physics class 9 chapter 8 please tell​ please

Answer:

The magnetic force acting on the proton is 1.68 x 10^-18 N.

Explanation:

magnetic field, B = 5 T

speed , v = 2.1 m/s

charge q = 1.6 x 10^-19 C

Angle, A = 90 degree

The magnetic force on the charge particle is given by

[tex]F = q v B sin A\\\\F = 1.6\times 10^{-19}\times 2.1\times 5\times sin 90\\\\F = 1.68\times 10^{-18} N[/tex]

Answer:

The answer is "physics ".

Explanation:

Physics is the branch of science that addresses the properties of crystalline and its interaction with the fundamental elements of the universe. It covers subjects ranging in quantum mechanics with extremely little ones with quantum mechanics to the whole cosmos. You must be constant whether you like it or not, thus everyone must learn physics, irrespective of whether they're in Uganda, and plenty of other countries should have physics to dare study.

Mendeleev wrote down atomic mass

Answer:

A. Buoyancy.

hope it helps

Answer:

T₂ = 305.17 K

Explanation:

Given that,

Heat, Q = 6000 J

Mass, m = 200 gram

Initial temperature, T₁ = 25° C

We need to find its final temperature. Let it is T₂.

We know that,

[tex]Q=mc\Delta T[/tex]

Where

c is the specific heat of water, c = 4.18 J/g°C

So,

[tex]6000=200\times 4.18\times (T_2-298)\\\\\dfrac{6000}{200\times 4.18}=(T_2-298)\\\\7.17=(T_2-298)\\\\7.17+298=T_2\\\\T_2=305.17\ K[/tex]

So, the final temperature is equal to 305.17  K.

Answer:

Heat required = mass× latent heat Q = 0.15 × 871 ×

The heat required to evaporate 0.15 kg of lead at 1750°C will be 130,650 J.

The movement of energy from a hot to a cold item is characterized as heat. Heat energy flows from a hot material to a cold one.

This occurs because faster-vibrating molecules transmit their energy to slower-vibrating ones.

The given data in the problem is;

m is the mass of lead =  0.15 kg  

T is the temperature = 1750°C,

The latent heat of vaporization for lead is, [tex]\rm L_V[/tex] = 871 × 10³ J/kg.

The heat is found as;

[tex]\rm Q= m \times L_V \\\\ \rm Q= 0.15 \times 871 \times 10^3 \\\\ Q=130,650 \ J[/tex]

Hence the heat required to evaporate 0.15 kg of lead at 1750°C will be 130,650 J.

To learn more about the heat refer to the link;

brainly.com/question/1429452

Answer:

1. 36.35 g = 36.35E15 micrometer.

II. 36.35 g = 363.5 millimetre.

III. 36.35 g = 0.03635 kilogram.

Explanation:

Given the following data;

To convert the mass in grams (g) to;

I. Micrometer

Conversion:

1 g = 1 exp 15 um

36.35 g = X um

Cross-multiplying, we have;

X = 36.35 * 1 exp 15 = 36.35 exp 15 um

36.35 g = 36.35E15 micrometer

II. Millimetre

Conversion:

1 g = 1 milliliter

36.35 g = X milliliter

Cross-multiplying, we have;

X = 36.35 * 1 = 36.35 milliliter

Next, we would convert milliliter to millimetre;

1 milliliter = 10 millimetre

36.35 milliliter = X millimetre

Cross-multiplying, we have;

X = 36.35 * 10 = 363.5 millimetre

36.35 g = 363.5 millimetre

III. Kilogram

Conversion:

1000 grams = 1 kilogram

36.35 g = X kilogram

Cross-multiplying, we have;

X * 1000 = 36.35 * 1

Dividing both sides by 1000, we have;

X = 36.35/1000 = 0.03635 kilogram

36.35 g = 0.03635 kilogram

Note:

Answer:

Explanation:

The formula for this is

[tex]F_g=\frac{Gm_1m_2}{r^2}[/tex] where F is the gravitational force, G is the gravitational constant, m1 is the mass of one object and m2 is the mass of the other object. We are looking for r, the distance between the centers of their masses.

Filling in:

[tex]7.5*10^{-8}=\frac{6.67*10^{-11}(90.0)(30.0)}{r^2}[/tex] and moving things around to solve for r:

[tex]r=\sqrt{\frac{6.67*10^{-11}(90.0)(30.0)}{7.5*10^{-8}} }[/tex] Doing all that and rounding to the 3 sig fig's you need gives us a distance of 1.55 m

Answer:

Explanation:

First off, this lens is concave. Second, the image is obviously smaller, and third, the only thing that is NOT obvious, is the fact that real images are always upside down and virtual images are always right-side-up. So the choice you're looking for is D.

Answer:

D - A concave lens forms a smaller, virtual image

Explanation:

Answer:

mass= 0.1993 kg

Explanation:

Using the formula c = Q / (mΔT)

Answer:

(4) A = 3 A, A₂ = 11 A

(5) 7 A

Explanation:

(4)

From the diagram,

A = 3+6+2

A = 11 A

V = A₂R

A₂ = V/R₂............ Equation 1

Given: V = 12 V,  R₂ = 4 Ω

Substitute these values into equation 1

A₂ = 12/4

A₂ = 3 A

(5) Applying,

V = IR'

I = V/R'............ Equation 1

Where V = Voltage, I = cuurent, R' = total resistance.

But,

1/R' = (1/3)+(1/4)

1/R' = (3+4)/12

1/R' = 7/12

R' = 12/7 Ω

Given: V = 12 V

Substitute these values into equation 1

I = 12/(12/7)

I = 7 A

Therefore

A = 7 A

Answer:

[tex]V.E=498.02m/s^2[/tex]

Explanation:

From the question we are told that:

Radius [tex]r=301Km[/tex]

Gravitational acceleration [tex]g=0.412 m/s^2[/tex]

Generally the equation for Escape velocity is mathematically given by

 [tex]V.E^2=2gr[/tex]

 [tex]V.E^2=2*0.412m/s^2*301000[/tex]

 [tex]V.E^2=248024[/tex]

 [tex]V.E=\sqrt{248024}[/tex]

 [tex]V.E=498.02m/s^2[/tex]

Answer:

519.62 m/s

Explanation:

Applying,

v = √(T/m').............. Equation 1

Where v = velocity of the wave, T = Tension on the string, m' = mass per unit length of the string

From the question,

Given: T = 1350 N, m' = 0.005 kg/m

Substitute these values into equation 1

v = √(1350/0.005)

v = √(270000)

v = 519.62 m/s

Answer:

im pretty sure it should be 50.0

Answer:

It is proved.

Explanation:

PV = n R T

where,  P is the pressure, V is the volume, n is the number of moles, R is the gas constant, T is  the absolute temperature.

Unit of pressure is Pascal= Newton/square meter = [tex][ML^{-1}T^{-2}][/tex]

Unit of volume is cubic meter = [tex][L^{3}][/tex]

So, the unit of P V

[tex][ML^{-1}T^{-2}] \times [L^{3}]\\\\[ML^2T^{-2}}[/tex]

[tex][ML^{2}T^{-2}][/tex]

It is same as the unit of energy.

hence proved.  

Answer:

The correct option is a) 10 mph, 0 mph.

Explanation:

1. The average speed (S) is a magnitude given by:

[tex] S = \frac{D}{T} [/tex]  

Where:

D: is the total distance = 1 mi

T: is the total time = 6 min

[tex] S = \frac{D}{T} = \frac{1 mi}{6 min}*\frac{60 min}{1 h} = 10 mph [/tex]

Hence, the average speed is 10 mph.

2. The average velocity is a vector:

[tex] V = \frac{\Delta d}{\Delta t} = \frac{d_{f} - d_{i}}{t_{f} - t_{i}} [/tex]

Where:

[tex]d_{f}[/tex]: is the final distance                                    

[tex]d_{i}[/tex]: is the initial distance  

[tex]t_{f}[/tex]: is the final time                

[tex]t_{i}[/tex]: is the initial time

Since Juanita ran one mile around her school track, the final position is the same that the initial position, so the magnitude of the average velocity is zero.                                               

Therefore, the correct option is a) 10 mph, 0 mph.

I hope it helps you!          

Answer:

3420.39 N

Explanation:

Applying,

Fd = 1/2(mv²-mu²)................. Equation 1

Where F = force on the bumber, d = distance, m = mass of the car, v = final velocity, u = initial velocity.

make F the subject of the equation

F = (mv²-mu²)/2d............... Equation 2

From the question,

Given: m = 890 kg, v = 0 m/s (to rest), u = 1.4 m/s, d = 0.255 m

Substitute these values into equation 2

F = [(890×0²)-(890×1.4²)]/(2×0.255)

F = -1744.4/0.51

F = -3420.39 N

The negative sign denotes that the force in opposite direction to the motion of the car.

[tex]\bold{{Answer}}[/tex]

O physical, chemical, respectively

please buddy correct me if im wrong

The ability of a metal to react with air and water is a  chemical, chemical, respectively property

A chemical change occur when the substance's composition is changed or when bonds are broken and new ones are forms a chemical change occur .

When a metal burn in air or react with air it form metal oxide and when metal react with water it forms oxides or hydroxides and release hydrogen gas . In both the cases change in composition is taking place . hence , bot are chemical reactions

correct answer is O chemical, chemical, respectively

learn more about chemical change

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Answer:

English only

Explanation:

When solving problems related to Electric Fields, care must be taken about symmetries. In our particular case when we take a look to at the drawings of the attached file, we realize:

1.-By symmetry each dx associated at a, has an opposite dx with point b as reference. The respective dE ( the charge is uniform ) is the same, as the charge of the wire is positive the force and the Field on a test charge (+) located at h will be upward, therefore the components dEx will cancel each other and the Electric Field becomes E = Ey = ∫ 2×dE× cosθ

The solutions:

A) Ey = 4623 N/C

B) Ey = 19.34 N/C

E = Ey = ∫ 2×dE× cosθ

Here     cosθ   = h/ d   ⇒  cosθ = h/√h² + x²      dE = K× dQ / d²

d² = h² + x²

k = 8.9 ×10⁹ Nm²C⁻²  ;   dQ = λ×dx     λ = 150×10⁻⁹ C    h = 0.08 m

Then by substitution

Ey =  2 ∫[K× λ×dx/ (h² + x²) ] × h / √h² + x²

reordering that equation:

Ey = 2×K×λ×h ∫ dx / [√ ( h² + x² ) ]³          (2)

To solve the integral we make use of a change of variables

x = h × tanα     then   x² = h² ×tan²α   and  dx = h× sec²α dα

plugging that values in equation (2)

Ey  =  2×K×λ×h ∫  h× sec²α× dα / [√ ( h² + h²tan²α)]³

Ey  = 2×K×λ×h² ∫ sec²α× dα / [ h × √ (1 + tan²α)]³            1 + tan²α = sec²α

Ey = 2×K×λ×h²× ∫ (sec²α / h³× sec³α )×dα

Ey = 2×K×λ/h × ∫ ( 1 / secα dα

Ey = 2×K×λ/h × sinα             now we αneed to come back to our original variables:

as   x = h × tanα         tanα = x/h   then x is the opposite leg in a right triangle  and h the adjacent one then the hypothenuse is √ (h² + x²)         then    sin α = x/ √ (h² + x²)      

Ey = 2×K×λ/h × x/ √ (h² + x²) |₀⁰°⁰⁵

Ey  = 2×8.9×10⁹× 150×10⁻⁹× 5×10⁻²/8× 10⁻²× √ 10⁻² ( 8 + 5 )   N/C

Ey = 4623 N/C

To answer the second question again we will make use of symmetries if you look at drawing ( Figure 2 ) you see that again the components in direction of x-axis cancel each other and the components in y-axis direction will add. Then

Ey = ∫ dE× cosθ

following the same procedure  we will find:

Ey = ∫ [K×λ × dl/d²] × h/ d

The importan point here is that the radius of the circle is

2×π×r = 0.01      ( the length of the wire)  ⇒  r = 0.16×10⁻² m

And we need to take into account that the integration is over the circle and the length of the circle is 0.01 m or ××2×π×r. All other factors are constant. Then by substitution

Ey = [K×λ ×h×  / ( √ r² + h²)³ ] × 10⁻²    N/C

Ey = 8.9 × 10⁹ × 150× 10⁻⁹ × 6× 10⁻² × 10⁻² / √ 10⁻² ( 0.16 + 6)

Ey = 0.8 × 10² / 6

Ey = 19.34 N/C

Answer:

the efficiency of the machine is 100%

Explanation:

Given;

load, L = 750 N

length of the lever, L = 2 m

effort applied, E = 250 N

Position of the load from the fulcrum,  = 50 cm

                                     50cm

     0↓--------------------------Δ-------------------------------↓---------200 cm

    750 N                                       x cm                 250 N

Apply the principle of moment;

750(50) = 250(x)

x = (750 x 50) / (250)

x = 150 cm

the distance of the effort = 150 cm = 1.5 m

the distance of the load = 50 cm = 0.5 m

The velocity ratio of the machine = 1.5/0.5

                                                        = 3

The mechanical advantage of the machine is calculated as;

M.A = Load/effort

M.A = 750 / 250

M.A = 3

The efficiency of the machine is calculated as;

E = (M.A / V.R) x 100%

E = (3/3) x 100% = 100%

Therefore, the efficiency of the machine is 100%

Answer:

C. 28.2 deg

Explanation:

The horizontal range of a projectile is given as:

[tex]R = \frac{v^2Sin2\theta}{g}[/tex]

where,

R = Range

v = speed

θ = angle of launch

g = acceleration due to gravity = 9.81 m/s²

First, we will find the launch speed (v) by using the initial conditions:

R = 120 m

θ = 45°

Therefore,

[tex]120\ m = \frac{v^2Sin 90^o}{9.81\ m/s^2}\\\\v = \sqrt{(120\ m)(9.81\ m/s^2)}\\\\v = 34.31\ m/s[/tex]

Now, consider the second scenario to hit the target:

R = 100 m

Therefore,

[tex]100\ m = \frac{(34.31\ m/s)^2Sin2\theta}{9.81\ m/s^2}\\\\Sin2\theta = \frac{(100\ m)(9.81\ m/s^2)}{(34.31\ m/s)^2}\\\\2\theta = Sin^{-1}(0.833)\\\\\theta = \frac{56.44^o}{2}\\\theta = 28.22^o[/tex]

Hence, the correct option is:

C. 28.2 deg

Answer:

345 K

Explanation:

Temperature can be defined as a measure of the degree of coldness or hotness of a physical object.

Generally, it is measured with a thermometer and its units are Celsius (°C), Kelvin (K) and Fahrenheit (°F).

Given the following data;

To convert the temperature in degree Celsius to Kelvin, we would use the following mathematical expression;

Kelvin = 273 + °C

Substituting into the formula, we have;

Kelvin = 273 + 72

Kelvin = 345 K

Therefore, the temperature of 72°C will be equivalent to 345 K on the Kelvin scale.

Answer:

$900

Explanation:

Step 1: Our output value is 9000.

Step 2: We represent the unknown value with x.

Step 3: From step 1 above,$9000=100\%$

Step 4: Similarly, x=10%

Step 5: This results in a pair of simple equations:

$9000=100

Step 6: By dividing equation 1 by equation 2 and noting that both the RHS (right hand side) of both

equations have the same unit (%); we have

\frac{9000}{x}=\frac{100\%}{10\%}

Step 7: Again, the reciprocal of both sides gives

\frac{x}{9000}=\frac{10}{100}$

\Rightarrow x=900$

Therefore, $10\%$ of $9000$ is $900$

unknown value with x

step 1 above,$9900=100%

similarly ,x=10%

Answer: hi your question is incomplete  attached below is the complete question

answer :

magnitude of acceleration :  | a | = g = 9.81 m/s^2

direction : a = - g j

Explanation:

Neglecting Air resistance

magnitude of acceleration :

| a | = g = 9.81 m/s^2

Direction of acceleration

a = - g j  ( given that the direction of acceleration is against the acceleration due to gravity i.e. in the opposite direction )

Answer:

1.the friction of air, gravity2.gravity

Answer:

The softball experiences an applied force as a result of Amy’s throw. As the ball moves, it experiences drag from the air it passes through. It also experiences a downward pull because of gravity.

Explanation:

Plato

Answer:

i) a₁ = -g (sin θ + μ cos θ), x = v₀² / 2a₁

ii) W = mg L sin  θ ,  iii)     Wₙ = 0

iv)  W = - μ m g  L cos  θ x

Explanation:

With a drawing this exercise would be clearer, I understand that you have a block on a ramp and it is subjected to some force that makes it rise, for example the tension created by a descending block.

The movement is that when the system is released, the tension forces are greater than the friction and the component of the weight and therefore the block rises up the ramp

At some point the tension must become zero, when the hanging block reaches the ground, as the block has a velocity it rises with a negative acceleration to a point and stops where the friction force and the weight component would be in equilibrium along the way. along the plane

i) Let's use Newton's second law

the reference system is with the x axis parallel to the ramp

Axis y

      N - W cos θ = 0

X axis

      T - W sin θ - fr = ma

the friction force is

      fr = μ N

      fr = μ mg cos θ

we substitute

      T - m g sin sin θ - μ mg cos θ = m a

      a = T / m - g (sin θ + μ cos θ)

With this acceleration we can find the height that the block reaches, this implies that at some point the tension becomes zero, possibly when a hanging block reaches the floor.

      T = 0

       a₁ = -g (sin θ + μ cos θ)

       v² = v₀² - 2a1 x

       v = 0       at the highest point

       x = v₀² / 2a₁

ii) the work of the gravitational force is

       W = F .d

       W = mg sin  θ   L

iii) the work of the normal force

the force has 90º with respect to the displacement so cos 90 = 0

         Wₙ = 0

iv) friction force work

friction force always opposes displacement

         W = - fr d

         W = - μ m g cos  θ L

Answer:

Friction always acts opposite to the motion.

Answer:

Refraction of waves involves a change in the direction of waves as they pass from one medium to another. Refraction, or the bending of the path of the waves, is accompanied by a change in speed and wavelength of the waves.