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aniked [119]
3 years ago
13

A projectile is fired straight upward from the Earth's surface at the South Pole with an initial speed equal to one third the es

cape speed. (The radius of the Earth is 6.38 106 m.)
Physics
1 answer:
7nadin3 [17]3 years ago
3 0
What is the question? if you are looking for the initial velocity of the projectile, then V escape of Earth = 11.2 km / s. divided by three
Vi = 3.73 km/s
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Rate of change of velocity is acceleration
8 0
3 years ago
uniform ladder of length 6.0 m and weight 300 N leans against a frictionless vertical wall. The foot of the ladder isplaced 3.0
olganol [36]

Answer:

Fx1 (6 m) sin 60 = 300 (3 m) cos 60  balancing torques about floor

Fx1 = 900 * 1/2 / 5.20 = 86.6 N  this is the horizontal force that must be supplied by the wall to balance torques about the floor

This is also equal to the static force of friction that must be applied at the point of contact with the floor to balance forces in the x-direction.

Fx1 = Fx2 = 86.6 N

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3 years ago
WHAT SHOULD I NAME MY DEAD RAT?
olga_2 [115]

Answer:

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6 0
3 years ago
You are standing at the top of a cliff that has a stairstep configuration. There is a vertical drop of 7 m at your feet, then a
Zolol [24]

Answer:

1. v = 6.67 m/s

2. d = 9.54 m

Explanation:

1. To find the horizontal velocity of the rock we need to use the following equation:

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

<u>Where</u>:

d: is the distance traveled by the rock

t: is the time

The time can be calculated as follows:

t = \sqrt{\frac{2d}{g}}

<u>Where:</u>

g: is gravity = 9.8 m/s²

t = \sqrt{\frac{2d}{g}} = \sqrt{\frac{2*7 m}{9.8 m/s^{2}}} = 1.20 s

Now, the horizontal velocity of the rock is:

v = \frac{d}{t} = \frac{8 m}{1.20 s} = 6.67 m/s      

Hence, the initial velocity required to barely reach the edge of the shell below you is 6.67 m/s.          

2. To calculate the distance at which the projectile will land, first, we need to find the time:

t = \sqrt{\frac{2d}{g}} = \sqrt{\frac{2*(7 m + 3 m)}{9.8 m/s^{2}}} = 1.43 s

So, the distance is:

d = v*t = 6.67 m/s*1.43 s = 9.54 m    

Therefore, the projectile will land at 9.54 m of the second cliff.

I hope it helps you!        

7 0
4 years ago
Jazz is a 172 lb athlete who exercised at 7.6 METs. At this workload, what is his energy expenditure in kcals/min.? Round to the
dezoksy [38]

Answer:

Energy expenditure in K cals/min = 10 K cals /min (approximately)

Explanation:

As we know

Energy expenditure in Kcal/min=  METs x 3.5 x Body weight (kg) / 200

Given is METs=7.6

Weight of Jazz= 172lb=78.02kg

putting the values in formula,

Energy expenditure in K cals/min=  7.6 x 3.5 x 78.02 / 200

                                                       =10.38 K cals /min

                                                       =10 K cals /min (approximately)

Therefore, Energy expenditure in K cals/min by Jazz will be approximately 10 K cals /min

8 0
3 years ago
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