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Anika [276]
3 years ago
7

Chemical compounds are named using an orderly set of rules called _____.

Physics
1 answer:
Vesna [10]3 years ago
6 0
Chemical nomenclature .... I believe this is right
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A solid of density 8000 kgm.. weighs 0.8 kgf in air. When it is completely
ad-work [718]

Answer:<em> 0.5kg</em>

Explanation:

5 0
2 years ago
Rearrange the momentum equation to solve for speed. Show all work.
nasty-shy [4]

Definition:                   Momentum  =  (mass) x (speed)


OK. Here we go.
Watch closely:


Divide each side        
by  'mass' :                  <span>Momentum / mass  =  Speed </span>


Did you follow that ?

3 0
3 years ago
A mass hanging from a spring is set in motion and its ensuing velocity is given by v (t )equals 2 pi cosine pi t for tgreater th
lianna [129]

Answer:

2(maximum), -2(minimum), -2(maximum).

Explanation:

V(t)= 2πcos πt--------------------------------------------------------------------------------(1).

Therefore, there is a need to integrate v(t) to get S(t).

S(t)= 2×sinπt + C ------------------------------------------------------------------------------(2).

Applying the condition given, we have s(0)= 0.

S(0)= 2sin ×π(0) + C.

Which means that; 0+C= 0. That is; C=0.

S(t)= 2 sin πt.

The mass moves to its highest positions at time,t=half(1/2=.5) and time,t=2.5.

Take note that;  sin(π/2) = sin(5π/2) = 1 .

Also, the mass moves to its lowest position at time,t=(3/2); also, sin(3π/2) = -1.

Therefore, we have that 2 maximum; -2 minimum and -2 maximum.

7 0
2 years ago
An ocean wave has an amplitude of 2.5 m. Weather conditions suddenly change such that the wave has an
Dovator [93]

Answer:

2.5

Explanation:2.5 +2.5 = 5.0

8 0
3 years ago
A 3.00 kg object is moving in the XY plane, with its x and y coordinates given by x = 5t³ !1 and y = 3t ² + 2, where x and y are
Hatshy [7]

Answer:

The net force acting on this object is 180.89 N.

Explanation:

Given that,

Mass = 3.00 kg

Coordinate of position of x= 5t^3+1

Coordinate of position of y=3t^2+2

Time = 2.00 s

We need to calculate the acceleration

a = \dfrac{d^2x}{dt^2}

For x coordinates

x=5t^3+1

On differentiate w.r.to t

\dfrac{dx}{dt}=15t^2+0

On differentiate again w.r.to t

\dfrac{d^2x}{dt^2}=30t

The acceleration in x axis at 2 sec

a = 60i

For y coordinates

y=3t^2+2

On differentiate w.r.to t

\dfrac{dy}{dt}=6t+0

On differentiate again w.r.to t

\dfrac{d^2y}{dt^2}=6

The acceleration in y axis at 2 sec

a = 6j

The acceleration is

a=60i+6j

We need to calculate the net force

F = ma

F = 3.00\times(60i+6j)

F=180i+18j

The magnitude of the force

|F|=\sqrt{(180)^2+(18)^2}

|F|=180.89\ N

Hence, The net force acting on this object is 180.89 N.

3 0
3 years ago
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