Answer: positively charged
Explanation: basically when a neutral atom gets some electrons it becomes negatively charged and when a neutral atom loses elections it's positively charged Ik science is weird lol.
I hope this was the answer you were looking for.
Place a small object on the number line below at the position marked zero. Draw a circle around the object. Mark the center of this circle with the symbol for “initial position”. Move the object 5.0cm to the right and stop. Label this circle with the correct symbol for “final position.”

(A) What was the initial position of the object?
(B) What is the final position of the object?
(C) What is the distance traveled by the object?
(D) What is the displacement of the object?
(E) Of the three underlined quantities, which are numerically equal?
A hockey player whacks a 162-g puck with her stick, applying a 171-N force that accelerates it to 42.3 m/s. A. If the puck was initially at rest, for how much time did the acceleration last? B. The puck then hits the curved corner boards, which exert a 151-N force on the puck to keep it in its circular path. What’s the radius of the curve?
Given parameters:
Mass of puck = 162g = 0.162kg (1000g = 1kg)
Force exerted on puck = 171N
Final velocity = 42.3m/s
Unknown
A. time of the acceleration
B. radius of the curve?
Solution:
A. time of the acceleration
the initial velocity of the puck = 0m/s
We know that;
Force = mass x acceleration
Acceleration = [tex]\frac{Final velocity - Initial velocity}{time taken}[/tex]
Acceleration = [tex]\frac{42.3 - 0}{t}[/tex]
So force = mass x [tex]\frac{42.3 }{t}[/tex]
Input the parameters and solve for time;
171 = 0.162 x [tex]\frac{42.3 }{t}[/tex]
171 = [tex]\frac{6.85}{t}[/tex]
t = [tex]\frac{6.85}{171}[/tex] = 0.04s
The time of acceleration is 0.04s
B. radius of the curve;
to solve this, we apply the centripetal force formula;
F = [tex]\frac{mv^{2} }{r}[/tex]
where;
F is the centripetal force
m is the mass
v is the velocity
r is the radius
Since the force exerted on the puck is 151;
input the parameters and solve for r²;
151 = [tex]\frac{0.162 x 42.3^{2} }{r}[/tex]
151r = 0.162 x 42.3²
r = 1.92m
The radius of the circular curve is 1.92m
Why after a certain time bouncing a ball does it stop?
Answer:
If you define "bouncing" as leaving the ground for any amount of time, the ball stops bouncing when the elastic energy stored in the compression phase of the bounce is not enough to overcome the weight of the ball. ... Some energy is dissipated in the compression and decompression phases.
Explanation:
what’s the best answer , a , b , c or d
Answer:
The Basic Concept
When we look at an object at rest, we say that it is in equilibrium since all the forces being applied on it, cancel out
now, if one of the force was slightly more than the other one in the same case. The object will start to move and NOT be in equilibrium
BUT
if an object is in a vacuum and moving on a frictionless surface, the object will attain equilibrium after some seconds since it will be moving with constant speed and all the forces acting on it will be equal
Hence, if the object is accelerating. we can say with surety that the object is not is equilibrium since from the second law of motion,
F = ma ; when a is a non-zero value, there is definitely some net force being applied on the object
Looking at the given case
in the question, we are given that the object is 'accelerating' upwards
we proved above that if an object is accelerating, there is some net force on that object and hence the object is NOT in equilibrium
Since the object is accelerating, from the second law of motion:
F = ma; m cannot be zero and if a is a non-zero value as given in the question, there is definitely some net force on the object
Since there is some force being applied on the object, the object is NOT in equilibrium
Conclusion
Since we found that the object is NOT in equilibrium and that there is some net force on the object,
The first option is correct
A 54.0-kg projectile is fired at an angle of 30.0° above the horizontal with an initial speed of 126 m/s from the top of a cliff 132 m above level ground, where the ground is taken to be y = 0. (a) What is the initial total mechanical energy of the projectile? (Give your answer to at least three significant figures.) J (b) Suppose the projectile is traveling 89.3 m/s at its maximum height of y = 297 m. How much work has been done on the projectile by air friction? J (c) What is the speed of the projectile immediately before it hits the ground if air friction does one and a half times as much work on the projectile when it is going down as it did when it was going up?
Answer:
a) The initial total mechanical energy of the projectile is 498556.296 joules.
b) The work done on the projectile by air friction is 125960.4 joules.
c) The speed of the projectile immediately before it hits the ground is approximately 82.475 meters per second.
Explanation:
a) The system Earth-projectile is represented by the Principle of Energy Conservation, the initial total mechanical energy ([tex]E[/tex]) of the project is equal to the sum of gravitational potential energy ([tex]U_{g}[/tex]) and translational kinetic energy ([tex]K[/tex]), all measured in joules:
[tex]E = U_{g} + K[/tex] (Eq. 1)
We expand this expression by using the definitions of gravitational potential energy and translational kinetic energy:
[tex]E = m\cdot g\cdot y + \frac{1}{2}\cdot m\cdot v^{2}[/tex] (Eq. 1b)
Where:
[tex]m[/tex] - Mass of the projectile, measured in kilograms.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]y[/tex] - Initial height of the projectile above ground, measured in meters.
[tex]v[/tex] - Initial speed of the projectile, measured in meters per second.
If we know that [tex]m = 54\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]y = 132\,m[/tex] and [tex]v = 126\,\frac{m}{s}[/tex], the initial mechanical energy of the earth-projectile system is:
[tex]E = (54\,kg)\cdot \left(9.807\,\frac{m}{s^{2}}\right)\cdot (132\,m)+\frac{1}{2}\cdot (54\,kg)\cdot \left(126\,\frac{m}{s} \right)^{2}[/tex]
[tex]E = 498556.296\,J[/tex]
The initial total mechanical energy of the projectile is 498556.296 joules.
b) According to this statement, air friction diminishes the total mechanical energy of the projectile by the Work-Energy Theorem. That is:
[tex]W_{loss} = E_{o}-E_{1}[/tex] (Eq. 2)
Where:
[tex]E_{o}[/tex] - Initial total mechanical energy, measured in joules.
[tex]E_{1}[/tex] - FInal total mechanical energy, measured in joules.
[tex]W_{loss}[/tex] - Work losses due to air friction, measured in joules.
We expand this expression by using the definitions of gravitational potential energy and translational kinetic energy:
[tex]W_{loss} = E_{o}-K_{1}-U_{g,1}[/tex]
[tex]W_{loss} = E_{o} -\frac{1}{2}\cdot m\cdot v_{1}^{2}-m\cdot g\cdot y_{1}[/tex] (Eq. 2b)
Where:
[tex]m[/tex] - Mass of the projectile, measured in kilograms.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]y_{1}[/tex] - Maximum height of the projectile above ground, measured in meters.
[tex]v_{1}[/tex] - Current speed of the projectile, measured in meters per second.
If we know that [tex]E_{o} = 498556.296\,J[/tex], [tex]m = 54\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]y_{1} = 297\,m[/tex] and [tex]v_{1} = 89.3\,\frac{m}{s}[/tex], the work losses due to air friction are:
[tex]W_{loss} = 498556.296\,J -\frac{1}{2}\cdot (54\,kg)\cdot \left(89.3\,\frac{m}{s} \right)^{2} -(54\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (297\,m)[/tex]
[tex]W_{loss} = 125960.4\,J[/tex]
The work done on the projectile by air friction is 125960.4 joules.
c) From the Principle of Energy Conservation and Work-Energy Theorem, we construct the following model to calculate speed of the projectile before it hits the ground:
[tex]E_{1} = U_{g,2}+K_{2}+1.5\cdot W_{loss}[/tex] (Eq. 3)
[tex]K_{2} = E_{1}-U_{g,2}-1.5\cdot W_{loss}[/tex]
Where:
[tex]E_{1}[/tex] - Total mechanical energy of the projectile at maximum height, measured in joules.
[tex]U_{g,2}[/tex] - Potential gravitational energy of the projectile, measured in joules.
[tex]K_{2}[/tex] - Kinetic energy of the projectile, measured in joules.
[tex]W_{loss}[/tex] - Work losses due to air friction during the upward movement, measured in joules.
We expand this expression by using the definitions of gravitational potential energy and translational kinetic energy:
[tex]\frac{1}{2}\cdot m \cdot v_{2}^{2} = E_{1}-m\cdot g\cdot y_{2}-1.5\cdot W_{loss}[/tex] (Eq. 3b)
[tex]m\cdot v_{2}^{2} = 2\cdot E_{1}-2\cdot m \cdot g \cdot y_{2}-3\cdot W_{loss}[/tex]
[tex]v_{2}^{2} = 2\cdot \frac{E_{1}}{m}-2\cdot g\cdot y_{2}-3\cdot \frac{W_{loss}}{m}[/tex]
[tex]v_{2} = \sqrt{2\cdot \frac{E_{1}}{m}-2\cdot g\cdot y_{2}-3\cdot \frac{W_{loss}}{m} }[/tex]
If we know that [tex]E_{1} = 372595.896\,J[/tex], [tex]m = 54\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]y_{2} =0\,m[/tex] and [tex]W_{loss} = 125960.4\,J[/tex], the final speed of the projectile is:
[tex]v_{2} =\sqrt{2\cdot \left(\frac{372595.896\,J}{54\,kg}\right)-2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0\,m)-3\cdot \left(\frac{125960.4\,J}{54\,kg}\right) }[/tex]
[tex]v_{2} \approx 82.475\,\frac{m}{s}[/tex]
The speed of the projectile immediately before it hits the ground is approximately 82.475 meters per second.
A 30-cm-diameter, 4-m-high cylindrical column of a house made of concrete ( k = 0.79 W/m⋅K, α = 5.94 × 10 −7 m2/s, rho = 1600kg/ m 3 , and c p = 0.84kJ/kg⋅K ) cooled to 14° C during a cold night is heated again during the day by being exposed to ambient air at an average temperature of 28° C with an aver-age heat transfer coefficient of 14 W/ m 2 ⋅K. Using the analyti-cal one-term approximation method, determine (a) how long it will take for the column surface temperature to rise to 27° C, (b) the amount of heat transfer until the center temperature reaches to 28° C, and (c) the amount of heat transfer until the surface temperature reaches 27° C.
Answer:
a) Time it will taken for the column surface temperature to rise to 27°C is
17.1 hours
b) Amount of heat transfer is 5320 kJ
c) Amount of heat transfer until the surface temperature reaches 27°C is 4660 kJ
Explanation:
Given that;
Diameter D = 30 cm
Height H = 4m
heat transfer coeff h = 14 W/m².°C
thermal conductivity k = 0.79 W/m.°C
thermal diffusivity α = 5.94 × 10⁻⁷ m²/s
Density p = 1600 kh/m³
specific heat Cp = 0.84 Kj/kg.°C
a)
the Biot number is
Bi = hr₀ / k
we substitute
Bi = (14 W/m².°C × 0.15m) / 0.79 W/m.°C
Bi = 2.658
From the coefficient for one term approximate of transient one dimensional heat conduction The constants λ₁ and A₁ corresponding to this Biot number are,
λ₁ = 1.7240
A₁ = 1.3915
Once the constant J₀ = 0.3841 is determined from corresponding to the constant λ₁
the Fourier number is determined to be
[ T(r₀, t) -T∞ ] / [ Ti - T∞] = A₁e^(-λ₁²t') J₀ (λ₁r₀ / r₀)
(27 - 28) / (14 - 28) = (1.3915)e^-(17240)²t (0.3841)
t' = 0.6771
Which is above the value of 0.2. Therefore, the one-term approximate solution (or the transient temperature charts) can be used. Then the time it will take for the column surface temperature to rise to 27°C becomes
t = t'r₀² / ₐ
= (0.6771 × 0.15 m)² / (5.94 x 10⁻⁷ m²/s)
= 23,650 s
= 7.1 hours
Time it will taken for the column surface temperature to rise to 27°C is
17.1 hours
b)
The heat transfer to the column will stop when the center temperature of column reaches to the ambient temperature, which is 28°C.
Maximum heat transfer between the ambient air and the column is
m = pV
= pπr₀²L
= (1600 kg/m³ × π × (0.15 m)² × (4 m)
= 452.389 kg
Qin = mCp [T∞ - Ti ]
= (452.389 kg) (0.84 kJ/kg.°C) (28 - 14)°C
= 5320 kJ
Amount of heat transfer is 5320 kJ
(c)
the amount of heat transfer until the surface temperature reaches to 27°C is
(T(0,t) - T∞) / Ti - T∞ = A₁e^(-λ₁²t')
= (1.3915)e^-(1.7240)² (0.6771)
= 0.1860
Once the constant J₁ = 0.5787 is determined from Table corresponding to the constant λ₁, the amount of heat transfer becomes
(Q/Qmax)cyl = 1 - 2((T₀ - T∞) / ( Ti - T∞)) ((J₁(λ₁)) / λ₁)
= 1 - 2 × 0.1860 × (0.5787 / 1.7240)
= 0.875
Q = 0.875Qmax
Q = 0.875(5320 kJ)
Q = 4660 kJ
Amount of heat transfer until the surface temperature reaches 27°C is 4660 kJ
when water in a brook or system of pipes flows from a wide region to a narrow region, the speed of water in the narrow region is
Answer:
more
hope this helps
plz mark brainliest
Ray runs 78 feet north, then 61 feet west. Calculate the total displacement traveled by in feet
Answer:
hdhshsisjsbrtheisebvrtctvsjusyevevrvrg eggs haushehegehs
What happens to the oceans tides if the ocean if the Earth spins slower?
Answer:
As the earth rotates, it tries to drag/bring the tidal bulges with it. When a large amount of friction is applied, the earth spin will gradually and slow down but not all the way down.
Answer:
If the Earth spun slower the rate of tides will be higher because the moon will start revolving faster tan the Earth, creating more tides as the moon will revolve more around the Earth in a month.
If my answer helped, kindly mark me as the brainliest !!
Thank You!!
The perception of an image first, followed by noticing individual pieces of the
image, can be described as:
A. sensation.
B. perceptual processing.
C. top-down processing.
D. bottom-up processing.
SUBMIT
Answer:
The answer is Top-Down processing
Explanation:
I had this question on a apex quiz and i got it correct.
What is the weight of a person whose mass at sea level is 72 kg ?
Answer:
158LB
Explanation:
The person's weight is equal to 705.6 N when the mass at sea level is 72 kg.
What is the weight of the object?The weight of a body can be defined as the force acting on the body due to gravity. Weight is a vector quantity if the gravitational force is working on the object.
The unit of the weight is that of force, which (S.I. unit) is Newton. A body with a mass of one Kg has a weight of 9.8 N on the surface of the Earth or on the seal level.
The mathematical equation to determine the weight of an object is written as follows:
W = mg
Given the mass of the person, m = 72 Kg
The acceleration due to gravity on the person, g = 9.8 m/s²
The weight of the person on the earth will be equal to:
W = 72 Kg × 9.8 m/s²
W = 705.6 N
Therefore, the weight of the person is 705.6 N.
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Which of the following elements are most likely to have similar chemical properties?
Answer:
transition metal, and inner transition metals groups are numbered 1-18 from left to right
is a guideline to help an individual write and achieve well-specified goals.
Answer:
is a guideline to help an individual write and achieve well-specified goals.
Explanation:
An action plan is a guideline to help an individual write and achieve well-specified goals.
Answer:
yes, the guideline is to help an individual write and achieve well-specified goals.
Explanation:
increased force will increase acceleration true or false.
A magnet is moved toward a coil or wire to induce an electric current. What will happen if the magnet is reversed and moved toward the coil of wire again? (Answer choices attached in photo)
Answer: the direction of the induced current will change :)
Explanation:
An object falls freely from rest on a planet
where the acceleration due to gravity is
29 m/s ^2
After 3.8 s, what will be its speed?
Answer in units of m/s.
Answer:
v=u+gt , initially u=0 and g acting in the direction of movement of body.
v=0+9.8×2
v=19.6m/s
Explanation:
sorry i dont have exact answer but hope this above equation will help you ....♡
100 POINTS HELP!!!!!!!!!!!!!
Answer:
it is whatever the temperature is at 5(I cant seem to see it clearly)
Explanation:
what happens to the matter that makes up a liquid when sound travels through it
Explanation:
The matter passes in the directions of the noise and flows from the source to a receiver like sound flows through a substance. As the sound flows through a fluid, the material is disrupted for an amount of time, but after the sound leaves, it restored to its normal location.
What is the elapsed time between the 0-m mark and the 40-m mark
Answer:
a) 4.0 s
b) 16 m/s
c) The distance covered between 4 - 5 s is four times the distance covered between 4 - 5 s
d) Equal distance are covered between 0 - 4 s and 4 -5 s
Note: The question is incomplete. The complete question is as follows;
Refer to the chart below that has data about a moving object to answer the following questions.
Time Elapsed 0.0s 1.0s 2.0s 3.0s 4.0s 5.0 s
Distance Traveled 0.0m 10.0m 20.0m 30.0m 40.0m 80.0 m
a. What is the elapsed time between the 0-m mark and the 40-m mark?
b. How large is the average velocity of the object for the interval from 0-5 s ?
c. How does the interval of 3-4 s compare with the interval from 4-5 s in terms of distance?
d. How does the interval of 0-4 s compare with the interval from 4-5 s in terms of distance?
Explanation:
a. From the data provide, time elapsed between the 0 m - 40 m mark is 4.0 s
b. Average velocity = total distance/ total time
average velocity = 80 m/ 5.0 s = 16.0 m/s
c. Distance covered between 3 - 4 s = 40 m - 30 m = 10 m
Distance covered between 4 - 5 s = 80 m - 40 m = 40 m
The distance covered between 4 - 5 s is four times the distance covered between 4 - 5 s
d. Distance covered between 0 - 4 s = 40 m - 0 m = 40 m
Distance covered between 4 - 5 s = 80 m - 40 m = 40 m
Equal distance are covered between 0 - 4 s and 4 -5 s
This glass of lemonade is sitting in the hot summer sun. As time passes, in which direction will heat transfer take place?
The heat transfer takes place from the Ice to lemonade (ice → lemonade).
What is heat transfer?The term “heat transfer” refers to the movement of heat. The flow of heat across a system's boundary is due to a temperature differential between the system and its surroundings.
When a temperature difference exists between states of matter, heat transfer happens solely in the direction of decreasing temperature, that is, from a hot object to a cold item.
The temperature of ice is increasing while the lemonade is decreasing. Heat transfer happens solely in the direction of decreasing temperature,
Hence, the heat transfer takes place from the Ice to lemonade (ice → lemonade.
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Answer: C
Explanation:
1. What happens to the current in a series circuit as it moves through each component? a. The current stays the same throughout the circuit.
b. The current will increase or decrease depending on the resistance.
c. The current decreases with each component it goes through.
d. The current increases with each component it goes through.
2. A 10-volt power supply is placed in series with two 5-ohm resistors. What is the current in the circuit after it passes through each of the two resistors?(1 point)
a. The current will stay the same at 1 amp after passing through both resistors.
b. The current will drop to 2 amps after the first resistor and then to 1 amp after the second resistor.
c. The current will stay the same at 2 amps after passing through both resistors.
d. The current will drop to 1 amp after the first resistor and then to 0 amps after the second resistor.
3. What is the voltage that passes through R1 and R2?
a. R1: 12 V, R2: 24 V
b. R1: 8 V, R2: 4 V
c. R1: 12 V, R2: 12 V
d. R1: 6 V, R2: 6 V
4. Which of the following correctly describes the magnitude of currents I1 and I2 ?
a. I1 is equal to I2
b. I1 and I2 approach zero
c. I1 is greater than I2
d. I1 is less than I2
5. If the energy of an electric charge flowing in a circuit is conserved, which of the following obeys the Kirchhoff junction rule?
a. The sum of the current flowing in is greater than the sum of current flowing out.
b. The sum of the current flowing in is less than the sum of the current flowing out.
c. The sum of the current flowing in is equal to the sum of current flowing out.
d. The sum of the current flowing in is zero and the sum of the current flowing out is greater than zero.
Answer: sorry here’s the answers, I didn’t feel like typing it all
Explanation:
The correct answer to the 5 questions are;
1) Option A; The current stays the same throughout the circuit.
2) Option A; The current will stay the same at 1 amp after passing through both resistors.
3) Option C; R1: 12 V, R2: 12 V
4) Option A; I1 is equal to I2
5) Option C; The sum of the current flowing in is equal to the sum of current flowing out.
1) In an electrical circuit, usually as current moves through each component, it stays the same. Thus, option A is correct.
2) Formula for current is;
I = V/R
We are told that there are two resistors in series each having a resistance of 5Ω. Thus; Total resistance = 5 + 5 = 10 Ω.
Thus; Current = 10/10 = 1 A.
The current will stay same at 1 A after passing through both resistors.
3) From the circuit we are given, we see that the Voltage is 12 V. Now, the same voltage would be transmitted through both resistor R1 and R2.
Option C is correct
4) The current splits upon passing resistor 1 and as such it means the current I2 going through the second resistor would be the same. Thus; I1 = I2.
5) Kirchoff's junction rule states that all the incoming currents to a particular junction must be equal to sum of all currents going out of that same junction. Thus, option C is correct.
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A solid nonconducting sphere of radius R carries a charge Q distributed uniformly throughout its volume. At a certain distance rl (r
(A) E/8
(B) E 78.
(C) E/2
(D) 2E
(E) 8E
Answer:
E ’= E / 8
therefore the correct answer is A
Explanation:
Let's calculate the electric field in an insulating sphere with a radius r <R, let's use Gauus's law, with a spherical Gaussian surface
Фi = ∫ E. dA = [tex]q_{int}[/tex] /ε₀
E (4πr²) = q_{int} / ε₀
density is
ρ = q_{int} / V
q_{int} = ρ V = ρ 4/3 π r³
we substitute
E (4π r²) = ρ 4/3 π r³ /ε₀
E = 1 /3ε₀ ρ r
let's change the density by
ρ = Q / V = Q / (4/3 π R³)
E = 1 / 4πε₀ Q r / R³
if we now distribute the same charge on a sphere of radius R' = 2R
E ’= 1 / 4pieo Q r / (2R)³
E ’= 1 / 4ft Qr / R³ ⅛
E ’= E / 8
therefore the correct answer is A
If you have a density of 100kg/L and a mass of 1000 units, tell me the following: second what is the volume
Answer:
volume is 0.1 L
Explanation:
you can use the equation density=mass/volume
100 = 1000 / v
divide by 1000 on both sides
0.1 = v
What is the momentum of a 20.0 kg scooter traveling at 5.00 m/s?
Answer:
The answer is 100 kgm/sExplanation:
To find the momentum of an object given it's mass and velocity we use the formula
momentum = mass × velocityFrom the question
mass = 20 kg
velocity = 5 m/s
We have
momentum = 20 × 5
We have the final answer as
100 kgm/sHope this helps you
Pest describes the act of using senses or tools to gather information?
ating a hypothesis
king an observation
mmarizing the results
ording the measurements
Answer:
Making an observation
Explanation:
The use of senses as a tool to gather information is described as making an observation.
While making an observation, the senses must be at alert.
Observation making is paramount to the scientific method. It is from observations that questions are asked and then hypothesis which can be tested are formulated. Observation can be carried out using the eyes, nose, feeling e.t.c. Nowadays, observation can also be carried out using some sophisticated equipment in the laboratory. This necessary for phenomenon the eludes our natural senses. A scientist must be a keen observer and their senses must be sharp.Therefore, the act of using senses or tools to gather information is called making an observation.
A mass of 15 kg is resting on a horizontal, frictionless surface. Force 1 of 206 N is applied to it at some angle above the horizontal, force 2 has a magnitude of 144 N and is applied vertically downward, force 3 has a magnitude of 5 N and is applied vertically upwards, and force 4 has a magnitude of 42 N and is applied in the -x direction to the object. When these forces are applied to the object, the object is moving at 20 m/s in the x direction in a time of 3 seconds. What is the normal force acting on the mass in Newtons
Answer:
N = 136.77 N
Explanation:
This is an exercise in Newton's second law, let's set a reference frame with the horizontal x-axis and the vertical y-axis. In the attachment we can see the applied forces.
Let's use trigonometry to decompose the force F1
cos θ = F₁ₓ / F₁
sin θ = F_{1y} / F₁
F₁ₓ = F₁ cos θ
F_{1y} = F₁ sin θ
now let's apply Newton's second law to each axis
X axis
F₁ₓ - F4 = m a
Y axis
N + F3 + F_{1y} -F₂ -W = 0
the acceleration can be calculated with kinematics
v = v₀ + a t
since the object starts from rest, the initial velocity is zero v₀ = 0
a = v / t
a = 20/3
a = 6.667 m / s²
we substitute in the equation
F₁ₓ = F₄ + m a
F₁ₓ = 42 + 15 6,667
F₁ₓ = 142 N
F₁ cos θ = 142
cos θ = 142/206 = 0.6893
θ = cos⁻¹ 0.6893
θ = 46.42º
now let's work the y axis
N = W + F₂ - F₃ - F_{1y}
N = 15 9.8 + 144 -5 - 206 sin 46.42
N = 286 - 149.23
N = 136.77 N
a torque of 100Nm is required to open a door. WHAT IS the minimum distance of the handle fromt he hinge. if the door is to be pulled open wth a force at handle not greater than 50N?
Answer:
At least [tex]2\; \rm m[/tex].
Explanation:
The torque [tex]\tau[/tex] that a force exerts on a lever is equal the product of the following:
[tex]F[/tex], the size of that force,[tex]r[/tex], the distance between the fulcrum and the point where that force is applied, and[tex]\sin\theta[/tex], the sine of the angle between the force and the lever.[tex]\tau = F\cdot r \cdot \sin\theta[/tex].
The force in this question is (at most) [tex]50\; \rm N[/tex]. That is: [tex]F = 50\; \rm N[/tex].
[tex]\sin \theta[/tex] is maximized when [tex]\theta = 90^\circ[/tex]. In other words, the force on the door gives the largest-possible torque when that force is applied perpendicular to the door. When [tex]\theta = 90^\circ\![/tex], [tex]\sin \theta =1[/tex].
If the force here is applied at a distance of [tex]r[/tex] meters away from the hinge (the fulcrum of this door,) the torque generated would be:
[tex]\begin{aligned}\tau &= F \cdot r \cdot \sin \theta \\ &= (50\, r)\; \rm N \cdot m\end{aligned}[/tex].
That torque is supposed to be at least [tex]100\; \rm N\cdot m[/tex]. That is:
[tex]50\, r \ge 100[/tex].
[tex]r \ge 2[/tex].
In other words, the force needs to be applied at a point a minimum distance of [tex]2\; \rm m[/tex] away from the hinge of this door.
Which technology collects data for weather climate and environmental environmental monitoring from space
Answer:B) satellites
Explanation:
I just took the tests
Find the velocity of the car after 6.9 s if its acceleration is 1.5 m/s² due south.
The velocity of the car after 6.9 s if its acceleration is 1.5 m/s² due south would be 10.35 meters / seconds.
What are the three equations of motion?There are three equations of motion given by Newton
v = u + at
S = ut + 1/2×a×t²
v² - u² = 2×a×s
Note that these equations are only valid for a uniform acceleration.
As given in the problem we have to find the velocity of the car we have to find the velocity of the car after 6.9 s if its acceleration is 1.5 m/s² due south,
The acceleration of the car = 1.5 m/s²
The time taken by the car = 6.9 seconds
By using the first equation of the motion,
v = u + at
v = 0 + 1.5*6.9
v = 10.35 meters / seconds
Thus, the velocity of the car after 6.9 s, if its acceleration is 1.5 m/s² due south, would be 10.35 meters / seconds.
To learn more about equations of motion from here, refer to the link;
brainly.com/question/5955789
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procedure for determining the thermal conductivity of a solid involves embedding thermocouple in a thick slab of the material and measuring the response to a prescribed change in temperature at one surface. Consider an arrangement for which the thermocouple is embedded 10 mm from a surface that is suddenly brought to a temperature of 100degreeC by exposure to boiling water. If the initial temperature of the slab was 30degreeC and the thermocouple measures a temperature of 65degreeC, 2 minutes after the surface is brought to 100degreeC, what is the thermal conductivity. The density of the material is 2200 kg/m3 and the specific heat is 700 J/M- Find: What is the thermal conductivity of the material
Answer:
The thermal conductivity [tex]k = 1.4094 W/ m\cdot K[/tex]
Explanation:
From the question we are told that
The depth of the thermocouple from the surface is x = 10 mm = 0.01 m
The temperature is [tex]T_f = 100 ^o C[/tex]
The initial temperature is [tex]T_i = 30 ^o C[/tex]
The temperature of the thermocouple after t = 2 minutes( 2 * 60 = 120 \ seconds) is [tex]T_t = 65 ^o C[/tex]
The density of the material is [tex]\rho = 2200 kg/m^3[/tex]
The specific heat of the solid [tex]c_s = 700 J/kg \cdot K[/tex]
Generally the equation for semi -infinite medium is mathematically as
[tex]\frac{T_s - T }{T_i - T} = erf [\frac{x}{2 \sqrt{\alpha * t} } ][/tex]
[tex]\frac{65 - 100 }{30 - 100} = erf [\frac{x}{2 \sqrt{\alpha * t} } ][/tex]
[tex]0.5 = erf [\frac{0.01}{2 \sqrt{\alpha * 120} } ][/tex]
Here [tex]\alpha[/tex] is a constant with unit [tex]m^2 /s[/tex]
[tex]\frac{0.01}{ 2 (\sqrt{\alpha * 120 } )}[/tex] this is from the Gaussian function table
[tex]0.0 1 = 0.954 * (\sqrt{\alpha * 120 } )[/tex]
=> [tex]\sqrt{\alpha * 120 } = \frac{0.01 }{0.954 }[/tex]
=> [tex]\alpha = 9.1525 *10^{-7} \ m^2 /s[/tex]
Generally the thermal conductivity is mathematically represented as
[tex]k = \alpha * \rho * c_s[/tex]
[tex]k = 9.1525 *10^{-7} * 2200 * 700[/tex]
[tex]k = 1.4094 W/ m\cdot K[/tex]