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SSSSS [86.1K]
3 years ago
14

Complete each statement with the word that makes it true.

Physics
1 answer:
Nuetrik [128]3 years ago
5 0

Answer:

reactive

halogens

nonmetals

Explanation:

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Arrange the examples in order, starting with the object that has the least amount of energy. In each case, assume there’s no fri
Artemon [7]
First example: book, m= 0.75 kg, h=1.5 m, g= 9.8 m/s², it has only potential energy Ep,

Ep=m*g*h=0.75*9.8*1.5=11.025 J

Second example: brick, m=2.5 kg, v=10 m/s, h=4 m, it has potential energy Ep and kinetic energy Ek,

E=Ep+Ek=m*g*h + (1/2)*m*v²=98 J + 125 J= 223 J

Third example: ball, m=0.25 kg, v= 10 m/s, it has only kinetic energy Ek

Ek=(1/2)*m*v²=12.5 J.

Fourth example: stone, m=0.7 kg, h=7 m, it has only potential energy Ep,

Ep=m*g*h=0.7*9.8*7=48.02 J

The order of examples starting with the lowest energy:

1. book, 2. ball, 3. stone, 4. brick 


4 0
3 years ago
For the potential , determine the number of bound states?A)6 b)4 c) 3 d) 1
s2008m [1.1K]

Do you see that blank, open space after the word "potential ..." ?

There's supposed to be a number there that actually tells us the value of the potential.  Without that number ... and a lot more description of the whole scenario here ... there's no possible answer to the question.

7 0
3 years ago
A point charge of -4.28 pC is fixed on the y-axis, 2.79 mm from the origin. What is the electric field produced by this charge a
makkiz [27]

Answer:

E = (-3.61^i+1.02^j) N/C

magnitude E = 3.75N/C

Explanation:

In order to calculate the electric field at the point P, you use the following formula, which takes into account the components of the electric field vector:

\vec{E}=-k\frac{q}{r^2}cos\theta\ \hat{i}+k\frac{q}{r^2}sin\theta\ \hat{j}\\\\\vec{E}=k\frac{q^2}{r}[-cos\theta\ \hat{i}+sin\theta\ \hat{j}]              (1)

Where the minus sign means that the electric field point to the charge.

k: Coulomb's constant = 8.98*10^9Nm^2/C^2

q = -4.28 pC = -4.28*10^-12C

r: distance to the charge from the point P

The point P is at the point (0,9.83mm)

θ: angle between the electric field vector and the x-axis

The angle is calculated as follow:

\theta=tan^{-1}(\frac{2.79mm}{9.83mm})=74.15\°

The distance r is:

r=\sqrt{(2.79mm)^2+(9.83mm)^2}=10.21mm=10.21*10^{-3}m

You replace the values of all parameters in the equation (1):

\vec{E}=(8.98*10^9Nm^2/C^2)\frac{4.28*10^{-12}C}{(10.21*10^{-3}m)}[-cos(15.84\°)\hat{i}+sin(15.84\°)\hat{j}]\\\\\vec{E}=(-3.61\hat{i}+1.02\hat{j})\frac{N}{C}\\\\|\vec{E}|=\sqrt{(3.61)^2+(1.02)^2}\frac{N}{C}=3.75\frac{N}{C}

The electric field is E = (-3.61^i+1.02^j) N/C with a a magnitude of 3.75N/C

8 0
3 years ago
Two flywheels of negligible mass and different radii are bonded together and rotate about a common axis (see below). The smaller
jekas [21]

Answer:

F_2 = 29.54 N

Explanation:

As we know that the combination is maintained at rest position

So we will take net torque on the system to be ZERO

so we know that

\tau = \vec r \times \vec F

here we will have

\vec r_1 \times F_1 = \vec r_2 \times F_2

so we have

13 \times 50 = 22 \times F_2

so we have

F_2 = \frac{13 \times 50}{22}

F_2 = 29.54 N

8 0
3 years ago
If tides occur twice a day in most places. How much time is there between a morning high tide and the next high tide?
algol13

About 12 hours is the time between a morning high tide and the next high tide

Explanation:

The Earth’s rotation happens between two tidal bulges  

The “periodic rise and fall” of the surface water levels of the ocean is called tides. The gravitational action and interaction on the earth by the sun and the moon causes these tides. Different regions of the World experiences different patterns of tides like the diurnal, semi-diurnal etc.

When there is one high and one low tide occurring on a lunar day, then it is diurnal pattern. Semi-diurnal pattern occurs when there are two equal high and low tides on a single lunar day.

Since the Earth’s rotation happens between two tidal “bulges” on each lunar day, the coastal areas can experience two high and two low tides in every 24 hours plus 50 minutes.

Accordingly the time between two high tides would be 12 hours plus 25 minutes. Similarly, the time gap between a high to low tide would be 6 hours plus 12.5 minutes.

8 0
3 years ago
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