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2 years ago

Which is better? ( Gifting or Receiving )

1 answer:

3 0

Honestly, giving. I know that receiving may feel very good, and you get that new thing that you wanted but if you give, you feel really good when you see that smile on someone else’s face.

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26. A 40 kg boy jumps from a height of 4m onto a plate-form mounted on springs. As the

denpristay [2]

Answer:

c. 1600J

Explanation:

The loss in potential energy of the boy is given by:

$U=mg \Delta h$

where

m = 40 kg is the mass of the boy

g = 9.8 m/s^2 is the acceleration of gravity

$\Delta h = 4 m + 0.02 m = 4.02 m$ is the total change in the height of the boy (4 metres + 2 cm due to the compression of the spring)

Substituting, we find

$\Delta U = (40 kg)(9.8 m/s^2)(4.02 m) = 1577 J \sim 1600 J$

4 0

2 years ago

Tectonic plate movement is the reason why northern California has a very different landscape than southern California. Two diffe

Studentka2010 [4]

Answer:

I don't know the answer I need points before I fail my math test

3 0

1 year ago

2. What will be the extension of this spring if the load is a) 4N and b) 75 g?

Furkat [3]

Answer:

Explanation:

just add

7 0

2 years ago

Suppose that a teacher driving a 1972 LeMans zooms out of a darkened tunnel at 34.5 m/s. He is momentarily blinded by the sunshi

valkas [14]

Answer:

489.19m

Explanation:

To find the minimum distance you first calculate the time in which the teacher stops:

$v=v_o-at\\\\t=\frac{v_o-v}{a}=\frac{34.5m/s-0m/s}{2.5m/s^2}=13.8s$

however, the reaction of the teacher is 0.31s later, then you use

t=13.8-0.31s=13.49s

during this time the camper has traveled a distance of:

$x=vt=(15.1m/s)(13.49s)=203.69m$ (1)

Next you calculate the distance that teacher has traveled for 13.6s:

$x=x_o+v_ot+\frac{1}{2}at^2\\\\x=0m+34.5m/s(13.49s)+\frac{1}{2}(2.5m/s^2)(13.49s)^2=692.88m$ (2)

The minimum distance between the driver and the camper will be the difference between (2) and (1):

$x_{min}=692.88m-203.69m=489.19m$

5 0

2 years ago

estimate the force that a cyclist travelling at 6m/s needs to apply to the brakes to bring the bicycle to a stop in a distance o

RSB [31]

v1 = 6m/s

v2 = 0

∆v = v1 - v2 = 6m\s

s = t * v = 15m

t = s\v1 = 15(m) \ 6(m\s) = 2.5s

a = ∆v\t = 6(m\s) \ 2.5s = 2.4m\s2

a = F\m = 2.4m\s2

F = a * m = 2.4m\s2 * ?kg

I can't tell you this because I don't know the mass of this cyclist

4 0

2 years ago