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4 years ago

14

Which are examples of projectile motion? Check all that apply. a kicker punting a football a car traveling along a highway a fre

e throw in basketball a ball rolling downhill a person throwing darts

2 answers:

o-na [289]4 years ago

8 0

A projectile is any object that is cast, fired, flung, heaved, hurled, pitched, tossed, or thrown<span>. </span>

Andrej [43]4 years ago

8 0

The answer would be A, C, and E

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8. Approximately how many hours a day are spent in school and doing homework?

Genrish500 [490]

Answer:

There would be 9 b/c there is so many hours of school and the homework depends on how old you are so i am pretty sure it is C.

Explanation:

3 0

3 years ago

A woman flies 300 kilometers north, 200 kilometers west, and 500 kilometers south. How far is she from her starting point?

maria [59]

She is
$\sqrt{200^2+200^2}= ~282.84$ kilometers away from her starting point

5 0

4 years ago

Two rather unequal cars are competing in a drag race. For simplicity, let's assume that the accelerations of the cars are consta

MA_775_DIABLO [31]

Answer:

car1: a=3.1m/s^2 , car2: a=6.1m/s^2

(a) 1/2*3.1*t^2= 1/2*6.1*(t-0.9)^2

1.55t^2= 3.05(t^2-1.8t+0.81)= 3.05t^2-5.49t+2.4705

1.5t^2-5.49t+2.4705= 0

t= 3.13457 = 3.14[s] after.

(b) d= 1/2*3.1*3.13457^2= 15.23[m] approx.

(c) car1: v=at = 3.1*3.13457= 9.717m/s

car2: v=at = 6.1*(3.13457-0.9)= 13.631m/s

13.631-9.717= 3.914 = 3.91[m/s] faster than car1.

6 0

4 years ago

In a cell, the amount nutrition coming in equals the amount of waste going out. This is an example of _____.

babymother [125]

The answer is B) <span>equilibrium
hope this helps!=-)</span>

5 0

4 years ago

You are climbing in the high sierra when you suddenly find yourself at the edge of a fog-shrouded cliff. to find the height of t

sergey [27]

The total time it takes the sound to reach our on top of the cliff (10.0 s) is actually sum of two times:
- the time it takes the rock to hit the ground starting from the top of the cliff, t1
- the time it takes the sound to reach the top of the cliff starting from the ground, t2
1) The rock moves by uniformly accelerated motion, with constant acceleration $g=9.81 m/s^2$ . Its law of motion is given by $y(t)=h-v_0t-\frac{1}{2}gt^2$ , where $v_0=0$ is the initial velocity of the rock, and h is the height of the cliff. The time t1 is the time at which the rock reaches the ground, so that y(t1)=0, and the equation becomes
$0=h-\frac{1}{2}gt_1^2$
2) The sound moves from the ground to the top of the cliff by uniform motion, with constant speed v=343 m/s. Therefore, the sound covers the distance h (the height of the cliff) in a time t2 given by
$h=vt_2$
3) If we rewrite h in both equations, we can write:
$=\frac{1}{2}gt_1^2=vt_2$ (1)
4) We also know that the sum of t1 and t2 is equal to 10 seconds:
$t_1+t_2=10$
from which we find
$t_2=10-t_1$
if we substitute this into eq.(1), we get
$\frac{1}{2}gt_1^2+vt_1-10v=0$
Numerically:
$4.9t_1^2 + 343 t_1 - 3430=0$
Solving the equation, we find the solution $t_1=8.87 s$ (the other solution is negative, so it does not have physical meaning). As a consequence,
$t_2 = 10-t_1 = 1.13 s$
and the height of the cliff is given by
<span> $h=vt_2=(343 m/s)(1.13 s)=388 m$ </span>

3 0

3 years ago

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