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zavuch27 [327]
3 years ago
14

Pls help my science teacher is a horrible teacher...

Physics
1 answer:
Anettt [7]3 years ago
4 0

First of all, looks like your teacher is indeed pretty horrible. Secondly, the constraints to consider would be proper weight distribution, methods to minimize excessive motion of the building structure, and quantities such as volume and density, which would help in determining the optimal structure. Keeping the frequency of oscillation for a building low in case of an earthquake or natural disaster would also be a priority.

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___ is produced when grains of quartz in sandstone are welded together when the rock is subjected to high temperatures.
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3 years ago
When a magnet is dropped why does it lose its magnetic properties
Veronika [31]
When a magnet is dropped, the domain become unaligned. 
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A child bounces a 52 g superball on the sidewalk. The velocity change of the superball is from 20 m/s downward to 14 m/s upward.
marusya05 [52]

Answer:

Force on superball will be =9.8222\times 10^{-4}N

Explanation:

We have given mass of superball m = 52 gram = 0.052 kg

Velocity change from 20 m/sec downward to 14 m/sec upward

Let downward velocity is positive then upward velocity is negative

So downward velocity is + 20 m/sec and upward velocity is -14 m/sec

Time is given as 1800 sec

We know that acceleration is rate of change of velocity

So a=\frac{20-(-14)}{1800}=0.0188m/sec^2

According to newton second law

Force = ma = 0.052×0.0188 =9.8222\times 10^{-4}N

6 0
4 years ago
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A particle moving along the x-axis has its velocity described by the function vx =2t2m/s, where t is in s. its initial position
Nesterboy [21]

The position of the object at time t =2.0 s is <u>6.4 m.</u>

Velocity vₓ of a body is the rate at which the position x of the object changes with time.

Therefore,

v_x= \frac{dx}{dt}

Write an equation for x.

dx=v_xdt\\ x=\int v_xdt

Substitute the equation for vₓ =2t² in the integral.

x=\int v_xdt\\ =\int2t^2dt\\ =\frac{2t^3}{3} +C

Here, the constant of integration is C and it is determined by applying initial conditions.

When t =0, x = 1. 1m

x= \frac{2t^3}{3} +C\\ x_0=1.1\\ x= (\frac{2t^3}{3} +1.1)m

Substitute 2.0s for t.

x= (\frac{2t^3}{3} +1.1)m\\ =\frac{2(2.0)^3}{3} +1.1\\ =6.43 m

The position of the particle at t =2.0 s is <u>6.4m</u>




5 0
3 years ago
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You can't see beyond a blind turn, so a mirror would allow you to see around the corner.
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