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Mnenie [13.5K]
3 years ago
12

20 points please answer

Physics
1 answer:
Anna [14]3 years ago
6 0
Another way to test your question is to build your own miniature buildings. Depending on how in-depth you go, building could get a little pricey, but if you keep it basic there shouldn't be a problem. Decide on a certain number of foundations to test [maybe 3 or so] and try simulating an earthquake. 
<span>Hope this helps! </span>
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A boy throws a ball and accidentally breaks a window. The momentum of the ball and all the pieces of glass taken together after
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When a boy throws a ball and accidentally breaks a window, the momentum of the ball and all the pieces of glass taken together after the collision is THE SAME as the momentum of the ball before the collision

hope this helps
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The correct answer is A. Solid Rock
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A dart is thrown horizontally with an initial speed of 10 m/s toward point P, the bull's-eye on a dart board. It hits at point Q
Kryger [21]
D=s(t) so it would be d=10(.19) d=.19 FOR BITH SNDWERS
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3 years ago
Technician A says that a tire will increase or decrease approximately 1 psi for each 10°F change of temperature. Technician B sa
bearhunter [10]

Answer:

Technician A

Explanation:

It is seen that a tire pressure will increase or decrease 1 psi for each 10^{\circ} F  change in temperature.

For Technician B vehicle pressure should not be adjusted after tire has been warmed as the warm air may increase the pressure but it will be auto adjusted as  the temperature falls to normal .          

8 0
3 years ago
The rocket's acceleration has components \(a_{x}(t)= \alpha t^{2}\) and \(a_{y}(t)= \beta - \gamma t\), where \(\alpha = 2.50 {\
lbvjy [14]
 it is just a matter of integration and using initial conditions since in general dv/dt = a it implies v = integral a dt 
v(t)_x = integral a_{x}(t) dt = alpha t^3/3 + c the integration constant c can be found out since we know v(t)_x at t =0 is v_{0x} so substitute this in the equation to get v(t)_x = alpha t^3 / 3 + v_{0x} 
similarly v(t)_y = integral a_{y}(t) dt = integral beta - gamma t dt = beta t - gamma t^2 / 2 + c this constant c use at t = 0 v(t)_y = v_{0y} v(t)_y = beta t - gamma t^2 / 2 + v_{0y} 
so the velocity vector as a function of time vec{v}(t) in terms of components as[ alpha t^3 / 3 + v_{0x} , beta t - gamma t^2 / 2 + v_{0y} ] 
similarly you should integrate to find position vector since dr/dt = v r = integral of v dt 
r(t)_x = alpha t^4 / 12 + + v_{0x}t + c let us assume the initial position vector is at origin so x and y initial position vector is zero and hence c = 0 in both cases 
r(t)_y = beta t^2/2 - gamma t^3/6 + v_{0y} t + c here c = 0 since it is at 0 when t = 0 we assume 
r(t)_vec = [ r(t)_x , r(t)_y ] = [ alpha t^4 / 12 + + v_{0x}t , beta t^2/2 - gamma t^3/6 + v_{0y} t ] 
5 0
3 years ago
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