(a) The normal force on the sled is 490 N.
( b ) The sleds acceleration in the horizontal direction is 0.52 m/s².
What is the normal force on the sled?
The normal force on the sled is the upward force on the sled due to weight of the sled acting downwards.
The normal force on the sled is calculated as follows;
N = mg
where;
m is the mass of the sledg is acceleration due to gravityN = 50 kg x 9.8 m/s²
N = 490 N
The acceleration of the sled is calculated by applying Newton's second law of motion as shown below;
Fx = ma
F cosθ = ma
a = Fcosθ / m
where;
θ is the inclination of the applied forcea = ( 30 N x cos30 ) / ( 50 )
a = 0.52 m/s²
Thus, the acceleration of the sled is a function of net force and mass of the sled.
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Which one of the statements below is true about mechanical waves?
They must travel in empty space.
They can travel in a vacuum.
Both sound and light are examples of mechanical waves.
They require a medium to travel through.
Fluid mechanics questions and answers
Answer:
Fluid mechanics is considered one of the toughest subdisciplines within mechanical and aerospace engineering. It is unique from almost any other field an undergraduate engineer will encounter. It requires viewing physics in a new light, and that's not always an easy jump to make.
Two points on a progressive wave are out of phase by 0.41 rad.
What is this phase difference?
[1 mark]
A 23° [e]
B 47° [e]
C 74° [e]
D 148° [e]
The phase difference between two points on a progressive wave are out of phase by 0.41 rad is 23°.
What is an equation?An equation is an expression that shows the relationship between two or more numbers and variables.
The two points on a progressive wave are out of phase by 0.41 rad.
Hence, Phase difference = 0.41 rad
But:
Rad to degree = (rad * 180/π)°
Hence:
0.41 rad = (0.41 rad * 180/π) = 23°
The phase difference between two points on a progressive wave are out of phase by 0.41 rad is 23°.
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A wave has a frequency of 2 Hz. Find its period
It takes 20 seconds to fill a two-liter bottle with water from your kitchen faucet. What is the mass flow rate from the faucet if water has a density of 1000 fraction numerator k g over denominator m cubed end fraction?
a. 0.1kg/sec.
b. 0.01kg/sec.
c. 1g/sec.
d. 1kg/sec.
Answer:
0.1 kg/s.
Explanation:
The density of water, d = 1000 kg/m³
Volume, V = 2 L
Time, t = 20 s
We need to find the mass flow rate from the faucet. We know that the density of an object is given by :
[tex]d=\dfrac{m}{V}\\\\m=d\times V\\\\\dfrac{m}{t}=\dfrac{dV}{t}\\\\\dfrac{m}{t}=\dfrac{1000\times 0.002}{20}\\\\=0.1\ kg/s[/tex]
So, the mass flow rate is equal to 0.1 kg/s.
Jovian planets have rings because:__________.
a. their thick gaseous atmospheres would disintegrate any small rock that enter them
b. there is too much material to have fit into the ball of each planet
c. tidal forces prevent the material in rings from forming into moons
d. Jovian planets rotate very rapidly, and some material near the equator of these planets was flung outward, forming the rings
e. tidal forces cause volcanic eruptions on some moons, and part of this material subsequently escaped the gravity of the moons, forming the rings.
Answer:
E
Explanation:
Planets after Mars in our solar system are called Jovian planets. Therefore, Jupiter, Saturn, Uranus and Neptune are Jovian planets. The specialty of these planets is that they mostly made of gases and have ring around them.
They have rings around them because tidal forces cause volcanic eruptions on some moons, and part of this material subsequently escaped the gravity of the moons, forming the rings.
Can anybody help in number 6? <3
Answer:
5.4 will be the weight in illustrate form
The steering wheel of a car has a radius of 0.19 m, and the steering wheel of a truck has a radius of 0.25 m. The same force is applied in the same direction to each steering wheel. What is the ratio of the torque produced by this force in the truck to the torque produced in the car
Answer:
[tex]\frac{T_t}{T_c} = 1.32[/tex]
Explanation:
The torque applied on an object can be calculated by the following formula:
[tex]T = Fr[/tex]
where,
T = Torque
F = Applied Force
r = radius of the wheel
For car wheel:
[tex]T_c = Fr_c\\[/tex]
For truck wheel:
[tex]T_t = Fr_t[/tex]
Dividing both:
[tex]\frac{T_t}{T_c} = \frac{Fr_t}{Fr_c}[/tex]
for the same force applied on both wheels:
[tex]\frac{T_t}{T_c} = \frac{r_t}{r_c} \\[/tex]
where,
rt = radius of the truck steering wheel = 0.25 m
rc = radius of the car steering wheel = 0.19 m
Therefore,
[tex]\frac{T_t}{T_c} = \frac{0.25\ m}{0.19\ m} \\[/tex]
[tex]\frac{T_t}{T_c} = 1.32[/tex]
A 1.8 kg book has been dropped from the top of the football stadium. Its speed is 4.8 m/s when it is 2.9 m above the ground. How high is the stadium?
Answer:
the height of the stadium is 4 m
Explanation:
The computation of the height of the stadium is shown below:
but before that total mechanic energy should be determined
E = PE + KE
where
PE = mgh
and, KE = 1 ÷2 mv^2
Now
E = mgh + 1 ÷2 mv^2
= (1.8) (9.8) (2.9) + 1 ÷ 2 (1.8) (4.8)^2
= 71.9J
= 72J
Now the height of the stadium is
TE = mgh
72 = (1.8) × (9.8) × h
So, h = 4 m
Hence, the height of the stadium is 4 m
Coherent monochromatic light of wavelength l passes through a narrow slit of width a, and a diffraction pattern is observed on a screen that is a distance x from the slit. On the screen, the width w of the central diffraction maximum is twice the distance x. What is the ratio a>l of the width of the slit to the wavelength of the light
Answer:
λ = a
Explanation:
This is a diffraction exercise that is described by the expression
a sin θ = m λ
sin θ = m λ/ a
the first zero of the diffraction occurs for m = 1
sin θ = λ / a
angles are generally very small and are measured in radians
sin θ = θ = y / x
we substitute
[tex]\frac{y}{x} = \frac{\lambda}{a}[/tex]
the width of the central maximum is twice the distance to zero
w = 2y
in the exercise indicate that this width is equal to twice the distance to the screen (2x)
W = 2x
2y = 2x
we substitute
1 = λ/ a
λ = a
we see that the width of the slit is equal to the wavelength used.
A turntable, with a mass of 1.5 kg and diameter of 20 cm, rotates at 70 rpm on frictionless bearings. Two 540 g blocks fall from above, hit the turntable simultaneously at opposite ends of a diameter, and stick to it.
Required:
What is the turntable's angular speed, in rpm, just after this event?
Answer:
The turntable's angular speed after the event is 28.687 revolutions per minute.
Explanation:
The system formed by the turntable and the two block are not under the effect of any external force, so we can apply the Principle of Conservation of Angular Momentum, which states that:
[tex]I_{T}\cdot \omega_{o} = (2\cdot r^{2}\cdot m +I_{T})\cdot \omega_{f}[/tex] (1)
Where:
[tex]I_{T}[/tex] - Moment of inertia of the turntable, in kilogram-square meters.
[tex]r[/tex] - Distance of the block regarding the center of the turntable, in meters.
[tex]m[/tex] - Mass of the object, in kilograms.
[tex]\omega_{o}[/tex] - Initial angular speed of the turntable, in radians per second.
[tex]\omega_{f}[/tex] - Final angular speed of the turntable-objects system, in radians per second.
In addition, the momentum of inertia of the turntable is determined by following formula:
[tex]I_{T} = \frac{1}{2}\cdot M\cdot r^{2}[/tex] (2)
Where [tex]M[/tex] is the mass of the turntable, in kilograms.
If we know that [tex]\omega_{o} \approx 7.330\,\frac{rad}{s}[/tex], [tex]M = 1.5\,kg[/tex], [tex]m = 0.54\,kg[/tex] and [tex]r = 0.1\,m[/tex], then the angular speed of the turntable after the event is:
[tex]I_{T} = \frac{1}{2}\cdot M\cdot r^{2}[/tex]
[tex]I_{T} = 7.5\times 10^{-3}\,kg\cdot m^{2}[/tex]
[tex]I_{T}\cdot \omega_{o} = (2\cdot r^{2}\cdot m +I_{T})\cdot \omega_{f}[/tex]
[tex]\omega_{f} = \frac{I_{T}\cdot \omega_{o}}{2\cdot r^{2}\cdot m +I_{T}}[/tex]
[tex]\omega_{T} = 3.004\,\frac{rad}{s}[/tex] ([tex]28.687\,\frac{rev}{min}[/tex])
The turntable's angular speed after the event is 28.687 revolutions per minute.
1.
is the rate that velocity changes
O Acceleration
O
Time
O
Distance
O
Mass
Answer:
a) acceleration
Explanation:
Acceleration is, by definition, the change of an object's velocity.
How much mechanical work is required to catch a 14.715N ball traveling at a velocity of 37.5m/s?
To catch a 14.715N ball traveling at a velocity of 37.5m/s, required mechanical work is 1056.10 joule.
What is work?Physics' definition of work makes clear how it is related to energy: anytime work is performed, energy is transferred.
In a scientific sense, a work requires the application of a force and a displacement in the force's direction. Given this, we can state that
Work is the product of the component of the force acting in the displacement's direction and its magnitude.
Weight of the ball = 14.715 N.
Mass of the ball = 14.715 N ÷ (9.8 m/s²) = 1.502 kg.
Velocity of the ball = 37.5 m/s
Kinetic energy of the ball = 1/2 × 1.502 × 37.5² Joule = 1056.10 Joule.
Hence, to catch a 14.715N ball traveling at a velocity of 37.5m/s, required work is 1056.10 joule.
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Which of the following is true of the
thermocline layer of the ocean?
A. rapidly decreases in temperature
B. warmest and least dense of the ocean layers
C. is the bottom layer of the ocean
D. is the top layer of the ocean
Answer:
d is trueeeeeeeeeeeeeeeee
Describe the formation of the land, the atmosphere, and the oceans of earth
What happens when a neutral atom gains an electrons?
Answer:
The neutral atom becomes an anion.
Explanation:
When a neutral atom gains an electron (e−), the number of protons (p+) in the nucleus remains the same, resulting in the atom becoming an anion (an ion with a net negative charge).
I WILL REPORT YOU IF YOU DON'T ANSWER QUESTION OR POST OTHER LINKS
Rank these objects in order of their resistance to change in motion (from greatest to least).
A. pyramid, cone, sphere, cube
B. cube, pyramid, cone, sphere
C. sphere, cone, pyramid, cube
D. None of the above
Answer:
pyramid, cone, sphere, cube
A light bulb has a resistance of 360 . What is the current in the bulb when it has a potential difference of 120 V across it? 0.33 A 3 A 480 A 43,200 A
Answer:
I=R×V
360×120 V
=43,200 A
Answer:
A.)0.33 A
Explanation:
i just took the quiz 2021
what happens during subduction
Answer:
Subduction , Latin for "carried under," is a term used for a specific type of plate interaction. It happens when one lithospheric plate meets another—that is, in convergent zones —and the denser plate sinks down into the mantle.
What is the index of refraction of a refractive medium if the angle of incidence in air is 40 degrees and the angle of refraction is 29 degrees?
The index of refraction is 1.33.
To find the answer, we need to know about index of refraction.
What is index of refraction?
The indication of the light bending ability of a medium is the refractive index of that medium.It determines how much the path of light is bent or refracted.It's a dimensionless number.What is the mathematical expression of refractive index?
Mathematically, the refractive index(n) isn= sin∅₁ / sin∅₂
where, ∅₁= angle of incident
∅₂= angle of refraction
What is the refractive index, if the angle of incidence in air is 40 degrees and the angle of refraction is 29 degrees?
n= sin(40) /sin(24) = 1.33Thus, we can conclude that the index of refraction is 1.33.
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The length of stereocilia actually vary from 10 to 50 micrometers. Again, assuming that they behave like simple pendula, over what frequency range of sound waves would they resonate. (The actual frequency range of human hearing is 20 Hz - 20,000 Hz, so there certainly must be other mechanisms involved in determining the frequency of these being the pendular model is rather oversimplified.)
a. About 70 Hz -160 Hz.
b. About 440 Hz - 1000 Hz.
c. About 20 Hz - 50 Hz.
d. About 0.07 to 0.16 Hz.
Answer:
the frequency range of sound waves is about 70 Hz - 160 Hz
Hence, Option a) About 70 Hz -160 Hz is the correct answer
Explanation:
Given the data in the question;
L₁ = 10 micrometers = 0.00001 m
so
T₁ = 2π√(L/g)
g = 9.8 m/s
so we substitute
T₁ = 2π√(0.00001 /9.8) = 0.006347
⇒ f₁ = 1 / T₁ = 1 / 0.006347 = 157.55 ≈ 160 Hz
L₂ = 50 micrometers = 0.00005 m
T₂ = 2π√(L/g)
g = 9.8 m/s
so we substitute
T₂ = 2π√(0.00005 /9.8) = 0.0142
f₂ = 1 / T₂ = 1 / 0.0142 = 70.422 ≈ 70 Hz
Therefore, the frequency range of sound waves is about 70 Hz - 160 Hz
Hence, Option a) About 70 Hz -160 Hz is the correct answer
A charged particle (charge 1.6x10-19 C and mass 1.67x10-27 kg) is initially moving with a velocity of 2x105 m/s and then moves into a region having a magnetic field and an electric field (6x104 V/m). The direction of initial velocity is perpendicular to the electric field and magnetic field. If the charged particle keep moving in the original direction without being deflected, what is the magnitude of the magnetic field
Answer:
[tex]0.3\ \text{T}[/tex]
Explanation:
q = Charge = [tex]1.6\times 10^{-19}\ \text{C}[/tex]
m = Mass of particle = [tex]1.67\times 10^{-27}\ \text{kg}[/tex]
v = Velocity = [tex]2\times 10^5\ \text{m/s}[/tex]
E = Electric field = [tex]6\times 10^4\ \text{V/m}[/tex]
B = Magnetic field
Magnetic field is given by
[tex]B=\dfrac{E}{v}\\\Rightarrow B=\dfrac{6\times 10^4}{2\times 10^5}\\\Rightarrow B=0.3\ \text{T}[/tex]
The magnitude of magnetic field is [tex]0.3\ \text{T}[/tex]
Which of the following is NOT true about
newspaper weather maps?
A. They report the temperature of the area in degrees
Fahrenheit.
B. They let you know how much precipitation to expect.
C. They provide more detail than weather service maps.
D. They tell you about the warm and cold air fronts.
Suppose we replace both hover pucks with pucks that are the same size as the originals but twice as massive. Otherwise, we keep the experiment the same. Compared to the pucks in the video, this pair of pucks will rotate View Available Hint(s) Suppose we replace both hover pucks with pucks that are the same size as the originals but twice as massive. Otherwise, we keep the experiment the same. Compared to the pucks in the video, this pair of pucks will rotate four times as fast. at the same rate. one-fourth as fast. twice as fast. one-half as fast.
Answer:
w = w₀ / 2 the angular velocity is half the initial value.
Explanation:
We can analyze this exercise as if we added another disk to obtain a disk with twice the mass, for which if the system is two disks, the angular tidal wave is conserved
initial instant.
L₀ = I₀ w₀
final moment
L_f = I w
the moment is preserved
L₀ = L_f
I₀ w₀ = I w
the moment of inertia of a disk is
I = ½ m R²
we substitute
½ m R² w₀ = ½ (2m) R² w
w = w₀ / 2
for the case of a disk with twice the mass, the angular velocity is half the initial value.
question is included in the picture!!! PUT REAL ANSWERS OR I WILL REPORT YOU
Answer:
Explanation:
this is like rubbing a balloon on your head to make your hair stand up. Do that to the can. The balloon is filled , ofc, and then just rub the balloon on the can. This will charge the can with static electricity. :P
A student hangs a block from a light string that is attached to a massive pulley of unknown radius R, as shown in the figure. The student allows the block to fall from rest to the floor. Which two of the following sets of data that could be measured or determined should the student use together to determine the final angular velocity of the pulley just before the block hits the floor? Select two answers. Justify your selections.
Answer:
The mass of the block, the distance of the block above the floor, and the time it takes the block to reach the floor, because these quantities can be used to determine the acceleration of the block.
The radius and the mass of the pulley, because these quantities can be used together to determine the rotational inertia of the pulley.
Explanation:
If the motion starts from rest, the initial angular velocity will be zero and the final angular velocity can be determined with the product of angular acceleration and time of motion of the pulley.
Angular velocity is defined as the change in the angular displacement per change in time of motion. This can be expressed mathematically as follows;
[tex]\omega = \frac{\Delta \theta}{\Delta t} = vr[/tex]
where;
Ф is the angular displacementt is the time of the motionv is the linear velocityr is the radius of the circular path.In a circular motion that starts from rest and ends with final velocity, the equation is given as;
[tex]\omega_f =\omega_i + \alpha t[/tex]
Where;
[tex]\omega_f[/tex] is the final angular velocity[tex]\omega_i[/tex] is the initial angular velocity[tex]\alpha[/tex] is the angular accelerationThus, if the motion starts from rest, the initial angular velocity will be zero and the final angular velocity can be determined with the product of angular acceleration and time of motion of the pulley.
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A 101 kg basketball player crouches down 0.380 m while waiting to jump. After exerting a force on the floor through this 0.380 m, his feet leave the floor and his center of gravity rises 0.920 m above its normal standing erect position. (a) Using energy considerations, calculate his velocity (in m/s) when he leaves the floor. m/s (b) What average force (in N) did he exert on the floor
Answer:
[tex]4.25\ \text{m/s}[/tex]
[tex]3391.22\ \text{N}[/tex]
Explanation:
y = Height of compression = 0.38 m
m = Mass of basketball player = 101 kg
h = Height of center of gravity after jump = 0.92 m
g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]
Energy balance of the system is given by
[tex]mgh=\dfrac{1}{2}mv^2\\\Rightarrow v=\sqrt{2gh}\\\Rightarrow v=\sqrt{2\times 9.81\times 0.92}\\\Rightarrow v=4.25\ \text{m/s}[/tex]
The velocity of the player when he leaves the floor is [tex]4.25\ \text{m/s}[/tex]
[tex]Fy=mgy+\dfrac{1}{2}mv^2\\\Rightarrow F=\dfrac{mgy+\dfrac{1}{2}mv^2}{y}\\\Rightarrow F=\dfrac{101\times 9.81\times 0.38+\dfrac{1}{2}\times 101\times 4.25^2}{0.38}\\\Rightarrow F=3391.22\ \text{N}[/tex]
The force exerted on the floor is [tex]3391.22\ \text{N}[/tex].
The driver of a car wishes to pass a truck that is traveling at a constant speed of 19.3 m/s . Initially, the car is also traveling at a speed 19.3m/s and its front bumper is a distance 25.0m behind the truck's rear bumper. The car begins accelerating at a constant acceleration 0.560m/s^2 , then pulls back into the truck's lane when the rear of the car is a distance 26.5m ahead of the front of the truck. The car is of length 4.50m and the truck is of length 20.7m .
Part A) How much time is required for the car to pass the truck?
Part B ) What distance does the car travel during this time?
Part C) What is the final speed of the car?
Answer:
A) t = 10.56 s, B) x = 235 m, C) v = 25.2 m / s
Explanation:
A) We can solve this problem using kinematics expressions.
The distance traveled by the truck is
x_c = v_c t
Distance traveled by the car.
The car must travel the distance that separates them from the truck x₀=25.0. Return to the lane at x₁ = 26.5 m. the length of the truck x₂=20.7m and the length of the car x₃ = 2 4.5 = 9 m, therefore the total length traveled by the car is
x_t = x₁ + x₂ + x₃
x_t = 26.5 + 20.7 +9 = 56.2 m
the distance traveled by the car when it returns to the lane is
x_c + x_t = x₀ + v₀ t + ½ a t²
when the car passes the car the distance traveled by the two vehicles is the same, we substitute
v_c t + x_t = x₀ + v₀ t + ½ a t²
½ a t² + t (v₀ -v_c) + (x₀ - x_t) = 0
we substitute the values
½ 0.560 t² + t (19.3 -19.3) + (25.0 - 56.2) =
0.28 t² -31.2 = 0
t = [tex]\sqrt{ \frac{31.2}{0.28} }[/tex]
t = 10.56 s
This is the time it takes for the car to pass the truck and back into the lane.
B) the distance traveled is
x = v₀ t + ½ a t²
x = 19.3 10.56 + ½ 0.560 10.56²
x = 235 m
C) the final velocity is
v = v₀ + a t
v = 19.3 + 0.560 10.56
v = 25.2 m / s
a student practicing for a track meet ran 250 m in 30 seconds. What was her average speed?
Answer:
8.33 meters/sec.
time = 30 sec. 30 sec. = 8.33 meters/sec.
I need help please someone !!!!! Would appreciate it
Answer:
Yes, it would make it back up.
Explanation:
If it has 100,000 Joules of gravitational potential energy at the top of the hill, by the time the cart gets to the bottom, it will become PE = 0, KE = 90,000 since 10% of 100,000 is 10,000. The cart only requires 80,000J to climb back up so it should easily do so.
I didn't quite understand if the 10% energy loss is total, or every time it goes up or down, but it isn't a problem because 10% of 90,000 is 9,000, which means it would have 81,000J of energy on the way back up IF it loses energy due to friction on the way back up also.
The only physical law you need to prove this is the Law of Conservation of Energy: no energy is lost, only transformed; 10% of the energy becomes heat, the rest remains mechanical energy, which is the reason why the reasoning above works.