = 30\sin 40^\circ = 19.3\frac{m}{s}

Horizontal:

v_x = 30\frac{m}{s}\cos 40^\circ = 23.0 \frac{m}{s}

(c) Use the kinematic equation for distance. Calculate only the vertical component (horizontal is irrelevant):

s_y = -\frac{1}{2}gt^2 +vt\,,\,\,\,s_y=0\\0 = -\frac{1}{2}gt^2 +vt\\\frac{1}{2}gt =v\\t = \frac{2v}{g}= \frac{2\cdot30\sin 40^\circ\frac{m}{s}}{9.8\frac{m}{s^2}}=3.9s

The ball will be in the air for about 3.9 s.

(d) The range is the horizontal distance traveled. We know the ball is in the air for 3.9s and it moves with a horizontal velocity of 23 m/s. So:

s_x = 23\frac{m}{s}\cdot 3.9 s = 89.7m

The range is 89.7 meters.

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kinetic energy = 97200Joules
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A car is traveling south is 200 kilometers from it’s starting point after 2 hours. What is the average velocity of the car
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Answer:

100

Explanation:

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A student adds ice to 150 mL of water. She observes that the temperature of the water decreases after 10 minutes. Which statemen
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Mohammad has been saving 8 each week. Today he spent 142 of the savings, and he now has 50 left. For how many weeks has he been
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3 0
3 years ago
Two particles P and Q each moves towards ther along Straight Line (M)(N), 51m long. P starts from M with velocity 5m/s and const
bagirrra123 [75]

The time required by the particles are as follows:

a. t = 1.5 seconds

b. t = 3 seconds

c. t = 0.4 seconds

<h3>What is the time required?</h3>

The time required for the particles to be at several distances apart is calculated using the equation of motion given below:

S = ut + \frac{1}{2}at^{2}

a) Time required to be 30 m apart:

Assuming the distance covered by P is S1 and distance covered by Q is S2.

S1 + S2 = 51 - 30

S1 + S2 = 21

Substituting the values of velocity and acceleration in the equation of motion above:

5t + 1/2t^2 + 6t + 3t^2 = 21

2t^2 + 11t - 21 = 0

Solving for time, t by factorization, t = 1.5 seconds

b) Time required to meet:

Assuming the distance covered by P is S1 and distance covered by Q is S2.

S1 + S2 = 51

Substituting the values of velocity and acceleration in the equation of motion above:

5t + 1/2t^2 + 6t + 3t^2 = 51

2t^2 + 11t - 51 = 0

Solving for time, t by factorization, t = 3 seconds

c) Time required for velocity of P is ¾ of the velocity of Q:

Using the equation of motion: V = u + at

Vp = 3/4 Vq

4Vp= 3Vq

Substituting the values:

4(5 + t) = 3(6 + 3t)

5t = 2

t = 0.4 seconds

Learn more about distance, velocity and acceleration at: brainly.com/question/14344386

#SPJ1

7 0
2 years ago
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