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makkiz [27]
3 years ago
12

Which statements accurately describe elements? Check all that apply

Physics
1 answer:
GalinKa [24]3 years ago
7 0

Answer:

Elements are made up of only one type of atom. Each element has a unique chemical symbol. Elements can be identified by their atomic number.

You might be interested in
In an inelastic collision, as compared to an elastic collision, what is to be expected?.
motikmotik
The loss or conservation of kinetic energy is the difference between an elastic and an inelastic collision. Kinetic energy is not preserved in an inelastic collision, and it will change forms into sound, heat, radiation, or another form. The kinetic energy in an elastic collision is preserved and does not change forms.
3 0
2 years ago
A weight suspended from a spring is seen to bob up and down over a distance of 20 cm triply each second. What is the period? Wha
Kisachek [45]

Answer:

1) Hence, the period is 0.33 s.

2) The amplitude is 10 cm.

Explanation:

1) The period is given by:

T = \frac{1}{f}

Where:

f: is the frequency = 3 bob up and down each second = 3 s⁻¹ = 3 Hz

T = \frac{1}{f} = \frac{1}{3 Hz} = 0.33 s

Hence, the period is 0.33 s.

2) The amplitude is the distance between the equilibrium position and the maximum position traveled by the spring. Since the spring is moving up and down over a distance of 20 cm, then the amplitude is:          

A = \frac{20 cm}{2} = 10 cm  

Therefore, the amplitude is 10 cm.          

I hope it helps you!                    

5 0
3 years ago
if you crash your car how could you decrease the damage to you or the car using the concept of impulse
kotykmax [81]

Explanation:

Crumple zones are sections in cars that are designed to crumple up when the car encounters a collision. Crumple zones minimize the effect of the force in an automobile collision in two ways. By crumpling, the car is less likely to rebound upon impact, thus minimizing the momentum change and the impulse.

3 0
3 years ago
A 55 kg person falling with a velocity of 0.6
siniylev [52]

Answer:

What are we supposed to find, if it is kinetic energy then this is the solution.

K.E=1/2mv^2

K.E= kinetic energy

M=mass

V=velocity

K.E =0.5*55*0.6^2

K.E=9.9J

Explanation:

3 0
2 years ago
A solid sphere of radius R is placed at a height of 30 cm on a15 degree slope. It is released and rolls, without slipping, to th
photoshop1234 [79]

Answer:

The height is  h_c = 42.857

A circular hoop of different diameter cannot be released from a height 30cm and match the sphere speed because from the conservation relation the speed of the hoop is independent of the radius (Hence also the diameter )

Explanation:

   From the question we are told that

           The height is h_s = 30 \ cm

            The angle of the slope is \theta = 15^o

According to the law of conservation of energy

     The potential energy of the sphere at the top of the slope = Rotational kinetic energy + the linear kinetic energy

                          mgh = \frac{1}{2} I w^2 + \frac{1}{2}mv^2

Where I is the moment of inertia which is mathematically represented as this for  a sphere

                    I = \frac{2}{5} mr^2

  The angular velocity w is mathematically represented as

                         w = \frac{v}{r}

So the equation for conservation of energy becomes

               mgh_s = \frac{1}{2} [\frac{2}{5} mr^2 ][\frac{v}{r} ]^2 + \frac{1}{2}mv^2

              mgh_s = \frac{1}{2} mv^2 [\frac{2}{5} +1 ]

             mgh_s = \frac{1}{2} mv^2 [\frac{7}{5} ]

            gh_s =[\frac{7}{10} ] v^2

              v^2 = \frac{10gh_s}{7}

Considering a circular hoop

   The moment of inertial is different for circle and it is mathematically represented as

             I = mr^2

Substituting this into the conservation equation above

              mgh_c = \frac{1}{2} (mr^2)[\frac{v}{r} ] ^2 + \frac{1}{2} mv^2

Where h_c is the height where the circular hoop would be released to equal the speed of the sphere at the bottom

                 mgh_c  = mv^2

                     gh_c = v^2

                     h_c = \frac{v^2}{g}

Recall that   v^2 = \frac{10gh_s}{7}

                    h_c= \frac{\frac{10gh_s}{7} }{g}

                      = \frac{10h_s}{7}

            Substituting values

                   h_c = \frac{10(30)}{7}

                       h_c = 42.86 \ cm    

       

     

                         

8 0
3 years ago
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