A horizontal rectangular surface has dimensions 3.00 cm by 4.20 cm and is in a uniform magnetic field that is directed at an angle of 32.0 ∘ above the horizontal.

What must the magnitude of the magnetic field be to produce a flux of 3.60 × 10 −4 Wb through the surface?

The acceleration due to gravity g at height 7 times the radius of Earth above the surface of the Earth is  8.933 m/s².

It is the acceleration caused by any object when placed on the surface on the Earth due to its weight.

The acceleration due to gravity is related to the distance as

g =GM/R²

where G is the gravitational constant = 6.67 x 10⁻¹¹ N.m²/kg²

Given is the height R = 7 x 6.4 x 10⁶ m and mass of  Earth M = 6 x 10²⁴ kg

Substitute the values into the expression to find the acceleration due to gravity.

g = ( 6.67 x 10⁻¹¹ x 6 x 10²⁴)/ ( 7 x 6.4 x 10⁶)²

g = 8.933 m/s²

Thus,  the acceleration due to gravity g at height 7 times the radius of Earth above the surface of the Earth is  8.933 m/s².

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Answer:

real depth / apparent depth = R.I.

Real depth = 1.5 x 1 = 1.5 m

a)A spherical convex mirror form the image at image distance is 8.57cm.

b) Mirror forms the image of height is 12.855 cm.

c)The magnification of the spherical convex lens is 0.4285.

The mirror which is a part of sphere with radius R is painted from inside forms the convex mirror.

The focal length of convex lens,  f = R/2

Given , the radius of mirror is R =30cm, the focal length will be

f =30cm/2 = 15cm

a) Object distance u =20 cm, object height =0.30m, then the image distance will be 1/f = 1/u + 1/v where v is the image distance from the lens.

1/15 = 1/20 + 1/v

v= 8.57 cm

Thus, the image distance is 8.57cm.

b) Ratio of image distance to the object distance is equal to the ratio of image height to the object height.

Substitute the values into the expression, we get the image height as

h(image) = 8.57/20 x 30

h(image) = 12.855 cm

Thus, the image height is 12.855 cm.

c) The magnification is the ratio of image height to the object height.

Putting the values, we have

= 12.855/30

= 0.4285

Thus, the magnification is 0.4285.

d) Size of object is large as compared to the size of image. Object distance is small as compared to the image distance. The image formed is virtual.

e)  The setup drawing is attached.

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The average speed is 18.73 km/hour.It is found as the ratio of the total distance and the total time taken.

The total distance traveled by an object divided by the total time taken is the average speed.

The average speed of an object indicates the pace at which it will traverse a distance. The metric unit of speed is the meter per second.

The given data in the problem is;

Distance (d₁) = 18 Kilometers

Time (t₁) =  2.5 hours

Distance (d₂) = 150 Km

Time (t₂)= 4 hours

Distance (d₃) =10 Km

Time (t₃)=3 hours.

The average speed is found as;

[tex]\rm v_{avg}= \frac{Total \ distance }{Total \ time } \\\\ V_{avg}=\frac{18+150+10}{2.5+4+3} \\\\ v_{avg}= 18.73 \ km/hours[/tex]

Hence, the average speed is 18.73 km/hour.

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Answer:

acts as a one-way switch for current

[tex]\huge{\color{red}{\fbox{\textsf{\textbf{Answer}}}}} [/tex]

94.6J

Explanation:

Force (F) = 22N

mass (m) = 12kg

distance (S) = 4.3m

While

[tex] \sf Work \: done = Force \times distance[/tex]

This means

W = FS

while mass has no role in equation

Now substituting the required values

[tex] \sf W = 4.3 \times 22 \\ \\ \sf W = 94.6J[/tex]

Answer:

Explanation:
Answer: B mechanical waves.

Waves that do require a medium are called mechanical waves.

Answer:

0.5

Explanation:

I think it is this beacause if u have the answer you have that

Answer:

false

Explanation:

The average reaction time for humans is 0.25 seconds to a visual stimulus, 0.17 for an audio stimulus, and 0.15 seconds for a touch stimulus.

Consider,

[tex]{:\implies \quad \displaystyle \sf \langle p\rangle =m\langle v(t)\rangle=m\int_{-\infty}^{\infty}x\bigg\{\dfrac{\partial \Psi^{*}(x,t)}{\partial t}\Psi (x,t)+\Psi^{*}(x,t)\dfrac{\partial \Psi (x,t)}{\partial t}\bigg\}dx}[/tex]

Multiply both sides by ih and simplification will yield

[tex]{:\implies \quad \displaystyle \sf ih\langle p\rangle =m\int_{-\infty}^{\infty}x\bigg[\Psi (x,t)\bigg\{\dfrac{h^2}{2m}\dfrac{\partial^{2}\Psi^{*}(x,t)}{\partial x^2}-V(x)\Psi^{*}(x,t)\bigg\}+\Psi^{*}(x,t)\bigg\{V(x)\Psi (x,t)-\dfrac{h^2}{2m}\dfrac{\partial^{2}\Psi (x,t)}{\partial x^2}\bigg\}\bigg]dx}[/tex]

Some simplification, Then Integrate by parts and then knowing the fact that the wave function vanishes for [tex]{\bf x\to \pm \infty}[/tex] will yield:

[tex]{:\implies \quad \displaystyle \sf \langle p\rangle =\dfrac{ih}{2}\int_{-\infty}^{\infty}\bigg\{\dfrac{\partial \Psi^{*}(x,t)}{\partial x}\Psi (x,t)-\Psi^{*}(x,t)\dfrac{\partial \Psi (x,t)}{\partial x}\bigg\}dx}[/tex]

Integrating by parts and knowing the same fact by some simplification will yield:

[tex]{:\implies \quad \displaystyle \sf \langle p\rangle =-ih\int_{-\infty}^{\infty}\Psi^{*}(x,t)\dfrac{\partial \Psi (x,t)}{\partial x}dx}[/tex]

The momentum is thus contained within the wave function, so we can then deduce that:

[tex]{:\implies \quad \sf p\rightarrow -ih\dfrac{\partial}{\partial x}}[/tex]

[tex]{:\implies \quad \sf p^{n}\rightarrow \bigg(-ih\dfrac{\partial}{\partial x}\bigg)^{n}}[/tex]

[tex]{:\implies \therefore \quad \displaystyle \sf \langle p^{2}\rangle =-h^{2}\int_{-\infty}^{\infty}\Psi^{*}(x,t)\dfrac{\partial^{2}\Psi (x,t)}{\partial x^2}dx}[/tex]

Now the kinetic energy

[tex]{:\implies \quad \displaystyle \sf \langle K\rangle =\dfrac{\langle p^{2}\rangle}{2m}=\dfrac{-h^2}{2m}\int_{-\infty}^{\infty}\Psi^{*}(x,t)\dfrac{\partial^{2}\Psi (x,t)}{\partial x^2}dx}[/tex]

The classical formula for the total energy

[tex]{:\implies \quad \sf \dfrac{p^2}{2m}+V(x)=E}[/tex]

Multiplying this equation by [tex]{\sf \Psi (x,t)=\psi (x)exp\bigg(\dfrac{-iEt}{h}\bigg)}[/tex] and use the above equations and simplify it we will be having

[tex]{:\implies \quad \boxed{\bf{\dfrac{-h^2}{2m}\dfrac{d^{2}\psi (x)}{dx^{2}}+V(x)\psi (x)=E\psi (x)}}}[/tex]

This is the Famous Time-Independent Schrödinger wave equation

Note:- If I write all the explanation then the Answer box willn't allow me to submit the answer

the car take to catch up to the truck =  same distance

d truck = d car

vt = vot+1/2 at² (vo=0, from rest)

vt = 1/2at²

15t=1.5t²

1.5t²-15t=0

t(1.5t-15)=0

t = 0 or

1.5t=15⇒ t=10 s

Explanation:

Standard one kilogram is the mass of a platinium- iridium cylinder having equal diameter at the height kept at 0⁰C at the International bureau of Weights and Measures, held in paris in France. The accurate copies of atandard one kilogram are prepared to Measures mass in different countries.....

[tex]....[/tex]

It is defined as the mass of a particular international prototype made of platinum-iridium and kept at the International Bureau of Weights and Measures. It was originally defined as the mass of one liter (10-3 cubic meter) of pure water. At the Earth's surface, a mass of 1 kg weighs approximately 2.20 pounds (lb).

Answer:

the first one, The object moving through air

Answer:

1.48kg

Explanation:

Here,

potential energy (P.E) = 29j

height (h) = 2m

acceleration due to gravity(g) =

[tex]9.8m {s}^{ - 2} [/tex]

mass(m) = ?

we know,

P.E = mgh

or, 29 = m×9.8×2

or, 29/19.6 = m

or,m = 1.48kg

F=m.a

a = F/m

a = 90/65

a = 1.385 m/s²

Answer:

solution

here,

Explanation:

weight = 90N

mass=65kg

acceleration due to gravity (g)=??

Now,

w=m×g

90=65×g

90/65=g

:g=1.38m/s^2

sine's law

h/sin 30 = 50/sin 60

h/0.5 = 50/(0.5√3)

h=28.9 m

In the text "White House Press Release Announcing the Bombing of Hiroshima August 6, 1945" by President Truman, the atomic bomb was developed by the United States. Therefore, the correct answer is B. The United States.

During World War II, a top-secret project known as the Manhattan Project was conducted by the United States. The goal of this project was to develop an atomic bomb. The research and development efforts were led by American scientists, engineers, and military personnel. The project involved numerous facilities across the United States, including Los Alamos, New Mexico, where the final design and testing of the atomic bomb took place.

President Truman's press release announcing the bombing of Hiroshima on August 6, 1945, indicates that the United States successfully developed and deployed the atomic bomb. This historical event marked the first use of atomic weapons in warfare and had a significant impact on the outcome of World War II.

Therefore, based on the context of President Truman's press release and the historical record, it is clear that the atomic bomb was developed by the United States. Therefore, the correct answer is B.

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The best estimate for the ratio of Power radiated per unit Area on Earth and Power radiated per unit Area on Venus is 1/16. The correct option is D.

Power radiated is directly proportional to the fourth power of Temperature at the surface and also to the cross-section area.

P = σ A T⁴

For 600 K at Venus, σ A T⁴ = σ A (600)⁴

For 300 K at Earth , σ A T⁴ = σ A (300)⁴

Comparing, we get

[σ A (300)⁴/ σ A (600)⁴ ] = 1/16

Thus, the the best estimate for the ratio is 1/16. The correct option is D.

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Answer:

The charge is believed to be from the charge of the quarks that make up the nucleons (protons and neutrons). A proton is made of two Up quarks, with 2/3 positive charge each and one Down Quark with a negative 1/3 charge (2/3 + 2/3 + -1/3 = 1).

Explanation:

Hope this helps :)

Answer:

Resistors connected in series

That states that the blackhole ends are working like bridges between one space to another.

This may be possible

Answer:

thermometer A

Explanation:

The range of a thermometer is the difference between the minimum and maximum temperatures that the thermometer can read.

             ∴ Range of thermometer A = 110° - (-10)° = 120°

                Range of thermometer B = 50° - (-10)° = 60°

∴ Range of thermometer A is greatest.

Hope this helps!

By applying the principles of relativity and uncertainty, rather than Newtonian physics,  the position of an electron is best determined. The correct option is D.

It states that position and the velocity of an object can't be measured exactly at the same time. One of them may have the approximate value, using the principle of uncertainty.

Every object is moving relative to some frame of reference. So, exact velocity and at what position an object is, can't be find by theory of Newtonian physics. Speed of a truck, length of object at rest and position of a flying bird flying overhead can be found using the Newtonian physics.

An electron is very minute particle whose velocity is unpredictable at any location.  So, the position of an electron is best determined by the theory of relativity and uncertainty.

Thus, The correct option is D.

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Answer:

2

Explanation:

there are two lonely pairs

2 hydrogens

Explanation:

oxygen has 3 lone pairs of electrons which do not take part in the bonding leaving two electrons in the outermost shell to be part of it, hydrogen has only one electron in it's outermost shell it needs one electron to be stable so two atoms of hydrogen bonds with one oxygen atom to to share there electrons to be stable.

Answer:

613.5 pm

Explanation:

We know that de Broglie wavelength of an electron :

λ = 1.227 / √E nm

λ = 1.227 / √400 nm

λ = 1.227 / 20 nm

λ = 0.6135 nm

λ = 0.6135 x 10⁻⁹ m

λ = 613.5 x 10⁻¹² m

λ = 613.5 pm

Wavelength of election having kinetic energy of 400 ev.

Answer: Limestone Caves

Explanation: The most common feature that can be caused purely by chemical weathering is Karst Landscape, which can lead to caverns and sinkholes.

The rate at which energy is lost from the whale's core from the core/blubber interface through the blubber to the skin is 3,162.5 W.

The rate of conduction of core/blubber interface is calculated as follows;

[tex]\frac{Q}{t} = \frac{kA\Delta T}{d}[/tex]

where;

[tex]\frac{Q}{t} = \frac{kA\Delta T}{d} = \frac{kA}{d} (\Delta Ta \ +\Delta Tb)\\\\\frac{Q}{t} = \frac{kA}{d} ((T_1- T_2 \ +\ T_2-T_3)\\\\\frac{Q}{t} = \frac{0.25 \times 22}{0.04} ( 35 - 29 \ + \ 29 - 12)\\\\\frac{Q}{t} = 137.5( 6 + 17)\\\\\frac{Q}{t} = 3,162.5 \ W[/tex]

Thus, the rate at which energy is lost from the whale's core from the core/blubber interface through the blubber to the skin is 3,162.5 W.

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