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gladu [14]
3 years ago
5

What is the slowest speed (in mph) that Mr. P can travel and still jump across the canyon? (I have attached a picture) Hint: The

answer is between 100-150 mph.

Physics
1 answer:
Alex17521 [72]3 years ago
7 0
might be 140mph, so that is a guess that i just made so plz let me know if im wrong or correct
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Explanation:

m=2xKE / v²

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A model rocket is shot directly upward, rises to its maximum height and then returns to its launch position in 10.0 s. Assuming
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d. 49.0 m/s

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3 0
3 years ago
Both objects A and D represent fixed, negatively-charged particles of equal magnitude and Object B represents a fixed, positivel
Kipish [7]

Answer:

Option A = 1.

Explanation:

So, in order to solve this question we are given the Important infomation or data or parameters in the question above as;

(1). First, Both objects A and D represent fixed.

(2).  Both objects A and D are negatively-charged particles of equal magnitude.

(3). "Object B represents a fixed, positively-charged particle (equal, but opposite charge from A and D)."

(4). "Object C shows a moving, positively-charged particle."

So, our mission is to determine the arrow that would correctly show the force of attraction or repulsion on object C caused by the other two objects.

We can do that by drawing out the forces of attraction and the resultants. Therefore, CHECK THE ATTACHED FILE/PICTURE FOR THE DRAWINGS.

The forces of attraction due to objects A and B on on object C will be towards themselves. Hence, the resultant is ONE(1).

6 0
3 years ago
brick, dry soil, paper, and water. If all four substances were exposed to sunlight for the same amount of time, which substance
denis-greek [22]
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4 0
3 years ago
A basketball player makes a jump shot. The 0.599 kg ball is released at a height of 2.18 m above the floor with a speed of 7.05
7nadin3 [17]

Answer:

W_{drag} = 4.223\,J

Explanation:

The situation can be described by the Principle of Energy Conservation and the Work-Energy Theorem:

U_{g,A}+K_{A} = U_{g,B} + K_{B} + W_{drag}

The work done on the ball due to drag is:

W_{drag} = (U_{g,A}-U_{g,B})+(K_{A}-K_{B})

W_{drag} = m\cdot g\cdot (h_{A}-h_{B})+ \frac{1}{2}\cdot m \cdot (v_{A}^{2}-v_{B}^{2})

W_{drag} = (0.599\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot (2.18\,m-3.10\,m)+\frac{1}{2}\cdot (0.599\,kg)\cdot [(7.05\,\frac{m}{s} )^{2}-(4.19\,\frac{m}{s} )^{2}]

W_{drag} = 4.223\,J

7 0
3 years ago
Read 2 more answers
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