What is the most important safety rule to remember during lab activities


Answer 1
Answer: To follow instructions

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The 2-Mg truck is traveling at 15 m/s when the brakes on all its wheels are applied, causing it to skid for 10 m before coming to rest. The total mass of the boat and trailer is 1 Mg. Determine the constant horizontal force developed in the coupling C, and the friction force developed between the tires of the truck and the road during this time.
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Given a double slit apparatus with slit distance 1 mm, what is the theoretical maximum number of bright spots that I would see when I shine light with a wavelength 400 nm on the slits



The maximum number of bright spot is n_(max) =5001


From the question we are told that

     The  slit distance is d =  1 \ mm  =  0.001 \ m

      The  wavelength is  \lambda =  400 \ nm  =  400*10^(-9 ) \ m


Generally the condition for interference is  

        n *  \lambda  =  d  * sin \theta

Where n is the number of fringe(bright spots) for the number of bright spots to be maximum  \theta =  90

=>     sin( 90   )=  1


     n  =  (d   )/(\lambda )

substituting values

     n  =  ( 1 *10^(-3) )/( 400 *10^(-9) )

     n  =  2500

given there are two sides when it comes to the double slit apparatus which implies that the fringe would appear on two sides so the maximum number of bright spots is mathematically evaluated as

        n_(max) = 2 *  n  + 1

The  1  here represented the central bright spot


      n_(max) = 2 *  2500  + 1

     n_(max) =5001      


HELP ME PLS!!!!Find the location of beryllium (Be) on the periodic table. What type of ion will
beryllium form?
A. An ion with a -2 charge
B. An ion with a +6 charge
C. An ion with a +2 charge
D. An ion with a -6 charge


The Beryllium (Be) has an atomic number of 4 and belongs to Group-2 elements. The Beryllium will form a divalent cation (+2). Thus, option C is correct.

What are cations and anions?

In an atom, the number of electrons equals the number of protons. If the electrons are removed from the atom or the electrons are added to the atom, the atom has an excessive positive or negative charge.

This excessive of electrons or lack of electrons forms Ions. The excess of electrons has a negative charge or anions and the lack of electrons has a positive charge or cations.

Beryllium has 4 electrons. Two electrons are occupied in the valence shell of beryllium. Group 2 elements always form the positive ions or cations, to become stable ions.

The outermost shell of beryllium has two electrons. In order to form a stable ion, beryllium should lose its two electrons or gain six electrons. Beryllium belongs to the Group-2 element, it always loses two electrons and forms Be²⁺, to form a stable ion.

Hence, Beryllium forms an ion with a +2 charge. Thus, the correct option is C.

To learn more about the Cations and Anions:




the answer is c which is a+2 charge


Beryllium is in group 2A. It's nearest noble gas is Helium, which is 2 elements behind Beryllium. ThBeryllium wants to lose two electrons. When it does that, Beryllium will have a positive chargeof two, and it will be stated as B-e two plus.

A celestial body moving in an ellipical orbit around a star


A celestial body moving in an elliptical orbit around a star is a planet.

Depending on its size, composition, and the eccentricity of its orbit, that scanty description could apply to a planet, an asteroid, a comet, a meteoroid, or another star.

A small space probe of mass 170 kg is launched from a spacecraft near Mars. It travels toward the surface of Mars, where it will eventually land. At a time 22.9 seconds after it is launched, the probe is at location <5600, 7200, 0> m, and at this same instant its momentum is <51000, -7000, 0> kg·m/s. At this instant, the net force on the probe due to the gravitational pull of Mars plus the air resistance acting on the probe is <-4000, -780, 0> N.Assuming that the net force on the probe is approximately constant during this time interval, what is the change of the momentum of the probe in the time interval from 22.6 seconds after the probe is launched to 22.9 seconds after the launch?



The change  in momentum is  \Delta p = <-1200 , -234  , 0>  kg \cdot m/s      


From the question we are told that  

       The mass of the probe is  m = 170 kg

       The location of the prob at time t = 22.9 s is  A  = <5600, 7200,0>

       The  momentum at time  t = 22.9 s is  p = < 51000, -7000, 0> kg m/s

        The net force on the probe is  F = <-4000 , -780 , 0> N

Generally the change in momentum is mathematically represented as

              \Delta p = F * \Delta t

The initial time is   22.6 s

 The final time  is  22.9 s

             Substituting values  

           \Delta p = <-4000 , -780 , 0> * (22.9 - 22.6)

            \Delta p = <-4000 , -780 , 0> * (0.3)  

              \Delta p = <-1200 , -234  , 0>  kg \cdot m/s        


A 1 225.0 kg car traveling initially with a speed of 25.000 m/s in an easterly direction crashes into the back of a 9 700.0 kg truck moving in the same direction at 20.000 m/s. The velocity of the car right after the collision is 18.000 m/s to the east.(a) What is the velocity of the truck right after the collision? (Give your answer to five significant figures.) m/s east (b) What is the change in mechanical energy of the cartruck system in the collision? J (c) Account for this change in mechanical energy.



The answers to the questions are;

(a) The velocity of the truck right after the collision is 20.884 m/s

(b) The change in mechanical energy of the car truck system in the collision is -9076.4384 J

(c) The change in mechanical energy is due to energy consumed by the collision process.


(a) From the principle of conservation of linear momentum, we have

m₁·v₁+m₂·v₂ = m₁·v₃ + m₂·v₄


m₁ = Mass of the car = 1225.0 kg

m₂ = Mass of the truck = 9700.0 kg

v₁ = Initial velocity of the car = 25.000 m/s

v₂ = Initial velocity of the truck = 20.000 m/s

v₃ = Final velocity of the car right after collision = 18.000 m/s

v₄ = Final velocity of the truck right after collision


1225.0 kg × 25.000 m/s  +  9700.0 kg × 20.000 m/s = 1225.0 kg × 18.000 m/s  + 9700.0 kg × v₄

That is 30625 kg·m/s + 194000 kg·m/s = 22050 kg·m/s + 9700.0 kg × v₄

Making v₄ the subject of the formula yields

v₄ = (202575 kg·m/s)÷9700.0 kg = 20.884 m/s

The velocity of the truck right after the collision to five significant figures = 20.884 m/s

(b) The change in mechanical energy of the car truck system in the collision can be found by

The change in kinetic energy of the car truck system

Change in kinetic energy, ΔK.E. = Sum of final kinetic energy - Sum of initial kinetic energy

That is ΔK.E. = ∑ Final K.E -∑ Initial K.E.

ΔK.E. = ((1)/(2) m_1v_3^(2)+(1)/(2) m_2v_4^(2)) - ((1)/(2) m_1v_1^(2) +(1)/(2) m_2v_2^(2) )

         = ((1)/(2)·1225·18²+ (1)/(2)·9700·20.884²) - ((1)/(2)·1225·25²+(1)/(2)·9700·20²)

         = 2313736.0616 kg·m²/s² - 2322812.5 kg·m²/s² =  -9076.4384 kg·m²/s²

1 kg·m²/s² = 1 J ∴ -9076.4384 kg·m²/s² = -9076.4384 J

(c) The energy given off by way of the 9076.4384 J is energy transformed into other forms including

1) Frictional resistance between the tires and the road for the truck and car

2) Frictional resistance in the transmission system of the truck to increase its velocity

3) Sound energy, loud sound heard during the collision

4) Energy absorbed when the car and the truck outer frames are crushed

5) Heat energy in the form of raised temperatures at the collision points of the car and the truck.

6) Energy required to change the velocity of the car over a short distance.

1. A surfboarder rides a wave for 23.7 m at a constant rate of 4.1 m/s. How long did his triptake?



His trip took 5.78 seconds


23.7m divided by 4.1m/s = 5.78048780488