The velocity of the tennis ball when it hits the ground is calculated as 4.43m/s. Details about velocity can be found below.
How to calculate velocity?The velocity of a moving object can be calculated using the equation of motion as follows:
v² = u² + 2as
Where;
v = final velocityu = initial velocitya = accelerations = distancev² = 0² + 2(9.8) (1)
v² = 19.6
v = ✓19.6
v = 4.43m/s
Therefore, the velocity of the tennis ball when it hits the ground is calculated as 4.43m/s.
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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.
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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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!
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]
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:
nitrogen dioxide is a major component of smog which type of chemical reaction creates this product?
a. synthesis reaction
b. replacement reactions
c. combustion reactions
d. decomposition reaction
Answer:
A.synrhesis reaction
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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]
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]
which of the following are features of the nucleus of an atom
finding evidence to support or dispove is how science advanes
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. ❞
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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:
please help me with this
Answer:
>400N is needed to balance that lever
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:
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)
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:
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
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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Assume ideal behavior. How many moles of are needed to fill a 2000L weather balloon at 210K and 1atm pressure? Round your answer to the nearest mole (use R = 0.082057 L atm mol-1K-1)
Considering the ideal gas law, 116.06 moles of are needed to fill a 2000 L weather balloon at 210 K and 1 atm pressure.
An ideal gas is a theoretical gas that is considered to be composed of randomly moving point particles that do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.
The pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:
P×V = n×R×T
where P is the gas pressure, V is the volume that occupies, T is its temperature, R is the ideal gas constant, and n is the number of moles of the gas. The universal constant of ideal gases R has the same value for all gaseous substances.
In this case, you know:
P= 1 atmV=2000 Ln= ?R= 0.082057 [tex]\frac{atmL}{molK}[/tex]T= 210 KReplacing in the ideal gas law:
1 atm× 2000 L = n× 0.082057 [tex]\frac{atmL}{molK}[/tex]× 210 K
Solving:
[tex]n=\frac{1 atmx 2000 L}{0.082057 \frac{atmL}{molK}x 210 K}[/tex]
n=116.06 moles
Finally, 116.06 moles of are needed to fill a 2000 L weather balloon at 210 K and 1 atm pressure.
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the impulse-momentum relationship is a direct result of
Newton's second law.
__________________
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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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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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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:
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 ❤.
Stacey runs around a 400 meter track 2 times in 5 minutes. How does her displacement differ from her distance? *
Answer:
Distance is 800 m and Displacement is 0 m
Explanation:
Total Distance
= 400(2)
= 800 m
Total Displacement
= 0 m since she returns to the same spot
Liquids and solids are:
a- compressible and expandable.
b- incompressible and non-expandable.
c- incompressible and expandable.
Answer:
incompressible and expandable
. 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:
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
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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How is the independent variable affected by the dependent variable
Answer:
A dependent variable is a variable that is tested in an experiment. An independent variable is that can be modified. Depending on what you are testing, the dependent variable will change accordingly to the dependent variable.
- I'm reading this back and it doesn't make much sense, if you want me to reword this I can
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