Answer:
4.3 m/s^2
Explanation:
a = vf - vi / t
26 - 20 / 1.4
= 4.28 ( round your numbers)
If you drag a 50kg block across the floor which has a coefficient of friction of .30, what is
the force needed to accelerated it at 2.0m/s^2 ?
Answer:
750 Newton
Explanation:
force=mass(acceleration)
I need help in question 7, a and b.
Answer:
The graph for 7a is shown in the attachment. For question 7b she walks a distance of 16 meters. (m)
Explanation:
A particular roller coaster has a mass of 3500 kg, a height of 4.0, and a velocity of 12m/s. What is the potential energy? If needed, use g=10.m/s^2
[tex]▪▪▪▪▪▪▪▪▪▪▪▪▪ {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪[/tex]
Gravitational Potential Energy of an object is calculated by formula ~
[tex] \large\boxed{\sf P = mgh}[/tex]
where,
m = mass of the object = 3500 kgg = Acceleration due to gravity = 12 m/s²h = height attained by the object = 4 mNow, let's calculate its potential energy ~
[tex]3500 \times 10 \times 4[/tex][tex]140000 \: \: joules[/tex][tex]140 \: \: kj[/tex]Answer:
Potential Energy of an object is calculated by formula:
Potential Energy (P.E)=m×g×hWhere,
m=mass of bodyg=acceleration due to gravityh=height from the earth surfaceNow, let's solve the question.
Given,
mass(m)=3500 kgheight (h)=4mvelocity (v)=10m/s²Now,
We know that,
Potential Energy (P.E)=mgh
[tex] = 3500 \times 10 \times 4[/tex]
[tex] = 3500 0\times 4[/tex]
[tex] =140000 joules [/tex]
[tex]\mathfrak{\blue{DisneyPrincess29}}[/tex]
Se aplican dos fuerzas concurrentes a un objeto de 4N a la derecha y 5N a la izquierda. ¿Hacia donde se movió y con cuanta fuerza?
The forces move strongly towards the left by 1N
Given the following
Force towards the right = 4N
Force towards the left = 5N
Note that the force acting towards the left is negative, hence the force acting towards the left is -5N
Take the sum of force
Resultant force = -5N + 4N
Resultant force = -1N
This shows that the forces move strongly towards the left by 1N
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two objects are sitting 6m apart, one object has a mass of 100kg and the other has a mass of 200kg. what is the gravitational attraction between them
PLEASE ANSWER NEED HELP!!!!!!!! PLEASE THE CORRECT ANSWER!!!!!!
Evonne does not have a lot of time to work out during the week. She wonders if she spent half a day on the weekend getting is as many different types of exercise as possible if that would help. What is the BEST advice for Evonne?
A.
She has the right idea since variety is more important than intensity.
B.
She can do that, but should have a clear goal in mind to define her exercise.
C.
She should only focus on one type of exercise since she can do it once a week.
D.
She would be better off fitting in 15-20 minutes of exercise several times a week.
PLEASE NO LINKS
She would be better off fitting in 15-20 minutes of exercise several times a week. This t is the BEST advice for Evonne.
What are the benefits of regular exercise?Regular physical activity and exercise have many positive health effects that are difficult to deny. Everyone, regardless of age, sex, or physical ability, benefits from exercise.
Exercise can assist sustain weight loss or prevent excessive weight gain. Calorie burn occurs during physical exertion. You burn more calories when you engage in more vigorous exercise.
Numerous health issues and concerns, such as stroke, the metabolic syndrome, high blood pressure, and type 2 diabetes, can be prevented or managed with regular exercise. It can also help with cognitive improvement and reduce the danger of dying from any cause.
Exercise helps your circulatory system function more effectively and distributes oxygen and nutrients to your tissues. Additionally, you have greater energy to complete daily tasks as your heart and lung health improves.
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7. A bus covers a certain distance in 60 minutes if it runs at a speed of 60 km/hr.
What must be the speed of the bus in order to reduce the time of journey by 40
minutes?
Answer:
90 Km/h
Explanation:
60 mins is 1 hr
so in an hour bus covers 60 Km
so new speed:
d/t
(60km/40mins)*60mins/h
90Km/h
A bob attached to a string of length L = 1.25 m, initially found at the equilibrium
position, is given an initial velocity v = 0.8 m/s. The maximum displacement angle,
Omax, is: (Take g=10 m/s)
17
3.2
13
10.2
6.4
Clear selection
A hoh attached to a string of length 2 m is displaced by an angle of 8º and then
The maximum displacement angle of the bob is 13⁰.
The given parameters;
Length of the pendulum, L = 1.25 mInitial velocity of the bob, v = 0.8 m/sThe maximum displacement of the bob is calculated by applying the principle of conservation of energy;
[tex]P.E = K.E\\\\mgh = \frac{1}{2} mv^2\\\\h = \frac{v^2}{2g} \\\\h = \frac{0.8^2}{2\times 10} \\\\h = 0.032 \ m[/tex]
The maximum displacement angle is calculated as follows;
[tex]cos \theta = \frac{L-h}{L} \\\\cos \theta = \frac{1.25 - 0.032}{1.25} \\\\\cos \theta = \frac{1.218}{1.25} \\\\cos \theta = 0.9744\\\\\theta = cos^{-1}(0.9744)\\\\\theta = 13\ ^0[/tex]
Thus, the maximum displacement angle of the bob is 13⁰.
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If you double the pressure of a constant amount of gas at a constant temperature, what happens to the volume
Answer: if you double the pressure, you will halve the volume. If you increase the pressure 10 times, the volume will decrease 10 times.
Explanation:
A very light rod 40cm long is pivoted at the centre. A weight of 50N is placed at one end. Where is the place to put a weight of 200N in order that the rod is in equilibrium?
Hi there!
We can go about this problem using a summation of torques.
In order to ensure the rod is in equilibrium, we must satisfy the condition:
Στ = 0
Since the rod is "very light", we can disregard its mass.
The equation for torque is:
τ = rFsinθ
In this instance, the torques are the weights of the objects and their distance from the pivot.
As the rod is 40 cm, the pivot is at 20 cm. Also, the torques must sum up to 0, so:
0 = rF1 - rF2
r1F1 = r2F2
0.20(50) = r2(200)
Solve:
10 = r2(200)
r2 = 0.05 m = 5 cm
The 200N weight must be put at a distance of 5 cm from the OTHER SIDE of the pivot in order to balance the rod.
Calculate torque using angular momentum
Answer:
The equation net τ=ΔLΔt net τ = Δ L Δ t gives the relationship between torque and the angular momentum produced.
if you have 3 moles of iron, how many grams of iron do you have?
Answer:
55.84 look at periodic table
so 55.84*3= 167.52gFe
how have astronomers interpreted the unexpectedly fast rotation of galaxies
Answer:
There must be a lot of dark matter that can be felt but not seen
The astronomers interpreted the unexpectedly fast rotation of galaxies that there must be a large quantity of dark energy whose gravity is detectable yet invisible.
What is a galaxy?Any system of stars plus interstellar material that makes up the cosmos is referred to as a galaxy. Such assemblages are common, and many of them are so massive that they hold tens of trillions of stars.
A vast variety of galaxies, from dim, hazy dwarf objects to spectacular spiral-shaped giants, have been created by nature. Almost all galaxies seem to have formed shortly after the universe started, and they are everywhere in space, even at the farthest limits of the universe that can be seen by the most advanced telescopes.
The majority of galaxies are found in clusters, many of which are further organized into clusters that span hundreds of billions of light-years.
Since there are almost empty spaces between these so-called super clusters, the universe's overall structure resembles a network of sheets or chains of galaxies.
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please help me solve all of them ( a, b, c and d ) thankiew !!
I’m also kind of in a rush
Answer:
a-
V= IR
9V = I ×( 12+6)
I = 9/ 18 A = 0.5 A
b
V=IR
240 = 6 A ×( 20 + R)
40 = 20 + R
R = 20 ohm
c
resultant resistance of the 2 parallel resistances= Ro
1/Ro = 1/ 5 + 1/ 20
1/Ro =( 20+5)/100
= 1/Ro = 1/4
Ro= 4 ohm
V=IR
V = 2A × ( 1+ 4 OHM)
V = 10V
d
equivalent resistance = Ro
1/Ro = 1/(2+8) + 1/(5+5)
1/Ro = 1/10 +1/10
2/10 = 1/ Ro
Ro= 10/2 = 5 ohm
V = IR
12V = I × 5Ohm
I=2.4 A
The wave height is equal to twice the _____.wave period wave height wave amplitude wave velocity wavelength
The wave height is equal to twice the amplitude of the wave.
The wave height of a wave of given wave with amplitude, period and wavelength is equal to twice the amplitude of the wave.
The amplitude of a wave is the maximum displacement of the wave, starting from the zero position of the wave. The wave height measures twice the maximum displacement of the wave.
Thus, we can conclude that the wave height is equal to twice the amplitude of the wave.
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Virtual image formed by concave mirror
Answer:
yes
Explanation:
because it is a diverging mirror
3.
If the specific heat of iron = 0.46 J/g Cº, how much heat is needed to warm 50 g of
iron from 20° C to 100° C?
Answer:
1.84 kJ (kilojoules)
Explanation:
A specific heat of 0.46 J/g Cº means that it takes 0.46 Joules of energy to raise the temperature of 1 gram of iron by 1 Cº.
If we want to heat 50 g of iron from 20° C to 100° C, we can make the following calculation:
Heat = (specific heat)*(mass)*(temp change)
Heat = (0.46 J/g Cº)*(50g)*(100° C - 20° C)
[Note how the units cancel to yield just Joules]
Heat = 1840 Joules, or 1.84 kJ
[Note that the number is positive: Energy is added to the system. If we used cold iron to cool 50g of 100° C water, the temperature change would be (Final - Initial) or (20° C - 100° C). The number is -1.84 kJ: the negative means heat was removed from the system (the iron).
Compare the amplitude of these waves.
help?
Answer:
The greater the amplitude of a wave then the more energy it is carrying.
Explanation:
E is carrying the most energy which means that it has the highest amplitude
A has the least energy because it has the lowest amplitude
a gas increases in pressure from 2.00 atm to 6.00 atm at a constant volume of 1.00 m3 and then expands at constant pressure to a volume of 3.00 m3 before returning to its initial state as shown in the figure below. how much work is done in one cycle?
The work done on the given gas in one cycle is -405.3 kJ.
The given parameters;
initial pressure of the gas, P₁ = 2 atmfinal pressure of the gas, P₂ = 6 atminitial volume of the gas, V₁ = 1 m³final volume of the gas, V₂ = 3 m³Convert the pressure to Pascal (N/m²);
1 atm = 101325
The work done in one cycle is the area of the triangle and it is calculated as follows
[tex]Area = \frac{1}{2} \times base \times height\\\\Area = \frac{1}{2} \times (3 \ m^3\ -\ 1 \ m^3)\times (6 \ atm \ - \ 2 \ atm)\\\\Area = 4 \ atm -m^3\\\\Area = 4 \ atm -m^3 \ \times \frac{101325 \ Pa}{1 \ atm} \\\\Area = 405,300 \ m^3.Pa\\\\Area = 405,300 \ m^3. (N/m^2)\\\\Area = 405,300 \ Nm\\\\Area = 405,300 \ J\\\\Area = 405.3 \ kJ[/tex]
the net work done on the gas = - 405.3 kJ
Thus, the work done on the given gas in one cycle is -405.3 kJ.
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The mercury in a barometer of a cross-sectional area 1 cm² stands at 75 cm, and the space above it is 9 cm in length. What
volume of air, measured at 3. The mercury in a barometer of a cross-sectional area 1 cm² stands at 75 cm, and the space above it is 9 cm in length. What
volume of air, measured at atmospheric pressure, would have to be admitted into the space to cause the column of the mercury
to drop to 59 cm?
The ideal gas equation and the pressure in barometer allows us to find the amount of air that we must introduce into the barometer for the change in height of the mercury column is:
The variation of the volume is: ΔV = 7.67 cm³
Pressure is defined by the relationship between force and area.
P = F / A
The ideal gas equation establishes a relationship between pressure, volume, and temperature of an ideal gas.
PV = nR T
Where P is pressure, V is volume, and T is temperature.
Let's write this equation for two points assuming that the temperature has not changed.
P₀ V₀ = P₁ V₁
V₁ = [tex]\frac{P_o}{P_1} \ V_o[/tex] (1)
The subscript "o" is used for the start point and the subscript "1" for the end point.
The pressure in a barometer is:
P = ρ g y
They indicate the initial height of the barometer y₀=75 cm, the distance from empty space y'₀ = 9 cm and the final height of the barometer y₁ = 59 cm.
The volume of the cylinder is
V = π r² y
Let's calculate the initial volume.
V₀ = π 1 9
V₀ = 28.27 cm³
We substitute in equation 1.
V₁ = [tex]\frac{\rho \ g \ y_o}{\rho \ g \ y_1} \ V_o[/tex]
V₁ = [tex]\frac{y_o}{y_1} \ V_o[/tex]
Let's calculate.
V₁ = [tex]\frac{75}{59} \ 27.27[/tex]
V₁ = 35.94 cm³
The volume to be incremented is
ΔV = V₁ - V₀
ΔV = 35.94 - 28.27
ΔV = 7.67 cm³
Using the ideal gas equation and the pressure in barometer we can find the amount of air that we must introduce into the barometer for the change in height of the mercury column is:
The change of the volume is: ΔV = 7.67 cm³
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How long would it take a 500 w electric motor to do 150,000 j of work
Explanation:
300 second = 5 minutes
details
500 watts = 500 joules per second
Time = (1.50 x 10^5 joules)/(5 x 10^2 joules per second) = .3 x 10^3 seconds =300 second
The efficiency of a motor is 15%. The student calculated the useful output energy transfer is 1.20J. Calculate the total input energy transfer.
Answer:
[tex]8.0\; \rm J[/tex].
Explanation:
The efficiency of a machine is the ratio between the useful output and the energy input:
[tex]\begin{aligned}\text{efficiency} &= \frac{\text{useful output}}{\text{energy input}} \times 100\%\end{aligned}[/tex]
Rearrange this equation to find energy input in terms of efficiency and useful output:
[tex]\displaystyle \text{energy input} = \frac{\text{useful output}}{\text{efficiency} / (100\%)}[/tex].
Substitute in the values: [tex]\text{useful output} = 1.2\; \rm J[/tex] and [tex]\text{efficiency} = 15\%[/tex]. Evaluate to find the value of [tex]\text{energy input}[/tex]:
[tex]\begin{aligned} \text{energy input} &= \frac{\text{useful output}}{\text{efficiency} / (100\%)} \\ &= \frac{1.20\; \rm J}{15\% / (100\%)} \\ &= 8.0\; \rm J\end{aligned}[/tex].
(Rounded to two significant figures as in the value of efficiency.)
A catamaran with a mass of 5.44×10^3 kg is moving at 12 knots. How much work is required to increase the speed to 16 knots? (One knot = 0.51 m/s.)
show all steps
The work that is required to increase the speed to 16 knots is 14,176.47 Joules
If a catamaran with a mass of 5.44×10^3 kg is moving at 12 knots, hence;
5.44×10^3 kg = 12 knots
For an increased speed to 16knots, we will have:
x = 16knots
Divide both expressions
[tex]\frac{5.44 \times 10^3}{x} = \frac{12}{16}\\12x = 16 \times 5.44 \times 10^3\\x = 7.23\times 10^3kg\\[/tex]
To get the required work done, we will divide the mass by the speed of one knot to have:
[tex]w=\frac{7230}{0.51}\\w= 14,176.47Joules[/tex]
Hence the work that is required to increase the speed to 16 knots is 14,176.47 Joules
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Why are people so rude???
Answer:
It could be low self-esteem
Explanation:
There probably insecure have problems and issues of their own could be that were abused growing up and weren’t treated right who knows.
definiciones de estrés
Answer:
dame la marca más inteligente, por favor
Explanation:
En el campo de la medicina, es la respuesta del cuerpo a una presión física, mental o emocional. El estrés produce cambios químicos que elevan la presión arterial, la frecuencia cardíaca y las concentraciones de azúcar en la sangre. También suele producir sentimientos de frustración, ansiedad, enojo o depresión.
Answer:
El estrés es la reacción del cuerpo al sentirse amenazado o bajo presión.
Explanation:
A falling object with a weight of 15N is moving downwards towards earth at a steady speed. What is the size of the drag force on this object?
Answer:
15 N :D
Explanation:
The size of the drag force on this object falling at a constant speed is determined as 15 N.
Drag force on the objectThe drag force on the object is determined by calculating the net force on the object;
F(net) = W - D
where;
W is weight of the objectD is the drag force of airma = W - D
at constant speed, acceleration, a = 0
0 = W - D
W = D
15 N = D
Thus, the size of the drag force on this object falling at a constant speed is determined as 15 N.
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a block weighing (Fg) 500 N is resting on a steel table ( us=0.74) the minimum force start this block moving is?
The minimum force required to start this block moving is 370 N.
The given parameters;
weight of the block, W = 500 Ncoefficient of static friction, [tex]\mu_s[/tex] = 0.74The minimum force required to start this block moving is calculated as follows;
[tex]F= \mu_s F_n[/tex]
where;
[tex]F_n[/tex] is the normal force on the block which is equal to the weight of the block
[tex]F= \mu_s F_n \\\\F= \mu_s W\\\\F = 0.74 \times 500 \\\\F= 370 \ N[/tex]
Thus, the minimum force required to start this block moving is 370 N.
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easy one - giving brainly if correct.
Gas.
HOPE YOU GET 100!
when a mechanical wave travels through a medium, how does the wave affect the medium's particles?
thanks ;w;
Answer:
PLEASE MAARK ME BRAINLEST!!!
Explanation:
The particles of the medium just vibrate in place. As they vibrate, they pass the energy of the disturbance to the particles next to them, which pass the energy to the particles next to them, and so on.
A car traveling at 35.6m/s crashes into a concert barrier and comes to a stop in 0.35 seconds. Calculate the average force applied to the 75kg driver.A 3.2kg steel ball traveling at 4.1m/s strikes a second ball of a mass 2.3kg Initially at rest. Calculate the velocity of the second ball when the first one continues traveling in the same direction with a speed of 1.5m/s2 balls of putty are shot towards one another. Ball 1 has a mass of 4.3kg and is moving at 18.6m/s . Ball 2 has a mass of 5.8kg and is moving at 9.5m/s. They collide and stick together. Calculate their final combine velocity.I really appreciate those attempting the problems. I do know the answers but I’m unaware of the steps to get there. Please include all formulas in your response and steps so I can learn and understand.Check your answer:7629N3.6m/s2.46m/sThank you all!
The force on the driver is 7629 N. The velocity of the second ball is 3.6 m/s. The combined velocity of the balls is 13.37 m/s.
We have to find the acceleration using;
v = u - at
v = final velocity = 0 m/s
u = initial velocity = 35.6m/s
a = acceleration = ?
t = time = 0.35 s
u = at
a = u/t = 35.6m/s / 0.35 s
a = 101.7 ms-2
The force on the driver = 75kg × 101.7 ms-2 = 7629 N
Using the principle of conservation of momentum;
Momentum before collision = momentum after collision
m1u1 +m2u2 = m1v1 + m2v2
Hence
(3.2 × 4.1) + 0 = (3.2 × 1.5) + 2.3v2
13.12 = 4.8 + 2.3v2
13.12 - 4.8 = 2.3v2
v2 = 13.12 - 4.8/2.3
v2 = 3.6 m/s
Using the principle of conservation of linear momentum;
m1u1 + m2u2 = m1v1 + m2v2
(4.3 × 18.6) + (5.8 × 9.5) = (4.3 + 5.8) v
v = 79.98 + 55.1/10.1
v = 13.37 m/s
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