Answer:
yes broHope it helps youWhat is the force on a 1000 kg elevator that is free falling at 9.8 m/s/s?
Answer:
Explanation:
Force = (mass) · (acceleration)
= (1,000 kg) · (9.8 m/s²)
= 9,800 newtons
pls I will give brainliest for correct answer
find the distance travelled im the first 8 hours
Answer
I think it is 112 havent done this in a while but im pretty sure thats it Good luck
Explanation:
If an elevator is moving at a speed of 5 m/s and the person in it is 82 kg what is the person’s kinetic energy?
Answer:
KE 1025J
[tex] \frac{1}{2} m {v}^{2} \\ [/tex]
Explanation:
[tex] \frac{1}{2} \times 82 \times {5}^{2} [/tex]
1025J
How will you show that metals form basic oxides?
Answer:
when Metallic oxide reacts with water they form metallic hydroxide which are basic in nature
A population is known to have a standard deviation of 26.1. A sample space of 35 items has a mean of 562. Construct a 90% confidence interval estimate of the name of the population.
Answer:
553.74≤x≤578.26
Explanation:
The formula for calculating the confidence interval is expressed as
CI = xbar±(z×s/√n)
xbar is the mean = 562
s is the standard deviation = 26.1
n is the sample size = 35
z is the z score at 90% CI = 1.645
Substitute the values into the formula
CI = xbar±(z×s/√n)
CI = 561±(1.645×26.1/√35)
CI = 561±(1.645×4.412)
CI = 561±7.257
CI = {561-7.257, 571+7.257}
CI = {553.74, 578.26}
Hence the 90% confidence interval estimate of the name of the population is 553.74≤x≤578.26
The most common elements in the Earth’s crust are rarely found on their own. They are usually found combined. Suggest why this is.
Answer:
Because there are other elements that get in touch with it. And over time with, evolution, erosion and temperature their particles tend to mix. So what once was very common became mixed over time.
The most common element throughout the Earth’s crust that are rarely found on their own will be "Astatine".
Astatine seems to be the very most uncommon or infrequent natural substance throughout Earth's crust that occurs just under 1 gram at any time.It's indeed exceedingly impossible to make Astatine-210, particularly in perhaps the most useful state, don't just in less and less astatine present in the fundamental natural world.
Thus the above answer is correct.
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HELPP WILL MARK BRAINLIEST!!!
Answer: the answer is c
Choose the element that has the larger atomic radius: francium or sodium.
Answer:
Francium.
I hope it will be useful.
I'll give brainliest....
. a train starts from rest and moves with a uniform acceleration of x metre per second square for 5 minutes determine the value of x if it covers a distance of 5.8 km during the journey
Answer:
x = ~0.1289 m/s^2
Explanation:
We can use one of the kinematic formulas. We see that the equation doesn’t contain the final velocity, so we use the kinematic equation d = v_0t + (1/2)at^2.
Since the train started from rest, the initial velocity, v_0, is 0. Thus, the equation becomes:
d = (1/2)at^2
Note that 5 minutes is 300 seconds and 5.8 km is 5800 m. Substituting the values we are given, we get:
5800 m = (1/2)(x m/s^2)(300 s)^2 = (45000 s^2)(x m/s^2)
Dividing by 45,000 on both sides, we get:
x = ~0.1289 m/s^2
I hope this helps! :)
12. A rolling ball has 18 J of kinetic energy and is rolling 3.0 m/s. Find its mass.
Answer:
4g
I hope it will be useful.
Explanation:
KE = (1/2)*m*v^2
What is horizontal motion
Horizontal motion is defined as a projectile motion in a horizontal plane depending upon the force acting on it. For a short distance, the vertical and horizontal components of a projectile are perpendicular and independent of each other.
Answer:
A projectile moves along its path with a constant horizontal velocity
A 0N
B 6N
C 10 N
D 12 N
Answer:
The net force acting on the object is 0 N
Explanation:
Newton's Second Law of Forces
The net force acting on a body is proportional to the mass of the object and its acceleration.
The net force can be calculated as the sum of all the force vectors in each rectangular coordinate separately.
The image shows a free body diagram where four forces are acting: two in the vertical direction and two in the horizontal direction.
Note the forces in the vertical direction have the same magnitude and opposite directions, thus the net force is zero in that direction.
Since we are given the acceleration a =0, the net force is also 0, thus the horizontal forces should be in equilibrium.
The applied force of Fapp=10 N is compensated by the friction force whose value is, necessarily Fr=10 N in the opposite direction.
The net force acting on the object is 0 N
Write skeleton equations for the following reaction. Write the correct formula in each blank. You do not need to make the subscripts small. For example for carbon dioxide, write CO2
barium chloride ____ + potassium sulfate ____ → barium sulfate _____ + potassium chloride _____ sodium+iron(III) chloride→iron+sodium chloride
Please help me!!
1. BaCl2+K2SO4⇒BaSO4+KCl
2. Na+FeCl3⇒Fe+NaCl
Further explanationGiven
Word equation
Required
Skeleton equation
Solution
A chemical reaction can be expressed in terms of:
words equation the skeleton equation balanced equationThe skeleton equation is still in unbalanced form
So only chemical formulas of reactants and products and the relative amounts of the reactants and products are not included
For example :
• word equation
Ethane + Oxygen ⇒ Carbon dioxide + Water
• skeleton equation
C₂H₆ + O₂ ⇒ CO₂ + H₂O
• balanced equation
2C₂H₆ + 7O₂ ⇒ 4CO₂ + 6H₂O
1.
word equation
barium chloride ____ + potassium sulfate ____ → barium sulfate _____ + potassium chloride _____
skeleton equation
BaCl2+K2SO4⇒BaSO4+KCl
2.
word equation
sodium+iron(III) chloride→iron+sodium chloride
skeleton equation
Na+FeCl3⇒Fe+NaCl
1. Which of the following is an example of projectile motion?
An elevator
Ajet taking off
O A football flying through the air
Dropping an aluminum can into the garbage
Answer:
A football flying through the air
Explanation:
when an object has projectile motion it is moving through the air
From the given options, the football flying through the air is the prime example of projectile motion. (option d)
Projectile motion refers to the motion of an object that is launched into the air and moves under the influence of gravity alone. It follows a curved path known as a projectile trajectory. Among the options provided, the example that best represents projectile motion is a football flying through the air.
When a football is kicked or thrown, it experiences an initial force that propels it forward. However, as soon as it leaves the kicker's or thrower's hand or foot, the only force acting on it is gravity. Gravity pulls the football downward, causing it to follow a curved path. The shape of the path depends on the angle at which the football is launched and its initial velocity.
During its flight, the football follows a parabolic trajectory, moving both horizontally and vertically. The horizontal component of its motion remains constant, while the vertical component experiences a downward acceleration due to gravity. This combination of horizontal and vertical motion creates the characteristic curved path of projectile motion.
Therefore, option (d) is correct.
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If red and blue light rays fall with the same angle of incidence on the separating
surface between two different transparent media, then the ratio between the refraction
angle of the red light and the refraction angle of the blue light (.) is
a) greater than 1
b)equal to 1
c )indeterminable
d)less than 1
Answer:
I'm gonna say it's D
Explanation:
but when u do the experiment on in u head you'll actually find out that it is actually , indeterminable
Which of the following is true about thoughts
Answer:
they can be unpredictable until you think it
Explanation:
Standing waves are created in the four strings shown in Figure 25. All strings have the same mass per unit length and are under the same tension The lengths of the strings are given. Rank the frequencies of the oscillations, from largest to smallest
Answer:
The rank of the frequencies from largest to smallest is
The largest frequency of oscillation is given by the string in option D
The second largest frequency of oscillation is given by the string in option B
The third largest frequency of oscillation is given by the string in option A
The smallest frequency of oscillation is given by the string in option C
Explanation:
The given parameters are;
The mass per unit length of all string, m/L = Constant
The tension of all the string, T = Constant
The frequency of oscillation, f, of a string is given as follows;
[tex]f = \dfrac{(n + 1) \times \sqrt{\dfrac{T}{m/L} } }{2 \cdot L}[/tex]
Where;
T = The tension in the string
m = The mass of the string
L = The length of the string
n = The number of overtones
[tex]Therefore, \ {\sqrt{\dfrac{T}{m/L} } } = Constant \ for \ all \ strings = K[/tex]
For the string in option A, the length, L = 27 cm, n = 3 we have;
[tex]f_A = \dfrac{(n + 1) \times \sqrt{\dfrac{T}{m/L} } }{2 \cdot L} = \dfrac{(3 + 1) \times K }{2 \times 27} = \dfrac{2 \times K}{27} \approx 0.07407 \cdot K[/tex]
For the string in option B, the length, L = 30 cm, n = 4 we have;
[tex]f_B = \dfrac{(n + 1) \times \sqrt{\dfrac{T}{m/L} } }{2 \cdot L} = \dfrac{(4 + 1) \times K }{2 \times 30} = \dfrac{ K}{12} \approx 0.08 \overline 3\cdot K[/tex]
For the string in option C, the length, L = 30 cm, n = 3 we have;
[tex]f_C = \dfrac{(n + 1) \times \sqrt{\dfrac{T}{m/L} } }{2 \cdot L} = \dfrac{(3 + 1) \times K }{2 \times 30} = \dfrac{K}{15} \approx 0.0 \overline 6 \cdot K[/tex]
For the string in option D, the length, L = 24 cm, n = 4 we have;
[tex]f_D = \dfrac{(n + 1) \times \sqrt{\dfrac{T}{m/L} } }{2 \cdot L} = \dfrac{(4 + 1) \times K }{2 \times 24} = \dfrac{5 \times K}{48} \approx 0.1041 \overline 6 \cdot K[/tex]
Therefore, we have the rank of the frequency of oscillations of th strings from largest to smallest given as follows;
1 ) [tex]f_D[/tex] 2) [tex]f_B[/tex] 3) [tex]f_A[/tex] 4) [tex]f_C[/tex]
The order of the frequencies is [tex]f_D>f_B>f_A>f_C[/tex]
Standing waves:The frequency of the standing wave in a string tied at both ends is given by:
[tex]f=\frac{nv}{2L}[/tex]
where n is the mode of frequency
v is the velocity of the wave
and L is the length of the string.
Now the velocity of a wave in a string tied at both ends is given by
[tex]v=\sqrt{\frac{T}{\mu}}[/tex]
where T is the tension and μ is the mass per unit length.
Since T and μ are the same for all the strings, velocity [tex]v[/tex] will be the same for all.
Now to find the mode of frequency we can calculate the number of nodes (including the nodes at the ends) in the given figure and subtract by 1. Nodes are the point where the amplitude of the wave is zero.
[tex]f_A=\frac{3v}{2\times27}=\frac{v}{18}\;s^{-1}\\\\f_B=\frac{4v}{2\times30}=\frac{v}{15}\;s^{-1}\\\\f_C=\frac{3v}{2\times30}=\frac{v}{20}\;s^{-1}\\\\f_D=\frac{4v}{2\times24}= \frac{v}{12}\;s^{-1}[/tex]
Hence, [tex]f_D>f_B>f_A>f_C[/tex]
Learn more about standing waves:
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The diagram shows a state of matter in a closed system before and after undergoing a
change.
System before change
System after change
SEE
Which statement best explains the change in the system?
Particle motion decreases when thermal energy is added to the system
O Particle motion increases when solid particles are added to the system.
O Particle motion increases when thermal energy is added to the system.
O Particle motion decreases when gas particles are added to the system.
Answer:
the second one
Explanation:
Particle motion increases when solid particles are added to the system.
concave lens has negative focal length why
Free electromagnetic oscillations occur in an electrical circuit. Knowing the characteristic dimensions of the elements, L=20mH și C=0,005mF, determine: a)The period of free electromagnetic oscillations: b)Frequency of free electromagnetic oscillations: c)Electromagnetic energy in the circuit when the maximum value of the current intensity is equal to 100 mA.
Answer:
a. 0.199 ms b. 5.03 kHz c. 0.1 mJ
Explanation:
a. The period of oscillation of an L-C circuit is T = 2π√(LC) where L = inductance = 20 mH = 20 × 10⁻³ H and C = capacitance = 0.005 mF = 5 × 10⁻⁶ F.
So, T = 2π√(LC)
= 2π√(20 × 10⁻³ H × 5 × 10⁻⁶ F)
= 2π√(100 × 10⁻¹¹)
= 2π√(10 × 10⁻¹⁰)
= 2π(3.16 × 10⁻⁵)
= 19.87 × 10⁻⁵
= 1.987 × 10⁻⁴ s
= 1.99 × 10⁻⁴ s
= 0.199 × 10⁻³ s
= 0.199 ms
b. frequency , f = 1/T where T = period = 0.199 × 10⁻³ s.
So, f = 1/0.199 × 10⁻³ s
= 5.03 × 10³ Hz
= 5.03 kHz
c. The electromagnetic energy E = 1/2Li² where L = inductance = 20 × 10⁻³ H and i = current = 100 mA = 0.1 A
So, E = 1/2Li²
= 1/2 × 20 × 10⁻³ H × (0.1 A)²
= 0.1 × 10⁻³ J
= 0.1 mJ
When the mallet hits the ball with an action force, the ball exerts a reaction 1 force on the mallet as explained by: 1) Newton's first law 2) Newton's second law 3) Newton's third law 4) all three of Newton's laws
Answer:
It's Newton's third law. For every known reaction, there is a equal and opposite reaction.
Explanation:
The mallet which hits the ball with an action force, the ball exerts a reaction force on the mallet as explained by the Newton's third law. Thus, the correct option is 3.
What is the Newton's third law?Newton's third law states that for every action (force) in nature there is an equal and opposite force or reaction. An example of third law of motion is that the mallet hits the ball with an action force, then the ball exerts a reaction.
If an object A exerts a force on the object B, then object B must also exert a force of equal magnitude and in the opposite direction back on object A. The law represents a certain symmetry in the nature that the forces always occur in pairs, and one body cannot exert a force on another without experiencing any kind of force itself.
Therefore, the correct option is 3.
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calculate the magnitude of the net force on a 4.5 kg mass if the mass moves straight up at constant speed
Answer:
0
Explanation:
as Newton 1st Law said that every object will remain at rest or in uniform motion (constant speed) in a straight line unless compelled to change its state by the action of an external force
Net Force = 0
A cart is moving at 55 m/s at an angle of 25° to the ground. Determine the horizontal component.
Answer:
The horizontal component of the velocity is 49.85 m/s.
Explanation:
Rectangular Components of a Vector
A 2D vector can be expressed in several forms. The rectangular form gives its two components, one for each axis (x,y). The polar form gives the components as the pair (r,θ) being r the magnitude and θ the angle.
When the magnitude and angle of the vector are given, the rectangular components are calculated as follows:
[tex]v_x=v\cos\theta[/tex]
[tex]v_y=v\sin\theta[/tex]
Where v is the magnitude of the vector and θ is the angle with respect to the x positive direction.
The cart is moving at v=55 m/s at θ=25°, thus:
[tex]v_x=55\cos 25^\circ[/tex]
[tex]v_x=49.85\ m/s[/tex]
The horizontal component of the velocity is 49.85 m/s.
Refresh the simulation, and then add six protons and six electrons to make a neutral carbon atom. the Mass Number window. What's the mass number of this carbon atom?
Answer:
The mass number of this carbon atom is 12.
Explanation:
Mass number = Number of protons + Number of neutrons = 6 + 6 = 12
Brainlist pls!
Answer:
6
Explanation:
if a truck has the mass of 2,000 kilometres and a velocity of 35 m/s, what is the momentum
Answer:
70,000 kg.m/sExplanation:
The momentum of an object can be found by using the formula
momentum = mass × velocity
From the question we have
momentum = 2000 × 35
We have the final answer as
70,000 kg.m/sHope this helps you
Was the cannonball able to hit its target of 50 meters when the initial velocity was 20 m/s? Why?/Why Not?
(was not able to hit the target, but I don't know the reason)
Answer:
The cannon ball was not able to hit the target because the target is located at a height of 50 m whereas the cannon ball was only above to get to a height of 20 m.
Explanation:
From the question given above, the following data were obtained:
Height to which the target is located = 50 m
Initial velocity (u) = 20 m/s
To know whether or not the cannon ball is able to hit the target, we shall determine the maximum height to which the cannon ball attained. This can be obtained as follow:
Initial velocity (u) = 20 m/s
Final velocity (v) = 0 (at maximum height)
Acceleration due to gravity (g) = 10 m/s²
Maximum height (h) =?
v² = u² – 2gh (since the ball is going against gravity)
0² = 20² – (2 × 10 × h)
0 = 400 – 20h
Collect like terms
0 – 400 = – 20h
– 400 = – 20h
Divide both side by – 20
h = – 400 / – 20
h = 20 m
Thus, the the maximum height to which the cannon ball attained is 20 m.
From the calculations made above, we can conclude that the cannon ball was not able to hit the target because the target is located at a height of 50 m whereas the cannon ball was only above to get to a height of 20 m.
marked as brainiest if correct
Answer:
I think the the answer is creating and layout and template style ( C )
Explanation:
I did it before and i was checking my notes and i wrote that down , Hope this Helps :)
A repeating disturbance that transfers energy through matter or space is called a _____.
Answer:
Collateral Damage
Explanation:
Answer:
The answer is Wave
Explanation:
calculate the resistance of a wire 150cm long and diameter 2.0mm constructed from an alloy of resistivity 44*10-⁸Ωm
Answer:
R = 0.21 Ω
Explanation:
the formula:
R = r x l/A
R = (44 x 10-⁸ Ωm) x 1.5 / (π x (1 x 10-³ m)²)
R = 6.6 x 10-⁷ / 3.14 x 10-⁶
R = 0.21 Ω
Chef Andy tosses an orange in the air, then catches it again at the same height. The orange is in the air for 0.75\,\text s0.75s0, point, 75, start text, s, end text. We can ignore air resistance.
What was the orange's velocity at the moment it was tossed into the air?
Answer:
90.82%digrash silsho