Answer:
[tex]\dfrac{1}{64}[/tex] as loud.
Explanation:
The sound intensity varies inversely as the square of the distance from the sound source. Mathematically,
[tex]I=\dfrac{k}{d^2}[/tex]
You change your seat to one that is eight times as far from the speakers, d' = 8d.
New sound intensity is :
[tex]I'=\dfrac{k}{d'^2}\\\\I'=\dfrac{k}{(8d)^2}\\\\I'=\dfrac{1}{64}\times \dfrac{k}{d^2}\\\\I'=\dfrac{1}{64}\times I[/tex]
So, the new intensity is (1/64) times of the initial sound intensity.
does velocity, speed, and rate acceleration change if it was dropped from a higher building?
Answer:
Your answer is correct, the rates would stay the same
Explanation:
An oxygen-16 ion with a mass of perpendicular to a 1.20-T magnetic field, which makes it move in a circular arc with a 0.231-m radius.
Required:
a. What positive charge is on the ion?
b. What is the ratio of this charge to the charge of an electron?
c. Discuss why the ratio found in (b) should be an integer.
Answer:
a) 4.8*10^-19 C
b) 3
Explanation:
We know that
F = ma
Also, recall that a = v²/r and F = qvB
Substituting this in the first equation, we have
qvB = mv²/r, since we're looking for q, we make it the subject of the formula and solve for it.
q = mv²/rvB
q = mv/rB
q = [2.66*10^-26 * (5*10^6)] / 0.231 * 1.2
q = 1.33*10^-19 / 0.2772
q = 4.8*10^-19 C
Ratio between the charge and the electron is q/c =
4.8*10^-19 / 1.6*10^-19 = 3
Therefore, the ratio between the charge and electron is 3
A car accelerates from a standstill to 60 km/hr in 10.0 seconds, what is its acceleration?
a= vf - vi/t
a= 60-0/10
a= 60/10
a= 6 m/s^2
Answer:
6 km/hr/s
Explanation:
A 95 N net force is applied to an ice block with a mass of 24 kg. Find the acceleration of the block if it moves on a smooth horizontal surface
Answer:
The answer is 3.96 m/s²Explanation:
The acceleration of an object given it's mass and the force acting on it can be found by using the formula
[tex]a = \frac{f}{m} \\ [/tex]
f is the force
m is the mass
From the question we have
[tex]a = \frac{95}{24} \\ = 3.95833333333...[/tex]
We have the final answer as
3.96 m/s²Hope this helps you
5n right 5n left I don’t know this please help!
Answer:
this is my exception I am not sure but I think its help full 5Nr+5NL so it will be 0N which is equilibrium
This problem relates to a design competition in which a 1-kg battery-powered vehicle must be propelled up a 0.9-m tall ramp within a 15-second time interval. a. For a run time of 9 seconds, how much power must the battery supply to the vehicle
Answer:
P = 0.98 W
Explanation:
Given that,
Mass, m = 1 kg
Height, h = 0.9 m
(a) We need to find power must the battery supply to the vehicle for a run time of 9 seconds. Power is given by work done divided by time. It can be given by :
[tex]P=\dfrac{W}{t}\\\\P=\dfrac{mgh}{t}\\\\P=\dfrac{1\times 9.8\times 0.9}{9}\\\\P=0.98\ W[/tex]
So, the required power is 0.98 W.
A 872 g wooden block is initially at rest on a rough horizontal surface when a 15.6 g bullet is fired horizontally into (but does not go through) it. After the impact, the block-bullet combination slides 6.50 m before coming to rest. If the coefficient of kinetic friction between block and surface is 0.750, determine the speed of the bullet immediately before impact.
Answer:
The speed of the bullet before the impact is 556.5 m/s
Explanation:
Given;
mass of the wooden block, m₁ = 872 g = 0.872 kg
initial velocity of the wooden block, u = 0
mass of the bullet, m₂ = 15.6 g = 0.0156 kg
distance traveled by the block-bullet system after impact, d = 6.5 m
coefficient of kinetic friction, μ = 0.75
Work done by friction on the system after impact is given by;
W = (m₁ + m₂)μg x d
kinetic energy of the system after impact is given by;
K.E = ¹/₂(m₁ + m₂)v²
Apply work energy theorem;
(m₁ + m₂)μg x d = ¹/₂(m₁ + m₂)v²
μg x d = ¹/₂v²
v² = 2μg x d
v = √(2μg x d)
v = √(2 x 0.75 x 9.8 x 6.5)
v = 9.78 m/s
The speed of the bullet before impact is calculated by applying principle of conservation of linear momentum;
m₁u₁ + m₂u₂ = v(m₁ + m₂)
0.872(0) + 0.0156u₂ = 9.78(0.872 + 0.0156)
0 + 0.0156u₂ = 8.681
u₂ = 8.681 / 0.0156
u₂ = 556.5 m/s
Therefore, the speed of the bullet before the impact is 556.5 m/s
A ship maneuvers to within 2.47×10^3 m of an island’s 1.77 × 10^3 m high mountain peak and fires a projectile at an enemy ship 6.02 × 10^2 m on the other side of the peak, as illustrated. The ship shoots the projectile with an initial velocity of 2.55×10^2 m/s at an angle of 72.4. The acceleration of gravity is 9.81 m/s 2. How close (vertically) does the projectile
come to the peak?
I've attached a sketch of what I interpret the situation to be. It seems to me that you have to find the height of the projectile above the mountain's peak, assuming it actually does hit the ship on the other side. (Note that the sketch is not drawn to-scale. I don't mean to suggest that the projectile reaches its maximum height directly above the mountain.)
Compute the horizontal and vertical components of the velocity:
[tex]v_{i,x}[/tex] = (2.55 x 10² m/s) cos(72.4º) ≈ 77.1 m/s
[tex]v_{i,y}[/tex] = (2.55 x 10² m/s) sin(72.4º) ≈ 243 m/s
The projectile's horizontal and vertical positions at time t from its launching point are
x = [tex]v_{i,x}[/tex] t
y = [tex]v_{i,y}[/tex] t - 1/2 (9.81 m/s²) t²
Find the time it takes for the projectile to travel the distance from the first ship to the mountain's peak:
2.47 x 10³ m = (77.1 m/s) t
t = (2.47 x 10³ m) / (77.1 m/s)
t ≈ 32.0 s
Find the vertical position at this time:
y = (243 m/s) (32.0 s) - 1/2 (9.81 m/s²) (32.0 s)²
y ≈ 2750 m = 2.75 x 10³ m
Take the difference of this height and the height of the mountain:
2.75 x 10³ m - 1.77 x 10³ m = 0.980 x 10³ m = 9.80 x 10² m
An astronaut on the surface of Mars standing on the edge of a 73.3 meter tall cliff throws a golf ball
down with an initial speed of 4.5 m/s. The ball hits the ground 5.2 seconds later. Find the
acceleration due to gravity on Mars.
Answer:
3.69m/s²
Explanation:
Using the formula;
S = ut + 1/2gt²
Where;
S = distance (m)
u = initial velocity (m/s)
t = time (s)
g = accelerate due to gravity (m/s)
Based on the information provided, S = 73.3m, u = 4.5m/s, t = 5.2s, g= ?
73.3 = (4.5 × 5.2) + 1/2 (g × 5.2²)
73.3 = 23.4 + 1/2 (27.04g)
73.3 = 23.4 + 13.52g
73.3 - 23.4 = 13.52g
49.9 = 13.52g
g = 49.9/13.52
g = 3.6908
g = 3.69m/s²
A train travels due south at 60 m/s. It reverses its direction and travels due north at 60 m/s. What is the change in velocity of the train
The change in velocity of the train is 120 m/sec.
What is Velocity?The rate at which an object's position changes when observed from a specific point of view and when measured against a specific unit of time is known as its velocity. Unit of velocity is m/sec.
Given in the question train travels due south at 60 m/s. It reverses its direction and travels due north at 60 m/s,
We see that the change in direction of the train is 180 degree,
the net change in velocity is given as,
v = 60 m/s + 60 m/s
v = 120 m/s
A train travels due south at 60 m/s. It reverses its direction and travels due north at 60 m/s the change in velocity of the train is 120 m/sec.
To learn more about velocity refer to the link:
brainly.com/question/18084516
#SPJ1
Answer the following: What are the units of mass and the units of weight for the metric system? What are the units of mass and the units of weight for the English system?
Answer:
Explanation:
Units are measure of a quantity. There are two types of units,
The fundamental units
The derived units
The fundamental units are units that are independent i.e they can stand alone while derived units are derived from fundamental units (They are dependent)
The fundamental units are the units of fundamental quantities (Length, mass and time)
Length is measured in metres (m)
Mass is measured in kilograms (kg)
Time is measured in seconds(s)
This are called the standard units where other units are derived.
Other units of mass are grams
Weight is the product of mass and acceleration due to gravity
Weight = mass×acc due to gravity
W = mg
Using units
W = kg × m.s²
Since acceleration is measured in m/s²
W = kgm/s²
Hence the metric unit of weight is kgm/s² or Newtons
1. Imagine that you take a 200.0 km trip. You travel the first half of the trip at 100.0 km/hr. Your speed for the second half is 50.0 km/hr. How much time does the entire trip take?
2. In the above problem, what is your average speed in km/hr and m/s
Answer:
Explanation:
Speed = Distance/Time
Given
Distance = 200km
Total speed = 100+50 = 150km/hr
Time = Distance/Speed
Time = 200/150
Time = 1.33hr
b) Average speed in km/hr is 150km/hr
Convert to m/s
150km/hr = 150 * 1000 m/1 * 3600s
150km/hr = 150,000/360
150km/hr = 416.67m/s
David tosses an egg horizontally from the top of a building, with an initial velocity of 6.5 m/s. The building is 60m high. The time takes the egg to hit the ground is
Answer: 3.5 seconds
Explanation:
Why do objects have specific colors?
Answer:
The color of an object is the wavelength of light that it reflects.
A marble rolls off a ledge with an initial velocity of (1.125m/s, 0o). The marble lands 0.45m from the base of the ledge. Calculate the magnitude of the velocity, in m/s, of the marble just before it lands.
I don't want the answer but can someone please describe to me how I'm supposed to solve this and the variables I need.
Answer:
3.2m/s
Explanation:
Given parameters:
Initial velocity = 1.125m/s
Height of fall = 0.45m
Unknown:
Final velocity = ?
Solution:
To solve this problem, we have to look at our given parameters and compare to the appropriate motion equation.
we have been given; u (initial velocity) and height of fall (h)
Since this body falls under acceleration due to gravity, g = 9.8m/s²
Now, we use;
V² = U² + 2gh
where V is the final velocity
Input the parameters and solve;
V² = 1.125² + (2 x 9.8 x 0.45)
V = 3.2m/s
During which stage of sleep, also known as delta sleep, would a person have slow breathing and be difficult to wake.
A.
between stages 1 and 2 of non–rapid-eye-movement sleep
B.
in stage 4 of non–rapid-eye-movement sleep
C.
near the end of stage 3 of rapid-eye-movement sleep
D.
in stage 3 of non–rapid-eye-movement sleep
Answer:
In stage 3 so the answer is D
Answer:
It's D. in stage 3 of non-rapid-eye-movement sleep
Explanation:
**PLATO** "NREM-3 is when a person is in deep sleep and it lasts about 30 minutes. The person experiences slow breathing and pulse rates and limp muscles, and is difficult to wake. This stage is sometimes called delta sleep, because the brain produces slow delta waves."
Based on the periodic Table, how many neutrons are most likely in a neutral atom of potassium(K)?
a model rocket of mass 0.5 kg is launched straight up. At an altitude of 140 m, when its speed is 90 m/s, it spontaneously blows apart into two equal pieces. One piece continues upward at a speed of 10 m/s. How fast is the other piece moving
Answer:
The other piece is moving at 170 m/s.
Explanation:
To find the speed of the second piece can be found by conservation of linear momentum:
[tex] p_{i} = p_{f} [/tex]
[tex] m_{1}v_{1} = m_{2}v_{2} + m_{3}v_{3} [/tex]
Where:
m₁: is the rocket's mass = 0.5 kg
m₂ = m₃: is the mass of the two equal pieces = 0.25 kg
v₁: is the rocket speed = 90 m/s
v₂: is the speed of the first piece = 10 m/s
v₃: is the speed of the second piece =?
[tex] 0.5*90 = 0.25(10 + v_{3}) [/tex]
[tex] v_{3} = \frac{45}{0.25} - 10 = 170 m/s [/tex]
Therefore, the other piece is moving at 170 m/s.
I hope it helps you!
In a grocery store, you push a 15.5 kg shopping cart with a force of 10.5 N. If the cart starts at rest, how far does it move in 3.00 s?
Answer:
The distance moved by the cart is 3.05 m
Explanation:
Given;
mass of the cart, m = 15.5 kg
applied force, f = 10.5 N
initial velocity, u = 0
time of motion, t = 3s
The final velocity of the cart is given by;
[tex]F = ma\\\\F = m(\frac{v-u}{t})\\\\ m(v-u) = Ft\\\\m(v-0) = Ft\\\\mv = Ft\\\\v = \frac{Ft}{m}\\\\ v = \frac{10.5*3}{15.5}\\\\ v = 2.032 \ m/s[/tex]
The acceleration of the cart is given by;
F = ma
a = F/m
a = 10.5 / 15.5
a = 0.677 m/s²
The horizontal distance moved by the cart is given by;
s = ut + ¹/₂at²
s = 0 + ¹/₂ (0.677)(3)²
s = 3.05 m
Therefore, the distance moved by the cart is 3.05 m
PLEASE HELP
You are walking to school one day and can't wait to give Mr. DeBaz the homework you finished. As you are walking down the street, a zombie comes out from behind the Dunkin' Donuts. He looks at you, growls, and starts coming after you. You are being chased by a zombie and you run 2 m/s. You are trying to get to the school, for safety, and it is 250 m away. How long will it take you to get there?
A billiard ball moving at 5.30 m/s strikes a stationary ball of the same mass. After the collision, the first ball moves at 4.72 m/s, at an angle of θ = 27.0° with respect to the original line of motion. Assuming an elastic collision (and ignoring friction and rotational motion), find the struck ball's velocity after the collision.
Answer:
The velocity of the struck billiard ball after collision is 2.41 m/s at an angle of 62.8⁰.
Explanation:
Given;
initial velocity of the billiard ball, u₁ = 5.3 m/s
initial velocity of the second ball, u₂ = 0
final velocity of the first ball, v₁ = 4.72 m/s at θ = 27.0°
final velocity of the second ball, v₂ = ?
Apply the principle of conservation of kinetic energy;
¹/₂m₁u₁² + ¹/₂m₂u₂² = ¹/₂m₁v₁² + ¹/₂m₂v₂²
m₁u₁² + m₂u₂² = m₁v₁² + m₂v₂²
m₁u₁² + m₂(0) = m₁v₁² + m₂v₂²
m₁u₁² = m₁v₁² + m₂v₂²
m₁ = m₂
u₁² = v₁² + v₂²
v₂² = u₁² - v₁²
v₂² = (5.3)² - (4.72)²
v₂² = 5.812
v₂ = √5.812
v₂ = 2.41 m/s
The direction of the second ball is given by;
v₂sinθ = v₁sin27
sinθ = (v₁sin27) / (v₂)
sinθ = (4.72 x 0.454) / (2.41)
sinθ = 0.8892
θ = sin⁻¹ (0.8892)
θ = 62.8⁰
Thus, the velocity of the struck billiard ball after collision is 2.41 m/s at an angle of 62.8⁰.
need help with momentum!
Answer:
The total kinetic energy must be greater than the momentum of either ball
Explanation:
In order to solve this problem, we must clarify the following conditions, the balls are in motion that is, there is kinetic energy, the kinetic energy is a scalar magnitude, therefore this energy is greater than zero.
But however momentum is equal to zero, momentum is a vector quantity, that is, these velocities have a direction, momentum is defined as the product of mass by Velocity.
P = m*v (momentum lineal)
where:
P = lineal momentum [kg*m/s]
m = mass [kg]
v = velocity [m/s]
Since the masses of the balls are equal, we have:
m1 = m2
therefore:
m1*v1 + (m2*v2) = 0 (total momentum)
For the sum to be equal to zero, besides the equivalation of the masses, the velocities must also be equal, but the velocities must be of opposite sign, to meet equality to zero.
m1*v1 - m2*v2 = 0
And the kinetic energy of a body is defined by means of the following equation
Ek = 0.5*m*v²
Since the magnitude of the velocities is greater than zero, so that there is movement, we can say that the kinetic energy is much greater than zero, as well as greater than the total linear momentum
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Answer:Never smoke ciga If you are a smoker, throw them away now! Smoking is the number one cause of preventable death. It causes many cancers, heart attacks, strokes, lung diseases, and other nonfatal problems, like stained teeth, poor athletic performance, erectile dysfunction, and bad breath. There is help available to quit, including free counseling at 1-800-QUIT-NOW and medications to ease the cravings. Ask your doctor for help. Stopping is the biggest favor you can do for yourself.
A spring stretches by 21.0 cm when a 135 N object is attached. What is the weight of a fish that would stretch the spring by 31 cm?
Answer:
199.28 N
Explanation:
It is given that,
A spring stretches by 21.0 cm when a 135 N object is attached.
We need to find the weight of a fish that would stretch the spring by 31 cm.
We know that, the force in a spring is given by as per Hooke's law as follows :
F = kx
Where k is spring constant
[tex]\dfrac{F_1}{F_2}=\dfrac{x_1}{x_2}\\\\\dfrac{135}{F_2}=\dfrac{21}{31}\\\\F_2=\dfrac{135\times 31}{21}\\\\F_2=199.28\ N[/tex]
So, the required weight of a fish is 199.28 N.
Consider a rocket (initial mass m o) accelerating from rest in free space. At first, as it speeds up, its momentum p increases, but as its mass m decreases p eventually begins to decrease. For what value of m is p maximum
Answer:
m = m_o/e
Explanation:
The rocket equation is given as;
m(dv/dt) = -v_ex(dm/dt)
v_ex is the exhaust velocity
Now, formula for momentum is;
p = mv
Differentiating with respect to time, we have;
dp/dt = m'v + mv'
Where;
m' is mass rate
v' is rate of change in velocity
Since we are dealing with exhaust velocity and momentum(p) is Maximum when v = v_ex then mv' can also be written as: -m'v_ex.
Thus;
dp/dt = m'v - m'v_ex
dp/dt = m'(v - v_ex)
Now, from the rocket equation we dt will cancel out to give;
-dv/v_ex = dm/m
Integrating both sides;
-∫dv/v_ex = ∫dm/m
This gives;
-v/v_ex = In(m/m_o)
Since momentum(p) is Maximum when v = v_ex
Thus;
-v/v = In(m/m_o)
-1 = In(m/m_o)
e^(-1) = m/m_o
m = m_o(e^(-1))
m = m_o/e
On June 17th 2003, in Santa Barbara, California, the morning low tide was measured at -0.365 m. High tide was measured at 1.12 . Calculate the tidal range between these tides.
A)there was no tidal range.
b)0.755 m
c)1.485 m
This cost a lot of points pls give me an amswer before 10:50 its k if its past 10:50
Answer:
I am not sure but I think it's choice c.
c)1.485 m
Explanation:
I hope this helps.
Why does the man fly out of the car?
A. Newtons 1st Law
B. Newtons 2nd Law
C. Newtons 3rd Law
what capacitance is required to store an energy of 10kw.h at a potential difference of 1000v?
Two pendulum bobs of unequal mass are suspended from the same fixed point by strings of equal length. The lighter bob is drawn aside and then released so that it collides with the other bob on reaching the vertical position. The collision is elastic. What quantities are conserved in the collision?
Answer:
"Both kinetic energy and angular moment of the system" is the correct solution.
Explanation:
The kinetic energy is:
[tex]KE=\frac{1}{2}m_{1}v_{1}^2 + \frac{1}{2}m_{2}v_{2}^2[/tex]
Its conversion is:
⇒ [tex](KE)_{1}=(KE)_{f}[/tex]
Angular momentum is:
[tex]L=I_{1}\omega_{1}+I_{2}\omega_{2}[/tex]
Its conversion is:
⇒ [tex]L_{i}=L_{f}[/tex]
So that both "KE" and "L" of the system conserved.
A ball is thrown straight up in the air with a velocity of 12 m/s. Find the maximum height the ball
reaches.
Answer:
The maximum height is 7.34 meters.
Explanation: