Answer:
Hopefully this is correct:
Given: m=40kg θ=60 to the horizontal
Ff to floor is 100N and Tension=300N
Find Work:
Given:
[tex]W=fd[/tex]
Where f= net force applied parallel to the block: Use cos(60)
=300cos(60)-100=50(4)=200J work
Calculate the speed of your cat as it runs towards its food bowl 14.7m away in 4.5 s. Give answer in mph.
Answer:
3.26mph
Explanation:
To calculate speed, use the formula distance/time. In this case, just divide 14.7 by 4.5.
The speed with which the cat runs towards its food bowl is 7.3 Miles per hour.
Given the data in the question
Distance of cat from it's food bowl; [tex]s = 14.7m[/tex]Time taken for the cat to reach the food bowl; [tex]t = 4.5s[/tex]Speed of the cat; [tex]v = \ ?[/tex]
Speed is the time rate at which an object is moving along a path
Speed is the time rate at which an object is displaced.
Speed = Distance / Time
[tex]v = s / t[/tex]
We substitute our values into the equation
[tex]v = \frac{14.7m}{4.5s}\\\\v = 3.267 m/s\\\\v = 7.3mph[/tex]
Therefore, the speed with which the cat runs towards its food bowl is 7.3 Miles per hour.
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These waves are most harmful for living things.
What waves are most harmful for living things?
A radio
B microwaves
C infrared
D visible
E ultraviolet
F x-rays
G gamma rays
Explanation:
G. gamma rays because they are produced by the hottest and most energetic objects in the universe, such as neutron stars and pulsars, supernova explosions, and regions around black holes.
hope this helps you
have a nice day:)
A 10kg block is Pulled along a horizontal
Surface by a force
of 50N at an angles
of 37° with the horizontal If the
coefficient of sliding friction b/n the
block and the surface is o.2
(g=10m/s^2 Sin 37=O.6 and cos 37 = 0.8)
A, what frictional forces acting on the block?
B,what is the acceleration of the block?
Answer:
hope u can understand the method
The density of table sugar is 1.59g/cm3 what is the volume of 7.85g of sugar?
Answer: 4.94cm³
Explanation:
Data;
ρ = 1.59g/cm³
mass = 7.85g
volume = ?
density = mass / volume
ρ = m / v
v = m / ρ
v = 7.85 / 1.59
v = 4.94cm³
On a distance-time graph, what is shown when the curve is flat going from left to the right?
A. a negative speed
B. no speed
C. a positive speed
D. It does not mean anything.
Please help me !!im on a test
girl is sitting on a tire swing that is attached with a rope that is 2.1 m in length. Her dad pushes
her with a speed of 3.0 m/s. If the centripetal force is 88 N, what is the mass of the girl?
pls help
Answer:
m=29.6kg
Explanation:
your welcome:) have a great day
Which is the best way to become familiar with your company's policies and procedures?
O
A. ask the person who hired you
O
B. look in the employee handbook
C. tell your supervisor you need help
D. visit the company's website
what unit is used to measure resistance
Answer:
Ohms
Explanation:
The unit of resistance is ohms.
When a particular hanging mass is suspended from the string, a standing wave with two segments is formed. When the weight is reduced by 2.2 kg, a standing wave with five segments is formed. What is the linear density of the string
Solution :
Mass is varied keeping frequency constant.
Wavelength, λ [tex]$=\frac{2l}{n}$[/tex]
where length of spring = l
number of segments = n
Velocity, v = λ x f
= [tex]$\sqrt{\frac{T}{\mu}}$[/tex]
[tex]$\mu $[/tex] = mass density, T = tension in string
[tex]$T=\frac{4 \mu l^2f^2}{n^2}$[/tex]
[tex]$T=mg = \frac{4 \mu l^2f^2}{n^2}$[/tex] , n = 2
[tex]$T = (m-2.2)g = \frac{4 \mu l^2f^2}{n^2}, n = 5$[/tex]
[tex]$\Rightarrow \frac{m}{m-2.2}=\frac{25}{4}$[/tex]
[tex]$\Rightarrow m = 2.619\ kg$[/tex]
Therefore, μ = 0.002785 kg/ m
Frequency is varied keeping T constant
[tex]$T=\frac{4 \mu l^2f^2}{n^2}, f=60 , \ \ n = 2$[/tex]
[tex]$T=\frac{4 \mu l^2f^2}{n^2}, f=? , \ \ n = 7$[/tex]
[tex]$\Rightarrow \frac{60^2}{4}=\frac{f^2}{49}$[/tex]
f = 210 Hz
A block of wood has mass 4kg, what is the density of the wood if the mass of water is 3kg and its volume is half that of the block of the block of wood ?
Se transmiten ondas transversales en una cuerda tensada orientada sobre el eje x. La función de ondas correspondientes es y= 5.00 sen(6.05x + 5.19t + 1.57). Donde y y x están en metros y t en segundos. Cual es la rapidez de las ondas que se transmiten en dicha cuerda?
Answer:
Explanation:
Whats the entire question?
g You drop a 3.6-kg ball from a height of 3.5 m above one end of a uniform bar that pivots at its center. The bar has mass 9.9 kg and is 4.2 m in length. At the other end of the bar sits another 3.6-kg ball, unattached to the bar. The dropped ball sticks to the bar after the collision. Assume that the bar is horizontal when the dropped ball hits it. How high (in meters) will the other ball go after the collision
Answer:
h = 3.5 m
Explanation:
First, we will calculate the final speed of the ball when it collides with a seesaw. Using the third equation of motion:
[tex]2gh = v_f^2 - v_i^2\\[/tex]
where,
g = acceleration due to gravity = 9.81 m/s²
h = height = 3.5 m
vf = final speed = ?
vi = initial speed = 0 m/s
Therefore,
[tex](2)(9.81\ m/s^2)(3.5\ m) = v_f^2 - (0\ m/s)^2\\v_f = \sqrt{68.67\ m^2/s^2}\\v_f = 8.3\ m/s[/tex]
Now, we will apply the law of conservation of momentum:
[tex]m_1v_1 = m_2v_2[/tex]
where,
m₁ = mass of colliding ball = 3.6 kg
m₂ = mass of ball on the other end = 3.6 kg
v₁ = vf = final velocity of ball while collision = 8.3 m/s
v₂ = vi = initial velocity of other end ball = ?
Therefore,
[tex](3.6\ kg)(8.3\ m/s)=(3.6\ kg)(v_i)\\v_i = 8.3\ m/s[/tex]
Now, we again use the third equation of motion for the upward motion of the ball:
[tex]2gh = v_f^2 - v_i^2\\[/tex]
where,
g = acceleration due to gravity = -9.81 m/s² (negative for upward motion)
h = height = ?
vf = final speed = 0 m/s
vi = initial speed = 8.3 m/s
Therefore,
[tex](2)(9.81\ m/s^2)h = (0\ m/s)^2-(8.3\ m/s)^2\\[/tex]
h = 3.5 m
Hypothesis what is it
Explanation:
hope it helps.
stay safe healthy and happy.Calculate the potential difference across the 8 ohm resistor
Explanation:
if the current is 1A
V=iR
V= 1 × 8
V = 8volts
In the physics lab, a block of mass M slides down a frictionless incline from a height of 35cm. At the bottom of the incline it elastically strikes another block that is only one-half its mass. Find the velocity of block M at the bottom of the incline before the collision with the small block m.
Solution :
Given :
M = 0.35 kg
[tex]$m=\frac{M}{2}=0.175 \ kg$[/tex]
Total mechanical energy = constant
or [tex]$K.E._{top}+P.E._{top} = K.E._{bottom}+P.E._{bottom}$[/tex]
But [tex]$K.E._{top} = 0$[/tex] and [tex]$P.E._{bottom} = 0$[/tex]
Therefore, potential energy at the top = kinetic energy at the bottom
[tex]$\Rightarrow mgh = \frac{1}{2}mv^2$[/tex]
[tex]$\Rightarrow v = \sqrt{2gh}$[/tex]
[tex]$=\sqrt{2 \times 9.8 \times 0.35}$[/tex] (h = 35 cm = 0.35 m)
= 2.62 m/s
It is the velocity of M just before collision of 'm' at the bottom.
We know that in elastic collision velocity after collision is given by :
[tex]$v_1=\frac{m_1-m_2}{m_1+m_2}v_1+ \frac{2m_2v_2}{m_1+m_2}$[/tex]
here, [tex]$m_1=M, m_2 = m, v_1 = 2.62 m/s, v_2 = 0$[/tex]
∴ [tex]$v_1=\frac{0.35-0.175}{0.5250}+\frac{2 \times 0.175 \times 0}{0.525}[/tex]
[tex]$=\frac{0.175}{0.525}+0$[/tex]
= 0.33 m/s
Therefore, velocity after the collision of mass M = 0.33 m/s
A solenoid that is 66.2 cm long has a cross-sectional area of 18.0 cm2. There are 1300 turns of wire carrying a current of 8.15 A. (a) Calculate the energy density of the magnetic field inside the solenoid. (b) Find the total energy in joules stored in the magnetic field there (neglect end effects).
Answer:
(a) Energy Density = 160.94 J/m³
(b) Energy Stored = 0.192 J
Explanation:
(a)
The energy density of the magnetic field inside the solenoid is given by the following formula:
[tex]Energy\ Denisty = \frac{B^2}{2\mu_o}\\[/tex]
where,
B = magnetic field strength of solenoid = [tex]\frac{\mu_oNI}{l}[/tex]
Therefore,
[tex]Energy\ Density = \frac{\mu_oN^2I^2}{2l^2}[/tex]
where,
μ₀ = permeability of free space = 4π x 10⁻⁷ N/A²
N = No. of turns = 1300
I = current = 8.15 A
L = length = 66.2 cm = 0.662 m
Therefore,
[tex]Energy\ Density = \frac{(4\pi\ x\ 10^{-7}\ N/A^2)(1300)^2(8.15\ A)^2}{2(0.662\ m)^2}[/tex]
Energy Density = 160.94 J/m³
(b)
Energy Stored = (Energy Density)(Volume)
Energy Stored = (Energy Density)(Area)(L)
Energy Stored = (160.94 J/m³)(0.0018 m²)(0.662 m)
Energy Stored = 0.192 J
Types of government
Answer:
Explanation:
The different types of government include a direct democracy, a representative democracy, socialism, communism, a monarchy, an oligarchy, and an autocracy
oscillating spring mass systems can be used to experimentally determine an unknown mass without using a mass balance. a student observes that a particular spring-mass system has a frequency of oscillation of 10 Hz. the spring constant of the spring is 250 N/m. what is the mass?
Answer:
0.063 Kg
Explanation:
From the question given above, the following data were obtained:
Frequency (f) = 10 Hz
Spring constant (K) = 250 N/m
Mass (m) =?
Next, we shall determine the period of oscillation. This can be obtained as follow:
Frequency (f) = 10 Hz
Period (T) =?
T = 1/f
T = 1/10
T = 0.1 s
Finally, we shall determine the mass of the spring. This can be obtained as follow:
Spring constant (K) = 250 N/m
Period (T) = 0.1 s
Pi (π) = 3.14
Mass (m) =?
T = 2π√(m/K)
0.1 = 2 × 3.14 × √(m/250)
0.1 = 6.28 × √(m/250)
Divide both side by 6.28
0.1 / 6.28 = √(m/250)
Take the square of both side.
(0.1 / 6.28)² = m/250
Cross multiply
m = (0.1 / 6.28)² × 250
m = 0.063 Kg
Therefore, the mass of the spring is 0.063 Kg.
A soccer ball is released from rest at the top of a grassy incline. After 6.2 seconds, the ball travels 47 meters. One second later, the ball reaches the bottom of the incline.
(a) What was the balls acceleration?(assume that the acceleration was constant).
(b) How long was the incline?
Answer:
(a) a = 2.44 m/s²
(b) s = 63.24 m
Explanation:
(a)
We will use the second equation of motion here:
[tex]s = v_it+\frac{1}{2}at^2[/tex]
where,
s = distance covered = 47 m
vi = initial speed = 0 m/s
t = time taken = 6.2 s
a = acceleration = ?
Therefore,
[tex]47\ m = (0\ m/s)(6.2\ s)+\frac{1}{2}a(6.2\ s)^2\\\\a = \frac{2(47\ m)}{(6.2\ s)^2}[/tex]
a = 2.44 m/s²
(b)
Now, we will again use the second equation of motion for the complete length of the inclined plane:
[tex]s = v_it+\frac{1}{2}at^2[/tex]
where,
s = distance covered = ?
vi = initial speed = 0 m/s
t = time taken = 7.2 s
a = acceleration = 2.44 m/s²
Therefore,
[tex]s = (0\ m/s)(6.2\ s)+\frac{1}{2}(2.44\ m/s^2)(7.2\ s)^2\\\\[/tex]
s = 63.24 m
The table provided shows the masses and volumes of four liquids. A student has a mystery liquid with a density of 3 g/mL. Which liquid does the student have?
Liquid 1
Liquid 2
Liquid 3
Liquid 4
Answer: Liquid 2, if it's 3 g/mL and liquid 2 has 6gs and 2 mLs than it's 3g/mL
what is the maximum distance we can shoot a dart,from ground level provided our toy dart gun gives a maximum initial velocity of 2.7m/s and air resistance is negligible
Answer:
R = v^2 sin 2 theta / g
The range provides the distance a projectile can travel
R(max) = v^2 / g if theta = 45 deg
R = 2.7^2 / 9.8 = .74 m
A pendulum swings past it's pole 20 times in 18 seconds, what is the period of the pendulum?
Answer: 0.9
Explanation:
If frequency equals the amount of oscillations in one second then 20/18 = f = 1.11 and to find period (T) then we use the equation T= 1/f or in this case 1/1.11 = 0.9
An ocean thermal energy conversion system is being proposed for electric power generation. Such a system is based on the standard power cycle for which the working fluid is evaporated, passed through a turbine, and subsequently condensed. The system is to be used in very special locations for which the oceanic water temperature near the surface is approximately 300 K, while the temperature at reasonable depths is approximately 280 K. The warmer water is
Answer:
Explanation:
Dear Student, this question is incomplete, and to attempt this question, we have attached the complete copy of the question in the image below. Please, Kindly refer to it when going through the solution to the question.
To objective is to find the:
(i) required heat exchanger area.
(ii) flow rate to be maintained in the evaporator.
Given that:
water temperature = 300 K
At a reasonable depth, the water is cold and its temperature = 280 K
The power output W = 2 MW
Efficiency [tex]\zeta[/tex] = 3%
where;
[tex]\zeta = \dfrac{W_{out}}{Q_{supplied }}[/tex]
[tex]Q_{supplied } = \dfrac{2}{0.03} \ MW[/tex]
[tex]Q_{supplied } = 66.66 \ MW[/tex]
However, from the evaporator, the heat transfer Q can be determined by using the formula:
Q = UA(L MTD)
where;
[tex]LMTD = \dfrac{\Delta T_1 - \Delta T_2}{In (\dfrac{\Delta T_1}{\Delta T_2} )}[/tex]
Also;
[tex]\Delta T_1 = T_{h_{in}}- T_{c_{out}} \\ \\ \Delta T_1 = 300 -290 \\ \\ \Delta T_1 = 10 \ K[/tex]
[tex]\Delta T_2 = T_{h_{in}}- T_{c_{out}} \\ \\ \Delta T_2 = 292 -290 \\ \\ \Delta T_2 = 2\ K[/tex]
[tex]LMTD = \dfrac{10 -2}{In (\dfrac{10}{2} )}[/tex]
[tex]LMTD = \dfrac{8}{In (5)}[/tex]
LMTD = 4.97
Thus, the required heat exchanger area A is calculated by using the formula:
[tex]Q_H = UA (LMTD)[/tex]
where;
U = overall heat coefficient given as 1200 W/m².K
[tex]66.667 \times 10^6 = 1200 \times A \times 4.97 \\ \\ A= \dfrac{66.667 \times 10^6}{1200 \times 4.97} \\ \\ \mathbf{A = 11178.236 \ m^2}[/tex]
The mass flow rate:
[tex]Q_{H} = mC_p(T_{in} -T_{out} ) \\ \\ 66.667 \times 10^6= m \times 4.18 (300 -292) \\ \\ m = \dfrac{ 66.667 \times 10^6}{4.18 \times 8} \\ \\ \mathbf{m = 1993630.383 \ kg/s}[/tex]
If you dropped a ball off the roof, and could neglect air resistance, the impact velocity of the ball on the sidewalk would be 102 miles per hour [mph]. Calculate the height of the building in units of feet.
Answer:
h = 105.98 m
Explanation:
We will use the third equation of motion here:
[tex]2gh= v_f^2-v_i^2[/tex]
where,
g = acceleration due to gravity = 9.81 m/s²
h = height = ?
vf = final speed = (102 mi/h)(1 h/3600 s)(1609.34 m/1 mi) = 45.6 m/s
vi = initial speed = 0 m/s
Therefore,
[tex]2(9.81\ m/s^2)h = (45.6\ m/s)^2-(0\ m/s)^2\\\\h = \frac{ (45.6\ m/s)^2}{2(9.81\ m/s^2)}[/tex]
h = 105.98 m
How much heat is required to raise the temperature of 50 grams of water from 30 °C to 90 °C? C of water 4186 J / kg C.
12558 J
12558000 J
125580 J
1255800 J
Answer:
12558 J
Explanation:
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Please help in one minute will mark brainlist
Answer: Different types of telescopes usually don't take simultaneous readings. Space is a dynamic system, so an image taken at one time is not necessarily the precise equivalent of an image of the same phenomena taken at a later time. And often, there is barely enough time for one kind of telescope to observe extremely short-lived phenomena like gamma-ray bursts. By the time other telescopes point to the object, it has grown too faint to be detected.
Explanation: Trust me
A 4.00-kg block hangs by a light string that passes over a massless, frictionless pulley and is connected to a 6.00-kg block that rests on a frictionless shelf. The 6.00-kg block is pushed against a spring to which it is not attached. The spring has a spring constant of 180 N/m, and it is compressed by 30.0 cm. Find the speed of the block after the spring is released and the 4.00-kg block has fallen a distance of 40.0 cm.
Answer:
v = 2.82 m/s
Explanation:
For this exercise we can use the conservation of energy relations.
We place our reference system at the point where block 1 of m₁ = 4 kg
starting point. With the spring compressed
Em₀ = K_e + U₂ = ½ k x² + m₂ g y₂
final point. When block 1 has descended y = - 0.400 m
Em_f = K₂ + U₂ + U₁ = ½ m₂ v² + m₂ g y₂ + m₁ g y
as there is no friction, the energy is conserved
Em₀ = Em_f
½ k x² + m₂ g y₂ = ½ m₂ v² + m₂ g y₂ + m₁ g y
½ k x² - m₁ g y = ½ m₂ v²
v² = [tex]\frac{k}{m_2} x^2 - 2 \frac{m_1}{m_2} \ g y[/tex]
let's calculate
v² = [tex]\frac{180}{6.00} \ 0.300^2 - 2 \ \frac{4.00}{6.00} \ 9.8 \ (- 0.400)[/tex]
v² = 2.7 + 5.23
v = √7.927
v = 2,815 m / s
using of significant figures
v = 2.82 m/s
Please select the word from the list that best fits the definition
location of most volcanoes
Island arc
Lava
Basalt
Magma
Hotspot
Plate boundaries
Answer:
The answer is E., or "Plate Boundaries".
The word best describes the definition of the location of volcanoes is "plate boundaries". Therefore, option (E) is correct.
What are plate boundaries?The lithosphere is broken into tectonic plates that undergo some large motions and the boundary regions between plates are called plate boundaries. Based on their motions there are three kinds of plate boundaries: divergent, convergent, and transform.
Divergent boundaries are moving away from one another and separate, they form a rift. As the gap widens, the underlying layer may be soft for molten lava underneath to push upward and resulting in the formation of volcanic islands.
Convergent boundaries move towards one another and collide, subduction usually takes place. The denser plate goes underneath the less dense one and the plate boundaries also experience buckling.
Transform boundaries slide alongside one another and are exposed to huge amounts of stress and strain and are momentarily held in place. The two plates succeed in moving with respect to one another and cause earthquakes.
Learn more about plate boundaries, here:
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what is the difference between kinetic and potential energy
Answer:
Kinetic energy is energy possessed by a body by virtue of its movement. Potential energy is the energy possessed by a body by virtue of its position or state. While kinetic energy of an object is relative to the state of other objects in its environment, potential energy is completely independent of its environment. Hence the acceleration of an object is not evident in the movement of one object, where other objects in the same environment are also in motion. For example, a bullet whizzing past a person who is standing possesses kinetic energy, but the bullet has no kinetic energy with respect to a train moving alongside.
I think this answer is helped you
Answer:
the what state does alduos is 500 stack of ml tayo the little man ♂️ jajao
kalokoha ya
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What is an appraisal of an employee's performance in the workplace?
A. consensus
B. evaluation
C. procedure
D. metacognition
Answer:
B. evaluation
Explanation:
An employee can be defined as an individual who is employed by an employer of labor to perform specific tasks, duties or functions in an organization.
Basically, an employee is saddled with the responsibility of providing specific services to the organization or company where he is currently employed while being paid a certain amount of money hourly, daily, weekly, or monthly depending on the contractual agreement between the two parties (employer and employee).
Hence, while an employer may be the owner of a business firm or company, an employee is a subordinate employed to provide unwavering services to the employer while also, being professional and diligent at all times.
Human resources management (HRM) can be defined as an art of managing, controlling and improving the number of people (employees or workers), functions, evaluation of employees, activities which are being used effectively and efficiently by an organization.
Hence, human resources managers are saddled with the responsibility of recruiting, evaluating, managing and improving the welfare and working conditions of the employees working in an organization.
Basically, the human resources manager carry out appraisals on the performance of the various employees working in an organization. These informations about employees are typically used for promotional purposes, allowances and other benefits or form of rewards.