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Fofino [41]
3 years ago
8

Which statements are true about the periodic table? Check all that apply. A. The periodic table is organized by atomic number. B

. There are two different ways to number the groups of the periodic table. C. Each box on the table represents one element. D. Elements in the same row of the periodic table have the same number of electron shells.
Physics
1 answer:
natulia [17]3 years ago
5 0
Let's evaluate each answer closely:
<span>A. The periodic table is organized by atomic number.
This statement is true the atomic number is what defines each element and is the number of protons and electrons in a neutral atom

B. There are two different ways to number the groups of the periodic table.
This is true.  The more modern method is to number the groups 1 - 18.  Earlier models used a Roman numeral and either an A or B to denote groups.
 
C. Each box on the table represents one element.
THis is also true.  Each box contains a unique element.

D. Elements in the same row of the periodic table have the same number of electron shells.
This is true.  The row that the element is on is also the number of electron shells that are in the element.
</span>
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Part a which of the following statements are true concerning electromagnetic induction?
maksim [4K]
I had to look for the options and here is my answer:

Based on the actual options attached to this question, the statements that are considered true about electromagnetic induction are the following:
-It is possible to induce a current in a closed loop of wire without the aid of a power supply or battery.
-It is possible to induce a current in a closed loop of wire by changing the strength of a magnetic field enclosed by the wire.
-It is possible to induce a current in a closed loop of wire by change the orientation of a magnetic field enclosed by the wire.
-It is possible to induce a current in a closed loop of wire located in a uniform magnetic field by either increasing or decreasing the <span>area enclosed by the loop.</span>
4 0
4 years ago
Item 4 Which conditions produce the smallest and largest ocean waves? Choose the two correct answers.
Darya [45]

Answer:

strong winds that blow for a long time over a great distance

weak winds that blow for short periods of time with a short fetch

Explanation:

When the winds are weak and blow for short periods, we experience the smallest ocean waves but when there are strong winds over a longer duration, the largest ocean waves are seen. Therefore, the conditions to produce the smallest and largest ocean waves are strong winds that blow for a long time over a great distance and weak winds that blow for short periods of time with a short fetch.

5 0
3 years ago
Read 2 more answers
What average braking force is required to stop a 1134-kg car traveling at a speed of 83 km/hr before it reaches a stop sign that
m_a_m_a [10]

Given:

m(mass of the car)=1134 Kg

u(Initial velocity)=83Km/HR=23m/s

s(distance traveled by the car)=98m

v(final velocity)=0(as it is given the car stops).

Now we know,

v=u+at

Where v is the final velocity

u is the initial velocity

a is the acceleration

t is the time

0=23+at

at=-23

Also

s=ut+1/2(at^2)

s is the distance covered by the car

u is the initial velocity

t is the time necessary for the car to cover a particular distance.

a is the acceleration

Now substituting these values we get

98=23t-1/2(23t)

98=23t-11.5t

11.5t=98

t=8.52secs

Now we have already derived

at=-23

ax8.52=-23

a=-23/8.52

a=-2.75 m/s^2

F=mxa

Where F is the force acting on the car.

m is the mass of the car.

a is the acceleration.

F=1134 x-2.75

F=-3119N

3 0
4 years ago
12.1 Following data are given for a direct shear test conducted on dry sand: Specimen dimensions: diameter= 63 mm; height= 25 mm
oksano4ka [1.4K]

Answer:

a) 30.58°

b) 367.55 N

Explanation:

Given:

Diameter of the specimen, D = 63mm = 0.063 m

Height of the specimen, H = 25 mm

Shear force at failure, {V_u}=276\ N=0.276 KN

Shear stress at failure, \tau_f=\frac{V_u}{Area\ of\ the\ specimen}

Area = \frac{\pi D^2}{4}

or

Area = \frac{\pi (0.063)^2}{4}=0.00311 m^2

thus,

Shear stress at failure, \tau_f=\frac{0.276}{0.00311}=88.745 KN/m^2

a)  \tau_f=C'+ \sigma'\tan\phi'    ............(1)

where,

C' = cohesion

for sand C' = 0

∅' = angle of friction

σ' = Normal stress

on substituting the values we get,

88.745=0+ 150\times\tan\phi'

or

\tan\phi'=0.591

or

\phi'=30.58^o

b) Shear force required at the time of failure with normal stress 200 KN/m² can be calculated by using the equation 1

\tau_f=C'+ \sigma'\tan\phi'  

on substituting the values, we get

\tau_f=0+ 200\tan30.58^o  

or

\tau_f=118.185 KN/m^2

shear force will be = 118.185 ×  area

or

shear force will be = 118.185 ×  0.00311 = 0.36755 KN = 367.55 KN

3 0
4 years ago
How can people reach their full potential?
PolarNik [594]
If they have self motivation or others motivation, they will show their full potential.
6 0
4 years ago
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