Answer:
suna bro i am nepali and nepali are pro
Can i use a 1000mAh battery instead of 600mAh Battery?
please tell me quickly
To deliver more power or voltage in the circuit use a battery with more current rating which is 1000 mAh.
Power generated in a circuitThe power generated in a circuit is directly proportional to the current flowing in the circuit.
P = IV
where;
I is currentV is voltageVoltage in the circuit also increases with increase in the current flowing in the circuit.
Thus, to deliver more power or voltage in the circuit use a battery with more current rating which is 1000 mAh.
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A uniform plate of height 1.870 m is cut in the form of a parabolic section.
The lower boundary of the plate is defined by: y = 0.500[tex]x^{2}[/tex]. Find the distance from the rounded tip of the plate to the center of mass.
The distance from the rounded tip of the plate to the center of mass is 1.87 m.
What is center of mass?The center of mass is a point inside or outside the mass where all of the mass is concentrated.
The y-coordinate of the centroid is given by the ratio of two definite integrals;
Yc = ∫ydm/∫dm,
where dm is a density function
For the uniform plate, δ does not change with position in the plate.
Yc = ∫yδdA/∫δdA
Yc = ∫ydA/∫dA.
dA is a horizontal slice of the plate with dimensions xdy.
Solving the parabola for x,
y = 0.5x²
x = ± √(y/0.50), where the negative value corresponds to the left half of the parabola and the positive to the right half.
dA = (√(y/0.50)
= √(y/0.50))dy
= 2(√(y/0.50))dy
The limits of integration are from zero to 1.870, the top of the plate.
∫ydA = ∫2y√(y/0.50)dy = 7.232 m³
∫dA = ∫2√(y/0.50)dy = 3.868 m²
∫ydA/∫dA = 7.232 m³/3.868 m²
∫ydA/∫dA = 1.869700 m
∫ydA/∫dA = 1.87 m
Thus, the distance from the rounded tip of the plate to the center of mass is 1.87 m.
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A baseball is thrown at angle of 65 degrees with an initial velocity of 45 m/s. How long will the baseball remain in the air?
Assume yi = yf =0 m
G= -9.81 m/s^2
A) 4.15 seconds
B) 8.31 seconds
C) 9.37 seconds
D) 1.44 seconds
The time for which the baseball remains in the air will be 9.37 sec. Option C is correct.
What is the maximum height achieved in projectile motion?It is the height achieved by the body when a body is thrown at the same angle and the body is attaining the projectile motion.
From Newton's first equation of motion;
[tex]\rm v_y= u_y+gt \\\\ 0 = usin \theta - gt \\\\ t = \frac{u sin\theta }{g}\\\\t = 4.68 \ sec[/tex]
The time the baseball remains in the air will be twice the time he travels for one side motion;
T' = 2t
T'=2 × 4.68 sec
T'= 9.37 sec
Hence for 9.37 sec, the baseball willl remains in the air. Option C is correct.
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What is the acceleration of Karla's I Phone being thrown from Mr. Higley's classroom at 0m/s if it hits the wall 1.2 seconds later going 35 m/s?
The acceleration of Karla's I IPhone being thrown from Mr. Higley's classroom at 0 m/s will be 29.16 m/s²
What is acceleration?The rate of velocity change concerning time is known as acceleration. According to Newton's second law, the eventual effect of all forces applied to a body is its acceleration.
The pace at which a body's velocity varies is represented by acceleration, which is a vector quantity.
The given data in the problem is given by ;
u is the initial speed = 0 m/sec
v is the final speed= 35 m/sec
t is the time interval= 1.2 second
a is the acceleration=? m/sec²
The formula for acceleration is;
[tex]\rm a=\frac{v-u}{t} \\\\ a= \frac{35-0}{1.2} \\\\ a= 29.16 \ m./s^2[/tex]
Hence, the acceleration of Karla's iPhone being thrown from Mr. Higley's classroom at 0 m/s will be 29.16 m/s²
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Please I need help fast. Assignment is due soon.
1) A ball is thrown from the top of a building with an initial speed of 8.0 m/s at an angle of 35° above the horizontal. The building is 18 m tall.
a) How long is the ball in the air?
b) How far from the building does the ball land?
c) What is its impact speed?
2) A 65-kg person driving a car hits the gas, accelerating the car at a rate of 3.9 m/s^2. Find the magnitude and direction of the force exerted by the seat on the person's body. Remember to include both the horizontal and the vertical components.
(a) The time spent in the air by the ball is 1.5 seconds.
(b) The horizontal distance of the ball is 9.83 m.
(c) The impact speed of the ball is 20.37 m/s.
(2) The magnitude of the force is 253.5 N in horizontal direction.
Time of motion of the ball
The time of motion of the ball is calculated as follows;
h = vt + ¹/₂gt²
18 = 8sin(35)t + (0.5)(9.8)t²
18 = 4.589t + 4.9t²
4.9t² + 4.589t - 18 = 0
solve the quadratic equation using formula method,
t = 1.5 s
Horizontal distance of the projectileX = vcosθ(t)
X = 8cos(35) x 1.5
X = 9.83 m
Impact speed of the projectilevyf = vyi + gt
vyf = 8sin(35) + 9.8(1.5)
vyf = 19.289 m/s
vxf = vxi
vxf = 8 x cos(35)
vxf = 6.55 m/s
Resultant speed = √(19.289² + 6.55²) = 20.37 m/s
Magnitude of the force exerted by the seat on the personF = ma
F = 65 x 3.9
F = 253.5 N
Since the motion is horizontal, the angle of the motion is zero.
Fx = 253.5cos(0) = 253.5 N
Fy = 253.5sin(0) = 0 N
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please help me in this questions
Answer:
Rainy daywindy daysunny daycloudy daysorry I don't know the answer of question 8.sunglassesumbrella:-(:-):-):-(day ☀️night day ☀️night Day ☀️Day ☀️night nightExplanation:
Hope I give all correct answer please mark as brainlest answer
Select. The star cycle that is accurate
One stellar-mass star, red giants, white dwarfs, or planetary nebulae will make up the correct star cycle. Option C is correct.
What is the solar system?The satellites of the planet, countless comets, asteroids, and meteoroids, as well as the interplanetary medium, make up the solar system.
The complete question is;
"Choose the correct star life cycle.
A. Supernova, star as well as red giant, the nebula is incorrect.
B. Nebula, white dwarf, planetary nebula That is incorrect.
C. Planetary nebula, red giant, white dwarf, and star with a single stellar mass That is true!
D. Nebula, a star of one stellar mass, is not correct. It is correct if the star had four or more stellar masses."
Star of one stellar mass, red giant, white dwarf, and planetary nebula make up the proper life cycle.
Hence option C is correct.
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Need help with this question
Answer:
1. 92, 138, 92
2. 7, 7, 7
3. 17, 18, 17
4. 29, 34, 29
Explanation:
the number on the bottom is always going to be the atomic number, or the number of protons.
the number on the top is going to be the atomic mass number, which is the sum of the number of protons and neutrons.
there is always going to be the same number of electrons as protons, because that was the overall charge of each atom is 0.
for example, for the first one:
there are 92 protons
there are 230 - 92 = 138 neutrons
there are 92 electrons
A resistor has a resistance of 8.2 kΩ. If a voltage of 15.9 V were placed across it, what would be the current, in mA? Give the answer to two decimal places; don't worry if the computer adds zeroes.
The value of the electric current for the given conditions willl be 1939.02 A.
What is ohm’s law?Ohm's law claims that the voltage across a conductor is directly proportional to the current flowing through it.
Ohm's law claims that the voltage across a conductor is direct to the current flowing through it. This current-voltage connection may be expressed mathematically as,
V=IR
15.9 V = I × 8.2 × 10³ Ω
I = 1939.02 A
Hence, the value of the electric current for the given conditions willl be 1939.02 A.
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Which has more mass - a 5-kg bag of feathers or a 5-kg
cannon ball?
Answer:equal mass
Explanation:the mass of both is 5kg
a 2kg aluminum block and a6kg copper block are connected by alight string over a frictionless pulley and fixed steel block of angle 30 degree .if the coefficient of friction on the surface is 0.2,find the acceleration of the two block and the tension in the string?
(a) The acceleration of the two block is determined as 0.93 m/s².
(b) The tension in the string is 5.78 N.
Net force on the aluminum block
The net force on the aluminum block is calculated as follows;
[tex]T - \mu_a m_a g = m_a a \ --- \ (1)[/tex]
Net force on the copper blockThe net force on the copper block is calculated as follows;
[tex]m_cg sin(30) - T -\mu _cm_c gcos(30) = m_c a --- (2)[/tex]
where;
T is tension in the stringma is mass of aluminummc is mass of copperg is acceleration due to gravitySolve for T using (1) and (2)
[tex]a = g(\frac{m_c sin30\ - \ \mu _c m_c cos30 \ - \mu_ a m_a}{ma_a + m_c} )\\\\a = 9.8(\frac{6 sin30\ - \ 0.2 (6) cos30 \ - 0.2 (6)}{2 +6} )\\\\a = 0.93 \ m/s^2[/tex]
Tension in the stringFrom equation (1);
T = μm_ag + m_aa
T = 0.2(2)(9.8) + 2(0.93)
T = 5.78 N
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The heat capacity of nickel is 0.444 J/(g · °C). Calculate the amount of heat needed to raise the temperature of 18 g of nickel from 20 °C to 66 °C. Now imagine those same joules were used instead to accelerate the same mass of nickel from rest. What would be the final speed, in m/s?
Hints:
The formula for kinetic energy is: Ek=1/2 mv^2
For joules needed for heating, use the mass in grams, but for kinetic energy, convert the mass to kilograms.
Round your answer to the nearest whole number.
The final speed of the nickel at the given quantity of heat is determined as 202.1 m/s.
Final speed of the nickelApply the principle of conservation of energy.
Q = mcΔθ
Q = (18)(0.444)(66 - 20)
Q = 367.63 J
Q = K.E = ¹/₂mv²
2K.E = mv²
v = √(2K.E/m)
where;
v is the final speedv = √(2 x 367.63)/(0.018))
v = 202.1 m/s
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A teacher wants to demonstrate that the radioactive source emits alpha beta and gamma radiation. Describe a method the teacher could use (IMPORTANT REPLAY ASAP) I will give Brainly
By using an electric field, it is feasible to differentiate between these different forms of radiation.
What is a radioactive source?A source that emits radiation like gamma, beta, and alpha rays is said to be radioactive. Using an electric field, we can discriminate between these different forms of radiation.
The field does not deflate the gamma rays, but it does deflate the alpha and beta rays, with the alpha being deflated to the field's negative portion and the beta to its positive part.
Hence, by using an electric field, it is feasible to differentiate between these different forms of radiation.
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It is important that a frame of reference
Answer:
enable us to take in a wide variety of information, and process it based on our past experience and values.
Explanation:
The following table contains financial information for Trumpter Inc. before closing entries:
Cash
Supplies
Prepaid Rent
Salaries Expense
Equipment
Service Revenue
Miscellaneous Expenses
Dividends
Accounts Payable
Common Stock
Retained Earnings
$13,100
6,000
3,800
5,000
65,900
28,000
20,100
4,000
3,100
66,100
20,700
What is the amount of Trumpter's total assets?
The Total assets based on the given table is $88,800.
What are assets?An asset refers to a resource owned by an individual or company which has tangible value and whose value is expected to rise in the future.
The Total assets based on the table = Cash + Supplies + Prepaid rent + Equipment
Total assets = 13,100 + 6,000 + 3,800 + 65,900
Total assets = $88,800
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The value of acceleration due to gravity (g) on a point 10,000 kilometers above sea level is about 1.49 meters/second2. How much will an object, which weighs 98 newtons on the surface of Earth, weigh on this point? The value of acceleration due to gravity on Earth is 9.8 meters/second2.
An object, which weighs 98 N on the surface of Earth, weigh on this point is 2.23 x 10²⁴ kg.
What is gravity?The force of attraction felt by a person at the center of a planet or Earth is called as the gravity.
Given, the Earth has the acceleration due to gravity, g = 9.81 m/s².
Force of gravity W = mass x acceleration due to gravity
98N = m x 9.8m/s²
m = 10 kg
The value of acceleration due to gravity (g) on a point 10,000 km above sea level is about 1.49 m/s².
The acceleration due to gravity and mass is related as
g = GM/R²
where G = gravitational constant = 6.67 x 10⁻¹¹ N.m²/kg² and R is the distance between two masses.
Substituting the values, we get
1.49 m/s² = 6.67 x 10⁻¹¹ N.m²/kg² x M/ (10000 x 10³ m)²
M = 2.23 x 10²⁴ kg
Therefore, an object will weigh on this point is 2.23 x 10²⁴ kg
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As a warehouse worker pushes a crate across a concrete floor, the force he
applies is not perfectly horizontal, as shown in the image below. If the
coefficient of kinetic friction between the crate and concrete floor is 0.5, what
is the net force on the crate?
A. 136 N
B. 99 N
C. 73 N
D. 112 N
The net force on the crate will be 99 N.Option B is correct.
What is the friction force?It is a type of opposition force acting on the surface of the body that tries to oppose the motion of the body. its unit is Newton (N). it is defined as the product of the coefficient of friction and normal reaction.
On resolving the given force and acceleration.Mathematically in the different components and balancing the equation gets.Components in the x-direction.
When all the forces get resolved the y-direction of forces are;
300 sin 10° +N = 445
N = 445-295.4
N = 392.9 N
The normal force is 392.9 N.
The net force on the crate is found by resolving the force in the x-direction;
F = 300 cos 10° - μN
F= 295.44-196.45
F= 98.99
F=99
Hence, the net force on the crate will be 99 N.Option B is correct.
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A hydrogen bond forms by the electrostatic interaction of opposite charges in two molecules. If
the bond length is 2x10 -10 m and the magnitude of the charges involved is
approximately 1.60x10 -20C, what is the force between the molecules involved in the bond?
The force between the molecules involved in the bond is 6. 426 *10^-11 Newton
How to determine the force
Using the formula:
F = K[q1 x q2]/D^2
where K is coulombs constant =9 *10 ^9 Nm^2/C^2.
q1 and q2 = charges = 1.60x10 -20C
d = distance between the charges = 2x10 -10 m
Substitute the values into the formula
F = [tex]9 * 10^9\frac{ 1.60*10^ -20 * 1.60 *10^ -20}{2x10^ -10^{2} }[/tex]
F = [tex]9 *10^9\frac{2. 856* 10^-40}{4* 10^-20}[/tex]
F = [tex]9* 10^9 * 7. 14* 10^-21[/tex]
F = [tex]6. 426 * 10^-11[/tex] Newton
Thus, the force between the molecules involved in the bond is 6. 426 *10^-11 Newton
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Two identical 3.0-kg cubes are placed on a horizontal surface in
contact with one another. The cubes are lined up from left to
right and a force F₁ is applied to the left side of the left cube
causing both cubes to move at a constant speed v. If the
coefficient of kinetic friction between the cubes and the
surface is 0.3, what is the magnitude of the force exerted by the
left cube on the right cube?
(3
The magnitude of the force exerted by the left cube on the right cube is 17.64N.
What is frictional force?When an object is moving on a rough surface, it experiences opposition. This opposing force is called the friction.
Two identical 3.0-kg cubes are placed on a horizontal surface in contact with one another. The cubes are lined up from left to right and a force F₁ is applied to the left side of the left cube causing both cubes to move at a constant speed v. The coefficient of kinetic friction between the cubes and the surface is 0.3.
From the equilibrium of forces in vertical direction
Normal force N= 2m x g
friction force f = μN =μ(2m)g
From the equilibrium of forces in horizontal direction
F₁ =ma =0
using Newton's third law of motion, we get
F₁ - f =0
F₁ =f = μ(2m)g
Put the values, we get
F₁ = 17.64N
Thus, magnitude of the force exerted by the left cube on the right cube is 17.64N.
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A contestant in a winter games event pushes a 54.0 kg block of ice across a frozen lake as shown in Figure 4.29(a). The coefficient of static friction is 0.1 and the coefficient of kinetic friction is 0.03.
a) Calculate the minimum force F he must exert to get the block moving.
N
(b) What is its acceleration once it starts to move, if that force is maintained?
m/s2
Answer:
54.0 x0.1=5.4 x0.03=0.162
kinetic force
If a contestant in a winter games event pushes a 54.0 kg block of ice across a frozen lake. The coefficient of static friction is 0.1 and the coefficient of kinetic friction is 0.03, then the minimum force F he must exert to get the block moving would be 5.2974 Nwrons.
What is friction?
Friction is a type of force that resists or prevents the relative motion of two physical objects when their surfaces come in contact.
As given in the problem, If a contestant in a winter games event pushes a 54.0 kg block of ice across a frozen lake. The coefficient of static friction is 0.1 and the coefficient of kinetic friction is 0.03,
The force required to get the block moving = μMg
= 0.01×54×9.81
= 5.2974 Newtons
Thus, the minimum force required to move the block would be 5.2974 Newtons.
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A sled is moving down a steep hill. The mass of the sled is 50 kg and the net force acting on it is 20 n. What must be done to find the acceleration of the sled? check all that apply.
The acceleration of this sled can be calculated by doing: option A, B, E and F.
How to calculate the acceleration?The acceleration of this sled can be calculated by using Newton's Second law of motion. Mathematically, the acceleration of an object is given by this formula:
Net force = Mass × acceleration
Deductively, the acceleration of this sled can be calculated by doing the following:
The force of gravity must be broken into its parallel and perpendicular components.Acceleration can be found by dividing the net force by mass.Trigonometry can be used to solve for the magnitude of the force components.Net force must be found before acceleration can be found.Read more on acceleration here: brainly.com/question/14344386
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Complete Question:
A sled is moving down a steep hill. The mass of the sled is 50 kg and the net force acting on it is 20 n. What must be done to find the acceleration of the sled? check all that apply.
A. The force of gravity must be broken into its parallel and perpendicular components.
B. Acceleration can be found by dividing the net force by mass.
C. Acceleration can be found by multiplying the net force by mass.
D. The magnitude of the force components must be multiplied by gravity.
E. Trigonometry can be used to solve for the magnitude of the force components.
F. Net force must be found before acceleration can be found.
Answer:
To find the acceleration, you do 20N/50kg = 0.4 m/s^2
Explanation:
example: a = F/m = 10/2 = 5 m/s2
Lucia wants to change the motion map shown so that it’s shows uniform circular motion. What change should Lucia make ?
Answer:
The last one - each vector pointing towards the center of the circle must be the same length for uniform circular motion
What quantity is represented by a unit called a newton (N)?
A. Inertia
B. Acceleration
C. Force
D. Mass
[tex]\textsf {C. Force}[/tex]
[tex]\textsf {The quantity which is represented by a unit called Newton (N)}\\\textsf {is Force.}[/tex]
A comet goes around the sun in an elliptical orbit. At its farthest point, 455 million miles from the sun, it is traveling with a speed of 15,135 mi/h. How fast is it traveling at its closest approach to the sun, at a distance of 116 million miles?
Answer:
A 2kg block is moved a long a level floor by a horizontal force of 20N if the coefficient of sliding friction is 0.4
what is the acceleration of the blove
Which refers to a material that causes a wave to bounce off of it?
A. absorber
B. reflector
C. conductor
D. insulator
Answer:
the answer to the question is a reflector
. Two planets, both of mass m, are separated by a distance d. Their relative velocity is negligible, and there is an inertial frame in which both planets are essentially at rest. The gravitational potential u(r) at the position r in the presence of the two planets, located at R1 and R2, is given as u(r) = − Gm R1 −r − Gm R2 −r . This problem takes place far out in space and there are no other massive objects in the vicinity of the two planets. (a) Draw a graph of u as a function of position (r) along the line between the two planets. (b) There are space stations Alpha and Beta located on the line between the planets. Both space stations are at rest with respect to the planets. Alpha is at distance d 4 from planet 1 and Beta is at distance d 3 from planet 2. A projectile of mass m is fired from station Alpha, with its velocity v pointing directly at planet 2. What is the minimum speed v which will permit the projectile to reach station Beta?
(a) A graph of u as a function of position (r) along the line between the two planets is attached below.
(b) The minimum speed v which will permit the projectile to reach station Beta is √ [Gm/3d]
What is gravitational potential energy?If an object is lifted, work is done against gravitational force. The object gains energy.
Given are two planets, both of mass m, are separated by a distance d. Their relative velocity is negligible, and there is an inertial frame in which both planets are essentially at rest. The gravitational potential u(r) at the position r in the presence of the two planets, located at R1 and R2, is given as
u(r) = − (Gm /R1 −r) − (Gm / R2 −r) .
This problem takes place far out in space and there are no other massive objects in the vicinity of the two planets
(a) A graph of u (r) versus position (r) along the line between the two planets is attached in answer.
(b) There are space stations Alpha and Beta located on the line between the planets. Both space stations are at rest with respect to the planets. Alpha is at distance d 4 from planet 1 and Beta is at distance d 3 from planet 2. A projectile of mass m is fired from station Alpha, with its velocity v pointing directly at planet 2.
The range of the projectile is given by R = v²sin2θ / g
g = gravitational acceleration of Earth
If g = g(p) for planet , range R = v²sin2θ / g(p)..................(1)
The gravitational force of attraction = weight force
Gm² /d² = m g(p)
g(p) = Gm/d².........................(2)
For R = d/3, from equation (1), we have
d/3 = v²sin2θ / g(p)
Plug the expression for g(p) , we get
v = √ [Gm/3dsin2θ ]
For velocity to be minimum, sin2θ =1
So, the minimum velocity will be
v = √ [Gm/3d]
Thus, the minimum speed v which will permit the projectile to reach station Beta is √ [Gm/3d]
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A uniform electric field exist between two parallel plate seperated by 1.2 cm. The intensity of the field is 23 KN/C. What is the potential difference between the plates
Explanation:
this is the answer 276 volts
Please help me!
If ball C is 3 times the volume of ball D and ball D has 1/3 the mass of ball C, which has the greater density?
A. Ball C
B. Ball D
C. The Densities are equal
C. The Densities are equal.
What is density?
Density is mass per unit volume or mass of a unit volume of a material substance.
If m1, V1 and D1 = mass, volume and density respectively of ball C
m2, V2 and D2 = mass, volume and density respectively of ball D
According to the Question ,
[tex]V_{1} = 3V_{2} , m_{2} = \frac{1}{3} (m_{1} ) \\ \\= m_{1} = 3m_{2}[/tex]
Therefore,
[tex]\frac{D_{1} }{D_{2} } = (\frac{m_{1} }{V_{1} } )* (\frac{m_{2} }{V_{2} } )\\ \\= (\frac{3m_{2} }{3V_{2} })*(\frac{V_{2} }{m_{2} }) \\\\= 1[/tex]
Hence, D1 = D2
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A snowboarder on a slope starts from rest and reaches a speed of 4.2 m/s after 7.3 s
The car's acceleration will be 0.575 m/s².The unit of acceleration is m/sec².
What is acceleration?The rate of velocity change concerning time is known as acceleration.
Given data;
Initial velocity, u= 0 m/s
Final velocity, v= 4.2 m/s
Time elapsed, t = 7.3 seconds.
To find ;
Acceleration, a
The acceleration when the change in velocity is observed by the formula as:
a= (v-u)/(t)
Substitute the given values:
a= (4.2-0)/(7.3)
a=(4.2)/(7.3)
a= 0.575 m/s²
Hence, the car's acceleration will be 0.575 m/s².
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A 545 N sled is pulled a distance of 385 m. The task is done by pulling on a rope with a force of 1225 N at an angle of 19° with the horizontal. How much work is done in pulling the sled? What is the acceleration in the x direction, assuming that friction is negligible? Assuming the sled started at rest, how long does it take to pull the sled 185 m?
The work done in pulling the sled is 445,930.2 Joules.
The acceleration in the x direction, assuming that friction is negligible, is 20.85 m/s².
Time taken to pull the sled 185 m is 4.21s.
What is work done?Work done is equal to product of force applied and distance moved.
Given is a 545 N sled is pulled a distance of 385 m. The task is done by pulling on a rope with a force of 1225 N at an angle of 19° with the horizontal.
Work = Force x Distance x cos(angle)
W= 1225 x 385 x cos 19°
W = 445,930.2 Joules
Thus, the work done in pulling the sled is 445,930.2 Joules
From the Newton's second law of motion, we have'
F = ma
acceleration, a = 1225cos19° / ( 545 /9.81)
a = 20.85 m/s²
Thus, the acceleration in the x direction is 20.85 m/s²
Using the second equation of motion, we get
s = ut+ 1/2 at²
Substitute the values, we have
185 m = 0 +1/2 x 20.85 x t²
t = 4.21 s
Thus, the time taken to pull the sled 185 is 4.21s.
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