Answer:
U = - ∫ F dx
U = - 3F₀
Explanation:
Force and potential energy are related
F = [tex]- \frac{dU}{dx}[/tex]
dU = - F dx
∫ dU = - ∫ F dx
U- U₀ = - ∫ F dx
to solve the problem we must know the form of the force, if we assume that F = F₀ and U₀ = 0 for x = 0
U = - F₀ 3
The boiling point of a substance is _72 degree Celsius. This temperature will be equivalent to Kelvin scale is-------.
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;
Boiling point = 72°CTo 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.
In which region is there most likely to be a volcano
A
B
C
D
A cannon sitting on level ground is aimed at 45.0 degrees relative to the horizontal. It fires a test shot at a target located 100.0 meters away from the cannon on the same level ground. The test overshoots the target by 20.0 meters. Which of the following angles can the cannon be adjusted to to hit the target. You may neglect air resistance and assume the cannon always delivers the same initial velocity to the cannonball .
A. 35.9 deg
B. 49.1 deg
C. 28.2 deg
D. 52.8 deg
E. 22.7 deg
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
(a) What is the escape speed on a spherical asteroid whose radius is 301 km and whose gravitational acceleration at the surface is 0.412 m/s2
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]
State any ten reasons why students in Uganda need to study physics.(use examples of the possible technological advancements which can be made using the knowledge of physics)
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.
A piano string of density 0.0050 kg/m is under a tension of 1,350 N. Find the velocity with which a wave travels on the string.
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
What is the magnitude of the electric field at a point 0.0055 m from a 0.0025
C charge?
kg
Use E = and k=9.00 x 10 N.m²/C2.
O A. 7.4 x 1011 N
O B. 2.0 x 1010 N
O C. 4.1 x 10°N
OD. 7.9 x 1012 N
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
Which wave interaction results in a change in the direction of the wave as it passes through one medium to another medium?
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.
please help me with this question
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
Rick works off commission. He earns 10 percent of all manufacturing equipment he sells. if he made a sale of $9,000 how much was his commission
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%
(12 points) Analysis from the point where the block is released to the point where it reaches the maximum height i) Calculate the highest height reached by the block (or the largest distance travelled along the ramp.) ii) Calculate the work done by the gravitational force. iii) Calculate the work done by the normal force. iv) Calculate the work done by the friction force.
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
a ball is thrown straight up into the air while the ball is traveleling upwards what are the magnitue and direction
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 )
20 swings takes 5 seconds in the pendulum. calculate the periodic time of the swing
Answer:
0.25s
Explanation:
5/20 = 0.25s
This might be correct
In an experiment the mass of a calorimeter is 36.35 g . Express in micrometer ,millimetre and kg.
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;
Mass of calorimeter = 36.35 gramsTo 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:
g is the symbol for grams.Exp (E) means exponential = 10um is the symbol for micrometer.Complete the sentence-
Friction always acts
1 along the direction of the motion.
2 opposite to the motion.
3 both of these.
4 none of these.
Answer:
Friction always acts opposite to the motion.
Un alambre de plástico, aislante y recto mide 10 cm de longitud y tiene una densidad de carga de +150 nC/m, distribuidos de manera uniforme por toda su longitud. Se encuentra sobre una mesa horizontal. A) Encuentre la magnitud y la dirección del campo eléctrico que produce este alambre en un punto que está 8 cm directamente arriba de su punto medio. B) Si el alambre ahora se dobla para formar un círculo que se coloca aplanado sobre la mesa, calcule la magnitud y la dirección del campo eléctrico que produce en un punto que se encuentra 6 cm directamente arriba de su centro.
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
show that pv=nrt has the same unit with energy
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.
How much heat is required to evaporate 0.15 kg of lead at 1750°C, the boiling point for lead? The heat of vaporization for lead is Lv = 871 × 103 J/kg.
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.
What is heat?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
for the equation BaCI2 + Na2SO4 > BaSO4 + 2NaCI
A. reactants: 1 ;products: 1
B. reactants: 1 ;products: 2
C. reactants: 2 ;products: 1
D. reactants: 2 ;products: 2
please help me with my question due tomorrow morning,
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
Amy throws a softball through the air. What are the different forces acting on the ball while it’s in the air?
The softball experiences
force as a result of Amy’s throw. As the ball moves, it experiences
from the air it passes through. It also experiences a downward pull because of
.
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
URGENT
A student runs at 4.5 m/s [27° S of W] for 3.0 minutes and then he turns and runs at 3.5 m/s [35° S of E] for 4.1 minutes.
a. What was his average speed?
b. What was his displacement?
PLEASE SHOW ALL WORK
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]
In an electromagnet, what produces the magnetic field?
A. Electric charges moving through a conducting wire.
B. Electric charges moving through the metal core.
C. The metal core within a coil of wire.
D. The magnetic field of a permanent magnet.
Which statement is the best interpretation of the ray diagram shown?
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:
Un proton penetra perpendiculares en un campo magnetico de 5 teslas con una velocidad de 2.10 m/s calcula
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]
Complete the sentence below using one of the following
words: equilibrium, flux, motion.
If supported at its center of gravity, an object will remain in ______ any position.
Answer:
equilibrium
Explanation:
acceleration will remain constant
Juanita ran one mile around her school track in six minutes. What is
her average speed, and what is the magnitude of her average velocity?
10 mph, 0 mph
6 mph, 0 mph
6 mph, 6 mph
10 mph, 10 mph
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!
If 6000 J of heat is added to 200 gm of water at 25° C. What will be its final
temperature?
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.
Q011) The Doppler effect a. occurs when the frequency of sound waves received is lower if the wave source is moving toward you than if it's moving away. b. occurs when the pitch of a sound gets lower if the source is receding. c. is the basic explanation for the blue shift of light in our Universe. d. can be applied only to sound waves.
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.
A uniform metre ruler scale balanced at 40 cm mark, when weight 25 gf and 10gf are suspended at 10cm mark and 75 cm mark respectively.Calculate the weight of the metre scale.
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.
