Explanation:
initial height, yo = 2 m
initial velocity, u = 20 m/s
angle of projection,θ = 5 degree
distance of net = 7 m
height of net = 1 m
Let it covers a vertical distance y in time t .
Use Second equation of motion for vertical motion
As it hits the ground in time t, so put y = 0
Taking positive sign, t = 0.84 s
The ball travels a horizontal distance x in time t
X = 20 Cos5 x t
X = 16.76 m
As this distance is more than the distance of net, so it clears the net.
Let t' be the time taken to travel a horizontal distance equal to the distance of net
7 = 20 cos5 x t'
t' = 0.35 s
Let the vertical distance traveled by the ball in time t' is y'.
So,
y' = 2.008 m
So, it clears the net which is 1 m high.
It clears the net by a vertical distance of 2.008 - 1 = 1.008 m and horizontal distance 16.76 - 7 = 9.76 m
your welcome, and have a great day.
48. Final Velocity Your sister drops your house keys down
to you from 4.3 m What is the velocity of the keys when you
catch them?
Answer:
There is a difference between the words 'drops' and 'throws'
If an object is dropped, it's initial velocity is 0 but if the same object is thrown, it has some initial velocity
Since 'the sister' DROPS the keys, the initial velocity will be 0 m/s
We are given:
u=0 m/s
s = 4.3 m
a = 9.8 m/s/s (gravity)
From the third equation of motion:
v² - u² = 2as
Replacing the known values
v² - (0)² = 2(9.8)(4.3)
v² = 84.28
v = 9.2 m/s (approx)
Hence, the final velocity of the keys is 9.2 m/s
When putting the ball on the tee you want half of the golf ball to _________.
Question 1 options:
be above the club
be below the club
be in front of the club
be behind the club
Answer:
be in front of the club
Explanation:
The ball should be highest off the ground for a driver. The general recommendation is that the bottom of the golf ball on a tee should be level with the top of the driver; for long and mid-irons, push the tee into the ground so that only about a quarter-inch is above ground.
When a light ray traveling in a higher index of refraction material passes into a lower index of refraction material, the light ray Group of answer choices Travels in a straight line without changing direction Bends toward the axis perpendicular to the surface between the materials Bends away from the axis perpendicular to the surface between the materials
The light ray ; ( C ) Bends away from the axis perpendicular to the surface between the materials
Snell's lawSnell's law states that the ratio of the sines of the angles of incidence is equal to the ratio of the refractive indexes of the materials surface through which the light rays pass through. therefore
As the light ray travels from a material with a higher index of refraction into a material with a lower index of refraction at the axis that is perpendicular to the surface which is in between the materials the light ray will bend away due to Snell's law .
Hence we can conclude that The light ray Bends away from the axis perpendicular to the surface between the materials.
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Calculate the wave speed (in m/s) for the following waves:
a) A sound wave in steel with a frequency of 500 Hz and a wavelength of 3.0 meters. (2pts)
b) a ripple on a pond with a frequency of 2 Hz and a wavelength of 0.4 meters. (2pts)
Calculate the wavelength (in meters) for the following waves:
A wave on a slinky spring with a frequency of 2 Hz travelling at 3 m/s. (2pts)
An ultrasound wave with a frequency 40,000 Hz travelling at 1450 m/s in fatty tissue. (2pts)
Calculate the frequency (in Hz) for the following waves:
A wave on the sea with a speed of 8 m/s and a wavelength of 20 meters. (2pts)
A microwave of wavelength 0.15 meters travelling through space at 300,000,000 m/s. (2pts)
Answer: A : 250 is the answer
B; The frequency of a wave is the number of complete oscillations (cycles) made by the wave in one second.
Instead, the wavelength is the distance between two consecutive crests (highest position) or 2 troughs (lowest position) of the wave.
In this problem, we are told that the leaf does two full up and down bobs: this means that it completes 2 full cycles in one second. Therefore, its frequency is
where is called Hertz (Hz). So, the correct answer is
Explanation:
#Wavespeed
#1
[tex]\\ \rm\Rrightarrow v=\nu\lambda=500(3)=1500m/s[/tex]
#2
[tex]\\ \rm\Rrightarrow v=2(0.4)=0.8m/s[/tex]
#Wavelength
#1
[tex]\\ \rm\Rrightarrow \lambda=\dfrac{v}{\nu}=\dfrac{3}{2}=1.5m[/tex]
#2
[tex]\\ \rm\Rrightarrow \lambda= \dfrac{1450}{40000}=0.03625m[/tex]
#Frequency
[tex]\\ \rm\Rrightarrow \nu=\dfrac{v}{\lambda}=\dfrac{8}{20}=0.4Hz[/tex]
#2
[tex]\\ \rm\Rrightarrow \nu=\dfrac{3\times 10^8}{15\times 10^{-2}}=0.2\timee 10^{10}=2\times 10^9Hz[/tex]
(based on 15-103 in the text) At an initial instant, a 4-lb ball B is traveling around in a circle of radius r1 = 3 ft with a speed (vB)1 = 4.8 ft/s. The attached cord is then pulled down through the hole with a constant speed vr = 2.2 ft/s. a. Determine the ball's speed at the instant r2 = 2 ft. Neglect friction and the size of the ball. Note that particle path is no longer of constant radius, and the particle has velocity components in both tangential and radial directions. b. How much work is done to pull down the cord from the initial instant to the instant when r2 = 2 ft? Neglect friction and the size o
Answer:
a
[tex]v_r =8.65 \ ft/s [/tex]
b
[tex] W_{1-2} = 3.24 \ ft \cdot lb[/tex]
Explanation:
From the question we are told that
The mass of the ball is [tex]m = 4 \ lb[/tex]
The radius is [tex]r= 3 \ ft[/tex]
The speed is [tex]v_B_1 = 4.8 \ ft /s[/tex]
The speed of the attached cord is [tex]v_c =2.2 \ ft[/tex]
The position that is been considered is [tex]r_1 = 2 \ ft[/tex]h
Generally according to the law of angular momentum conservation
[tex]L_a = L_b[/tex]
Here [tex]L_a[/tex] is the initial momentum of the ball which is mathematically represented as
[tex]L_a = m* v_B_1 * r[/tex]
while
[tex]L_b[/tex] is the momentum of the ball at r = 2 ft which is mathematically represented as
[tex]L_a = m* v_B_2 * r_1[/tex]
So
[tex]m* v_B_1 * r = m* v_B_2 * r_1[/tex]
=> [tex] 4.8 * 3 = v_B_2 * 2[/tex]
=> [tex] v_B_2 = 7.2 \ ft/s [/tex]
Generally the resultant velocity of the ball is
[tex]v_r = \sqrt{v_B_2^2 + v_B_1^2 }[/tex]
=> [tex]v_r = \sqrt{7.2^2 + 4.8^2 }[/tex]
=> [tex]v_r =8.65 \ ft/s [/tex]
Generally according to equation for principle of work and energy we have that
[tex]K_1 + \sum W_{1-2} = K_2[/tex]
Here [tex]K_1[/tex] is the initial kinetic energy of the ball which is mathematically represented as
[tex]K_1 = \frac{1}{2} * m* v_B_1^2[/tex]
While [tex]\sum W_{1-2}[/tex] is the sum of the total workdone by the ball
and [tex]K_2[/tex] is the final kinetic energy of the ball which is mathematically represented as [tex]K_2 = \frac{1}{2} * m* v_r^2[/tex]
So
[tex]\sum W_{1-2} = \frac{1}{2} * m (v_r^2 - v_B_1^2)[/tex]
Here m is the mass which is mathematically represented as
[tex]m = \frac{W}{g}[/tex] here W is the weight in lb and g is the acceleration due to gravity which is [tex]g = 32 \ ft/s^2[/tex]
So
[tex]\sum W_{1-2} = \frac{1}{2} * \frac{4}{32} * (8.65^2 - 4.8^2)[/tex]
=> [tex] W_{1-2} = 3.24 \ ft \cdot lb[/tex]
Please I need help with this :(
three charged particals are located at the corners of an equil triangle shown in the figure showing let (q 2.20 Uc) and L 0.650
I need to know what the answer is to this
Answer:
i think its the top one
Explanation:
pls tell me if im wrong
Describe the buoyant force and explain how
it relates to Archimedes principle.
Answer:
Archimedes' principle states that the upward buoyant force that is exerted on a body immersed in a fluid, whether fully or partially submerged, is equal to the weight of the fluid that the body displaces.
What is the mass of a toy car if it has 5 J of potential energy and is sitting on top of a track that has a height of 2m?
(PE= m x g x h) (hint g=9.8 m/s2)
Explanation:
PE=mgh
5=m(9.8)(2)
m=5/19.6
m=0.2251 kg
m=225.1 grams
A man slides on snow without friction starting at 8.96m/s at the top of an inclined plane with height 8.21m. What is his speed at the bottom of a plane?
Answer:
V2 = 15.53 [m/s]]
Explanation:
In order to solve this problem we must use the principle of energy conservation, where potential energy is transformed into kinetic energy. At the bottom is taken as a reference level of potential energy, where the value of this energy is equal to zero.
Above the inclined plane we have two energies, kinetics and potential. While when the sled is at the reference level all this energy will have been transformed into kinetic energy.
[tex]E_{1}=E_{2}\\ m*g*h+(\frac{1}{2} )*m*v_{1} ^{2}=\frac{1}{2}*m*v_{2} ^{2} \\(9.81*8.21)+(0.5*8.96^{2} )=(0.5*v_{2}^{2} )\\(0.5*v_{2}^{2} )=120.68\\v_{2} ^{2}=241.36\\v_{2} =\sqrt{241.36}\\ v_{2} =15.53[m/s][/tex]
g A 38-g ball at the end of a string is swung in a vertical circle with a radius of 21 cm. The tangential velocity is 200.0 cm/s. Find the tension in the string:
Answer:
It depends on the location of the ball during the motion. The string tension are approximately 3.82 N (at the lowest point), 3.06 N (at the highest point), and 3.44 N (at the horizontal point).
Explanation:
Tension in the String can be determined by the Newton's 1st Law of Motion (The ball shouldn't be escaped from the trajectory). The value of [tex]\theta[/tex] indicates the angle that is measured from the vertical lines and the rope of length R)
[tex]\sum F=0\rightarrow \frac{mv^{2}}{R}+mg\cos\theta-T=0[/tex]
[tex]T=mg\cos\theta+\frac{mv^{2}}{R}[/tex]
[tex]T=(38\times10^{-3})(10)(\cos\theta)+(38\times10^{-3})\frac{2^{2}}{0.21^{2}}=0.38\cos\theta+3.44[/tex]
When the ball is at the lowest point, the value of angle [tex]\theta=0[/tex], so the string tension is approximately 3.82 N. If the ball is at the highest point the value of [tex]\theta=180^{0}[/tex], so the string tension is approximately 3.06 N, and at the horizontal point [tex]\theta=90^{0}[/tex], so the string tension is approximately 3.44 N.
Can someone help pleaseeee
Answer:
Motion with constant velocity of magnitude 1 m/s (uniform motion) for 4 seconds in a positive direction and then for 2 seconds uniform motion with constant velocity of magnitude 3 m/s in reverse direction .
Explanation:
The graph shows a constant velocity of 1 m/s for 4 seconds in the positive direction. After that, between 4 seconds and 6 seconds, the object reverses its motion with constant velocity of magnitude 3m/s.
The knot at the junction is in equilibrium under the influence of four forces acting on it. The F force acts from above on the left at an angle of α with the horizontal. The 5.7 N force acts from above on the right at an angle of 50◦ with the horizontal. The 6.2 N force acts from below on the right at an angle of 44◦ with the horizontal. The 6.7 N force acts from below on the left at an angle of 43◦ with the horizontal.
1. What is the magnitude of the force F?
2. What is the angle a of the force F in the figure above?
The knot is in equilbrium, so there is no net force acting on it. Starting with the unknown force and going clockwise, denote each force by F₁, F₂, F₃, and F₄, respectively. We have
F₁ + F₂ + F₃ + F₄ = 0
Decomposing each force into horizontal and vertical components, we have
F cos(180º - α) + (5.7 N) cos(50º) + (6.2 N) cos(-44º) + (6.7 N) cos(-137º) = 0
F sin(180º - α) + (5.7 N) sin(50º) + (6.2 N) sin(-44º) + (6.7 N) sin(-137º) = 0
Recall that cos(180º - x) = - cos(x) and sin(180º - x) = sin(x), so these equations reduce to
F cos(α) ≈ - 3.22 N
F sin(α) ≈ 4.51 N
(1) Recall that for all x, sin²(x) + cos²(x) = 1. Use this identity to solve for F :
(F cos(α))² + (F sin(α))² = F ² ≈ 30.73 N² → F ≈ 5.5 N
(2) Use the definition of tangent to solve for α :
tan(α) = sin(α) / cos(α) ≈ 1.399 → α ≈ 126º
or about 54º from the horizontal from above on the left of the knot.
(a) The magnitude of the force F acting on the knot is 5.54 N.
(b) The angle α of the force F is 54.4⁰.
The given parameters:
F force at α 5.7 N force at 50⁰6.2 N force at 44⁰6.7 N force at 43⁰The net vertical force on the knot is calculated as follows;
[tex]F_y = Fsin(\alpha) + 5.7 sin(50) - 6.2 sin(44) - 6.7 sin(43)\\\\F_y = F sin(\alpha) -4.51\\\\Fsin(\alpha) = 4.51[/tex]
The net horizontal force on the knot is calculated as follows;
[tex]F_x = -F cos(\alpha) + 5.7 cos(50) + 6.2cos(44) - 6.7cos(43)\\\\F_x = -Fcos(\alpha) + 3.22\\\\Fcos(\alpha) = 3.22[/tex]
From the trig identity;
[tex]sin^2 \theta + cos^ 2 \theta = 1\\\\[/tex]
[tex](Fsin(\alpha))^2 + (Fcos(\alpha))^2 = (4.51)^2 + (3.22)^2\\\\F^2(sin^ 2\alpha + cos^2 \alpha) = 30.71\\\\F^2(1) = 30.71\\\\F = \sqrt{30.71} \\\\F = 5.54 \ N[/tex]
The angle α of the force F is calculated as follows;
[tex]Fsin(\alpha) = 4.51\\\\sin(\alpha) = \frac{4.51}{F} \\\\sin(\alpha ) = \frac{4.51}{5.54} \\\\sin(\alpha ) = 0.814\\\\\alpha = sin^{-1}(0.814)\\\\\alpha = 54.5 \ ^0[/tex]
Find the image uploaded for the complete question.
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Two hockey pucks are moving toward each other. The mass of the first puck is half the mass of the second puck. Initially, the first puck is going at a speed of 5.0 m/s to the right, while the second puck is going at a speed of 2.0 m/s. The first puck rebounds with a speed of 9.0 m/s. What is the final speed of the second puck
The final speed of the second puck is 5 m/s.
What is speed?Speed can be defined as the rate of change of distance with respect to time.
To calculate the final speed of the second puck, we use the formula below.
Formula:
mu+m'u' = mv+m'v'................ Equation 1Where:
m = mass of the first puckm' = mass of the second pucku = initial velocity of the first pucku = initial velocity of the second puckv = final velocity of the first puckv' = final velocity of the second puck.
Make v' the subject of the equation
v' = (mu+m'u'-mv)/m'.............. Equation 2Assuming,
The mass of the first puck is y The left direction is negative and the right direction is positiveFrom the question,
Given:
m = ym' = 2yu = 5 m/su' = -2 m/sv = -9 m/s ( rebounds)Substitute these values into equation 2
v' = [(y×5)+(2y×(-2))-(y×(-9))]/2yv' = (5y-4y+9y)/2yv' = 10y/2yv' = 5 m/sHence, The final speed of the second puck is 5 m/s.
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Which type of energy increases when you compress a spring?
Question 8 options:
radiant energy
kinetic energy
elastic potential energy
sound energy
Answer:
Elastic potential energy
Answer:
Elastic potential energy: increases when you compress a spring.
Explanation:
A campus shuttle bus makes 1 revolution per second round a circular track of radius 100 cm. Determine its periodic time.
The periodic time of the campus shuttle is = 1 sec
What is periodic time ?The periodic time is also known as revolution per second is the time it takes object in motion ( revolving ) to pass through a point twice ( usually the starting position ) of the object and it calculated as :
For a simple pendulum = 2π√L/g
But a campus shuttle bus in motion
since it takes
1 revolution = 1 sec therefore the time period is = 1 sec
Hence we can conclude that the periodic time of the campus shuttle = 1 second.
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this diagram shows the sectionary box sitting on the floor which statement about other force exerted on the box are true?
questions:
it's normal force exerted by the floor. it balances the gravitational force.
it's friction force exerted by the floor. it balances the gravitational force.
it's normal for exerted by the floor. does not balance the gravitational force.
it's an applied force exerted from the left. It doesn't balance the gravitational force.
Answer: A) It's a normal force exerted by the floor. It balances out the gravitational force.
The normal force pushes up which means the box doesn't go through the floor. The up and down forces are balance. If the normal force was larger than the gravitational force, then the box would be pushed up and float/jump in the air. But the box is stationary and not moving, so that's why we don't have any acceleration and the forces effectively cancel each other out. In a way you can think of it like a see-saw.
Friction would only come into play if you applied a force and were trying to move the box. Friction counteracts the applied force. If you push the box to the left, then the friction would push to the right. Friction slows down the box and it allows it to not slide forever.
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Is normal force equal to gravity?
The normal force on an object at rest on a flat surface is equal to the gravitational force on that object.
▀▄▀▄Is friction a gravitational force?
Friction is the resistance to motion of one object moving relative to another. It is not a fundamental force, like gravity or electromagnetism. Instead, scientists believe it is the result of the electromagnetic attraction between charged particles in two touching surfaces.
▀▄▀▄Can normal force be greater than gravity?
Can you imagine the situation when Normal force is greater than mg ? Yes. When an additional downward force F is applied to a mass m resting on a horizontal surface, the normal force is FN=F+mg. ... An example is the normal force on an incline plane with an angle of θ due to a mass m.▄▀▄▀▄▀▄
What is the direction of the force you applied?
An applied force is an interaction of one object on another that causes the second object to accelerate or change velocity or direction. The force can be a push, pull, or drag. The resulting direction of an object depends on the relative direction of the force on the object. A force equation is F = ma.
Harry is pushing a car down a level road at 2.0 m/s with a force of 243 N. The
total force acting on the car in the opposite direction, including road friction and
air resistance, is which of the following?
a. Slightly more than 243 N.
b. Exactly equal to 243 N.
c. Slightly less than 243 N.
Answer:
C, slightly less than 243 N
Explanation:
Road friction and air resistance aren't that much on a force. Try pushing something and see how much friction there is. Not that much.
a car moving at a speed shows that the force applied to the car is greater than the frictional force and air resistance
c. Slightly less than 243 N.
a 2 kg ball traveling at 5m/s collides with a 1 kg ball at rest. After the collision the 1 kg ball moves off with a speed of 7 m/s. Find the final speed of the 2 kg ball.
You measured the length, diameter and mass of two different cylinders. In both cases, you found that the length had 3 significant figures and that length was the measurement with the fewest number of significant digits. If you found the weight densities to be 38119 N/m^3 and 38081 N/m^3 and you round these values to the correct number of significant figures, can you conclude the two cylinders are made of the same material (do they have the same weight density)?
a. Not enough information given.
b. Yes.
c. No.
Answer:
Weight Density 1 = 38100 N/m³
Weight Density 2 = 38100 N/m³
b. Yes
Explanation:
The formula for volume of cylinder is:
V = πr²l
where,
V = Volume
r = radius
l = length of cylinder
So, if length has the 3 significant figures which is least in all values, Then the volume must also be in 3 significant figures. The formula for weight density is:
Weight Density = Weight/Volume
Here, the volume has the least significant figures of 3, therefore, the weight densities must also have 3 significant figures:
Weight Density 1 = 38119 N/m³
Weight Density 1 = 38120 N/m³
Weight Density 1 = 38100 N/m³
Weight Density 2 = 38119 N/m³
Weight Density 2 = 38120 N/m³
Weight Density 2 = 38100 N/m³
Hence, the answer is:
b. Yes
A, B, or C for this question?
Answer:
A. right before it hits the ground.
Explanation:
because all the way down, it is building kinetic energy.
Which information did the Glomar Challenger study in 1968?
the rate of seafloor spreading
the direction of seafloor spreading
the age of rocks in various places in the ocean
the contents of rocks in various places in the ocean
Explanation:
Glomar Challenger studies about the "age of rocks in various places in the ocean" in 1968. EXPLANATION: Glomar Challenger was a "deep sea research vessel" for marine geology and oceanography studies.
I hope this helps you :)
Answer:
c no cappp :)
Explanation:
During which time Interval does the object travel approximately 10 meters?
OA. O seconds to 3 seconds
OB. 3 seconds to 5 seconds
OC. 5 seconds to 7 seconds
OD. 7 seconds to 8 seconds
OE. 8 seconds to 10 seconds
Answer:
[tex]OA. \: O \: seconds \: to \: 3 \: seconds[/tex]
Explain why we need to study the climate.
Answer:The current focus of climate science involves carbon dioxide (CO2)emissions. Carbon dioxide in the atmosphere acts as a blanket over the planet by trapping long wave radiation, which would otherwise radiate heat away from the planet. As the amount of carbon dioxide increases, so will its warming effect.
Explanation:
A boy of mass 40kg eats bananas contains of 980 joule. If this energy is used to lift him up from ground,the height to which he can climb is
Answer:
h = 2.49 [m]
Explanation:
In order to solve this problem we must use the definition of potential energy, which tells us that energy is equal to the product of mass by gravity by height.
The potential energy can be calculated by means of this equation:
Ep = m*g*h
where:
Ep = potential energy = 980 [J]
m = mass = 40 [kg]
g = gravity acceleration = 9.81 [m/s^2]
h = elevation [m]
Now replacing:
980 = 40*9.81*h
h = 2.49 [m]
Kepler's work revealed that the Earth was at the center of the circular orbits of the planets at the center of the elliptical orbits of the planets orbiting the Sun in an elliptical orbit O at one focus of the elliptical orbit of the planets
Answer:
orbiting the Sun in an elliptical orbit
Explanation:
When orbits are circular, the center and the focus are the same point. For an elliptical orbit, the orbited object is at one focus of the ellipse. Kepler found planetary orbits to be elliptical with the sun at one focus.
__
As with a lot of scientific theories, it only explained some of the motion. As with a lot of scientific theories, explaining the discrepancies resulted in new discoveries.
Answer:
at one focus of the elliptical orbit of the planets.
A body, with a volume of 2 m3, weighs 40 kN. Determine its weight when
submerged in a liquid with SG = 1.59.
Answer:
8.8 kN
Explanation:
V = 2 m³, W = 40 kN, SG = 1.59
Bouyant force N = 1.59 * 1000 kg/m³ * 9.81 N/kg * 2 m³ = 31.2 kN
So the weight becomes 40 - 31.2 = 8.8 kN
how does gravity change as it nears an object
what is acceleration due to gravity?
The acceleration due to gravity is 9.8 m/s^2.
Which statements describe kinetic and potential energy? Check all that apply.
Energy can be stored in the position of an object.
Energy is not present in a moving object.
Energy can be stored in the position of the particles that make up a substance.
Energy exists as movement of the particles of a substance.
Energy is greater in faster-moving particles than in slower-moving particles.
Energy is lower in objects with greater mass than in objects with less mass.
Answer:
First option, third option, fourth option, and the fifth option.
Explanation:
Kinetic energy is energy an object has when it's motion, the greater the speed the greater the kinetic energy. For example, a car moving and increasing in speed is kinetic energy since the object is in motion. If the car stops and parks in a parking lot that is potential energy. Potential energy is the amount of energy an object has when it's at rest or not in motion.
So, the answer for this question is as followed first option or "energy can be stored in the position of an object." Third option or "Energy can be stored in the position of the particles that make up a substance." Fourth option or "Energy exists as movement of the particles of a substance." The last answer will be the fifth option or "Energy is greater in faster-moving particles than in slower-moving particles."
Hope this helps.