Answer:
v = 50.5 m/s
Explanation:
F = (m)(^v/^t)
115N = (0.04551kg)(v/(0.020s))
2,526.917161 m/s² = v/(0.020s)
v = 50.53834322 m/s
v = 50.5 m/s
ASAP!! Please help me out here
Answer:
Option: DExplanation:
As,
In Option,
A:
There are two same positive ions so they will move away from each other.
B:
There is one negative and positive ion so they will move towards each other.
C:
Again there is one negative and positive ion so they will move towards each other.
D:
Here, there is neutral ions so they will not move and its the correct option.
Two forces have magnitudes in the ratio
3 : 5 and the angle between their directions is 60°. If their resultant is 35 N,
what are their magnitudes ?
Answer:
F1=26N and F2=09N ..this is from the two simultaneously equations
What are the uses of magnetic force?
Answer:
Computer hard drives use magnetism to store the data on a rotating disk. More complex applications include: televisions, radios, microwave ovens, telephone systems, and computers. An industrial application of magnetic force is an electromagnetic crane that is used for lifting metal objects.
Answer:
Examples of magnetic force is a compass, a motor, the magnets that hold stuff on the refrigerator, train tracks, and new roller coasters. All moving charges give rise to a magnetic field and the charges that move through its regions, experience a force.
I Hope this will help you if not then sorry :)
Una carga q1 = - 45 µC esta colocada a 30 mm a la izquierda de una carga q2 = 25 µC . ¿Cuál es la fuerza resultante sobre una carga de q3 = 20 µC localizada exactamente 50 mm arriba de la carga de 25µC ?
Answer:
La fuerza resultante sobre q₃ es -1.2245 × 10⁻¹⁵ i + -0.24 × 10⁻¹⁵ j
La magnitud de la fuerza resultante sobre q₃ es aproximadamente 1.25 × 10⁻¹⁵ N
Explanation:
q₁ = -45 μC = -45 × 10⁻⁶ C
r₁₂ = 30 mm = 30 × 10⁻³ m
q₂ = 25 μC = 25 × 10⁻⁶ C
r₂₃ = 50 mm = 50 × 10⁻³ m
q₃ = 20 μC = 20 × 10⁻⁶ C
k = 9×10⁻⁹ N·m²/C²
Por lo tanto;
r₁₃ = √(50² + 30²) = 10·√(34)
F₁₂ = 9×10⁻⁹ × (-45 × 10⁻⁶)×(25 × 10⁻⁶)/(30 × 10⁻³)² = -1.125 × 10⁻¹⁴
F₁₂ = -1.125 × 10⁻¹⁴ N
F₂₃ = 9×10⁻⁹ × (20 × 10⁻⁶)×(25 × 10⁻⁶)/(50 × 10⁻³)² = 1.8 × 10⁻¹⁵ j
F₁₃ = 9×10⁻⁹ × (-45 × 10⁻⁶)×(20 × 10⁻⁶)/(10·√34 × 10⁻³)² = -2.38× 10⁻¹⁵
Los componentes de F₁₃ son;
-2,38 × 10⁻¹⁵ × cos (arctan (30/50)) = -2,04 × 10⁻¹⁵ j
-2,38 × 10⁻¹⁵ × sin (arctan (30/50)) = -1,2245 × 10⁻¹⁵ i
La fuerza resultante sobre la carga q₃, [tex]\left | \underset {F_3} \rightarrow \right |[/tex] = [tex]\underset{F_{13}}{\rightarrow}[/tex] + [tex]\underset{F_{23}}{\rightarrow}[/tex]
∴ [tex]\left | \underset {F_3} \rightarrow \right |[/tex] = 1.8 × 10⁻¹⁵ j + -1.2245 × 10⁻¹⁵ i + -2.04 × 10⁻¹⁵ j
La fuerza resultante sobre q₃ es [tex]\left | \underset {F_3} \rightarrow \right |[/tex] = -1.2245 × 10⁻¹⁵ i + -0.24 × 10⁻¹⁵ j
La magnitud de la fuerza resultante sobre q₃,
[tex]\left | F_3 \right |[/tex] = √((-1.2245 × 10⁻¹⁵)² + (-0.24 × 10⁻¹⁵)²) ≈ 1.25 × 10⁻¹⁵
La magnitud de la fuerza resultante sobre q₃, [tex]\left | F_3 \right |[/tex] ≈ 1.25 × 10⁻¹⁵ N.
If 20N force produces an acceleration of 5ms^-2 In a body then the mass of the body will be:
A.4kg
B.5kg
C.1/4kg
D.1/5kg
Calculate the acceleration of a train of mass 30 000 kg when driven by a force of 15000 N.
Answer:
Explanation:
F = ma, so filling in:
15000 = 30000a and
a = .50 m/s/s
The acceleration of a train is 0.50 [tex]m/s^{2}[/tex].
What is acceleration?Acceleration is the rate of change of the velocity of an object with respect to time.
F = ma
a = F/m
a = 15000/30000
a = 0.50 [tex]m/s^{2}[/tex]
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HELP‼️‼️ A car horn creates a 595 Hz tone at rest. Two cars pass on the street, each going 20.0 m/s; the first car honks. What frequency does the other car hear before they pass each other?
The frequency of the other car is 669HZ
Let start off by writing out the parameters given in the question
frequency= 595 HZ
Speed of sound(v)= 343m/s
Speed of the car(vs)= 20m/s
Speed of the second car(vo) = 20m/s
The frequency of the other car can be calculated as follows
(v+vo/v-vs) f
= (343+20/343-20)595
= (363/323)595
= 1.1238(595)
= 669 HZ
Hence the frequency of the other car is 669 HZ
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There is given an ideal capacitor with two plates at a distance of 3 mm. The capacitor is connected to a voltage source with 12 V until it is loaded completely. Then the capacitor is disconnected from the voltage source. After this the two plates of the capacitor are driven apart until their distance is 5 mm. Now a positive test charge of 1 nC is brought from the positively charged plate to the negatively charges plate. How large is the kinetic energy of the test charge? The test charge of 1 nC can be regarded to be so small that it does not influence the electric field between the two plates of the capacitor.
The kinematic energy of the positive charge is 2 10⁻⁸ J
This electrostatics exercise must be done in parts, the first part: let's start by finding the charge of the capacitor, the capacitance is defined by
C = [tex]\frac{Q}{\Delta V}[/tex]
C = ε₀ [tex]\frac{A}{d}[/tex]
we solve for the charge (Q)
[tex]\frac{Q}{\Delta V} = \epsilon_o \frac{A}{d}[/tex]
indicates that for the initial point d₁ = 3 mm = 0.003 m and the voltage is DV₁ = 12
Q = [tex]\epsilon_o \ \frac{A \ \Delta V_1 }{d_1}[/tex]
Now the voltage source is disconnected so the charge remains constant across the ideal capacitor.
For the second part, the condenser is separated at d₂ = 5mm = 0.005 m
Q = \epsilon_o \ \frac{A \ \Delta V_2 }{d_2}
we match the expressions of the charge and look for the voltage
[tex]\frac{\Delta V_1}{d_1} = \frac{\Delta V_2}{d_2}[/tex]
ΔV₂ = [tex]\frac{d_2}{d_1 } \ \Delta V_1[/tex]
The third part we use the concepts of conservation of energy
starting point. With the test load (q = 1 nC = 1 10⁻⁹ C) next to the left plate
Em₀ = U = q DV₂
Em₀ = q \frac{d_2}{d_1 } \ \Delta V_1
final point. Proof load on the right plate
Em_f = K
energy is conserved
Em₀ = em_f
q \frac{d_2}{d_1 } \ \Delta V_1 = K
we calculate
K = 1 10⁻⁹ 12 [tex]\frac{0.005}{0.003}[/tex]
K = 20 10⁻⁹ J
In this exercise, as the conditions at two different points of separation give, the area of the condenser is not necessary and with conservation of energy we find the final kinetic energy of 2 10⁻⁸ J
Which word describes the maximum a point moves from its rest position when a wave passes?
Answer:
Amplitude.
Explanation:
A wave can be defined as a disturbance in a medium that progressively transports energy from a source location to another location without the transportation of matter.
In Science, there are two (2) types of wave and these include;
I. Electromagnetic waves: it doesn't require a medium for its propagation and as such can travel through an empty space or vacuum. An example of an electromagnetic wave is light.
II. Mechanical waves: it requires a medium for its propagation and as such can't travel through an empty space or vacuum. An example of a mechanical wave is sound.
An amplitude can be defined as a waveform that's measured from the center line (its origin or equilibrium position) to the bottom of a trough or top of a crest.
Hence, an amplitude is a word that describes the maximum displacement a point moves from its rest position when a wave passes.
On a graph, the vertical axis (y-axis) is the amplitude of a waveform and this simply means that, it's measured vertically.
Mathematically, the amplitude of a wave is given by the formula;
x = Asin(ωt + ϕ)
Where;
x is displacement of the wave measured in meters.
A is the amplitude.
ω is the angular frequency measured in rad/s.
t is the time period measured in seconds.
ϕ is the phase angle.
Which type of reactions usually happens slowest?
Answer:
option b is correct..................
3kg of water at 80degree celcius is added to 8 kg of water at 25 degree celcius. find the temperature of final mixture provided there is no loss of heat in the surrounding. the specific heat capacity is 4200j/kg
Answer:
hope fully it help s
when you go out to your car one cold winter morning you discover a 0.80 cm thick layer of ice on the windshield which has an area of 1.4m square if the temperature of the ice is - 2.0 degrees celsius and its density is 917 kg/m to the third find the heat required to melt all the ice
Answer:
The total heat required to melt the ice is approximately 3.473 MJ
Explanation:
The given parameters for the layer of ice are;
The thickness of the layer of ice, t = 0.80 cm = 0.008 m
The area of the wind shield, A = 1.4 m²
The initial temperature of the ice, T₁ = -2.0 °C = 271.15 K
The density of the ice, ρ = 917 kg/m
The temperature at which ice melts, T₂ = 0 °C = 273.15 K
We have;
The mass of the ice, m = ρ × t × A
∴ m = 917 kg/m³ × 0.008 m × 1.4 m² = 10.2704 kg
The specific heat capacity of ice, c = 2,090 J/(kg·K)
∴ The equation for the heat capacity of the ice to melt, is given as follows;
ΔQ = m·c·ΔT
Where;
ΔT = T₂ - T₁
∴ ΔT = T₂ - T₁ = 273.15 K - 271.15 K = 2 K
ΔQ = 10.2704 kg × 2,090 J/(kg·K) × 2 K = 42.930272 kJ
The latent heat to melt the ice, Q = The latent heat of fusion of ice, L × Mass of ice, m
The latent heat of fusion of ice, L = 334 kJ/kg
∴ Q = 334 kJ/kg × 10.2704 kg = 3,430.3136 kJ
The total heat required to melt the ice, H = ΔQ + Q
∴ H = 42.930272 kJ + 3,430.3136 kJ = 3,473.243872 kJ ≈ 3.473 MJ.
A cricket ball of mass 800g has momentum of 20 kg m/sec . Calculate velocity of ball
Answer: 25 m/s
Explanation:
1. rearrange equation p = mv to find v
2. v = p/m - then convert 800 g into kg (= 0.8 kg)
3. plug in p and m - v = 20 / 0.8 = 25 m/s
Answer:
V= 25m/s
Explanation:
Mass=800g
Momentum=20kgm/s
V= ?
Convert 800g to kg
800/1000 = 0.8kg
Momentum=mass× Velocity
Momentum/Mass = Velocity
20/0.8 =25m/s
You toss an apple across the room to a friend. Which of the following statements is true about the apple at the top of its trajectory?
A. Its acceleration is zero.
B. The horizontal component of its velocity is zero.
C. The vertical component of its velocity is 9.8 m/s down.
D. Its acceleration is 9.8 m/s2 down.
i. The lift raises a car to a height of 1.8 m using a force of 5500 N. How much work does the lift
perform? (1 point)
Work = force x distance
Work = 5500 x 1.8
Work = 9900 N
The work measure of energy transfer that occurs when an object is moved over a distance by an external force at least part of which is applied in the direction of the displacement.
Work = force x distance
Work = 5500 x 1.8
Work = 9900 N
therefore, work does 9900 N
What is work?Work is force applied over distance. Examples of work include lifting an object against the Earth's gravitation, driving a car up a hill, and pulling down a captive helium balloon. Work is a mechanical manifestation of energy. The standard unit of work is the joule (J), equivalent to a newton - meter (N · m).
What is work and energy?Work is defined as transferring energy into an object so that there is some displacement. Energy is defined as the ability to do work. Work done is always the same. Energy can be of different types such as kinetic and potential energy.
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The refractive index of water is 1.33
and that of diamond is 2.42. Draw a labelled diagram to show how a light ray bends when it travels from water
into diamond.
Explanation:
light travel slower in daimond
what is the definition of unit?definition of unit?
Answer:
The Reference standard with which we carry out the measurement of any physical quantity of the same kind is know as unit .
I HOPE I HELP YOU.
Answer:
The standard quantity which is used for the comparison with an unknown quantity is called unit.
Unit is a known or standard quantity in terms of which other physical quantities are measured.
for examples;kilogram,metre,second,etc.
Explanation:
hope it helps you
calculate the potential energy stored in a meta ball of mass of 80 kg kept at a height of 15m from the earth surface . What will be the potential energy when the metal ball is kept on the earth surface Take (g- 9.8 m/s]
Answer:
11760 J
Explanation:
cuz potential engery is PE = MHG
SO 80×15×9.8= 11760 J
Answer:
11769j
Explanation:
here,
mass(m)=80 kg
height(h)=15m
acceleration due to gravity(g)=9.8m/s^2
now,
potential energy = m×g×h
= 80×9.8×15
= 11760j
why is it important to have regular super vision of the weights and the measurements in the market?
Answer:
Obeying to weights & measurement regulations in both national and international metrology legislation, standards and test procedures is a requirement to participate in any market because it's aimed to safeguard the consumers and promote fair competition, which provides efficiency and saves unnecessary costs to U.S. businesses and stakeholders.
Explanation:
In the U.S., the National Institute of Standards and Technology (NIST) has an Office of Weights and Measures (OWM) that represents the country in the International Organization of Legal Metrology (OIML)
A group of students are designing a field study to investigate the length of time a traffic light remains yellow.
Answer:
3 to 7 seconds.
Explanation:
According to the traffic engineers the length or the duration of the signal to turn form yellow or amber to green is more than 5 seconds. A time interval is needed to easy out the traffic flow.Can someone pls help, thank you in advance!
What is an example of a force applied at an angle to displacement
Answer:
an object sliding down hill
Explanation:
On a slope, the force applied is due to gravity. Its direction is straight down. If the object is sliding down the hill, its displacement is at an angle to the applied force. The angle of displacement will depend on the steepness of the hill.
The number of windings on the primary coil of a transformer
is 1.5 times greater than on the secondary coil. The primary
coil has a current of 3.0 A and a voltage of 12.0 V. Determine
the voltage and current on the secondary coil.
Answer:
I. Vs = 8.0 Volts.
II. Is = 4.5 Amperes.
Explanation:
Given the following data;
Np = 1.5Ns = [tex] \frac {N_{P}}{N_{S}} = 1.5 [/tex] ..... equation 1
Ip = 3.0 A
Vp = 12 V
To find the voltage and current on the secondary coil;
I. For the voltage in the secondary coil (Vs), we would use the following formula;
[tex] \frac {V_{P}}{V_{S}} = \frac {N_{P}}{N_{S}} [/tex] ...... equation 2.
Substituting eqn 1 into eqn 2, we have;
[tex] \frac {V_{P}}{V_{S}} = 1.5 [/tex]
[tex] \frac {12}{V_{S}} = 1.5 [/tex]
Cross-multiplying, we have;
[tex] V_{S} * 1.5 = 12 [/tex]
[tex] V_{S} = \frac {12}{1.5} [/tex]
Vs = 8.0 V
II. For the current in the secondary coil (Is), we would use the following formula;
[tex] \frac {I_{S}}{I_{P}} = \frac {N_{P}}{N_{S}} [/tex] .... equation 3
Substituting eqn 1 into eqn 3, we have;
[tex] \frac {I_{S}}{I_{P}} = 1.5 [/tex]
[tex] \frac {I_{S}}{3.0} = 1.5 [/tex]
Cross-multiplying, we have;
[tex] I_{S} = 1.5 * 3.0 [/tex]
Is = 4.5 A
Which statement about the ocean is true?
No evaporation or precipitation in the water cycle occurs over the ocean.
Most evaporation and precipitation in the water cycle occur over the ocean.
All evaporation and precipitation in the water cycle occur over the ocean.
Evaporation, but not precipitation, in the water cycle occurs over the ocean.
Answer:
most evaporation and precipitation in the water cycle occus over the ocean
The option Most evaporation and precipitation in the water cycle occur over the ocean. is the correct answer
I let go of a piece of bread from a balcony. A bird flying 5.0 m overhead sees me drop it, and starts to dive straight down towards the bread the instant that I release it. She catches it after it falls 3.0 m. Assuming she accelerates constantly from rest (v0 = 0) at the time I let go of the bread, what is her acceleration? Show your work
This question can be solved using the equations of motion. There are two scenarios where the equations of motion can be used. The first scenario is the free-fall motion of the piece of bread. The second scenario is the uniformly accelerated motion of the bird.
The acceleration of the bird is "a = 26.13 m/s²".
First, we will calculate the time taken by the bread to fall 3 m. Using the second equation of motion for this free-fall motion:
[tex]h = v_it + \frac{1}{2}gt^2[/tex]
where,
h = height fall = 3 m
vi = initial velocity = 0 m/s
g = acceleration due to gravity = 9.8 m/s²
t = time taken = ?
Therefore,
[tex]3\ m = (0\ m/s)t+\frac{1}{2}(9.8\ m/s^2)t^2\\t = \sqrt{\frac{(3\ m)(2)}{9.8\ m/s^2}}\\\\t = 0.78\ s[/tex]
The bird took the same time to catch the bread. Now applying the second equation of motion to the bird's motion:
[tex]s = v_it + \frac{1}{2}at^2[/tex]
where,
s = distance covered by the bird = 5 m + 3 m = 8 m
vi = initial velocity of the bird = 0 m/s
a = acceleration of the bird = ?
t = time taken = 0.78 s
Therefore, using these values we get:
[tex]8\ m = (0\ m/s)(0.78\ s)+\frac{1}{2}a(0.78\ s)^2\\\\a = \frac{16\ m}{(0.78\ s)^2}[/tex]
a = 26.13 m/s²
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A low-power laser used in a physics lab might have a power of 0.50 mW and a beam diameter of 3.0 mm. Calculate:a.The average light intensity of the laser beam.b. The intensity of a lightbulb producing 100-W light viewed from 2.0 m.c.Compare the intensity of the laser to the intensity of the lightbulb. Is it advisable to look directly at a laser
Answer:
A) I_laser = 70.74 W/m²
B) I_bulb = 1.989 W/m²
C) it is not advisable to look at the laser beam directly.
Explanation:
We are given;
Power; P = 0.50 mW = 0.5 × 10^(-3) W
Diameter; d = 3 mm = 0.003 m
Radius; r = d/2 = 0.003/2 = 0.0015 m
A) Area of beam; A = πr²
A = 0.0015²π
Now, formula for average intensity is;
I = P/A
I = (0.5 × 10^(-3))/0.0015²π
I = 70.74 W/m²
B) We are told to find the intensity of a lightbulb producing 100-W.
Thus, P = 100 W
A light bulb is spherical in shape. Thus;
Area; A = 4πr²
We are told it's 2 m away.
Thus; r = 2 m
A = 4π(2)²
A = 16π
Thus, I = P/A = 100/16π
I = 1.989 W/m²
C) The intensity of the laser beam is far greater than that of the light bulb. Thus, it is not advisable to look at the laser beam directly.
Which of the following changes would not lead to changes in the efficiency of
a heat engine?
A. Doubling the work done while keeping the heat flow into the
engine the same
B. Doubling the heat flow into the engine while halving the work done
C. Doubling both the work done and the heat flow into the engine
D. Doubling the heat flow into the engine while keeping the work
done the same
Explanation:
B. Doubling the heat flow into the engine while halving the work done
hope this helps you
have a nice day
Why aren’t magazine photos a good representation of what a healthy person looks like
A 0.0780 kg lemming runs off a
5.36 m high cliff at 4.84 m/s. What
is its potential energy (PE) when it
jumps?
Answer:
4.097 Joules
Explanation:
Applying,
P.E = mgh............... Equation 1
Where P.E = Potential energy, m = lemming mass, h = height, g = acceleration due to gravity
From the question,
Given: m = 0.0780 kg, h = 5.36 m
Constant: g = 9.8 m/s²
Substitute these values into equation 1
P.E = 0.0780(9.8)(5.36)
P.E = 4.097 Joules.
Hence the potential energy is 4.097 Joules
Answer:
4.097 J
Explanation:
The plane of a5cm*8cm rectangular loop of wire is parallel to a 0.19T magnetic field the loop carries a current of 6.2 A. What torque acts on aloop? What is the magnetic moment of the loop?
Answer:
Torque; τ = 4.712 × 10^(-3) J
Magnetic moment; M = 0.0248 J/T
Explanation:
Torque is gotten from the formula;
τ = BIA
Where;
B is magnetic field
I is current
A is area
We are given;
B = 0.19T
I = 6.2A
Rectangle dimensions = 5cm by 8cm = 0.05m by 0.08m
Thus;
Area; A = 0.05m × 0.08m = 0.004 m²
Thus;
τ = 0.19 × 6.2 × 0.004
τ = 4.712 × 10^(-3) J
Formula for the magnetic moment is given by;
M = IA
M = 6.2 × 0.004
M = 0.0248 J/T
Un cuerpo es saltado desde cierta altura , llegando al piso luego de 7 s. Determine de altura se le soltó . (g=10m/segundo al cuadrado )
Answer:
Maximum height, H = 35 meters
Explanation:
Given the following data;
Time = 7 seconds
Acceleration due to gravity, g = 10 m/s²
To find the maximum height;
Mathematically, the maximum height is given by the formula;
H = ½gt²
H = ½ * 10 * 7
H = 5 * 7
Maximum height, H = 35 meters
