Answer:
w = 0.319 rad / s
Explanation:
This is an angular momentum problem, let's form a system composed of the feeder and the squirrel, therefore the forces during the collision are internal and the angular momentum is conserved.
initial instant. Before the squirrel jumps
L₀ = m v r
final instant. After the trough and the squirrel are together
L_f = (I_fetter + I_ardilla) w
angular momentum is conserved
L₀ = L_f
m v r = (I_fetter + I_ardilla) w
w = [tex]\frac{mvr}{I_{fetter} + I_{ardilla} }[/tex]
the moment inercial ofbody is
I_thed = 2.00 kg m²
We approach the squirrel to a specific mass
I_ardilla = m r²
we substitute
w = m v r / ( I_[feefer + m r²)
let's calculate
w = 3 3.40 6.30 10⁻² / (2.00 + 3.00 (6.30 10-2)² )
w = 0.6426 / 2.0119
w = 0.319 rad / s
Classify each change (which can be manipulated within the green box) according to its effect on the wavelength.
a. Decrease frequency
b. Decrease damping
c. Decrease amplitude
d. Increase frequency
e. Increase amplitude
f. Increase damping
g. Shortens wavelength
Answer:
Explanation:
The classification will be made into 3 categories, which are
Ones that shortens wavelengths
Ones that lengthens wavelengths
Ones that has no effect on wavelengths
Shortens wavelengths -> Increase frequency
Lengthens wavelengths -> Decrease frequency
No effect -> Increase amplitude, decrease amplitude, increase damping, decrease damping.
can someone please take there time and answer this for me :)
Answer:
number 1
Explanation:
they have common ancestors
19. Using the formula V - IX R, find the voltage in an electrical component with a current of 2 A, and a resistance of 5 omega
O A 20 v
O 8.5v
O C. 2.5 V
O D. 10 v
Answer:
B
Explanation:
Answer:
B
Explanation:
BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB
red light from a He-Ne laser is at 590.5 nm in the air. it is fired at an angle of 31.0 to horizontal at a flat transparent crystal of calcite (n= 1.34 ar this frequency) .find the wavelength and frequency of the light inside the crystal and the angle from horizontal that it travels inside the calcite crystal.
Answer:
7374.4
Explanation:
I took the test
(filler so I can post)
A 4 kg box is at rest on a table. The static friction coefficient u, between the box and table is 0.30, and
the kinetic friction coefficient Hi is 0.10. Then, a 10 N horizontal force is applied to the box.
Answer:
The box will not move from its position.
Explanation:
First, we will calculate the static frictional force that is stopping the box to move from its position:
[tex]f = \mu R = \mu W=\mu mg[/tex]
where,
f = static frictional force = ?
μ = coefficient of static friction = 0.3
m = mass of box = 4 kg
g = acceleration due to gravity = 9.81 m/s²
Therefore,
[tex]f = (0.3)(4\ kg)(9.81\ m/s^2)\\f=11.77\ N[/tex]
Since the frictional force (11.77 N) is greater than the applied force (10 N).
Therefore, the box will not move from its position.
an object is moving at 60m/s and has a mass of 5 kg what is its momentum
Answer:
300
Explanation:
the momentum is 300
p=mv
p=5×60
5×60 =300
What type of electromagnetic radiation is being shown in the picture?
A. Gamma rays
B. Ultraviolet radiation
C. X-rays
D. Infrared radiation
Answer:
I think D. Infrared radiation.
Answer:
infrared radition
Explanation:
valid
mdjxjxjcjfkfjjdksklqlakzjxjxkkskakMmznxkxkdkd?
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
I hope this helped!+*
Consider a wheel (solid disk) of radius 1.12 m, mass 10 kg and moment of inertia 1 2 M R2 . The wheel rolls without slipping in a straight line in an uphill direction 37◦ above the horizontal. The wheel starts at angular speed 12.0536 rad/s but the rotation slows down as the wheel rolls uphill, and eventually the wheel comes to a stop and rolls back downhill. How far does the wheel roll in the uphill direction before it stops?
Answer:
d= 23.25 m
Explanation:
Assuming no other external forces acting on the disk, total mechanical energy must be conserved.Taking the initial height of the disk as the zero reference for the gravitational potential energy, initially. all the energy is kinetic.This kinetic energy is part translational kinetic energy, and part rotational kinetic energy, as follows:[tex]E_{o} = K_{transo} + K_{roto} (1)[/tex]
When the disk rolling uphill finally comes to an stop, its energy is completely gravitational potential energy, as follows:[tex]E_{f} = m*g*h (2)[/tex]
Since the angle with the horizontal of the track on which the disk is rolling, is 37º, we can express the height h in terms of the distance traveled d and the angle of 37º, as follows:[tex]h = d* sin 37 (3)[/tex]
Replacing (3) in (2):[tex]E_{f} = m*g* d * sin 37 (4)[/tex]
Since the wheel rolls without sleeping, this means that at any time there is a fixed relationship in the translational speed and the angular speed, as follows:[tex]v = \omega * R (5)[/tex]
For a solid disk, as mentioned in the question, the moment of inertia is just 1/2*M*R².The rotational kinetic energy of a rotating rigid body can be written as follows:[tex]K_{rot} = \frac{1}{2}* I * \omega^{2} (6)[/tex]
Replacing I from (6) and ω from (5), and remembering the definition of the translational kinetic energy, we can solve (1) in terms of v, m and r as follows:[tex]E_{o} = K_{transo} + K_{roto} = \frac{1}{2}* m* v^{2} +(\frac{1}{2}* \frac{1}{2}) *m*r^{2}*(\frac{v}{r}) ^{2} = \\ \frac{3}{4} * m * v^{2} (7)[/tex]
Since (4) and (7) must be equal each other, we can solve for d as follows:[tex]d =\frac{3}{4} * \frac{v^{2}}{g*sin37} = \frac{3}{4}*\frac{(\omega*r)^{2}}{g*sin 37} (8)[/tex]
Replacing by the values, we finally get:[tex]d =\frac{3}{4}*\frac{(\omega*r)^{2}}{g*sin 37} = \frac{3}{4} *\frac{(12.0536rad/sec*1.12m)^{2}}{9.8 m/s2*0.601} = 23. 25 m.[/tex]
11.Electricity is made at a power plant by huge
(C). Coil
(D). Wire
(A). Motor
(B). Generator
Scientists are constantly exploring the universe, looking for new planets that support life similar to the life on
Earth. A new planet that supports life would have all of the following characteristics except -
A. a gaseous atmosphere.
B. an orbiting moon.
C. liquid water.
D. protection from radiation.
A new planet that supports life would have all the following characteristics except an orbiting moon. Hence, option B is correct.
What is a Planet?An enormous, spherical celestial object known as a planet is neither a star nor its remains. The nebular hypothesis, which states how an interstellar cloud falls out of a nebula to produce a young protostar encircled by a protoplanetary disk, is now the best explanation for planet formation.
By gradually accumulating material under the influence of gravity, or accretion, planets develop in this disk.
The rocky planets Mercury, Venus, Earth, and Mars, as well as the giant planets Jupiter, Saturn, Uranus, and Neptune, make up the Solar System's minimum number of eight planets. These planets all revolve around axes that are inclined relative to their respective polar axes.
To know more about Planet:
https://brainly.com/question/14581221
#SPJ2
which of the following best defines spring constant ?
a. the amount of force needed to extend or compression og the spring.
b. the amount of force needed every 1 meter of stretch or compression of the spring.
c. the amount of energy needed to extend or compress a spring for every 1 kilogram of mass of the spring.
d. the amount of energy needed for every 1 meter of stretch or compression of the spring.
Answer:
A
Explanation:
How much work will a 500 watt motor do in 10 seconds?
Answer:
50j
Explanation:
Watts are units used to measure power. power can be defined as rate of energy transfer
500 watts means - 500 J of energy per second
in 1 second - 500 J of work is done
therefore within 10 seconds - 500 J/s x 10 s = 5000 J
work of 5000 J is carried out in 10 seconds
Answer:
Watts are units used to measure power. power can be defined as rate of energy transfer
500 watts means - 500 J of energy per second
in 1 second - 500 J of work is done
therefore within 10 seconds - 500 J/s x 10 s = 5000 J
work of 5000 J is carried out in 10 seconds
Explanation:
What does the Curl-up test assess?
O A.
Body composition
ОВ.
Muscular strength and endurance
O C. Flexibility
D.
Cardiovascular fitness
HURRRY
(it’s pe not physics)
Answer:
assesses C.) muscular endurance
Explanation:
The universe cooled after the Big Bang.At some point hydrogen atoms combined to form helium.What is this process called?
Answer:
Nuclear fusion
Explanation:
ILL GIVE BRAINLIST PLS In which circuit does charge reverse direction many times per second?
A. A DC circuit
B. A combined circuit
C. A parallel circuit D. An AC circuit
Answer: D. An AC circuit
Explanation:
I took it on a test and it was correct ; )
Bonnie has 16 coins in her pocket worth $1.50. What are two different combinations of coins she could have in her pocket?
A student must analyze data collected from an experiment in which a block of mass 2M traveling with a speed v0 collides with a block of mass M that is initially at rest. After the collision, the two blocks stick together. What applications of the equation for the conservation of momentum represent the initial and final momentum of the system for a completely inelastic collision between the blocks?
Solution :
In the question, it is given that the collision is inelastic and the blocks stick together.
In an inelastic collision, the linear momentum is conserved but the kinetic energy is not conserved.
The linear momentum is given by :
[tex]$\vec p = m \vec v$[/tex] (mass x velocity)
So according to the conservation of linear momentum,
[tex]$\vec p_{(\text{before collision})}=\vec p_{(\text{after collision})}$[/tex]
Let the velocity after the collision is [tex]$v_F$[/tex]
[tex]$m_1v_0+m_2 \times 0 = m_1v_F+m_2v_F$[/tex]
Putting the values of [tex]$m_1 \text{ and}\ m_2$[/tex]
[tex]$m_1=2M \text{ and}\ m_2=M$[/tex]
∴ [tex]$2Mv_0=2Mv_F+Mv_F$[/tex], as the blocks stick together after the collision.
and [tex]$2MV_0=3Mv_F$[/tex], as the blocks stick together after the collision.
what term is used to describe the block of organs (heart, lungs, liver, kidneys and spleen) that are removed during the autopsy?
The term used to describe the block of organs that are removed during the autopsy is Thoracic organs.
The organs that are removed during autopsy include:
Thoracic organs;Cervical organs, and Abdominal organsThe thoracic cavity contains organs and tissues that function in the respirator, cardiovascular, nervous and digestive system.
These thoracic organs include the following;
heart, lungs, liver, kidneys and spleen.Thus, we can conclude that the term used to describe the block of organs that are removed during the autopsy is Thoracic organs.
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03: A mass with a 60 g vibrate at the end of a spring. The amplitude of the motion is 0.394 ft
and a frequency is 0.59 HZ. Find the perind and spring constant, the maximum speed and
acceleration of the mass, the speed and acceleration when the displacement is 6 cm, compute the
kinetic and the potential energy when the position is 6 cm
Answer:
a) T = 1.69 s, b) k = 0.825 N / m, c) v = 1.46 feet/s, d) a = 5.41 ft / s²,
e) v = - 1,319 ft / s, a = - 2.70 ft / s², f) K = 4.8 10⁻³ J, U = 1.49 10⁻³ J
Explanation:
In a mass-spring system with simple harmonic motion, the angular velocity is
w = [tex]\sqrt{\frac{k}{m} }[/tex]
a) find the period
angular velocity, frequency, and period are related
w = 2π f = 2π / T
f = 1 / T
T = 1 / f
T = 1 / 0.59
T = 1.69 s
b) the spring constant
w = 2π f
w = 2π 0.59
w = 3.70 rad / s
w² = k / m
k = w² m
k = 3.70² 0.060
k = 0.825 N / m
c) the maximum speed
simple harmonic movement is described by the expression
x = A cos (wt + Ф)
speed is defined by
v =[tex]\frac{dx}{dt}[/tex]
v = -A w sin (wt + fi)
the speed is maximum when the cosine is ± 1
v = A w
v = 0.394 3.70
v = 1.46 feet/s
d) maximum acceleration
a = [tex]\frac{dv}{dt}[/tex]
a = - A w² cos wt + fi
the acceleration is maximum when the cosine is ±1
a = A w²
a = 0.394 3.70²
a = 5.41 ft / s²
e) velocity and acceleration for x = 6 cm
let's reduce the cm to feet
x = 6 cm (1 foot / 30.48 cm) = 0.1969 foot
Before doing this part we must find the phase angle (Ф), the most common way to start the movement is to move the spring a small distance and release it, so its initial speed is zero for t = 0 s
let's use the expression for the velocity
v = -A w sin (0 + Фi)
0 = - A w sin Ф
so sin Ф = 0 which implies that Фi = 0
the equation of motion is
x = A cos wt
x = 0.394 cos 3.70t
we substitute
0.1969 = 0.394 cos 370t
3.70 t = cos⁻¹ (0.1969 / 0.394)
let's not forget that the angle is in radians
3.70, t = 1.047
t = 1.047 / 3.70
t = 0.2826 s
we substitute this time in the equation for velocity and acceleration
v = - Aw sin wt
v = - 0.394 3.70 sin 3.70 0.2826
v = - 1,319 ft / s
a = - A w² cos wt
a = - 0.394 3.70² cos 3.70 0.2826
a = - 2.70 ft / s²
f) the kinetic and potential energy at this point
K = ½ m v²
let's slow down to the SI system
v = 1.319 ft / s (1 m / 3.28 ft) = 0.402 m / s
K = ½ 0.060 0.402²
K = 4.8 10⁻³ J
U = ½ k x²
U = ½ 0.825 0.06²
U = 1.49 10⁻³ J
Objects A and B, of mass M and 2M respectively, are each pushed a distance d straight up an inclined plane by a force F parallel to the plane. The coefficient of kinetic friction between each mass and the plane has the same value μ.k At the highest point is:______
a. KEA > KEB
b. KEA = KEB
c. KEA < KEB
d. The work done by F on A is greater than the work done by F on B.
e. The work done by F on A is less than the work done by F on B.
Answer:
The correct answer is option (A) that is KEA > KEB .
Explanation:
Let us calculate -
If the object is straighten up and inclined plane , the work done is
[tex]W=F_d- F_f_r_id-F_gh[/tex]
[tex]W=F_d-\mu_kmgdcos\theta-mgdsin\theta[/tex]
The change in kinetic energy is ,
[tex]\Delta K=\frac{1}{2}mv^2-\frac{1}{2}m\nu_0^2[/tex]
At the top of the inclined plane , the velocity is zero
So,
[tex]\Delta K=\frac{1}{2} m(0)^2-\frac{1}{2}m\nu_0^2[/tex]
[tex]\Delta KE=-\frac{1}{2}m\nu_0^2[/tex]
From the work energy theorem , we have [tex]W=-\Delta K[/tex] in case of friction , so
[tex]\frac{1}{2}m\nu_0^2=Fd-\mu_kmgdcos\theta-mgdsin\theta[/tex]
[tex]KE=Fd-\mu_kmgdcos\theta-mgdsin\theta[/tex]
For object A-
[tex]KE_A=Fd-\mu_kmgdcos\theta-mgdsin\theta[/tex]
For object B
[tex]KE_B= Fd -2\mu_kMgdcos\theta-2Mgdsin\theta[/tex]
[tex]KE_B= Fd -2(\mu_kMgdcos\theta-Mgdsin\theta)[/tex]
Thus , larger mass is going to mean less total work and a lower kinetic energy .
From the above results , we get
[tex]KE_A >KE_B[/tex]
Therefore , option A is correct .
A running Marites launched the egg she
stole as she was about to be caught with
a velocity of 25 m/s in a direction making
an angle of 20° upward with the
horizontal
a) What is the maximum height reached by
the egg?
b) What is the total flight time (between
launch and touching the ground) of the
egg?
c) What is the horizontal range (maximum
* above ground) of the egg?
d) What is the magnitude of the velocity
of the egg just before it hits the ground?
Answer:
a) y = 3.73 m, b) t = 1.74 s, c) R = 40.99 m,
d) vₓ = 23.49 m/s, v_y = -8.5 m / s
Explanation:
This is a projectile launching exercise, we start by breaking down the initial velocity
sin θ = v_{oy} / v₀
cos θ = v₀ₓ / v₀
v_{oy} = v₀ sin θ
v₀ₓ = v₀ cos θ
v_{oy} = 25 sin 20 = 8.55 m / s
v₀ₓ = 25 cos 20 = 23.49 m / s
a) when the egg reaches the maximum height its vertical speed is zero
v_y² = v_{oy}² - 2 g y
0 = v_[oy}² - 2g y
y = v_{oy}² / 2g
y = [tex]\frac{8.55^2}{2 \ 9.8 }[/tex]
y = 3.73 m
b) flight time
y = v_{oy} t - ½ g t²
the time of flight occurs when the body reaches the ground y = 0
0 = (v_{oy} - ½ g t) t
The results are
t₁ = 0s this time is for using the body star
v_{oy} - ½ g t = 0
t = [tex]\frac{2v_{oy}^2}{g}[/tex]
t = 2 8.55 / 9.8
t = 1.74 s
c) the range
R = v₀² sin 2θ / g
R = 25² sin (2 20) / 9.8
R = 40.99 m
d) speed at the point of arrival
horizontal speed is constant
vₓ = v₀ₓ = 23.49 m/s
vertical speed is
v_y = Iv_{oy} - g t
v_y = 8.55 - 9.8 1.74
v_y = -8.5 m / s
Which type of wires are ferromagnetic metals?
cooper
aluminum
string
Swordfish are capable of stunning output power for short bursts. A 650 kg swordfish has a cross-sectional area of 0.92 m2 and a drag coefficient of 0.0091- very low due to some evolutionary adaptations. Such a fish can sustain a speed of 30 m/s for a few seconds. Assume seawater has a density of 1026 kg/m3. a) How much power does the fish need to put out for motion at this high speed
Answer:
the required or need power is 115960.57 Watts
Explanation:
First of all, we take down the data we can find from the question, to make it easier when substituting values into formulas.
mass of swordfish m = 650 kg
Cross - sectional Area A = 0.92 m²
drag coefficient C[tex]_D[/tex] = 0.0091
speed v = 30 m/s
density p = 1026 kg/m³
Now, we determine our Drag force F[tex]_D[/tex]
Drag force F[tex]_D[/tex] = [tex]\frac{1}{2}[/tex] × C[tex]_D[/tex] × A × p × v²
Next, we substitute the values we have taken down, into the formula.
Drag force F[tex]_D[/tex] = [tex]\frac{1}{2}[/tex] × 0.0091 × 0.92 × 1026 × (30)²
Drag force F[tex]_D[/tex] = 4.294836 × 900
Drag force F[tex]_D[/tex] = 3865.3524
Now, we determine the power needed P[tex]_w[/tex]
P[tex]_w[/tex] = F[tex]_D[/tex] × v
we substitute
P[tex]_w[/tex] = 3865.3524 × 30
P[tex]_w[/tex] = 115960.57 Watts
Therefore, the required or need power is 115960.57 Watts
The nucleus of a certain type of uranium atom contains
92 protons and 143 neutrons. What is the total charge of
the nucleus?
Answer:
charge = electrons + protons
=92+92
=184
What is the magnitude of the gravitational force acting on a
1.0 kg object which is 1.0 m from another 1.0 kg object?
Ans[tex]^{}[/tex]wer and expl[tex]^{}[/tex]anation is in a fi[tex]^{}[/tex]le. Li[tex]^{}[/tex]nk below! Go[tex]^{}[/tex]od luck!
bit.[tex]^{}[/tex]ly/3a8Nt8n
Which statement best explains why objects are pulled toward Earth’s center?
Answer:
Earth has a much greater mass than objects on its surface
An object is placed 12.0 cm from a thin diverging lens with a focal length of 4 cm. Which one of the
following statements is true concerning the image?
A. The image is virtual and 3.0 cm from the lens.
B. The image is real and 6.0 cm from the lens.
C. The image is virtual and 12 cm from the lens.
D. The image is real and 12 cm from the lens.
Answer:
soluble soluble soluble soluble
Explanation:
solublesolublesolublesolublesolublesolublesoluble dguhjjewugbcsbdc csyuhjci
How much energy must the brakes absorb to bring a 1200kg car from 30m/s to 15 m/s?
Careful measurements reveal that a star maintains a steady apparent brightness at most times except that at precise intervals of 127 hours the star becomes dimmer for about 4 hours. The most likely explanation is that Careful measurements reveal that a star maintains a steady apparent brightness at most times except that at precise intervals of 127 hours the star becomes dimmer for about 4 hours. The most likely explanation is that:________
a. the star is a white dwarf.
b. the star is periodically ejecting gas into space, every 127 hours.
c. the star is a Cepheid variable.
d. the star is a member of an eclipsing binary star system.
Answer:
d. the star is a member and also a part of an eclipsing binary star system.
Explanation:
If any star happens to be brighter for an extended period of time, however, at some times, it becomes dimmer, is due to the fact that the star is being overshadowed (hiding behind another star that is known as eclipse).
The above-mentioned eclipsing binary star system is essentially what has been defined. It occurs when two stars' orbit planes are so similar that one star will obscure (the light) of the other.
Thus, option D is correct.
