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Strike441 [17]
3 years ago
7

Can someone help me with this I'm-- never mind I would just like help..

Physics
2 answers:
Alik [6]3 years ago
6 0
Silicon: Si
Iron: Fe
K: Potassium
Hg: Mercury
I: Iodine
Copper: Cu

A. Bromine
B.Beryllium
C. Gold
D. Tin

A. Alkaline Metals
B. Transition Metals
C. Reactive Nonmetals
D. Alkaline Earth Metals
serg [7]3 years ago
6 0
Dang it I love elements :(
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A force acting over a large area will exert less pressure per square inch than the same force acting over a smaller area.
babunello [35]

Answer:

True

Explanation:

Pressure is defined as:

p=\frac{F}{A}

where

F is the magnitude of the force perpendicular to the surface

A is the surface

Therefore, pressure is inversely proportional to the area of the surface:

p\propto \frac{1}{A}

this means that, assuming that the forces in the two situations (which have same magnitude) are both applied perpendicular to the surface, the force exerted over the smaller area will exert a greater pressure. Hence, the statement"

<em>"A force acting over a large area will exert less pressure per square inch than the same force acting over a smaller area"</em>

is true.

8 0
3 years ago
A reaction mixture in a 3.67 L flask at a certain temperature initially contains 0.763 g H2 and 96.9 g I2, At equilibrium, the f
Alexxandr [17]

Answer:

13 530 482

Explanation:

                            H2    +          I2     ------>      2HI

start (mol)             0.3785         0.3818                   0

change (mol)       -0.3534        -0.3534            +0.7067

equilibrium (mol)  0.0251         0.0284             0.7067

concentra (mol/L) 0.0068        0.0077              0.1926

K_{c} = \frac{0.1926^{2}}{0.0068^{2}*0.0077^{2} } = 13530482

7 0
3 years ago
An N-slit system has slit separation d and slit width a. Plane waves with intensity I and wavelength O are incident normally on
Leokris [45]

Answer:

E. d and O

Explanation:

"Light passing through a single slit forms a diffraction pattern somewhat different from those formed by double slits or diffraction gratings".

According to Huygens’s principle, "for each element of the wavefront in the slit emits wavelets. These are like rays that start out in phase and head in all directions. (Each ray is perpendicular to the wavefront of a wavelet.) Assuming the screen is very far away compared with the size of the slit, rays heading toward a common destination are nearly parallel".

The destructive interference for a single slit is given by:

d sin \theta = m\lambda , m=1,-1,2,-2,3,...

Where

d is the slit width

\lambda=O is the light's wavelength

\theta is the angle relative to the original direction of the light

m is the order od the minimum

I represent the intensity

When the intensity and the wavelength are incident normally the angular as we can see on the expression above the angular separation just depends of the distance d and the wavelength O.

7 0
3 years ago
Question 24 of 33 Which of the following is an example of uniform circular motion? A. A car speeding up as it goes around a curv
Fofino [41]

Answer:

Uniform Circular Motion is the Movement or Rotation of an Object along a circular Path at constant speed.

OPTION C IS YOUR ANSWER!.

8 0
3 years ago
A wire of length 6cm makes an angle of 20° with a 3 mT
Crazy boy [7]

Answer:

Approximately 7.3 \times 10^{-3}\; \rm A (approximately 7.3\; \rm mA) assuming that the magnetic field and the wire are both horizontal.

Explanation:

Let \theta denote the angle between the wire and the magnetic field.

Let B denote the magnitude of the magnetic field.

Let l denote the length of the wire.

Let I denote the current in this wire.

The magnetic force on the wire would be:

F = B \cdot l \cdot I \cdot \sin(\theta).

Because of the \sin(\theta) term, the magnetic force on the wire is maximized when the wire is perpendicular to the magnetic field (such that the angle between them is 90^\circ.)

In this question:

  • \theta = 20^\circ (or, equivalently, (\pi / 9) radians, if the calculator is in radian mode.)
  • B = 3\; \rm mT = 3 \times 10^{-3}\; \rm T.
  • l = 6\; \rm cm = 6 \times 10^{-2}\;\rm m.
  • F = 1.5\times 10^{-4}\; \rm N.

Rearrange the equation F = l \cdot I \cdot \sin(\theta) to find an expression for I, the current in this wire.

\begin{aligned} I &= \frac{F}{l \cdot \sin(\theta)} \\ &= \frac{3\times 10^{-3}\; \rm T}{6 \times 10^{-2}\; \rm m \times \sin \left(20^{\circ}\right)} \\ &\approx 7.3 \times 10^{-3}\; \rm A = 7.3 \; \rm mA\end{aligned}.

5 0
3 years ago
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