Answer:
Approximately [tex]10\; \rm m \cdot s^{-1}[/tex] at [tex]5.6^\circ[/tex] below the horizon.
Horizontal component of velocity: [tex]10\; \rm m \cdot s^{-1}[/tex].Vertical component of velocity: [tex]0.981\; \rm m\cdot s^{-1}[/tex] (downwards.)(Assumption: air resistance on the ball is negligible; [tex]g = 9.81\; \rm m \cdot s^{-2}[/tex].)
Explanation:
Assume that the air resistance on the ball is negligible. The horizontal component of the velocity of the ball would stay the same at [tex]10\; \rm m \cdot s^{-1}[/tex] until the ball reaches the ground.
On the other hand, the vertical component of the ball would increase (downwards) at a rate of [tex]g = 9.81\; \rm m\cdot s^{-2}[/tex] (where [tex]g[/tex] is the acceleration due to gravity.) In [tex]0.1\; \rm s[/tex], the vertical component of the velocity of this ball would have increased by [tex]9.81\; \rm m \cdot s^{-2} \times 0.10\; \rm s = 0.981\; \rm m \cdot s^{-1}[/tex].
However, right after the ball rolled off the edge of the table, the vertical component of the velocity of this ball was [tex]0\; \rm m\cdot s^{-1}[/tex]. Hence, [tex]0.10\; \rm s[/tex] after the ball rolled off the table, the vertical component of the velocity of this ball would be [tex]0\; \rm m \cdot s^{-1} + 0.981\; \rm m\cdot s^{-1} = 0.981\; \rm m \cdot s^{-1}[/tex].
Calculate the magnitude of the velocity of this ball. Let [tex]v_{x}[/tex] and [tex]v_{y}[/tex] and denote the horizontal and vertical component of the velocity of this ball, respectively. The magnitude of the velocity of this ball would be [tex]\displaystyle \sqrt{{v_x}^{2} + {v_y}^{2}}[/tex].
At [tex]0.10\; \rm s[/tex] after the ball rolled off the table, [tex]v_x = 10\; \rm m \cdot s^{-1}[/tex] while [tex]v_y = 0.981\; \rm m \cdot s^{-1}[/tex]. Calculate the magnitude of the velocity of the ball at this moment:
[tex]\begin{aligned} \| v \| &= \sqrt{{v_x}^{2} + {v_y}^{2}} \\ &= \sqrt{\left(10\; \rm m \cdot s^{-1}\right)^{2} + \left(0.981\; \rm m \cdot s^{-1}\right)^{2}} \\ &\approx 10.0\; \rm m\cdot s^{-1}\end{aligned}[/tex].
Calculate the angle between the horizon and the velocity of the ball (a vector) at that moment. Let [tex]\theta[/tex] denote that angle.
[tex]\displaystyle \tan \theta = \frac{\text{rise}}{\text{run}}[/tex].
For the vector representing the velocity of this ball:
[tex]\displaystyle \tan \theta = \frac{0.981\; \rm m \cdot s^{-1}}{10\; \rm m \cdot s^{-1}} = 0.0981[/tex].
Calculate the size of this angle:
[tex]\theta = \arctan 0.0981 \approx 5.62^\circ[/tex].
Notice that the vertical component of the velocity of this ball at that moment points downwards (towards the ground.) Hence, the corresponding velocity should point below the horizon.
An astronaut has a mass of 70kg.
The force of gravity on the Moon is approximately 1.5N/kg.
Calculate the weight of the astronaut when they are on the Moon.
Weight = .............................................................. N
Answer:
Weight = 105 Newton
Explanation:
Given the following data;
Mass = 70kg
Force of gravity = 1.5N/kg
To find the weight;
Weight = mass * force of gravity
Substituting into the equation, we have;
Weight = 70 * 1.5
Weight = 105 Newton
Therefore, the weight of the astronaut when they are on the Moon is 105 Newton.
Example 4:
Find the index of refraction for diamond if the speed of light in diamond is 1.24 x 10m/s
Use the proper FORMAT
Answer:
Given speed of light in diamond =1.24*10m/s
speed of light (c) =3*10^8m/s
refractive index of diamond =3*10^8/1.24*10=2.41*10^7m/s
A. 1 and 2 > 3 and 4 = 4 and 5
B. 1 and 2 > 3 and 4 > 4 and 5
C. 3 and 4 > 4 and 5 > 1 and 2
D. 1 and 2 > 4 and 5 > 3 and 4
E. 3 and 4 = 4 and 5 > 1 and 2
Answer:
1 and 2>3 and 4=4 and 5 that is the answer
A force of 5 N accelerates an object. The objects mass is 50 G. What is the acceleration of the object? (Formula: F=ma)
Answer:
100 m/s^2
Explanation:
F=ma
50 g = 0.05 kg
5 = 0.05a
a=100m/s^2
Answer:
100MS
Explanation:
I did the test and got 100%
1. What is the velocity of an 250-kg that has 6,250 J of energy?
Please help
Explanation:
KE = 1/2mv^2
6,250 J =1/2(250kg) v ^2
6250 J = 125kgv^2
125kg. 125kg
v =
[tex] \sqrt{50} [/tex]
Explain the benefits of understanding sound waves properties in real life.
Answer:
The vibrations from sound waves cause our ears to send signals to our brains to create sound. The speed of sound waves will determine the sound's pitch, or how high or low something sounds. Sound waves are important because they allow us to hear important messages and emergency signals to protect ourselves.
Explanation:
I hope this helps :)
Rank the items from slowest to faster
Answer:since time goes to the right you look at the lines that are most pointed to the right with less distance so that would be B, A, D, C
Explanation:
Which of the following shows a balanced nuclear reaction?
Answer:
No. 2
Explanation:
The balanced nuclear reaction is shown by the following: ²³⁷₉₃ Np[tex]\rightarrow[/tex]⁴ ₂He +²³³₉₅ Am nuclear fission reaction.
What are nuclear reactions?There are two types of nuclear reactions which are nuclear fusion and nuclear fission .They involve the combination and disintegration of the element's nucleus respectively.
In nuclear fission, the nucleus of the atom is bombarded with electrons of low energy which splits the nucleus in to two parts .Large amount of energy is released in the process.It is used in nuclear power reactors as it produces large amount of energy.
In nuclear fusion,on the other hand, is a reaction which occurs when two or more atoms combine to form a heavy nucleus.Large amount of energy is released in the process which is greater than that of the energy which is released in nuclear fission process.
Learn more about nuclear reactions,here:
https://brainly.com/question/12786977
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What is the energy transformation of home gas?
Answer:
The Sun transforms nuclear energy into heat and light energy.
Our bodies convert chemical energy in our food into mechanical energy for us to move.
An electric fan transforms electrical energy into kinetic energy.
explain how to perform the exercise leg raises
A lift that weighs 300 N moves a man with a weight of 950 N upwards by 12 m. How much work is done to move the lift this height?
The total work done to move the lift to this height if, A lift that weighs 300 N moves a man with a weight of 950 N upwards by 12 m, is 15000 Joules.
What is work done?When an object is moved over a distance by an external force, at least a portion of that force must be applied in the direction of the displacement. This is known as work in physics.
Given:
A lift that weighs 300 N moves a man with a weight of 950 N upwards by 12 m,
Calculate the total force as shown below,
Total force = Lift weight + man weight
Total force = 300 + 950
Total force = 1250 N
Calculate the work done as shown below,
The work done = Total force × displacement by total force
The work done = 1250 × 12
The work done = 15000 Joules
Thus, the total work done is 15000 joules.
To know more about work done:
https://brainly.com/question/29762840
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10 A ship is made of 4000 m' steel of density
7900 kgm^3
a. What volume of fresh water
1000 kg m) would the ship need to
displace in order for it to float?
b. Explain how the design of the ship enables
it to displace this much water.
Answer:
Explanation:
mass of ship = volume x density
= 4000 x 7900 = 316 x 10⁵ kg
The ship must displace water equal to weight of ship that is 316 x 10⁵ kg .
If volume of displaced water be V .
V x 10³ is mass of displaced water
V x 10³ = 316 x 10⁵
V = 31600 m³.
b ) Now the volume of steel making ship is only 4000 m³ so if it is in the form of a lump , it will displace only water equal to its volume ie 4000 m³ . So in order that it may displace water equal to volume 31600 m³ , it must be given shape so that it is hollow .
Two coulombs of charge pass through a wire in 5 seconds. Calculate the current in the wire. 1
Answer:
0.4A.
Explanation:
Current (A) = Charge (coulomb)/Time (secs)
2 coulombs/5 secs = 0.4A
A boy moves around a square path of side 15m. He completes 1 round in 40 sec. 1 Calculate the displacement of the boy at the end of 2min 20sec.
Plsss answer fast!
Answer:
Explanation:
2 min 20 sec = 140 sec .
In 40 sec one round
In 140 sec 140 / 40 round = 3.5 round .
That means the boy will be positioned at the diametrically opposite corner of the square path .
side of square path = 15 m
length of diameter = 15√2
= 21.21 m
Displacement of boy is 21.21 m .
Plz help in this question.
Answer:
The area of the liquid surface does not affect the pressure.
If you are startled by a sudden loud noise, your (____)
activated.
nervous system is
A. autonomic
B. parasympathetic
C. somatic
D. sympathetic
Answer:
D. sympathetic
Explanation:
Bill and Nageen have each built an electric motor and they are testing them by having them lift boxes. Bill's motor lifts a box 0.39 metres in 2 seconds. Nageen's motor lifts the same box 0.35 metres in 2 seconds. Whose motor has applied more power to the box?
Answer:
Bill's motor
Explanation:
Bill's motor lifts a box 0.39 metres in 2 seconds.
Nageen's motor lifts the same box 0.35 metres in 2 seconds.
Power = Work done/time
or
[tex]P=\dfrac{mg\times h}{t}[/tex]
Power of bill's motor,
[tex]P_1=\dfrac{m\times 9.8\times 0.39}{2}\\\\P_1=1.91m\ Watts[/tex]
Power of Nageen's motor,
[tex]P_2=\dfrac{m\times 9.8\times 0.35}{2}\\\\P_2=1.71m\ Watts[/tex]
So, Bill's motor applied more power to the box.
A baseball player (mass = 75 kg) is running north towards a base. In order to avoid
being tagged by the ball, the baseball player slides into the base. If his acceleration
in the slide is 5.0 m/s2 south, what is the kinetic frictional force between the
baseball player and the ground?
it's 1727 +822 just kidding
A baseball player slows down as he slides to base is he balanced or unbalanced?
Answer : this pretty much gives you the answer Lol
Explanation:
Fundamentals. When the forces acting upon an object are balanced, the object will maintain its state of motion; that is, its velocity remains constant. If the forces are unbalanced, the object will either speed up or SLOW DOWN. ... And if an object is speeding up or slowing down, the forces are unbalanced.
what are some activities you can do on a regular basis? Physical education
activities you can do on a regular basis for physical education?
If so
WalkingGardeningDancingBicyclingSwimmingYogaThe answer to anyones problems-
Answer:
POGGGGGGGGGGGGGGG
Explanation:
A student wants to make an analogy to help her remember the characteristics of the planets. Which analogy makes the most sense?
The inner planets are like cotton candy and the outer planets are like hard butterscotch candy.
The outer planets are like bowling balls and the inner planets are like basketballs.
The inner planets are like hard butterscotch candy and the outer planets are like cotton candy.
The outer planets are like roses and the inner planets are like daisies.
Answer:
The last one makes the most sense as they combine two like things that are easy to visualize
Explanation:
Answer:
C. The inner planets are like hard butterscotch candy and the outer planets are like cotton candy.
Explanation:
Right on ED2021, goodluck!
Essential Question: How do different forces affect our daily experience? Think about us walking, gravity, etc.
Answer:
dd
Explanation:
I dont know I dont know I dont know I dont know
1. Calculate the momentum of each car before the collision: SHOW YOUR WORK!
Pred p = mxv> (1 kg) x (+5) = +5 kg m/s
Pblue: P = mxv
(1 kg) x (0) = 0 kg m/s
Answer:
Momentum of red car = 5kgm/s
Momentum of blue car = 0kgm/s
Explanation:
Momentum = mass × velocity
For the red car
Mass = 1kg
Velocity = 5m/s
Momentum of the red car = 1kg × 5m/s
Momentum of the red car = 5kgm/s
For the blue car.
Mass = 1kg
Velocity = 0m/s(shows that the blue car is stationery)
Momentum = 1kg ×0m/s
Momentum of the blue car = 0kgm/s
Can someone answer these questions? Im really sorry for the long 15 questions but it would save me and basically my whole class.
Answer:
yes
Explanation:
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Now consider θc, the angle at which the blue refracted ray hits the bottom surface of the diamond. If θc is larger than the critical angle θcrit, the light will not be refracted out into the air, but instead it will be totally internally reflected back into the diamond. Find θcrit.
Answer:
θ_c = 24.4º
Explanation:
To find the critical angle, let's use the law of refraction where index 1 refers to the incident medium (diamond) and index 2 refers to the medium where it is to be refracted (air)
n₁ sin θ₁ = n₂ sin θ₂
for the critical angle the ray comes out refracted parallel to the surface, therefore the angle is
θ₂ = 90
n₁ sin θ₁ = n₂
θ_c = θ₁ = sin⁻¹ [tex]\frac{n_2}{n_1}[/tex]
the index of refraction of the diamond is tabulated
n₁ = 2.419
let's calculate
θ_c = sin⁻¹ ([tex]\frac{1}{2.419}[/tex])
θ_c = 24.4º
Is pulling a dog a contact force or a non contact force
How would you describe a crescent moon?
1.) A crescent moon is when we can view more than half of the illuminated
side of the moon.
2.) A crescent moon is when we can view less than half of the illuminated
side of the moon.
3.) A crescent moon is when we can view all of the illuminated side of the
moon.
4.) A crescent moon is when we can view none of the illuminated side of the
moon.
Which of the following would be the best conductor - iodine, silver, or sulfur? Explain
Answer:
Silver
Explanation:
Silver would be the best conductor.
Because silver is metal and iodine, sulfur are non-metals.Metals are good conductor as compared to non-metals.Answer:
silver
Explanation:
transition metal that has properties of a good conducfor.
Putting a beaker of water in a microwave and turning it on for a minute or two will increase the temperature of the water. Describe an experiment to investigate the relationship between the time the microwave is on and the increase in temperature of water in a beaker. In your answer, include: The equipment you will use The measurements you will take The safety precautions you will take
Answer:
The equipment to use is: a beaker, a fixed amount of water, a thermometer.
The mass of water, the time, the temperature for each time should be noted and a graph of Temperature versus time should be made
Explanation:
The design of an experiment is to place the beaker in the microwave, with a good amount of water (approximately ⅔ of its capacity) and turn it on for small periods of time, generally the minimum is 30 s, quickly open the microwave, place a thermometer or better yet an infrared thermometer to measure the temperature of the water; repeat this several times.
The advantage of the infrared thermometer is that it reduces the transfer of heat between the water and the thermometer.
The mass of water, the time, the temperature for each time should be noted and a graph of Temperature versus time should be made.
The equipment to use is: a beaker, a fixed amount of water, a thermometer.
The main precaution that must be taken is not to open the microwave while it is on.
