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GenaCL600 [577]
4 years ago
14

What two simple machines make a skateboard?

Physics
1 answer:
Andrews [41]4 years ago
6 0

Answer:

a saw and wheels

Explanation:

you need a saw for the wood and trucks and you need the wheels to ride the board

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1. Mike wears an 8.50 kg backpack while scaling a cliff. After 90 minutes, Mike is 11.2 m above the starting point.
svet-max [94.6K]

Answer:

a. 952Joules

b. 8008 joules

c. 1.48 watts

8 0
3 years ago
The strength of an electromagnet can be altered by
denpristay [2]
The strength of an electromagnet can be altered by increasing the number of coils around the core. The more times the coil is wrapped, the stronger the electromagnet is.

Your answer is: B) Increasing the number of coils around the core 

Have an amazing day and stay hopeful!

3 0
3 years ago
Read 2 more answers
Thiết bị trao đổi nhiệt ngược chiều, nhiệt độ chất lỏng nóng vào là 310°C, ra là 115°c. Nhiệt độ chất lỏng lạnh vào là 25°c và r
Vika [28.1K]

Answer: 129.5 m

Explanation:

310 + 115 + 25 + 68 = 518

518 / 4 = 129.5 m

i think. Sorry if this is wrong

:)

5 0
3 years ago
A process occurs in which a system's potential energy increases while the environment does work on the system.
iogann1982 [59]
Ok, now what do you want to know about it?
4 0
3 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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