Answer:
the mass of the box is 51.98 kg.
Explanation:
Given;
applied horizontal force, F = 450 N
coefficient of friction, μ = 0.795
constant velocity, v = 1.2 m/s
At constant velocity, the acceleration of the object is zero and the net force will be zero.
[tex]F_{Net} = F - F_k\\\\0 = F - F_k\\\\F_k = F\\\\\mu \ N = F\\\\\mu (mg) = F\\\\m = \frac{F}{\mu g} \\\\m = \frac{405}{0.795 \ \times \ 9.8} \\\\m = 51.98 \ kg[/tex]
Therefore, the mass of the box is 51.98 kg.
Scientists create models to better understand Earth. Which evidence has led scientists to conclude that there are different layers within Earth's interior?
A.analysis of seismic wave data
B.measurement of Earth's diameter
C.temperatures taken within each layer
D.rock samples taken from Earth's core
Answer:
it is A or D
Explanation:
Answer:
ANswer:A
Explanation:
Allen and Jason are chucking a speaker around. On one particular throw, Allen throws the speaker, which is playing a pure tone of frequency f, at a speed of 10 m/s directly towards Jason, but his aim is a bit off. As a result, Jason runs forward towards the speaker at a speed of 6 m/s before catching it. Then, the frequency that Jason hears while running can be written as (m/n)f Hz, where m and n are relatively prime positive integers. Compute m n.
Answer:
Explanation:
We shall apply Doppler's effect of sound .
speaker is the source , Jason is the observer . Source is moving at 10 m /s , observer is moving at 6 m/s .
apparent frequency = [tex]f_o\times\frac{V+v_o}{ V-v_s}[/tex]
V is velocity of sound , v₀ is velocity of observer and v_s is velocity of source and f_o is real frequency of source .
Here V = 340 m/s , v₀ is 6 m/s , v_s is 10 m/s . f_o = f
apparent frequency = [tex]f\times \frac{340+6}{340-10}[/tex]
= [tex]f\times \frac{346}{330}[/tex]
So m = 346 , n = 330 .
A 22.0 kg child is riding a playground merry-go- round that is rotating at 40.0 rev/min. What centripetal force must
Answer:
F = 482.51 N
Explanation:
Given that,
Mass of a child, m = 22 kg
Angular velocity of the merry-go-round, [tex]\omega=40\ rev/min[/tex]
Let the radius of the path, r = 1.25 m
We need to find the centripetal force acting on the child. The formula for the centripetal force is given by :
[tex]F=m\omega^2r\\\\=22\times (4.18879)^2\times 1.25\\\\=482.51\ N[/tex]
So, the required centripetal force is 482.51 N.
All of the following are ways in which sports psychologists help athletes except __________.
A.
staying motivated
B.
managing fear of failure
C.
improving performance
D.
enhancing memory
Please select the best answer from the choices provided
A
B
C
D
Answer:
D-Enhancing memory
Explanation:
A hammer strikes a nail with a 10 N force for 0.01 seconds. Calculate the impulse of the hammer.
Answer:
0.1Ns
Explanation:
Impulse is the product of Force and time
Impulse = Force * Time
Given
Force = 10N
Time = 0.01s
Substitute into the formula
Impulse = 10 * 0.01
Impulse = 10 * 1/100
Impulse = 10/100
Impulse = 0.1Ns
hence the impulse of the hammer is 0.1Ns
The human nervous system can propagate nerve impulses at about 102 m>s. Estimate the time it takes for a nerve impulse to travel 2 m from your toes to your brain.
Answer:
t = 0.196 s
Explanation:
The speed of a pulse is determined by the characteristics of the medium, its density and its resistance to stress, as long as these remain the speed will be constant for which we can use the kinetic expressions of the uniform movement
v = x / t
t = x / v
calculate
t = 2/102
t = 0.196 s
An inductor is connected to a 120-V, 60-Hz supply. The current in the circuit is 2.4 A. What is the inductive reactance
Answer:
Inductive reactance is 50.00 ohms
Explanation:
Given the following data;
Voltage = 120v
Frequency = 60Hz
Current = 2.4 A
To find the inductive reactance;
Inductive reactance, XL = V/I
Where;
XL represents the inductive reactance. V represents the voltage. I represents the current.Substituting into the equation, we have;
XL = 120/2.4
XL = 50.00 ohms
A small lead ball, attached to a 1.10-m rope, is being whirled in a circle that lies in the vertical plane. The ball is whirled at a constant rate of three revolutions per second and is released on the upward part of the circular motion when it is 1.3 m above the ground. The ball travels straight upward. In the absence of air resistance, to what maximum height above the ground does the ball rise
Answer:
1.84 m
Explanation:
For the small lead ball to be balanced at the tip of the vertical circle just before it is released, the reaction force , N equal the weight of the lead ball W + the centripetal force, F. This normal reaction ,N also equals the tension T in the string.
So, T = mg + mrω² = ma where m = mass of small lead ball, g = acceleration due to gravity = 9.8 m/s², r = length of rope = 1.10 m and ω = angular speed of lead ball = 3 rev/s = 3 × 2π rad/s = 6π rad/s = 18.85 rad/s and a = acceleration of normal force. So,
a = g + rω²
= 9.8 m/s² + 1.10 m × (18.85 rad/s)²
= 9.8 m/s² + 390.85 m/s²
= 400.65 m/s²
Now, using v² = u² + 2a(h₂ - h₁) where u = initial velocity of ball = rω = 1.10 m × 18.85 rad/s = 20.74 m/s, v = final velocity of ball at maximum height = 0 m/s (since the ball is stationary at maximum height), a = acceleration of small lead ball = -400.65 m/s² (negative since it is in the downward direction of the tension), h₁ = initial position of lead ball above the ground = 1.3 m and h₂ = final position of lead ball above the ground = unknown.
v² = u² + 2a(h₂ - h₁)
So, v² - u² = 2a(h₂ - h₁)
h₂ - h₁ = (v² - u²)/2a
h₂ = h₁ + (v² - u²)/2a
substituting the values of the variables into the equation, we have
h₂ = 1.3 m + ((0 m/s)² - (20.74 m/s)²)/2(-400.65 m/s²)
h₂ = 1.3 m + [-430.15 (m/s)²]/-801.3 m/s²
h₂ = 1.3 m + 0.54 m
h₂ = 1.84 m
A motorcycle moving 18.8 m/s has
57800 J of KE. What is its mass?
Answer:
m = 327.07 kg
Explanation:
Given that,
Kinetic energy of a motorcycle, E = 57800 J
Velocity of the motorcycle, v = 18.8 m/s
We need to find the mass of the motorcycle. The kinetic energy of an object is given by :
[tex]E=\dfrac{1}{2}mv^2[/tex]
m is mass
[tex]m=\dfrac{2E}{v^2}\\\\m=\dfrac{2\times 57800 }{(18.8)^2}\\\\m=327.07\ kg[/tex]
So, the mass of the motorcycle is 327.07 kg.
A block weighing 9.3 N requires a force of 3.7 N to push it along at constant velocity. What is the coefficient of friction for the surface
Answer:
0.398
Explanation:
According to friction, the frictional force is directly proportional to the normal reaction
Ff = nR
Ff is the frictional force
n is the coefficient of friction
R is the reaction
Reaction is equal to the weight
R= W = 9.3N
Fm = Ff = 3.7N
Fm is the moving force
Get the coefficient of friction
n = Ff/R
n = 3.7/9.3
n = 0.398
Hence the coefficient of friction for the surface is 0.398
A cart's initial velocity is +3.0 meters per second. What is its final velocity after accelerating at a rate of 1.5 m/s2 for 8.0 seconds?
A. 36 m/s
B. 9 m/s
C. 72 m/s
D. 15 m/s
Answer:
V = 15m/s
Explanation:
Given the following data;
Initial velocity = 3m/s
Time = 8secs
Acceleration = 1.5m/s²
To find the final velocity, we would use the first equation of motion;
V = U + at
Substituting into the equation, we have
V = 3 + 1.5*8
V = 3 + 12
V = 15m/s
Why could it be argued that the respiratory system is most critical to sustaining life?
Explanation:
Energy is the most important ingredient for life. Organisms use energy in diverse ways. Scientifically, energy is defined as the ability to do work. Without this ability, organisms would not exist.
So, the most important process is one that can furnish the body with energy.
The respiratory system happens to be the one that furnishes the body with energy. During respiration, the energy needs of the body is met by series of processes. Oxygen is taken in and use to liberate calories from chemical substances packed with energy. So, without respiration, the bodily energy demands will not be met.2. A 2500 kg car is slowed down uniformly from an initial velocity of 20.0 m/s to
the north by a 6250 N braking force acting opposite the car's motion. Use the
impulse-momentum theorem to answer the following questions:
a. What is the car's velocity after 2.50 s?
b. How far does the car move during 2.50 s?
c. How long does it take the car to come to a complete stop?
Answer:
13.75m/s; 42.2m; 8s
Explanation:
(a) the car's velocity after 2.50 s is 13.75 m/s
(b) The distance traveled by the car is 42.18 m
(c) the time taken for the car to come to complete stop is 8 s.
The given parameters;
mass of the car, m = 2500 kg
initial velocity of the car, u = 20 m/s
breaking applied on the car, f = 6250 N
The acceleration of the car is calculated as follows;
[tex]F = ma \\\\a = \frac{F}{m} = \frac{6250}{2500} = 2.5 \ m/s^2[/tex]
(a) Using impulse-momentum theorem, the car's velocity after 2.5 s is calculated as follows;
[tex]F = \frac{m(u-v)}{t} \\\\m(u-v) = Ft\\\\u-v = \frac{Ft}{m} \\\\v = u - \frac{Ft}{m} \\\\v = 20 - \frac{6250 \times 2.5}{2500} \\\\v = 13.75 \ m/s[/tex]
(b) The distance traveled by the car during the 2.5 s;
[tex]v^2 = u^2 - 2as\\\\2as = u^2 - v^2\\\\s = \frac{u^2 - v^2}{2a} \\\\s = \frac{20^2 - 13.75^2}{2\times 2.5} \\\\s = 42.18 \ m[/tex]
(c) The time taken for the car to come to a complete stop;
when the car stop's the final velocity, v = 0
v = u - at
0 = 20 - 2.5t
2.5t = 20
[tex]t = \frac{20}{2.5} \\\\t = 8 \ s[/tex]
Thus, the time taken for the car to come to complete stop is 8 s.
Learn more here: https://brainly.com/question/14559060
If you travel from Tucson to Argentina, you will see some different constellations in the night sky. true or false
Answer:
its true!!
Explanation: have a nice day !!
The resistance of a wire depends on its length i and on its cross sectional area A the resistance is
Answer:
The resistance of a wire is directly proportional to its length and inversely proportional to its cross-sectional area
Explanation:
If a man traveled to a different planet
Answer:
He would be in space.
Explanation:
Answer: What would happen if a human traveled to Uranus?
As an ice giant, Uranus doesn't have a true surface. The planet is mostly swirling fluids. While a spacecraft would have nowhere to land on Uranus, it wouldn't be able to fly through its atmosphere unscathed either. The extreme pressures and temperatures would destroy a metal spacecraft.
Explanation:
An energy source forces a constant current of 2A to flow through a light bulbfilament for twenty seconds. If 4.6 kJ is given off in the form of light and heatenergy, calculate the voltage drop across the bulb.
Answer:
The voltage drop across the bulb is 115 V
Explanation:
The voltage drop equation is given by:
[tex]V=\frac{\Delta W}{\Delta q}[/tex]
Where:
ΔW is the total work done (4.6kJ)
Δq is the total charge
We need to use the definition of electric current to find Δq
[tex]I=\frac{\Delta q}{\Delta t}[/tex]
Where:
I is the current (2 A)
Δt is the time (20 s)
[tex]2=\frac{\Delta q}{20}[/tex]
[tex]q=40 C[/tex]
Then, we can put this value of charge in the voltage equation.
[tex]V=\frac{4600}{40}=115 V[/tex]
Therefore, the voltage drop across the bulb is 115 V.
I hope it helps you!
An electron traverses a vacuum tube with a length of 2 m in 2 X 10- 4
sec. What is the average speed of the
electron during this time?
Answer:
Average speed = 10,000 m/s
Explanation:
Given the following data;
Distance = 2m
Time = 0.0002secs
To find the average speed;
Average speed = distance/time
Average speed = 2/0.0002
Average speed = 10,000 m/s
Therefore, the average speed of the
electron is 10,000 meters per seconds.
Which is greater, the energy of one photon of orange light or the energy of one quantum ofradiation having a wavelength of 3.36 * 10^-9
The question is incomplete, here is the complete question:
Which is greater, the energy of one photon of orange light or the energy of one quantum of radiation having a wavelength of [tex]3.36\times 10^{-9}m[/tex]
Answer: The energy of one quantum of radiation having wavelength [tex]3.36\times 10^{-9}m[/tex] is greater than the energy of 1 photon of orange light.
Explanation:
To calculate the energy of one photon, we use the Planck's equation:
[tex]E=\frac{N_Ahc}{\lambda}[/tex]
where,
E = energy of radiation
[tex]N_A[/tex] = Avogadro's number = [tex]6.022\times 10^{23}mol^{-1}[/tex]
h = Planck's constant = [tex]6.626\times 10^{-34}Js[/tex]
c = speed of light = [tex]3\times 10^8 m/s[/tex]
[tex]\lambda}[/tex] = wavelength of radiation
For orange light:For 1 photon, the term [tex]N_A[/tex] does not appear
[tex]\lambda}[/tex] = 620 nm = [tex]620\times 10^{-9}m[/tex] (Conversion factor: [tex]1nm=10^{-9}m[/tex] )
Putting values in above equation, we get:
[tex]E=\frac{6.626\times 10^{-34}Js\times 3\times 10^8m/s}{620\times 10^{-9}m}\\\\E=3.206\times 10^{-19}J[/tex]
For one quantum of radiation:[tex]\lambda}[/tex] = [tex]3.36\times 10^{-9}m[/tex]
Putting values in above equation, we get:
[tex]E=\frac{6.022\times 10^{23}mol^{-1}\times 6.626\times 10^{-34}Js\times 3\times 10^8m/s}{3.36\times 10^{-9}m}\\\\E=3.56\times 10^{7}J/mol[/tex]
Hence, the energy of one quantum of radiation having wavelength [tex]3.36\times 10^{-9}m[/tex] is greater than the energy of 1 photon of orange light.
In a particular crash test, an automobile of mass 1577 kg collides with a wall and bounces back off the wall. The x components of the initial and final speeds of the automobile are 17 m/s and 1.5 m/s, respectively. If the collision lasts for 0.18 s, find the magnitude of the impulse due to the collision. Answer in units of kg · m/s.
Answer:
Ft=24,443.5 kgm/s
Explanation:
Step one
Given data
Mass of automobile m=1577kg
Initial Velocity u=17m/s
Final Velocity v=1.5m/s
Time t=0.18s
Step two
From the impulse and momentum equation
Ft=mΔv
Substitute
Ft=1577*(17-1.5)
Ft=1577*15.5
Ft=24,443.5 kgm/s
Brainliest brainliest help help help mememememememme
Answer:
????????????????????,
Explanation:
I need points sorry
Answer:
honestly this was so long ago can i get brainliest i need 2 more until i am at expert level
Explanation:
Water runs out of a horizontal drainpipe at the rate of 135 kg/min. It falls 3.1 m to the ground. Assuming the water doesn't splash up, what average force does the water exert on the ground
Answer:
The average force exerted by the water on the ground is 17.53 N.
Explanation:
Given;
mass flow rate of the water, m' = 135 kg/min
height of fall of the water, h = 3.1 m
the time taken for the water to fall to the ground;
[tex]h = ut + \frac{1}{2}gt^2\\\\h = 0 + \frac{1}{2}gt^2\\\\t = \sqrt{\frac{2\times 3.1}{9.8} } \\\\t = 0.795 \ s[/tex]
mass of the water;
[tex]m = m't\\\\m = 135 \ \frac{kg}{min} \ \times \ 0.795 \ s \ \times \ \frac{1 \ \min}{60 \ s} \ = 1.789 \ kg[/tex]
the average force exerted by the water on the ground;
F = mg
F = 1.789 x 9.8
F = 17.53 N
Therefore, the average force exerted by the water on the ground is 17.53 N.
What can we conclude from observing an attractive force between a positively charged rod and some object
Answer: that the object is negatively charged.
Explanation:
We know that the force between objects that have the same type of charge is a repulsive force, while for objects with an opposite charge, the force is attractive.
In this case, we know that we have an attractive force between an object and a positively charged rod.
Then the only conclusion we can take in this situation is that the object is negatively charged.