Answers
Answer:
x = 10.75 m
Explanation:
For this problem we will solve it in two parts, the first using energy and the second with kinematics
Let's use the energy work relationship to find the velocity of the block as it exits the ramp
W = [tex]Em_{f}[/tex] - Em₀
Starting point. Higher
Em₀ = U = m g h
the height from the edge of the ramp of the graph has a value
h = 9-3 = 6 m
Final point. At the bottom of the ramp
Em_{f} = K = ½ m v²
Friction force work
W = - fr d
The friction force has the formula
fr = μ N
On the ramp, we can use Newton's second law
N - W cos θ = 0
N = W cos θ
where the angle is obtained from the graph
tan θ = (9-3) / (0.5-4) = -6 / 3.5
θ = tan⁻¹ (-1,714)
θ = -59.7º
the distance d is
d = √ (Δx² + Δy²)
d = √ [(0.5-4)² + (9-3)²]
d = 6.95 m
for which the work is
W = - μ mg cos 59.7 d
we substitute
W = Em_{f} -Em₀
- μ mg cos 59.7 d = ½ m v² - m g h
In the graph o text the value of the friction coefficient is not observed, suppose that it is μvery = 0.2
- μ g cos 59.7 d = ½ v² - g h
v² = 2g (h - very d coss 59.7)
let's calculate
v² = 2 9.8 (6 - 0.2 6.95 cos 59.7)
v = √ 103.8546
v = 10.19 m / s
in the same direction as the ramp
in the second part we use projectile launch kinematics
let's look for the components of velocity
v₀ₓ = vo cos -59.7
[tex]v_{oy}[/tex] = vo sin (-59,7)
v₀ₓ = 10.19 cos (-59.7) = 5.14 m / s
v_{oy} = 10.19 if (-59.7) = -8.798 m / s
Let's find the time to get to the floor (y = o)
y = y₀ + v_{oy} t - ½ g t²
to de groph y₀=3 m
0 = 3 - 8.798 t - ½ 9.8 t²
t² - 1.796 t - 0.612 = 0
we solve the quadratic equation
t = [1.796 ±√(1.796² + 4 0.612)] / 2
t = [1,795 ± 2,382] / 2
t₁ = 2.09 s
t₂ = -0.29 s
since time must be a positive quantity the correct value is t = 2.09 s
we calculate the horizontal displacement
x = v₀ₓ t
x = 5.14 2.09
x = 10.75 m
The motion of the box, after it exits the incline is the motion and trajectory
of a projectile.
Horizontal distance from the right-hand edge of the incline to the point of
contact with the floor is approximately 1.24613 m.
Reasons:
Mass of the block, m = 3.5 kg
Coefficient of kinetic friction, μ = 1.2
Location of the = 0.5 m from the origin
Required:
Horizontal distance between the block's point of contact with the floor and
the bottom right-hand edge of the incline.
Solution:
Let θ represent the angle the incline make with the horizontal.
The normal reaction of the incline on the block, [tex]F_N[/tex] = m·g·cos(θ)
Work done on friction = [tex]F_N[/tex]×μ×Length of the incline, L
Rise of the incline = 10 - 3 = 7
Run of the incline = 4
L = √(6.125² + 3.5²) = [tex]\dfrac{7 \times \sqrt{65} }{8}[/tex]
Let ΔP.E.₁ represent the potential energy transferred to kinetic energy
and work along the incline, we have;
Energy of the block at the bottom of the incline, M.E.₂, is found as follows;
K.E.₂ = mgh - m·g·μ·cos(θ)·L
[tex]K.E. =\frac{1}{2} \times 3.5 \times v^2 = 3.5 \times 9.81 \times 6.125 - 3.5 \times 9.81 \times 1.2 \times \dfrac{4}{\sqrt{65} } \times \dfrac{7 \times \sqrt{65} }{8}[/tex]
v ≈ 6.1456 m/s
The vertical component of the velocity is therefore;
[tex]v_y = v \cdot sin(\theta)[/tex]
[tex]v_y = 6.1456 \times \dfrac{7}{\sqrt{65} } \approx 5.33588[/tex]
From the equation, h = u·t + 0.5·g·t² derived from Newton's Laws of motion, we have;
ΔP.E.₁ = 3.5×9.81×7
3 = 5.33588·t + 0.5×9.81·t²
Factorizing, the above quadratic equation, we get;
The time it takes the block to reach the floor, t ≈ 0.40869 seconds
Horizontal component of the velocity is [tex]v_x \approx 6.1456 \times \dfrac{4}{\sqrt{65} } \approx 3.04908[/tex]
The horizontal distance, x = vₓ × t
∴ x = 3.04908 × 0.40869 ≈ 1.08194
Horizontal distance from the right-hand edge of the incline to the point of
contact with the floor, x ≈ 1.24613 m.
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Related Questions
What is potential energy? Use in your own words.
Answers
Answer:
Potential energy is the latent energy in an object at rest.
Explanation:
Hope this helps- this is how I would answer :)
A circular coil of wire 8.6 cm in diameter has 15 turns and carries a current of 2.7 A. The coil is in a region where the magnetic field is 0.56 T. (a) What orientation of the coil gives the maximum torque on the coil, and what is this maximum torque? (b) For what orientation of the coil is the magnitude of the torque 71% of the maximum found in part (a)?
Answers
Answer:
(a) (i) The orientation of the coil which gives maximum torque on the coil is 90⁰
(a)(ii) The maximum torque is 0.132 Nm
(b) The orientation of the coil is 45⁰
Explanation:
Given;
diameter of the circular wire, d = 8.6 cm = 0.086 m
radius of the wire, r = d /2 = 0.043 m
number of turns, N = 15 turns
magnetic field, B = 0.56 T
The torque on the wire is given by;
τ = NIABsinθ
where;
θ is the orientation of the wire
(a) maximum torque occurs when the orientation of the wire is at 90⁰
The maximum torque is given by;
τ = NIABsin(90⁰)
τ = NIAB
τ = (15)(2.7)(π x 0.043²)(0.56)
τ = 0.132 Nm
(b)
71% of 0.132 = 0.71 x 0.132 = 0.0937 Nm
[tex]\tau = NIAB sin\theta\\\\sin\theta = \frac{\tau}{NIAB }\\\\ sin\theta = \frac{0.0937}{(15)(2.7)(\pi *0.043^2)(0.56)} \\\\sin\theta = 0.7111\\\\\theta = sin^{-1}(0.7111)\\\\\theta =45.32\\\\\theta = 45^0[/tex]
Calculate the de Broglie wavelength of: a) A person running across the room (assume 180 kg at 1 m/s) b) A 5.0 MeV proton
Answers
Answer:
a
[tex]\lambda = 3.68 *10^{-36} \ m[/tex]
b
[tex]\lambda_p = 1.28*10^{-14} \ m[/tex]
Explanation:
From the question we are told that
The mass of the person is [tex]m = 180 \ kg[/tex]
The speed of the person is [tex]v = 1 \ m/s[/tex]
The energy of the proton is [tex]E_ p = 5 MeV = 5 *10^{6} eV = 5.0 *10^6 * 1.60 *10^{-19} = 8.0 *10^{-13} \ J[/tex]
Generally the de Broglie wavelength is mathematically represented as
[tex]\lambda = \frac{h}{m * v }[/tex]
Here h is the Planck constant with the value
[tex]h = 6.62607015 * 10^{-34} J \cdot s[/tex]
So
[tex]\lambda = \frac{6.62607015 * 10^{-34}}{ 180 * 1 }[/tex]
=> [tex]\lambda = 3.68 *10^{-36} \ m[/tex]
Generally the energy of the proton is mathematically represented as
[tex]E_p = \frac{1}{2} * m_p * v^2_p[/tex]
Here [tex]m_p[/tex] is the mass of proton with value [tex]m_p = 1.67 *10^{-27} \ kg[/tex]
=> [tex]8.0*10^{-13} = \frac{1}{2} * 1.67 *10^{-27} * v^2[/tex]
=> [tex]v _p= \sqrt{\frac{8.0 *10^{-13}}{ 0.5 * 1.67 *10^{-27}} }[/tex]
=> [tex]v = 3.09529 *10^{7} \ m/s[/tex]
So
[tex]\lambda_p = \frac{h}{m_p * v_p }[/tex]
so [tex]\lambda_p = \frac{6.62607015 * 10^{-34}}{1.67 *10^{-27} * 3.09529 *10^{7} }[/tex]
=> [tex]\lambda_p = 1.28*10^{-14} \ m[/tex]
(1.5 pts) A woman pushes on a box to the left. If the box is accelerating, what forces are working on the
Question 2:
box? (Draw both y and x forces)
Answers
Answer:
Nope
Explanation:
5. Peter had an inflated balloon that he released and flies across the room. The balloon slows down and then stops on top of the dinning table, As the balloon slows downs the force becomes
A.balanced
B.frictional
C.restricted
D.unbalanced
Answers
Ruby has an index of refraction of 1.7. Calculate the maximum angle with respect to the normal at which a light ray escapes a ruby if the ruby is in air.
a. 17°
b. 36°
c. 53°
d. 90°
e. Light escapes at all incident angles.
Answers
Answer:
Umm sorry just came for the points.
Sam heats an 8kg sample of sand, with a specific heat of 664 J/kg·C°, from 20° to 40°. What is the change in thermal energy?
Answers
Answer:
106.24 kJ.
Explanation:
Given that,
Mass of sample of sand, m = 8 kg
Specific heat of sand, c = 664 J/kg-°C
The temperature changes from 20° C to 40° C. We need to find the change in thermal energy. It is given by :
[tex]Q=mc\Delta T\\\\Q=8\times 664(40-20)\\\\=106240\ J\\\\=106.24\ kJ[/tex]
So, the change in thermal energy is 106.24 kJ.
How long does it take a vehicle to reach a velocity of 32 m/s if it accelerates from rest at a rate of 4.2 m/s^2?
What is the initial velocity of the vehicle?
What is the final velocity of the vehicle?
What is the acceleration of the vehicle?
Write the equation you will use to solve the problem.
How long does it take the vehicle to reach its final velocity?
0.13 seconds
18.1 seconds
7.62 seconds
134.4 seconds
Answers
Answer:
7.62
Explanation:
because you have to divide 32/4.2
and can you do a friend request so i can accept it
Staying with the dragster problem...two more... If the dragster did the amount of work you calculated in the previous problem in 4.38 seconds, how much power was generated in Watts?
Answers
Answer:
1,439,283.10Watts
Explanation:
The question looks incomplete. We need the workdone to get the power generated.
Power = Workdone/Time
Power = Force * Distance/Time
Let
Force = 24340 Newtons
Distance covered = 259meters
Given
Time = 4.38secs
Required
Power generated
Substitute the given parameters into the formula;
Power generated = 24340*259/4.38
Power generated = 6,304,060/4.38
Power generated = 1,439,283.10
Hence the amount of power generated is 1,439,283.10Watts
A boat travels 28 m while it reduces its velocity from 27.5 m/s to 14.5m/s. What is the boat’s acceleration while it travels that distance?
Answers
Answer:
9.75 m/s²
Explanation:
Given that,
v= 14.5 m/s
u = 27.5 m/s
s = 28 m
a = ?
v² = u² -2as [ minus sign due to deceleration]
14.5² = 27.5² - 2 × a × 28
210.25 - 756.25 = -56a
-56a = -546
a = 9.75 m/s²
A car has a mass of 850 kg. By pushing on the car, Evan increases its speed
from 3.5 m/s to 5 m/s. What impulse did Evan apply to the car?
A. 4250 kg•m/s
B. 1275 kg•m/s
C. 850 kg•m/s
D. 2975 kg•m/s
Answers
Answer:
B. 1275 kg*m/s
Explanation:
I = F(deltaT) = (deltaP) = mv2- mv1
Therefore,
I = mv2-mv1
m = 850 kg
v2 = 5 m/s
v1 = 3.5 m/s
I = (850)(5)-(850)(3.5)
I = 1275 kg* m/s
What is the difference between the two graphs
Answers
Answer:
one of the graph is postion-time graph while the other one is velocity-time graph
A pressure antinode in a sound wave is a region of high pressure, while a pressure node is a region of low pressure.
True
False
Answers
The answer is false
At a young age, who has the greatest influence on your choices?
Why do stretching exercises increase flexibility more than cardio exercises?
Answers
Stretching exercises increase flexibility more than cardio exercises because being flexible requires you to stretch parts of your body. For instance, when you touch your toes your being flexible by stretching your arms to touch your toes.
Answer:
Explanation:
No da esta aplicación y si para que usted vella
If I ride my bike at 10 mph and traveled 5 miles, how long did I ride in both hours and
minutes?
Answers
STATION 1
Jane moved a 800kg piano to the right across the
carpet with a coefficient of friction of 0.4. What is the
magnitude of the force of friction acting on the
piano?
If she moved it at a constant velocity what is the
applied force acting on the piano?
Answers
A footballer kicks a ball from rest. The foot is in contact with the ball for 0.30s and the final velocity of the ball is 15ms-1 .What is the average acceleration of the ball?
Answers
Answer:
50m/s^2Explanation:
Step one:
given data
initial velocity u= 0m/s since the ball is at rest
time of contact t= 0.3s
final velocity v=15m/s
Required
acceleration a
from the first law of motion
v=u+at
substitute our given data
15=0+a*0.3
15=0.3a
divide both sides by 0.3
a=15/0.3
a=50m/s
The average acceleration is 50m/s^2
In a certain time, light travels 3.50 km in a vacuum. During the same time, light travels only 2.35 km in a liquid. What is the refractive index of the liquid?
Answers
Answer:
1.45
Explanation:
Refractive index of the liquid is given as;
Refractive index = [tex]\frac{speed of light in vacuum}{speed of light in liquid}[/tex]
But,
speed = [tex]\frac{distance}{time}[/tex]
Since a certain light of specific wavelength was used during the same time, let the time be represented by t.
So that;
speed of light in vacuum = [tex]\frac{3500}{t}[/tex]
speed of light in the liquid = [tex]\frac{2350}{t}[/tex]
Refractive index = [tex]\frac{3500}{t}[/tex] ÷ [tex]\frac{2350}{t}[/tex]
= [tex]\frac{3500}{t}[/tex] x [tex]\frac{t}{2350}[/tex]
Refractive index = [tex]\frac{3500}{2350}[/tex]
= 1.4536
= 1.45
The refractive index of the liquid is 1.45.
A 27-g steel-jacketed bullet is fired with a velocity of 640 m/s toward a steel plate and ricochets along path CD with a velocity 500 m/s. Knowing that the bullet leaves a 50-mm scratch on the surface of the plate and assuming that it has an average speed of 600 m/s while in contact with the plate, determine the magnitude and direction of the impulsive force exerted by the plate on the bullet.
Answers
Answer:
F = - 3.56*10⁵ N
Explanation:
To attempt this question, we use the formula for the relationship between momentum and the amount of movement.
I = F t = Δp
Next, we try to find the time that the average speed in the contact is constant (v = 600m / s), so we say
v = d / t
t = d / v
Given that
m = 26 g = 26 10⁻³ kg
d = 50 mm = 50 10⁻³ m
t = d/v
t = 50 10⁻³ / 600
t = 8.33 10⁻⁵ s
F t = m v - m v₀
This is so, because the bullet bounces the speed sign after the crash is negative
F = m (v-vo) / t
F = 26*10⁻³ (-500 - 640) / 8.33*10⁻⁵
F = - 3.56*10⁵ N
The negative sign is as a result of the force exerted against the bullet
In your new job, you are the technical advisor for the writers of a gangster movie about Bonnie and Clyde. In one scene Bonnie and Clyde are being pursued by a police car. They are 750m from a level railroad crossing travelling at 100 km/hr. A train is 500 meters from the crossing travelling at 130 km/hr. The level crossing is on the state line, and so if they can beat the train, they could evade capture, at least for a while until they become Federal fugitives. They accelerate at a constant rate of 4 m/s2 toward the crossing. The writers want to know if Bonnie and Clyde make it across the crossing before the train.
Answers
Answer:
Explanation:
Velocity of train 130 km /hr
distance of crossing s = 500 m = .5 km
Time taken by train to reach crossing = .5 / 130 = .003846 hr = 13.85 s
Time taken by gangster to reach the crossing = t
initial velocity u = 100km/h = 27.77 m /s
distance of crossing s = 750 m
acceleration a = 4 m /s
s = ut + .5 a t²
750 = 27.77 t + .5 x 4 x t²
2 t² + 27.77 t - 750 = 0
t = - 27.77 ± √(27.77² + 4 x 2 x 750) / 2 x 2
= - 27.77 ±√ ( 771.17 + 6000) / 4
= - 27.77 ±82.28 / 4
= 13.62 s
So this time is less than time taken by train so it will be able to cross the crossing before train arrives .
5. How does the existence of humans compare with Earth's age?
Answers
Answer:
Human activity has fundamentally changed our planet. We live on every continent and have directly affected at least 83% of the planet’s viable land surface. Our influence has impacted everything from the makeup of ecosystems to the geochemistry of Earth, from the atmosphere to the ocean. Many scientists define this time in the planet’s history by the scale of human influence, and label it as a new geological epoch called the Anthropocene.
Explanation:
if a toy car has a centripetal acceleration of 50 m/s2 and was making the turn at 10 m/s. what was his radius
a 2m
ь 500m
C 5m
d.25m
Answers
Answer:
Explanation:
The centripetal acceleration is expressed as;
a = v²/r
a is the acceleration = 50m/s²
v is the velocity = 10m/s
r is the radius
To get the radius
r = v²/a
r = 10²/50
r = 100/50
r = 2m
Hence its radius is 2m
an object falls from a hovering helicopter and hits the ground at a speed of 30m per seconds. how long does it take the object to reach the ground and how far does it fall? sketch a velocity-time graph for the object ( ignore air resistance
Answers
Answer:
45.9m
Explanation:
Given parameters:
Final velocity = 30m/s
Initial velocity = 0m/s
Unknown:
Time it takes for the object of fall = ?
Height of fall = ?
Solution:
For the first problem, we use the equation below to solve for t;
V = U + gt
V is the final velocity
U is the initial velocity
g is the acceleration due to gravity
t is the time taken
30 = 0 + 9.8 x t
30 = 9.8t
t = [tex]\frac{30}{9.8}[/tex] = 3.1s
Now, height of fall;
V² = U² + 2gH
30² = 0² + 2 x 9.8 x H
900 = 19.6H
H = 45.9m
1
Betty is sitting on of her surfboard out in the ocean. She is waiting for the
perfect wave to come along so she can ride it in to shore. As she waits, she
notices that the waves roll by in patterns, or sets. As the top of each wave
passes by Betty, it pushes her up. Which part of the wave does this? *
Answers
Explanation:
yooooooooo
A car is traveling west for 12 s its speed is 36.12 m/s in the same direction find the total distance the car traveled
Answers
We are given:
constant speed of the car (u) = 36.12 m/s
time in question (t) = 12 seconds
Solving for the Distance and Displacement:
from the second equation of motion:
s = ut + 1/2 at^2
since we have 0 acceleration:
s = ut
replacing the variables
s = 36.12 * 12
s = 433.44 m
Since the car is travelling in a straight line towards the same direction, it's Distance will be equal to its Displacement
Hence, both the Displacement and Distance covered by the car is
433.44 m
but since Displacement also has a direction vector along with it,
the Displacement will be 433.44 m due west
Which element is used in the manufacture of mirrors and bronze?
Answers
Answer:
Silver
Explanation:
Silver is an important element in the manufacturing process of mirrors. Silver is used to make mirrors through the process we call "silvering". Silvering is a process in which a glass is coated with reflective substances so as to produce reflections, and then mirrors.
In Silvering process, Chlorine is also used. Stannous Chloride is the particular compound used to carry out the silvering, it has the chemical formula, SnCl₂
Using a maximum allowable shear stress of 70 MPa, find the shaft diameter needed to transmit 40 kW when (a) The shaft speed is 2500 rev/min. (b) The shaft speed is 250 rev/min
Answers
Answer:
a
[tex]d = 0.0223 \ m[/tex]
b
[tex]d = 0.0481 \ m[/tex]
Explanation:
From the question we are told that
The maximum allowable shear stress is [tex]\sigma = 70 MPa = 70 *10^{6} \ Pa[/tex]
The power is [tex]P = 40 \ kW = 40 *10^{3} \ W[/tex]
considering question a
The shaft speed is given as [tex]v = 2500\ rev/min[/tex]
Generally the torque experienced by the shaft is mathematically represented as
[tex]\tau = \frac{ 9.55 * P}{v}[/tex]
=> [tex]\tau = \frac{ 9.55 * 40 *10^{3}}{ 2500}[/tex]
=> [tex]\tau = 152.8 \ N \cdot m[/tex]
Generally the maximum torque experienced by the shaft is mathematically represented as
[tex]\tau_m = \frac{2 \tau }{ \pi r^2 }[/tex]
Generally diameter = 2 * radius (r)
So
[tex]\tau_m = \frac{2 \tau }{ \pi 4 d^2 }[/tex]
Generally the maximum allowable shear stress is mathematically represented as
[tex]\sigma = \frac{2 \tau }{ \pi 4 d^2 } * \frac{32}{d}[/tex]
=> [tex]\sigma = \frac{16 \tau }{ \pi d^3}[/tex]
=> [tex]d = \sqrt[3]{\frac{16 * \tau }{ \pi \sigma } }[/tex]
=> [tex]d = \sqrt[3]{\frac{16 * 152.8 }{ \pi * 70 *10^{6} } }[/tex]
=> [tex]d = 0.0223 \ m[/tex]
considering question b
The shaft speed is given as [tex]v = 250\ rev/min[/tex]
Generally the torque experienced by the shaft is mathematically represented as
[tex]\tau = \frac{ 9.55 * 40 *10^{3}}{250 }[/tex]
=> [tex]\tau = 1528 \ N \cdot m[/tex]
Generally the shaft diameter is mathematically represented as
[tex]d = \sqrt[3]{\frac{16 * \tau }{ \pi \sigma } }[/tex]
=>[tex]d = \sqrt[3]{\frac{16 * 1528 }{ 3.142 * 70 *10^{6} } }[/tex]
=>[tex]d = 0.0481 \ m[/tex]
Describe and give an example of mutualism.
Describe and give an example of commensalism.
Describe and give an example of parasitism.
Describe and give an example of competition.
Describe and give an example of predation.
Answers
Answer:
Mutualism - Bee to flower. Bee eats - flower reproduces
Commensalism - Tree Frog to plant or tree. Frog uses plant for protection.
Parasitism - Flea or tick to host. Parasite feeds off host.
Explanation:
Competition - relationship between organisms that strive for same resources. intraspecific and interspecific. ex) two males competing for mates.
predation - one organism kills and consumes another. wolf hunting moose, cat hunting mouse. venus fly trap killing insect
how does the uneaven heating of earths surface affects earths weather patterns
Answers
Answer: it causes some parts of the earth to get more radiation than others.
Explanation: earth rotates around the sun on a tilted axis so the Rays of the sun cause earth to have more radiation than it needs.
What horizontal speed must a pumpkin be thrown to hit a car 13.4 meters away from a building which stands 10.4 meters tall?
A) 1.5 m/s
B) 2.1 m/s
C)6.1 m/s
D) 8.9 m/s
Answers
Answer:
V₀ₓ = 9.2 m/s
Nearest answer:
D) 8.9 m/s
Explanation:
First we find the time taken by the pumpkin to hit the car. For that purpose we apply 2nd equation of motion to the pumpkin:
h = V₀y t + (1/2)gt²
where,
h = height of building = 10.4 m
V₀y = vertical component of initial speed = 0 m/s
t = time = ?
g = 9.8 m/s²
Therefore,
10.4 m = (0 m/s)(t) + (1/2)(9.8 m/s²)t²
t² = (10.4 m)(2)/(9.8 m/s²)
t = √[2.122 s²]
t = 1.45 s
Now, we analyze horizontal motion for horizontal component of initial velocity. We assume air friction to be zero so that the horizontal motion is uniform. Therefore,
s = V₀ₓ t
where,
s = horizontal distance between building and car = 13.4 m
V₀ₓ = Horizontal Component of Initial Velocity = ?
Therefore,
13.4 m = V₀ₓ(1.45 s)
V₀ₓ = 13.4 m/1.45 s
V₀ₓ = 9.2 m/s