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Anton [14]
3 years ago
15

Explain 5 things that could cause an incorrect mass when using the triple bean balance?

Physics
2 answers:
ludmilkaskok [199]3 years ago
5 0
If the scale is not "zeroed". If you do not use grams (g) to lable your products. If you do not unlock the balance. [that's about all I got doll]
Ierofanga [76]3 years ago
3 0
Incorrect mass labeling, Not starting at 0 and gradually increasing weights. Possibly weighing the wrong object (?) Maybe you keep your hand or another object on the scale... There are many things that can go wrong while usuing the triple beam scale.. Always make sure you take more then one measurement.
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What kind of quantity is displacement?
Serhud [2]

Answer:

calar quantity, length of path. displacement: vector quanity, "as the crow flies" difference between start and finish regardless of path taken. Term.

Explanation:

8 0
3 years ago
The position of a particle moving along the x-axis depends on the time according to the equation x = ct2 - bt3, where x is in me
Sav [38]

Answer:

(a):  \rm meter/ second^2.

(b):  \rm meter/ second^3.

(c):  \rm 2ct-3bt^2.

(d):  \rm 2c-6bt.

(e):  \rm t=\dfrac{2c}{3b}.

Explanation:

Given, the position of the particle along the x axis is

\rm x=ct^2-bt^3.

The units of terms \rm ct^2 and \rm bt^3 should also be same as that of x, i.e., meters.

The unit of t is seconds.

(a):

Unit of \rm ct^2=meter

Therefore, unit of \rm c= meter/ second^2.

(b):

Unit of \rm bt^3=meter

Therefore, unit of \rm b= meter/ second^3.

(c):

The velocity v and the position x of a particle are related as

\rm v=\dfrac{dx}{dt}\\=\dfrac{d}{dx}(ct^2-bt^3)\\=2ct-3bt^2.

(d):

The acceleration a and the velocity v of the particle is related as

\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(2ct-3bt^2)\\=2c-6bt.

(e):

The particle attains maximum x at, let's say, \rm t_o, when the following two conditions are fulfilled:

  1. \rm \left (\dfrac{dx}{dt}\right )_{t=t_o}=0.
  2. \rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}

Applying both these conditions,

\rm \left ( \dfrac{dx}{dt}\right )_{t=t_o}=0\\2ct_o-3bt_o^2=0\\t_o(2c-3bt_o)=0\\t_o=0\ \ \ \ \ or\ \ \ \ \ 2c=3bt_o\Rightarrow t_o = \dfrac{2c}{3b}.

For \rm t_o = 0,

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6\cdot 0=2c

Since, c is a positive constant therefore, for \rm t_o = 0,

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}>0

Thus, particle does not reach its maximum value at \rm t = 0\ s.

For \rm t_o = \dfrac{2c}{3b},

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6b\cdot \dfrac{2c}{3b}=2c-4c=-2c.

Here,

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}

Thus, the particle reach its maximum x value at time \rm t_o = \dfrac{2c}{3b}.

7 0
3 years ago
A brick is thrown vertically upward with an initial speed of 3.00 m/s from the roof of a building. If the building is 78.4 m tal
masya89 [10]

Answer:

0.918 sec

Explanation:

time of flight:

t=u²/g

=3²/9.8=0.918sec

6 0
3 years ago
Read 2 more answers
When its 75 kW (100 hp) engine is generating full power, a small single-engine airplane with mass 700 kg gains altitude at a rat
svlad2 [7]

Answer:

343/1500

Explanation:

Power: This can be defined as the product force and velocity. The S.I unit of power is Watt (w).

From the question,

P' = mg×v................. Equation 1

Where P' = power used to gain an altitude, m = mass of the engine, g = acceleration due to gravity of the engine, v = velocity of the engine.

Given: m = 700 kg, v = 2.5 m/s, g = 9.8 m/s²

Substitute into equation 1

P' = 700(2.5)(9.8)

P' = 17150 W.

If the full power generated by the engine = 75000 W

The fraction of the engine power used to make the climb = 17150/75000

= 343/1500

8 0
3 years ago
Which of the following is not a wave of the electromagnetic spectrum?
djyliett [7]
The correct answer is surface wave
6 0
2 years ago
Read 2 more answers
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