is acting on an object that is constrained to move along the x-axis. If the force moves the object 3m, from the origin, what is the potential energy difference due to this force acting on the object

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:

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:

[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:

0.25s

Explanation:

5/20 = 0.25s

This might be correct

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:

equilibrium

Explanation:

acceleration will remain constant

Answer:

Explanation:

The equation for the electric field is

[tex]E=\frac{kQ}{r^2}[/tex] so filling in:

[tex]E=\frac{9.00*10^9(.0025)}{(.0055)^2}[/tex] which in the end gives you

E = 7.4 × 10¹¹, choice A

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:

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:

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.

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 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: 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:

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:

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:

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.

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:

Option (c) is correct.

Explanation:

The apparent change in the frequency of light due to the relative motion between the source and the observer is called Doppler's effect.

When the source is moving towards the observer which is at rest, the apparent frequency increases and if the observer is moving away the frequency of sound decreases.

It occurs for both light and sound.  

So, to explain the blue shift of light in the universe is due to the Doppler's effect of light.

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:

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:

D)7 1/2 or 15/2

Explanation:

Let's calculate Combined resistance of the parallel first

1/Rt= 1/2+1/6=4/6

Rt=6/4 which is also equal with 3/2

Now let's add it with the series one

Rt= 6+3/2

=15/2 And when we put that un a mixed fraction 7 1/2

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:

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:

(a) 3.93 m/s

(b) 861.66 m

Explanation:

A = 4.5 m/s  [27° S of W] for 3.0 minutes

B = 3.5 m/s [35° S of E] for 4.1 minutes

Distance A = 4.5 x 3 x 60 = 810  m

Distance B = 3.5 x 4.1 x 60 = 861 m

(a) The average speed is defined as the ratio of the total distance to the total time.

Total distance, d = 810 + 861 = 1671 m

total time, t = 3 + 4.1 = 7.1 minutes = 7.1 x 60 = 426 seconds

The average speed is

[tex]v=\frac{1671}{426}=3.93 m/s[/tex]

(b)

[tex]\overrightarrow{A} = 810(- cos 27 \widehat{i} - sin 27 \widehat{j})=- 721.7 \widehat{i} - 367.7 \widehat{j}\\\\\overrightarrow{B} = 861( cos 35 \widehat{i} - sin 35 \widehat{j})= 705.3 \widehat{i} - 493.8 \widehat{j}\\\\\overrightarrow{C} = (- 721.7 + 705.3) \widehat{i} - (367.7 + 493.8) \widehat{j} \\\\\overrightarrow{C}= - 16.4 \widehat{i} - 861.5 \widehat{j}[/tex]

The magnitude is

[tex]C =\sqrt{16.4^2+861.5^2} = 861.66 m[/tex]

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:

40 gf.

Explanation:

Please see attached photo for diagram.

In the attached photo, W is the weight of metre rule.

The weight of the metre rule can be obtained as follow:

Clockwise moment = (W×10) + (10×35)

Clockwise moment = 10W + 350

Anticlock wise moment = 25 × 30

Anticlock wise moment = 750

Clockwise moment = Anticlock wise moment

10W + 350 = 750

Collect like terms

10W = 750 – 370

10W = 400

Divide both side by 10

W = 400/ 10

W = 40 gf

Thus, the weight of the metre rule is 40 gf

Answer:

40 gf

Explanation:

The balance point of the uniform meter rule with the suspended weights = 40 cm = The pivot point

The location where the 25 gf weight is suspended = 10 cm

The location where the 10 gf weight is suspended = 75 cm

Let W represent the weight of the meter rule.

We have that the location of the application of the weight of the meter rule is at the center, 50 cm mark, point

Given that the meter rule is balanced, and taking moment about the pivot point, we have;

The moment om the left hand side, LHS, of the pivot point = The moment on the right hand side, RHS, of the pivot point

The moment on the LHS = 25 gf × (40 cm - 10 cm) = 750 gf·cm

The moment on the RHS = W × (50 cm - 40 cm) + 10 gf × (75 cm - 40 cm)

The moment on the RHS = W·(10 cm)  + 350 gf·cm

∴ 750 gf·cm = W·(10 cm) + 350 gf·cm

W·(10 cm) = 750 gf·cm - 350 gf·cm = 400 gf·cm

W = (400 gf·cm)/(10 cm) = 40 gf

The weight of the meter scale (rule), W = 40 gf.