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natita [175]
3 years ago
14

Gladiators of Physics Plz Help On This Question. Steps will help! with formula

Physics
1 answer:
arsen [322]3 years ago
8 0
Simplified gravitational potential energy:
v = mgh

m mass
g gravitational acceleration, on earth = 9.81m/s²
h height

Solve for height.
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A school bus moves at speed of 35 mi/hr for 20 miles. How long will it take the bus to get to school
ludmilkaskok [199]

Answer:

Explanation:

Assuming school is at the end of the 20 mile route, then

20 mi / 35 mi/hr = 0.57142...hr

which is about 34 minutes 17 seconds

6 0
3 years ago
A roller coaster starts from rest at its highest point and then descends on its (frictionless) track. Its speed is 38 m/s when i
inysia [295]

Explanation:

look !

speed= 38m/s

start from rest= 0

5 0
3 years ago
A golf ball strikes a hard, smooth floor at an angle of 39.8 ° and, as the drawing shows, rebounds at the same angle. The mass o
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Answer:

The magnitude of the impulse is 1.33 kg m/s

Explanation:

please look at the solution in the attached Word file

Download docx
7 0
4 years ago
Read 2 more answers
The moon is always circling around the earth, in a stable orbit. This movement is the reason we see the different phases of the
Darina [25.2K]
The answer would be B. This is because all planets in our galaxy orbit the sun.
3 0
4 years ago
Riders in an amusement park ride shaped like a Viking ship hung from a large pivot are rotated back and forth like a rigid pendu
erica [24]

Answer:

a)  v = 16.57 m / s, b)  a = 19.6 m / s², d)    N = 1.76 10³ N,     N / W = 3

Explanation:

This exercise looks interesting, but I think you have some problem with the writing, the questions seem a bit disconnected from the initial text.

Let's answer the questions.

a) For this part we can use energy considerations.

Starting point. The upper part of the trajectory indicates that the arm is horizontally

          Em₀ = U = m g h

in this case h = r

Final point. For lower of the trajectory

          Em_f = K = ½ m v²

as they indicate that there is no friction

         Em₀ = em_f

         mgh = ½ m v²

         v = \sqrt{2gh}

let's calculate

        v = \sqrt{2 \ 9.8 \ 14.0}

         v = 16.57 m / s

b) the centripetal acceleration has the formula

           a = v² / r

           a = 16.57² / 14.0

           a = 19.6 m / s²

c) see attached where the diagram is

where N is the normal and w the weight

d) let's use Newton's second law

               N-W = m a

               N - mg = m ar

               N = m (g + a)

let's calculate

               N = 60.0 (9.8 + 19.6)

               N = 1.76 10³ N

the relationship with weight is

              N / W = 1.76 10³/( 60 9.8)

              N / W = 3

normal is three times greater than body weight

e) the answer is reasonable since by Newton's first law the body must continue in a straight line, therefore to change its trajectory a force must be applied to deflect it

6 0
3 years ago
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