Answer:
Sound waves bend around the corners of various obstacles.
Explanation:
I took the test
A 600g toy train completes 10 laps of its circular track in 1 min 20s. If the radius of the track is 1.2 m, Find the speed of the train in m/s.
Answer:
0.236 m/s² (rounded)
Explanation:
2. A nutty professor leaves a history classroom and walks 10. meters north to a drinking fountain. Then, in a daze, she turns and walks 30. meters south to an art classroom. (draw vectors to scale to show your work)
(a) What is the professor’s total displacement from the classroom to the art classroom?
(b) What was the total distance she traveled?
Total displacement:-
[tex]\\ \sf\Rrightarrow 10-30=20m[/tex]
Total distance
[tex]\\ \sf\Rrightarrow 10+30=40m[/tex]
HELPPP PLEASE
1) if the distance between two masses are 15 meters and the masses are 350kg and 492 kg
respectively; what is the magnitude of gravitational force?
Answer:
-8
5.092×10
Explanation:
given,
distance (d) =15m
mass (M1) =350kg
mass(M2) = 492kg
Gravitational constant (G)=6.67×10^-11
we know,
gravitational force=(Gm1m2)÷d^2
=(6.67×10^-11×350×492)÷15^2
= (1.14857×10^-5)÷225
= 5.092×10^-8
which of the following are features of the nucleus of an atom
finding evidence to support or dispove is how science advanes
What do you feel when you receive your homework?
The feeling can either be positive or negative.
How one feels depends on number of factorsA positive feeling can occur when one performs very well in the given home. Also, a positive feeling can come from a high level of satisfaction in the homework. When you complete your homework on time using the recommended steps, you will be sure to do well on the homework.
In other hand, a negative feeling may result from poor performance in the homework. In ability to complete the homework or missing some steps in the homework can increase your level of trepidation even before seeing your score.
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according to newton's third law, of a sledgehammer hits a wall with a force of 1000 n, how much force does the wall hit the sledgehammer? a - 1000 n b 1000 n c 0 n d 500 n
The correct option is A: the force exerted on the wall by the sledgehammer is - 1000 N
Newton's third lawNewton's third law of motion states that for every action or force, there is an equal but oppositely directed force.
Application of the newton's third lawForce with which the sledgehammer hits the wall = 1000 N
From the Newton's third law, an equal but oppositely directed force is exerted on the sledgehammer by the wall.Thus, the force exerted by the sledgehammer on the wall = -1000 N
The negative sign indicates that the force acts in a direction opposite to that of the sledgehammer.Thus, the correct option is A: the force exerted on the wall by the sledgehammer is - 1000 N
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what equation do you use to calculate force from work to distance
Answer:
Work can be calculated with this equation: Work = Force x Distance. The SI unit for work is the Newton meter (N m). One joule equals the amount of work that is done when 1 N of force moves an object over a distance of 1 m.
Explanation:
stella is driving down a steep hill. she should keep her car __________ to help _________.
Answer:
Stella is driving down a steep hill. She should keep her car in a lower gear to help slow her vehicle.
Explanation:
A wave has a frequency of 2700 Hz and λ of 6m. Calculate the speed of the wave.pls help
Answer:
16,200 m/sExplanation:
The speed of the wave given only it's frequency and wavelength can be found by using the formula
[tex]c = f \times \lambda[/tex]
where
c is the velocity of the wave in m/s
[tex] \lambda[/tex] is the wavelength in m
f is the frequency in Hz
From the question
c = 2700 × 6 = 16,200
We have the final answer as
16,200 m/sHope this helps you
General kinematics problem.
"Anna is returning from vacation driving down the Sunshine Highway. She travels for a stretch (SA) at a speed of 110 km/h. Then, due to an increase in traffic, Anna is forced to reduce her speed by travelling the second part of the journey (SB) at 60 km/h. The total distance and duration of the trip are 350 km and 4.0 hours respectively.
-> Calculate the time intervals needed to travel SA and SB
-> Calculate the distances covered in the two sections SA and SB".
The results I should get by doing the exercise are in the attached picture.
If you could explain the various steps in the problem you would be doing me a huge favor. Thanks!
Explanation:
What we know:
Sa + Sb = 350 km
Ta + Tb = 4h
S = Vt
With these equations above we can write a system of equations and find Tb, as you see in the picture(s).
After having Tb you can easily find Ta and then Sa and Sb. Hopefully it was clear.
If the western component of the wind is half of the south, what is the angle of the wind with the south?
Answer this question in details!
Answer is 26.57 or tan^-1 (1/2) or cot^-1 (2)
no links !
Answer:
This above a triangle that models our situation.
Explanation:
We have a two componens., since we have a western componet and southern component. One travel in a southern direction. and the other travel in the west.
Let the component that travel in the south be the length of a.
According to the problem, the westard component is half of that so let that length be a/2.
Now we must find the angle of the wind in the South.
This means that what is angle that is opposite of the western componet because that angle is the most southward angle. So know we apply the tan property.
[tex] \tan(x) = \frac{opp}{adj} [/tex]
Our side opposite of the angle we trying to find is the western component and the side adjacent to it is the southern component. Also remeber since western and Southern negative displacements, we have
[tex] \tan(x) = \frac{ - \frac{a}{2} }{ - a} [/tex]
[tex] \tan(x) = - \frac{a}{2} \times - \frac{1}{a} = \frac{1}{2} [/tex]
Now we take the arctan or inverse tan of 1/2.
[tex] \tan {}^{ - 1} ( \frac{1}{2} ) = 26.57[/tex]
Based on the information about vector components and graph provided, the angle of the wind with the south is 26.57°.
What are the values of the western and southern component of the wind?The western and southern component have a negative displacement as shown in the graph.
Let the southern component be -x
The western component is half of -x = -x/2
What is the angle of the wind with the South?Since the angle is with the south, the trigonometric ratio to be used to find the angle is:
Tan θ = opp/adjTan θ = -×/2 /-x
Tan θ = 1/2
θ = tan^-1(0.5)
θ = 26.6°
Therefore, angle of the wind with the south is 26.57°.
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. a. Calculate the work done while lifting 300 kg of wate through a vertical height of 6 m. (Assume g = 10 m a =
Answer:
potential energy = mgh = 300 × 10 × 6m = 18000 joule or 18 kilo joule.
Explanation:
Liquids and solids are:
a- compressible and expandable.
b- incompressible and non-expandable.
c- incompressible and expandable.
Answer:
incompressible and expandable
What are the rest of the blanks?
_____________ is the study of movement in athletes. A. Sports biomechanics B. Posture C. Dynamics D. Anatomy
Answer:
I believe its A: Sports biomechanics.
An automobile with a mass of 1250 kg accelerates at a rate of 4 m/sec2 in the forward direction. What is the net force acting on the automobile?
[tex]\large{\underline{\underline{\maltese{\pink{\pmb{\sf{ \; Question \; :- }}}}}}}[/tex]
An automobile with a mass of 1250 kg accelerates at a rate of 4 m/sec² in the forward direction. What is the net force acting on the automobile?
_________________________________________
[tex]\large{\underline{\underline{\maltese{\green{\pmb{\sf{ \; Given\; :- }}}}}}}[/tex]
➢ Mass of the automobile = 1250 kg➢ Acceleration of the body = 4m/sec²_________________________________________
[tex]\large{\underline{\underline{\maltese{\orange{\pmb{\sf{ \; To \: Find \; :- }}}}}}}[/tex]
Net force acting on the automobile = ?_________________________________________
[tex]\large{\underline{\underline{ \bigstar{{\pmb{\sf{ \; Concept \: and \: Formula \: Used \; :- }}}}}}}[/tex]
➯ Force is defined as the product of the mass of the body and its acceleration.
[tex] \bigstar \: \small{ \fbox {\textsf {\textbf{F = ma}}}}[/tex]
Where :
F = Force m = Mass of the bodya = acceleration due to gravity[tex]\large{\underline{\underline{ \bigstar{{\pmb{\sf{ \; Substituting \: the\: Given\: Values :- }}}}}}}[/tex]
[tex]⇝ \: \: \small \textsf {\textbf{ F = (1250 × 4) N }}[/tex]
[tex]⇝ \: \: \small \textsf {\textbf{ F = 5000 N }}[/tex]
[tex]\large{\underline{\underline{ \bigstar{{\pmb{\sf{ \; Therefore \; :- }}}}}}}[/tex]
❝ The net force acting on the automobile is 5000 N. ❞
▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬
How does NaOH form a basic solution when it dissolves in water
When NaOH is added to water, it separates into Na+ and OH-. The hydroxyls make it more basic.
What is basic solution?A basic solution is an aqueous solution containing more OH⁻ ions than H⁺ ions.
When NaOH as a solute is mixed into the solvent like water, it separates NaOH into Na+ and OH-. The sodium ions Na⁺ have little effect on the solution, but the hydroxyls OH⁻ ions makes the solution more basic.
Thus, NaOH form a basic solution by the presence of OH⁻ ions into it.
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please help me with this
Answer:
>400N is needed to balance that lever
a 10 mh inductor is connected in series with a 10 ohm resistor, a switch and a 6 v battery. what is the time constant of the circuit
Answer:
The inductive time constant is just T = L / R
T = 10E-3 henry/ 10 ohm = 10E-2 sec = .1 sec
(.01 / 10 = .1)
when a rigid body rotates about a fixed axis, all the points in the body have the same
Hi there!
[tex]\large\boxed{\text{Angular acceleration.}}[/tex]
For a rigid body rotating about a fixed axis, its angular characteristics ⇒ angular acceleration and velocity are constant throughout.
However, its LINEAR velocities/accelerations differ because of the following relationships:
v = ωr
a = αr
Thus, a point closer to the axis of rotation has a smaller linear velocity or acceleration compared to a point along the edge.
Which branch of science deals with the study of the structures shown here?
FISH
AMPHIBIAN
REPTILE
BIRD?
Answer:
Vertebrate zoology
Explanation:
Have a great day!
the impulse-momentum relationship is a direct result of
Newton's second law.
__________________
o0o0o0o0o0o0o0o0o0
The impulse-momentum relationship is a direct result of Newton's second law of motion.
What is the impulse-momentum?The impulse-momentum theorem states that the impulse applied to an object is equal to the change in its momentum. It proves that the change in momentum of an object depends not only on the amount of force applied but also on the duration of force applied.
Newton's second law states that the force acting on an object is equal to the rate of change of its momentum.
This means that a force applied to an object will cause a change in its momentum. The impulse-momentum relationship describes the relationship between the force applied to an object and the resulting change in its momentum.
The impulse-momentum relationship states that the impulse acting on an object is equal to the change in its momentum.
Impulse is defined as the force applied to an object over a period of time, while momentum is the product of an object's mass and velocity. Therefore, the impulse-momentum relationship can be expressed as:
Impulse = Change in momentum
This relationship is important in understanding the behavior of objects in motion, particularly in collisions or other situations where forces act over a period of time.
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Which of the following industries is the largest producer of primary air pollutants in the United States?
The largest producer of primary air pollution in the United States is what? electricity producing plants.
Answer:
electricity production
Explanation:
Connor rode an inner tube down a river. For 4.6 minutes, he moved downriver at 15 meters per minute. During this time, how far did he move?
Answer:
2 miles
Explanation:
What event will produce the greatest increase in the gravitational force between the two masses?
Question 5 options:
Doubling the distance between the masses
Reducing the small mass by half
Doubling the large mass
Reducing the distance between the masses by half
Answer:
Doubling the large mass
Explanation:
By doubling the mass you will get more gravational force
The following data were collected during a short race between two friends. Velocity (m/s) 0 0.5 1 1.5 2 2 4 6 2 0 Time (s) 0 2 4 6 8 10 12 14 16 18 a) Describe the different sections of the graph. b) Determine the acceleration over the first eight seconds. c) Determine the maximum acceleration. d) Using the graph calculate the displacement: i) over the first eight seconds ii) the total race. e) Find the maximum velocity reached by the runner.
The characteristics of the kinematics allow to find the results for the questions about the movement of the body are:
a) we have four sections;
0 to 8 s The body is accelerating. 8 to 10 s The body goes at a constant speed, the acceleration is zero. 10 to 14 Body accelerating. 14 to 18 Body slowing down.b) The acceleration is the first 8 s is: a = 0.25 m / s²
c) The maximum acceleration is: a = 1 m / s²
d) The displacement is: i) d₁ = 8m, ii) [tex]d_{total}[/tex]= 16 m
e) maximum speed is: v = 6 m / s
Kinematics studies the movement of bodies by finding relationships between the position, speed and acceleration of bodies.
v = v₀ + a t
y = v₀ t + ½ a t²
where v and v₀ is the current and initial velocity, respectively, a is the acceleration and t is time.
In many circumstances graphs are made for their analysis, in a graph of speed versus time when we have a horizontal line the speed is constant, the acceleration is zero and in the case of a slope there is an acceleration, we have two cases:
Positive slope the body is accelerating and the speed is increasing. Negative slope the body is stopping, the speed decreases.
Let's answer the different questions about the system.
a) in the attached we have a graph of the velocity versus time, each section corresponds to a change in the slope of the graph, we have four sections;
0 to 8 s The body is accelerating. 8 to 10 s The body goes at a constant speed, the acceleration is zero. 10 to 14 Body accelerating. 14 to 18 Body slowing down.
b) The acceleration is the first 8 s
v = v₀ + a t
[tex]a = \frac{v-v_o}{\Delta t}[/tex]
[tex]a = \frac{2-0}{8-0}[/tex]
a = 0.25 m / s²
c) The maximum acceleration is when the slope is maximum.
[tex]a = \frac{6-2}{ 14-10}[/tex]
a = 1 m / s²
Therefore the acceleration is maximum in the section between 10 and 14 s
d) The total displacement is the sum of the displacements of each section.
[tex]d_{total } = d_1 +d_2 + d_3 +d_4[/tex]
We look for every displacement.
d₁ = v₀ + ½ a₁ Δt²
d₁ = 0 + ½ 0.25 8²
d₁ = 8 m
In the second section the velocity is constant
d₂ = v₂ Δt₂
d₂ = 2 (10-8)
d₂ = 4 m
The third section.
d₃ = v₀ + ½ a t²
d₃ = 2 + ½ 1 (14-10) ²
d₃ = 10 m
The distance of the fourth section.
we look for acceleration
a₄ = [tex]\frac{v-v_o}{\Delta t}[/tex]
a₄ = [tex]\frac{0-6}{18-14}[/tex]
a₄ = -1.5 m / s²
d₄ = 6 + ½ (-1.5) (1814) ²
d₄ = -6 m
The total displacement is;
[tex]d_{total}[/tex] = 8 + 4 + 10 -6
[tex]d_{total}[/tex] = 16 m
e) The maximum speed is the highest point in the graph of speed versus time that in the attachment we can see corresponds to
v = 6 m / s
In conclusion using the characteristics of kinematics we can find the results for the questions about the motion of bodies are:
a) we have four sections;
0 to 8 s The body is accelerating. 8 to 10 s The body goes at a constant speed, the acceleration is zero. 10 to 14 Body accelerating. 14 to 18 Body slowing down.b) The acceleration is the first 8 s is: a = 0.25 m / s²
c) The maximum acceleration is: a = 1 m / s²
d) The displacement is: i) d₁ = 8m, ii) [tex]d_{total}[/tex]= 16 m
e) maximum speed is: v = 6 m / s
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Stewart (70 Kg) is attracted to Ms. Little (60 Kg) who sits 2 m away. What is the gravitational attraction between them? G=6.67×10^-11 (-11 is an exponent)
Happy Holidays!
We can use the following equation to solve for the gravitational force:
[tex]\large\boxed{F_g = G\frac{m_1m_2}{r^2}}[/tex]
Fg = force due to gravity (N)
G = Gravitational constant
m1,m2 = masses of the objects (kg)
r = distance between the objects (m)
Plug in the given values into the equation:
[tex]F_g = (6.67*10^{-11})\frac{(70)(60)}{(2)^2}} = \boxed{7.0 * 10^{-8}N}[/tex]
Annie comes to a sudden stop to avoid hitting a deer crossing the road at night. As she hits the breaks, her speed drops from 12 m/s to zero in 1.5 seconds.(p1) How far did she travel during that time? (p2)What was her average acceleration?
Distance(p1)
9 m
8 m
10 m
13.5 m
Acceleration(p2)
6 m/s2
8 m/s2
12 m/s2
16 m/s2
Answer:
1. around 22m
Explanation:
Annie traveled 9 meters during that time. Her average acceleration was -8 m/s².
Annie's initial speed was 12 m/s.
Her final speed was 0 m/s.
The time it took her to stop was 1.5 seconds.
Therefore, the distance she traveled during that time is:
distance = (initial speed + final speed)/2 * time
= (12 m/s + 0 m/s)/2 * 1.5 seconds
= 9 meters
Annie's average acceleration is calculated by dividing her change in velocity by time it took for that change to occur.
In this case, her change in velocity is from 12 m/s to 0 m/s, which is a decrease of 12 m/s.
The time it took for this change to occur is 1.5 seconds.
Therefore, Annie's average acceleration is:
acceleration = change in velocity / time
= -12 m/s / 1.5 seconds
= -8 m/s²
The negative sign indicates that Annie was slowing down.
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A 1500 kg car traveling 5.0 m/s collides head on with a 3000 kg truck traveling
7.0 m/s in the opposite direction. If the bumpers lock. What is the velocity of the
two vehicles together immediately following the collision?
Using law of conservation of momentum
[tex]\\ \sf\Rrightarrow m_1v_1-m_2v_2=(m1+m2)v_3[/tex]
[tex]\\ \sf\Rrightarrow 1500(5)-3000(7)=(1500+3000)v_3[/tex]
[tex]\\ \sf\Rrightarrow 7500-21000=4500v_3[/tex]
[tex]\\ \sf\Rrightarrow -13500=4500v_3[/tex]
[tex]\\ \sf\Rrightarrow v_3=-3m/s[/tex]
what is the electric field strength at a distance of .50 m from a 1.00x10^-6 c charge?
Answer:
e = k q / r² then,
e = 9×10^9 * 1×10^-6 / 50² = 3.6 N/C
hope this helps ❤.