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enot [183]
3 years ago
5

So i am doing a science fair project and its about earthquakes so im going to need different supports for my buildings. I found

springs but now i need 3 more. URGENT!!
Physics
1 answer:
In-s [12.5K]3 years ago
8 0
You could use clay, sticks, rocks, Legos,
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grandymaker [24]

Answer:

a) frequency = 0.1724 Hz

b) Period = 5.8 sec

c) speed = 7.04 m/s

d) acceleration = 7.62 m/s²

Explanation:

Given that;

radius = 6.5m

time period = 5.8 sec every circle

a)  the frequency

frequency is the number of rotation in unit time

frequency = 1 / time period = 1/5.8

frequency = 0.1724 Hz

b)  the period

period is time taken in one rotation

period = total time / rotation = 5.8 / 1

Period = 5.8 sec

c)  the speed

speed = distance/time = circumference/time period = 2πr / t = (2π×6.5) / 5.8

speed = 7.04 m/s

d) acceleration

To find the acceleration we take the linear velocity squared divided by the radius of the circle.

so

acceleration = v² / r = (7.04)² / 6.5 = 49.5616 / 6.5

acceleration = 7.62 m/s²

6 0
3 years ago
A 35 kg box rests on the back of a truck. The coefficient of static friction bet?005 (part 1 of 2)A 35 kg box rests on the back
elena-14-01-66 [18.8K]

Answer with Explanation:

We are given that

Mass of box=35 kg

Coefficient of static friction between box and truck bed=0.202

Acceleration due to gravity=9.8 m/s^2

a.We have to find the force by which the box accelerates forward.

Force by which box accelerates=\mu mg=0.202\times 9.8\times 35

Force by which box accelerates=62.286 N

b.We have to find the maximum acceleration can the truck have before the box slides.

Force =friction force

ma=\mu mg

a=\mu g=9.8\times 0.202=1.9796 m/s^2

Hence, the truck can have maximum acceleration before the box slide=1.9796 m/s^2

3 0
3 years ago
Estimate the magnitude of the electric field due to the proton in a hydrogen atom at a distance of 5.29 x 10-11 m, the expected
V125BC [204]
Would be A 1012 N/C because The magnitude of the electric field at distance r from a point charge q is E=k
e
​
q/r
2
, so
E=
(5.11×10
−11
m)
2

(8.99×10
9
N.m
2
/C
2
)(1.60×10
−19
C)
​

=5.51×10
11
N/C∼10
1
2N/C
making (e) the best choice for this question.
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Resistors tell me if im right

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