Answer:
is 6
Explanation:
h
 what is the difference between repelling and attracting
Answer:
Attracting means pulling toward you and repelling means pushing away
Explanation:
Answer: Repelling is when something will not connect with another object. The force will cause a repel between the two objects. Attracting is when something is attracted or being pulled to another object.
Explanation: Hope this helps!
5. A quarterback throws the football to a stationary receiver who is 31.5 m
down the field. If the football is thrown at an initial angle of 40.0° to the
ground, at what initial speed must the quarterback throw the ball for it
to reach the receiver? What is the ball's highest point during its flight?
The projectile launch equations allow to find the results for the questions about the movement of the ball are:
The initial velocity is: v₀ = 17.7 m / s. The maximum height is: y = 16 m.
Given parameters
Horizontal distance x = 31.5 m Launch angle tea = 40ºTo find
The initial speed. Maximum height.
Projectile launching is an application of kinematics, where on the x-axis there is no acceleration and on the y-axis is the gravity acceleration.
The range is the distance traveled for the same departure height, see attached.
.
R =[tex]\frac{v_o^2 \ sin 2\theta}{g}[/tex]
[tex]v_o^2 = \frac{ g R}{sin 2 \theta }[/tex]
Let's calculate.
v₀² = [tex]\frac{9.8 \ 31.5}{sin \ (2 \ 40)}[/tex]9.8 31.5 / sin (2 40.0)
[tex]v_o = \sqrt{313.46}[/tex]o = ra 313.46
v₀ = 17.7 m / s
At the point of maximum height the vertical speed is zero.
v² = v₀² - 2 g y
0 = v₀² - 2g y
y = [tex]\frac{v_o^2}{2g}[/tex]
Let's calculate.
y = [tex]\frac{17.7^2}{2 \ 9.8}[/tex]
y = 16 m
In conclusion, using the projectile launch equations we can find the results for the questions about the movement of the ball are:
The initial velocity is: v₀ = 17.7 m / s The maximum height is: y = 16 m.
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Alex is x years old . June is 7years older than Alex . their total combined ages is 29 years . find June,s age . show all work algebraically
Answer:
18 yearsExplanation:
Given,
Let Alex be = x years
Then June will be = (7 + x) years
We know that,
Their total combined age is 29 years
Therefore,
By the problem,
=> x + (7 + x) = 29
=> 2x = 29 - 7
=> 2x = 22
=> x = 22 ÷ 2
=> x = 11
So,
Required age of June is = (7 + x) years
= (7 + 11) years
= 18 years (Ans)
What is non examples of enlarged
reduction
deflation
these are examples of the opposite of enlarged to make something smaller is really the key thing here
Colette launches an air rocket in the upward, positive direction. It launches with an initial velocity of 25.5 m/s. It accelerates in the downward, negative direction at a rate of 9.81 m/s2. After 3.5 seconds, what is the magnitude of the rocket's displacement?
Answer:
Give me some hint please
Describe how you can determine:
a) Volume of an irregular body
b) Density of a liquid
Density of liquid try thank you so much
Answer:
a) measure the change in volume when the object is immersed; compute from range data
b) find the ratio of mass to volume for a measured mass and volume
Explanation:
a) The volume of a small enough irregular body can be found by measuring the difference in volume of the (semi-)fluid in which it is immersed, before and after immersion.
For irregular bodies for which that approach does not work, various 3D scanners are available for measuring volume and surface area. They may rely on optical (laser or camera), sonic, or radar measurements, and generally involve computation from distances to various points.
__
b) Density is the ratio of mass to volume. So, measurements of mass and volume of a liquid sample are sufficient to provide the basis for determining density.
Other methods include measuring buoyancy forces, and/or the depth of submersion of something that floats in the liquid. For specific liquids, hydrometers are available for measuring their density relative to that of water.
A mountain climber encounters a crevasse in an ice field. The opposite side of the crevasse is a height h lower, and is separated horizontally by a distance w. To cross the crevasse, the climber gets a running start and jumps in the horizontal direction. If the height of the crevasse increases but the width remains the same, then,
Select one:
O a. the minimum speed needed to cross the crevasse stays the same.
O b. the minimum speed needed to cross the crevasse will depend on the mass of the mountain climber.
O c. the minimum speed needed to cross the crevasse decreases.
O d. the minimum speed needed to cross the crevasse increases.
O e. the minimum speed needed to cross the crevasse will depend on the weight of the mountain climber.
A girl of mass m1=60.0 kilograms springs from a trampoline with an initial upward velocity of vi=8.00 meters per second. At height h=2.00 meters above the trampoline, the girl grabs a box of mass m2=15.0 kilograms. (Figure 1)
For this problem, use g=9.80 meters per second per second for the magnitude of the acceleration due to gravity.
What is the speed vbefore of the girl immediately before she grabs the box?
Express your answer numerically in meters per second.
What is the speed vafter of the girl immediately after she grabs the box?
Express your answer numerically in meters per second.
What is the maximum height hmax that the girl (with box) reaches? Measure hmax with respect to the top of the trampoline.
The conservation of momentum and energy allows to shorten the results for the movement of the girl on the trampoline holding the box are:
a) the girl's speed is v = 4.98 m / s
b) The speed of the girl + box system is: v_f = 0.996 m / s
c) the maximum height is: y = 2.05 m
Kinematics studies the movement of bodies, looking for relationships between the position, velocity and acceleration of bodies.
The momentum is defined by the product of mass and the velocity, when a system is isolated the momentum is conserved.
The mechanical energy is the sum of the kinetic energy plus the potential energies, when there is no friction in the system the mechanical energy is conserved.
Let's solve this exercise in parts:
a) Let's use kinematics to find the speed of the girl before she grabs the box
v² = v₀² - 2 g y₁
v² = 8² - 2 9.8 2.00
v = R 24.8 = 4.98 m / s
b) Let's use momentum conservation for when the speed of the girl and the box together. Let's write the moment in two moments.
Initial instant. Just before you grab the box.
p₀ = M v + 0
Final moment. Right after taking the box
[tex]p_f[/tex] = (m + M) [tex]v_f[/tex]
In system this form by the girl and the box therefore it is an isolated system and the momentum is conserved.
[tex]p_o = p_f[/tex]
mv = (m + M) [tex]v_f[/tex]
[tex]v_f = \frac{m}{m+M} \ v[/tex]
Let's calculate
[tex]v_f = \frac{15}{15+ 60} \ 4.98[/tex]
[tex]v_f[/tex] = 0.996 m / s
c) Now we use conservation of energy after the girl has the box.
Starting point. When the girl has the box
Em₀ = K + U
Em₀ = ½ (m + M) v² + (m + M) g y₁
Final point. At the highest point of the trajectory
[tex]Em_f[/tex] = U
[tex]Em_f[/tex] = (m + M) g y₁
As there is no friction, the energy is conserved.
[tex]Em_o = Em_f[/tex]
½ (m + M) v² + (m + M) g y₁ = (m + M) g y
y = [tex]\frac{v^2}{2g} + y_1[/tex]
Let's calculate
y = [tex]\frac{0.996^2}{2 \ 9.8} + 2.0[/tex]
y = 2.05 m
In conclusion using the conservation of momentum and energy we can shorten the results for the movement of the girl on the trampoline holding the box are:
a) the girl's speed is v = 4.98 m / s
b) The speed of the girl + box system is: v_f = 0.996 m / s
c) the maximum height is: y = 2.05 m
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Answer:
Vbefore = 4.98 m/s
Vafter = 3.98 m/s
Hmax = 2.81 m
50 points help
Column I Column II
______ 1. acceleration a. change in distance over time
______ 2. speed b. time interval
______ 3. velocity c. scalar
______ 4. Δt. d. change in position
______ 5. Magnitude only e. change in velocity over time
______ 6. Δx f. change in displacement over time
[tex]\\ \sf\longmapsto Acceleration\longrightarrow Change\:in\: Velocity\:over\:time[/tex]
[tex]\\ \sf\longmapsto Speed\longrightarrow Change\:in\: Distance\:over\:Time[/tex]
[tex]\\ \sf\longmapsto Velocity\longrightarrow Change\:in\: Displacement\: over\:time[/tex]
[tex]\\ \sf\longmapsto ∆t\longrightarrow Time\: interval [/tex]
[tex]\\ \sf\longmapsto Magnitude\:only\longrightarrow Scaler[/tex]
[tex]\\ \sf\longmapsto ∆x=Change\:in\: position [/tex]
The figure shows the light intensity on a screen behind a double slit. The slit spacing is 0.22 mm and the screen is 2.0 m behind the slits (Figure 1). What is the wavelength of the light?
The wavelength of the light is 550 nm
For a double slit interference pattern with slit spacing, d we have
dsinθ = mλ where d = slit spacing = 0.22 mm = 0.22 × 10⁻³ m, m = number of maximum fringe = 2(from the picture) and λ = wavelength of light.
Thus sinθ = mλ/d
Also, tanθ = L/D where L = distance between central maximum and fringe = 2.0 cm/2 = 1.0 cm = 1 × 10⁻² m and D = distance between slit and screen = 2.0 m
Since θ is small, sinθ ≅ tanθ
So, mλ/d = L/D
Making λ subject of the formula, we have
λ = dL/mD
Substituting the values of the variables into the equation, we have
λ = dL/mD
λ = 0.22 × 10⁻³ m × 1 × 10⁻² m/(2 × 2.0 m)
λ = 0.22 × 10⁻⁵ m²/4.0 m
λ = 0.055 × 10⁻⁵ m
λ = 0.55 × 10⁻⁶ m
λ = 550 × 10⁻⁹ m
λ = 550 nm
So, the wavelength of the light is 550 nm
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Define average atomic mass and explain how it is calculated
Answer:
The average atomic mass of an element is the sum of masses of it's isotopes
Each are multiplied by it's natural abundance
Explanation:
The average atomic mass of an element is the sum of masses of it's isotopes
Each are multiplied by it's natural abundance
Which statement about RNA polymerase is NOT true?
RNA polymerase reads a template strand of DNA 5' to 3'
RNA polymerase binds to a promoter region to initiate transcription
RNA polymerase adds a ribonucleotide to the 3' end of a growing RNA molecule
During transcription of a gene, RNA polymerase reads only one strand of DNA
Please explain!
Answer:
RNA polymerase reads a template strand of DNA 5' to 3'
Explanation:
RNA polymerase is the enzyme accountable for the transcription of the DNA. It shows one coast of the DNA and transcribed within the RNA. It combines the nucleotide to the 3' conclusion of the expanding series of the RNA. It commences the method by connecting to the promoter area of the gene. The enzyme increases nucleotide, not ribonucleotide.
Therefore, A. RNA polymerase reads a template strand of DNA 5' to 3' is false.
Answer:
RNA polymerase reads a template strand of DNA 5' to 3'.
Explanation:
RNA synthesis is less processive during elongation than during initiation. Synthesis begins with the polymerase binding two rNTP molecules. The polymerase frequently releases the nascent transcript before it reaches 8-10 nucleotides in length.
Required information
Medical testing has established that the maximum acceleration a pilot can be subjected to without losing consciousness is
approximately 5.00g. A pilot can avoid "blackout" at accelerations up to approximately 9.00g by wearing special "g-suits"
that help keep blood pressure in the brain at a sufficient level.
What is the minimum safe radius of curvature for an unprotected pilot flying an F-15 in a horizontal circular loop at 729 km/h?
Answer:
hi there is that OK for the weekend of the following week as well
Explanation:
6th of March is fine for me
The minimum safe radius of curvature for an unprotected pilot flying an F-15 in a horizontal circular loop at 729 km/h is approximately 838.1 meters.
To determine the minimum safe radius of curvature for an unprotected pilot flying an F-15 in a horizontal circular loop, we need to consider the maximum acceleration the pilot can withstand without losing consciousness.
Given:
Maximum acceleration without losing consciousness = 5.00g
Acceleration with g-suits to avoid blackout = 9.00g
First, we need to convert the speed of the F-15 from km/h to m/s:
Speed = 729 km/h = (729 * 1000) m/3600 s ≈ 202.5 m/s
Next, we'll calculate the acceleration experienced by the pilot in the circular loop. In a horizontal circular motion, the centripetal acceleration is given by:
Acceleration = ([tex]\rm Velocity^2[/tex]) / Radius
We can rearrange the equation to solve for the radius:
Radius = ([tex]\rm Velocity^2[/tex]) / Acceleration
Using the maximum acceleration of 5.00g, we convert it to [tex]\rm m/s^2[/tex]:
Maximum acceleration = 5.00g ≈ (5.00 * 9.8) [tex]\rm m/s^2[/tex] = 49 m/s^2
Now, we can calculate the minimum safe radius of curvature:
Radius = ([tex]\rm 202.5^2[/tex]) / 49 ≈ 838.1 meters
Therefore, the minimum safe radius of curvature for an unprotected pilot flying an F-15 in a horizontal circular loop at 729 km/h is approximately 838.1 meters.
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What on earth is equal to 9.8m/s/s
Answer:
Acceleration due to gravity
Which force, in real life, will have the least effect on a bowling ball rolling down a lane toward bowling pins?
A) magnetism
B) air resistance
C) gravity
D) friction
Answer:
Its Friction
Explanation:
the pins are not floating and they are not a magnet and not involved with air
The force, in real life, that will have the least effect on a bowling ball rolling down a lane toward bowling pins is magnetism. The correct option is A.
What is magnetism?Magnetism is basically the force which indeed magnets exert when they attract or even repel one another. The movement of electric charges resulting in magnetism.
Every substance is composed of tiny units referred to as atoms. Each atom contains electrons, which are charged particles.
To increase stability, the pins themselves have a low center of gravity. Because they are spherical in shape, they can roll and strike other pins in a variety of directions.
The force acting on the bowling ball is friction and air resistance. The friction force is equal to the friction coefficient multiplied by the normal force, and thus mass times acceleration.
Thus, the correct option is A.
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Acceleration of a car that speeds from 4.3 m/s to 12.9 m/s in 2 seconds
Explanation:
let v1 = 4.3 m/s
v2 = 12.9 m/s
t = 2 seconds
v2 = v1 + at
12.9 = 4.3 + a×2
2a = 12.9 - 4.3 = 8.6
a = 8.6/2
a = 4.3 m/s^2
A 770-kg two-stage rocket is traveling at a speed of 6.90×103 m/s away from Earth when a predesigned explosion separates the rocket into two sections of equal mass that then move with a speed of 2.60×103 m/s relative to each other along the original line of motion.What is the speed of each section (relative to Earth) after the explosion?How much energy was supplied by the explosion?
Answer:
Explanation:
Let's just have our reference frame travel along with the original un broken mass. This way the original velocity is not relevant.
Each half will have a mass of 770/2 = 385 kg
Each half will have the same magnitude of velocity (conservation of momentum) which will be 2.6 x 10³/2 = 1.30 x 10³ m/s
Now add back the reference frame velocity to get velocity relative to earth.
Section one will have velocity 6.90 x 10³ + 1.30 x 10³ = 8.2 x 10³ m/s
Section two will have velocity 6.90 x 10³ - 1.30 x 10³ = 5.6 x 10³ m/s
In the moving reference frame, each half will have kinetic energy which could only come from the explosion
KE = ½(385)(1.3 x 10³)² = 325,325,000 J
2(325,325,000) = 650,650,000 J released in the explosion.
Rounding to the three significant figures of the problem numerals
E = 6.50 x 10⁸ J or 650 MJ released
PLEASE HELP!!!
If a 40cm rope with a 220g bob can hold a maximum tension of 3N
a) what are the maximum angular velocity and inclination angle it can reach before the rope break?
b) Angle of inclination
Answer:
ω = 3.1 rad/s
θ = 36° from vertical
Explanation:
I will ASSUME that the bob and string is acting as a pendulum.
Please understand that the string will break when the bob is at the lowest point of the swing where the vectors of gravity and centripetal acceleration align. It will NOT break at the angle of maximum inclination measured from vertical. This angle is only a component of the maximum potential energy that gets converted to maximum kinetic energy at the lowest point of the swing.
At the bottom of the swing, the string must support the weight of the bob plus supply the required centripetal acceleration.
F = mg + mω²R
F/m = g + ω²R
F/m - g = ω²R
ω = √((F/m - g)/R)
ω = √((3/0.220 - 9.8)/0.40)
ω = 3.09691...
ω = 3.1 rad/s
Potential energy will convert to kinetic energy
mgh = ½mv²
h = v²/2g
R - Rcosθ = v²/2g
R(1 - cosθ) = v²/2g
1 - cosθ = v²/2gR
cosθ = 1 - v²/2gR
cosθ = 1 - (Rω)²/2gR
cosθ = 1 - Rω²/2g
cosθ = 1 - 0.40(3.1²)/(2(9.8))
cosθ = 0.804267
θ = 36.46045...
θ = 36°
Market researchers were interested in the relationship between the number of pieces in a brick-building set and the
cost of the set. Information was collected from a survey and was used to obtain the regression equation ý =
0.08x + 1.20, where x represents the number of pieces in a set and ŷ is the predicted price in dollars) of a set.
What is the predicted price of a set that has 500 pieces?
$40
$41.20
$600
$6,235
Which of the following correctly compares gravitational force and distance between two objects?
A. As the distance increases, the gravitational force decreases.
B. As the distance decreases, the gravitational force decreases.
C. There is no relationship between gravitational force and distance between two objects.
D. As the distance increases, the gravitational force increases.
Help pls…
Answer:
a. as the distances increases , the gravitational force decreases
When an elastic object is changed from its original shape:
A:Energy is released
B:Work is done
C:It is ruined
D:It makes a twanging sound
Answer:
deformation : elastic deformation is reversed when the force is removed. inelastic deformation is not fully reversed when the force is removed – there is a permanent change in shape.
Explanation:
tysm
hey I just wanted to know if any of the guys here are able to help answer my physics questions , it would be a great thankyou xoxoxoxo
what's your question?
A hot air balloon rising vertically is tracked by an observer located 2 miles from the lift-off point. At a certain moment, the angle between the observer's line-of-sight and the horizontal is , and it is changing at a rate of 0.1 rad/min. How fast is the balloon rising at this moment
We have that for the Question, it can be said that
the balloon rising at [tex]0.266miles/min[/tex]From the question we are told
An observer located 2 miles from the lift-off point. At a certain moment, the angle between the observer's line-of-sight and the horizontal is , and it is changing at a rate of 0.1 rad/min.From,
[tex]tan\theta = \frac{h}{2}[/tex]
differentiate with respect to h
[tex]sec^2\theta * \frac{do}{dz} = \frac{1}{2} * \frac{dh}{dz}\\\\\frac{dh}{dz} = 2 sec^\theta * \frac{d\theta}{dz}\\\\\theta = \frac{\pi}{6} and \frac{d\theta}{dz} = 0.1rad/min\\\\\frac{dh}{dz} = 2sec^2 (\frac{\pi}{6}) * (0.1)\\\\= 0.266miles/min[/tex]
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A Scooter has a mass of 250 kg. A constant force is exerted on it for 6.0 s. During the time the force is exerted, the scooter increases its speed from 6.00 m/s to 280 m/s. What is the magnitude of the force exerted on the scooter
Answer:
917 N
Explanation:
917 N, this is your answer!!
Glad to help.
Which scientist is credited with having the greatest contribution to early microscopy and was the first to observe and describe single-celled organisms?
Answer:
Antonie van Leeuwenhoek
Explanation:
1.25 is closer to 1.04 or not ?
plz heelp plz
Answer:
No, it is closer to 1.30
Explanation:
PLEASE PLEASE HELP!!
An arrow is shot straight up in the air from the ground with an initial velocity of 54.0 m/s. If on striking the ground it
embeds itself 15.0 cm into the ground, what is the acceleration required to stop the arrow when it hits the ground?
Answer:
you have patience the distance.
Explanation:
the train leaves at 6.30.
If you do 72 J of work in 1.2 seconds, how much power is produced
Answer:
Explanation:
Power is the rate of doing work
P = 72 / 1.2 = 60 Watts
which of the following is used to transport sound waves
A.medium
B.vacuum
C.mass
D.light
I think the answer is d.right?
A meter stick is attached to one end of a rigid rod with negligible mass of length l = 0.302 m. The other end of the light rod is suspended from a pivot point, as shown in the figure below. The entire system is pulled to a small angle and released from rest. It then begins to oscillate. A meter stick hung from a rod of length l. The rod is attached to the ceiling. The rod and meter stick extend downward in a straight line making a small angle with the vertical. (a) What is the period of oscillation of the system (in s)? (Round your answer to at least three decimal places.)
The period of oscillation of the system nearest to three decimal places
= 1.092 seconds
The period of an oscillation occurring in a system is the time taken to complete one cycle.
The formula that is used to calculate the period of oscillation (T) is
= 2π√[tex]\frac{l}{g}[/tex]
But,
π = 3.14159 (constant)
g= 10m/s² (acceleration due to gravity)
l = 0.302 m
Therefore T = 2 × 3.14159 × √[tex]\frac{0.302}{10}[/tex]
= 6.28318 x √0.0302
= 6.28318 x 0.17378
= 1.09189s
= 1.092 seconds ( to the nearest three decimal places)
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