Answer:
Being silent most of the time is a good virtue under certain circumstances and environment. It is always advisable to remain quite silent and not be too quick to respond to situations or issues so as to avoid making and saying wrong words.
The ethics in a workplace involves communicating with others with less amount of talking as possible and more of body languages and signs. This is because the workplace is meant to be a serene place.
If the velocity of a runner changes from -2 m/s to -4 m/s over a period of time, the
runner's kinetic energy will become:
(a) four times as great as it was.
(b) half the magnitude it was.
(c) energy is conserved.
(d) twice as great as it was.
(e) four times less than it was.
Answer:
It will be A. So since its 2 times more the kinetic energy. But then you have to square it 2^2 = 4
The average, year-after-year conditions of temperature, precipitation, winds, and cloud in an area are known as its
A.climate.
b.weather.
C. global warming
d. seasons
Answer:
a. global warming
Explanation:
that's the definitain of global warming
Answer:
A climate
Explanation:
World religions: Shinto
Most Shinto rituals are tied to
A) worshiping the kami.
B) the life-cycle of humans and the seasonal cycles of nature.
C) forgiveness of sins.
D) preparing for the afterlife.
Scenario 2: Use the following information to answer questions 3 and 4:
Your client, Jim, is interested in weight control. He weighs 75kg.
3. If Jim walks 3.3 mph (0% grade), how long must he walk to expend 300 kcal total?
A. 52 min
B. 42 min
C. 65 min
D. 99 min
4. If Jim exercises at an intensity of 6 kcal/min, what is the leg ergometer work rate?
A. 47 watts
B. 90 watts
C. 61 watts
D. 71 watts
Answer:
A. 52 min
.A. 47 watts
Explanation:
Given that;
jim weighs 75 kg
and he walks 3.3 mph; the objective here is to determine how long must he walk to expend 300 kcal.
Using the following relation to determine the amount of calories burned per minute while walking; we have:
[tex]\dfrac{MET*weight (kg)*3.5}{200}[/tex]
here;
MET = energy cost of a physical activity for a period of time
Obtaining the data for walking with a speed of 3.3 mph From the standard chart for MET, At 3.3 mph; we have our desired value to be 4.3
However;
the calories burned in a minute = [tex]\dfrac{4.3*75 (kg)*3.5}{200}[/tex]
= 5.644
Therefore, for walking for 52 mins; Jim burns approximately 293.475 kcal which is nearest to 300 kcal.
4.
Given that:
mass m = 75 kg
intensity = 6 kcal/min
The eg ergometer work rate = ??
Applying the formula:
[tex]V_O_2 ( intensity ) = ( \dfrac{W}{m}*1.8)+7[/tex]
where ;
[tex]V_O_2 ( intensity ) = \dfrac{1 \ kcal min^{-1}*10^{-3}}{5}[/tex]
[tex]V_O_2 ( intensity ) = \dfrac{6*1 \ kcal min^{-1}*10^{-3}}{5}[/tex]
[tex]V_O_2 ( intensity ) = 0.0012[/tex]
∴[tex]0.0012 = (\dfrac{W}{75}*1.8)+7 \\ \\ W = \dfrac{0.0012-7}{1.8}*75 \\ \\ W = \dfrac{7*75}{1.8} \\ \\ W = 291.66 \ kg m /min[/tex]
Converting to watts;
Since; 6.118kg-m/min is = 1 watt
Then 291.66 kgm /min will be equal to 47.67 watts
≅ 47 watts
Q) A particle in simple harmonic motion starts its motion from its mean position. If T be the time period, calculate the ratio of kinetic energy and potential energy of the particle at the instant when t = T/12.
t\12 and the parties are spreading ever
Explanation:
my point is that you can get sick if
you sont wash your ha
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save
Davina accelerates a box across a smooth frictionless horizontal surface over a displacement of 18.0 m with a constant 25.0 N force angled at 23.0° below the horizontal. How much work does she do on the box? A. 176 J B. 414 J C. 450 J D. 511 J Group of answer choices
Answer:
W = 414 J, correct is B
Explanation:
Work is defined by
W = ∫ F .dx
where F is the force, x is the displacement and the point represents the dot product
this expression can also be written with the explicit scalar product
W = ∫ F dx cos θ
where is the angle between force and displacement
for this case as the force is constant
W = F x cos θ
calculate
W = 25.0 18.0 cos (-23)
W = 414 J
the correct answer is B
Representar con una escala de 1cm = 10N dos fuerzas que tengan igual dirección, distinto sentido y sus intensidades son de 40n y 60n, respectivamente.
Alguien que me lo hagaaaaaaa
Answer:
To solve this problem we just need to graph two forces with same direction, pointing to different sides with intensities of 40 N and 60 N.
The image attached shows these forces.
Notice that the vectors are parallel, that's because they have the same direction, but they point to different sides, and their magnitudes have a difference of 20 N.
N capacitors are connected in parallel to form a "capacitor circuit". The capacitance of first capacitor is C, second one is C/2 and third one is C/4, forth one is C/8 and so on. Namely, capacitance of a capacitor is one-half of the previous one. What is the equivalent capacitance of this parallel combination when N goes to inifinity?
Answer:
2C
Explanation:
The equivalent capacitance of a parallel combination of capacitors is the sum of their capacitance.
So, if the capacitance of each capacitor is half the previous one, we have a geometric series with first term = C and rate = 0.5.
Using the formula for the sum of the infinite terms of a geometric series, we have:
Sum = First term / (1 - rate)
Sum = C / (1 - 0.5)
Sum = C / 0.5 = 2C
So the equivalent capacitance of this parallel connection is 2C.
A ball is thrown upward from the ground with an initial speed of 19.2 m/s; at the same instant, another ball is dropped from a building 18 m high. After how long will the balls be at the same height above the ground?
Answer:
0.938 seconds
Explanation:
For the ball thrown upwards, we use the formula below to solve it:
[tex]s = ut - \frac{1}{2}gt^2[/tex]
where s = distance moved
u = initial speed = 19.2 m/s
t = time taken
g = acceleration due to gravity = 9.8 [tex]m/s^2[/tex]
Let x be the height at which both balls are level, this means that:
=> [tex]x = 19.2t - 4.9t^2[/tex]________(1)
For the ball dropped downwards, we use the formula below:
[tex]s = ut + \frac{1}{2}gt^2[/tex]
u = 0 m/s
At the point where both balls are level:
s = 18 - x
=> [tex]18 - x = 0 + 4.9t^2[/tex]
=> [tex]x = 18 - 4.9t^2[/tex]__________(2)
Equating both (1) and (2):
[tex]19.2t - 4.9t^2 = 18 - 4.9t^2\\\\=> 19.2t = 18\\\\t = 18/19.2 = 0.938 secs[/tex]
They will be level after 0.938 seconds
How many ohms of resistance are in a 120–volt hair dryer that draws 7.6 amps of current?
From Ohm's law . . . Resistance = (voltage) / (current)
Resistance = (120 volts) / (7.6 Amperes)
Resistance = 15.8 Ω
The froghopper, a tiny insect, is a remarkable jumper. Suppose a colony of the little critters is raised on Rhea, a moon of Saturn, where the acceleration due to gravity is only 0.264 m/s2 , whereas gravity on Earth is =9.81 m/s2 . If on Earth a froghopper's maximum jump height is ℎ and its maximum horizontal jump range is R, what would its maximum jump height and range be on Rhea in terms of ℎ and R? Assume the froghopper's takeoff velocity is the same on Rhea and Earth.
Answer:
Maximum height of jump on Rhea is 37.16 times of that on Earth, i.e 37.16h
Maximum range of jump on Rhea is 37.16 of times that on Earth, i.e 37.16R
Explanation:
The acceleration due to gravity on Rhea = 0.264 m/s^2
Acceleration due to gravity on earth here = 9.81 m/s^2
this means that the acceleration due to gravity g on earth is 9.81/0.264 = 37.16 times that on Rhea.
maximum height that can be achieved by the froghopper is given by the equation;
h = [tex]\frac{u^{2}sin^{2} \alpha}{2g}[/tex]
let us put all the numerator of the equation as k, since the velocity of take off is the same for Earth and Rhea. The equation is simplified to
h = [tex]\frac{k}{2g}[/tex]
for earth,
h = [tex]\frac{k}{2*9.81}[/tex] = [tex]\frac{k}{19.62}[/tex]
for Rhea,
h = [tex]\frac{k}{2*0.264}[/tex] = [tex]\frac{k}{0.528}[/tex]
therefore,
h on Rhea is [tex]\frac{k}{0.528}[/tex] ÷ [tex]\frac{k}{19.62}[/tex] = 37.16 times of that on Earth, i.e 37.16h
Equation for range R is given as
R = [tex]\frac{u^{2}sin 2\alpha}{g}[/tex]
following the same approach as before,
R on Rhea will be [tex]\frac{k}{0.264}[/tex] ÷ [tex]\frac{k}{9.81}[/tex] = 37.16 of times that on Earth, i.e 37.16R
To study the properties of various particles, you can accelerate the particles with electric fields. A positron is a particle with the same mass as an electron but the opposite charge ( e). If a positron is accelerated by a constant electric field of magnitude 286 N/C, find the following.
(a) Find the acceleration of the positron. m/s2
(b) Find the positron's speed after 8.70 × 10-9 s. Assume that the positron started from rest. m/s
Answer:
a) a = 5.03x10¹³ m/s²
b) [tex]V_{f} = 4.4 \cdot 10^{5} m/s [/tex]
Explanation:
a) The acceleration of the positron can be found as follows:
[tex] F = q*E [/tex] (1)
Also,
[tex] F = ma [/tex] (2)
By entering equation (1) into (2), we have:
[tex] a = \frac{F}{m} = \frac{qE}{m} [/tex]
Where:
F: is the electric force
m: is the particle's mass = 9.1x10⁻³¹ kg
q: is the charge of the positron = 1.6x10⁻¹⁹ C
E: is the electric field = 286 N/C
[tex] a = \frac{qE}{m} = \frac{1.6 \cdot 10^{-19} C*286 N/C}{9.1 \cdot 10^{-31} kg} = 5.03 \cdot 10^{13} m/s^{2} [/tex]
b) The positron's speed can be calculated using the following equation:
[tex] V_{f} = V_{0} + at [/tex]
Where:
[tex]V_{f}[/tex]: is the final speed =?
[tex]V_{0}[/tex]: is the initial speed =0
t: is the time = 8.70x10⁻⁹ s
[tex] V_{f} = V_{0} + at = 0 + 5.03 \cdot 10^{13} m/s^{2}*8.70 \cdot 10^{-9} s = 4.4 \cdot 10^{5} m/s [/tex]
I hope it helps you!
A metal ring 4.60 cm in diameter is placed between the north and south poles of large magnets with the plane of its area perpendicular to the magnetic field. These magnets produce an initial uniform field of 1.12 T between them but are gradually pulled apart, causing this field to remain uniform but decrease steadily at 0.280 T/s.
A. What is the magnitude of the electric field induced in the ring?
B. In which direction (clockwise or counterclockwise) does the current flow as viewed by someone on the south pole of the magnet?1. Counterclockwise2. Clockwise
Answer:
A. Ein = 8.05*10^-4 V/m
B. Clockwise sense
Explanation:
A. the magnitude of the electric field induced in the ring is obtaind by using the following formula:
[tex]\int E_{in} \cdot ds=-\frac{d\Phi_B}{dt}[/tex] (1)
Ein: induced electric field
ds: differential of a path of the ring
ФB: magnetic flux in the ring
The Ein vector is parallel to ds in the complete ring. Furthermore, the area of the ring is constant, hence, you have in the equation (1):
[tex]\int E_{in}ds=E_{in}(2\pi r)=-A\frac{dB}{dt}\\\\E_{in}=-\frac{A}{2\pi r}\frac{dB}{dt}[/tex] (2)
dB/dt = -0.280T/s (it is decreasing)
A: area of the ring = π(r/2)^2= (π/4) r^2
r: radius of the ring = 4.60/2 = 2.30 cm
Then, you replace the values of all variables in the equation (2):
[tex]E_{in}=-\frac{(\pi/4)r^2}{2\pi r}\frac{dB}{dt}=\frac{r}{8}\frac{dB}{dt}\\\\E_{in}=-\frac{0.0230m}{8}(-0.280T)=8.05*10^{-4}\frac{V}{m}[/tex]
hence, the induced electric field is 8.05*10^-4 V/m
B. The induced current in the ring produced a magnetic field that is opposite to the magnetic field of the magnet. The, in this case you have that the induced current is in a clockwise sense.
assuming 100% efficient energy conversion how much water stored behind a 50 centimeter high hydroelectric dam would be required to charged the battery
Answer:
The amount of water that will power a battery with that rating = 7.35 m³
Explanation:
The power rating for the battery is missing from the question.
Complete Question
Assuming 100% efficient energy conversion how much water stored behind a 50 centimeter high hydroelectric dam would be required to charged the battery with power rating, 12 V, 50 Ampere-minutes
Solution
Potential energy possessed by water at that height = mgH
m = mass of the water = ρV
ρ = density of water = 1000 kg/m³
V = volume of water = ?
g = acceleration due to gravity = 9.8 m/s²
H = height of water = 50 cm = 0.5 m
Potential energy = ρVgH = 1000 × V × 9.8 × 0.5 = (4900V) J
Energy of the battery = qV
q = 50 A.h = 50 × 60 = 3,000 C
V = 12 V
qV = 3,000 × 12 = 36,000 J
Energy = 36,000 J
At a 100% conversion rate, the energy of the water totally powers the battery
(4900V) = (36,000)
4900V = 36,000
V = (36,000/4900)
V = 7.35 m³
Hope this Helps!!!
A man pushes a 25kg box up an incline 2.0m by applying a steady force of 95N parallel to the incline. The box moves up the incline at a steady speed. The incline makes an angle 15 degrees to the horizontal
a) What is the force of friction between the incline and the box
b)If the box is released at the top of the incline, what will its speed be at the bottom
Answer:
a) Ff = 19.29 N
b) v = 3.00 m/s
Explanation:
a) To calculate the friction force you use the second Newton Law in the incline plane, with an acceleration equal to zero, because the motion of the box has a constant velocity:
[tex]F-F_f-Wsin(\theta)=0\\\\[/tex] (1)
F: force applied by the man = 95N
Ff: friction force
W: weight of the box = Mg = (25kg)(9.8m/s^2) = 245N
θ: degree of the inclined plane = 15°
You solve the equation (1) for Ff and you replace the values of all variables in the equation (1):
[tex]F_f=-Wsin(\theta)+F\\\\F_f=-(245N)sin18\°+95N=19.29N[/tex]
b) To fins the velocity of the box at the bottom you use the following formula:
[tex]W_N=\Delta K[/tex] (2)
That is, the net work over the box is equal to the change in the kinetic energy of the box.
The net work is:
[tex]W_N=Mgsin(18\°)d-Ffd[/tex]
d: distance traveled by the box = 2.0m
[tex]W_N=245sin18\°(2.0m)N-19.29(2.0m)N=112.83J[/tex]
You use this value of the net work to find the final velocity of the box, by using the equation (2):
[tex]112.8J=\frac{1}{2}m[v^2-v_o^2]\\\\v_o=0m/s\\\\v=\sqrt{\frac{2(112.8J)}{m}}=\sqrt{\frac{225.67J}{25kg}}=3.00\frac{m}{s}[/tex]
The speed of the box, at the bottom of the incline plane is 3.00 m/s
a vector has components x=6 m and y=8 m. what is its magnitude and direction?
Answer: 10m
Explanation:
The magnitude of the vector would be 10
[tex]\sqrt{6^{2}+8^{2} } =10[/tex]
Refer to a situation where you exert a force F on a crate of mass M, moving it at a speed v a distance d across a floor in a time interval t. The quantity F d/t is?
a.) kinetic energy of the crate
b.) potential energy of the crate
c.) linear momentum of the crate
d.) work you do on the crate
e.) power you supply to the crate
Answer:
e.) power you supply to the crate
Explanation:
According to given data, we have:
F = Force exerted on the crate
M = Mass of the crate
v = Speed of motion of the crate
d = Distance traveled by the crate across the floor
t = Time interval passed
Now, we try to analyze the given quantity:
=> F d/t
=> (Force)(Displacement)/(Time)
but, (Force)(Displacement) = Work Done
Therefore,
=> Work Done/Time
but, Work Done/Time = Power
Therefore,
=> Power
Hence, the quantity F d/t is:
e.) power you supply to the crate
Block 1, of mass m1 = 2.50 kg , moves along a frictionless air track with speed v1 = 27.0 m/s. It collides with block 2, of mass m2 = 33.0 kg , which was initially at rest. The blocks stick together after the collision.A. Find the magnitude pi of the total initial momentum of the two-block system. Express your answer numerically.B. Find vf, the magnitude of the final velocity of the two-block system. Express your answer numerically.C. what is the change deltaK= Kfinal- K initial in the two block systems kinetic energy due to the collision ? Express your answer numerically in joules.
Answer:
a
The total initial momentum of the two-block system is [tex]p_t = 67.5 \ kg \cdot m/s^2[/tex]
b
The magnitude of the final velocity of the two-block system [tex]v_f = 1.9014 \ m/s[/tex]
c
the change ΔK=Kfinal−Kinitial in the two-block system's kinetic energy due to the collision is
[tex]\Delta KE =- 847.08 \ J[/tex]
Explanation:
From the question we are told that
The mass of first block is [tex]m_1 = 2.50 \ kg[/tex]
The initial velocity of first block is [tex]u_1 = 27.0 \ m/s[/tex]
The mass of second block is [tex]m_2 = 33.0\ kg[/tex]
initial velocity of second block is [tex]u_2 = 0 \ m/s[/tex]
The magnitude of the of the total initial momentum of the two-block system is mathematically repented as
[tex]p_i = (m_1 * u_1 ) + (m_2 * u_2)[/tex]
substituting values
[tex]p_i = (2.50* 27 ) + (33 * 0)[/tex]
[tex]p_t = 67.5 \ kg \cdot m/s^2[/tex]
According to the law of linear momentum conservation
[tex]p_i = p_f[/tex]
Where [tex]p_f[/tex] is the total final momentum of the system which is mathematically represented as
[tex]p_f = (m_+m_2) * v_f[/tex]
Where [tex]v_f[/tex] is the final velocity of the system
[tex]p_i = (m_1 +m_2 ) v_f[/tex]
substituting values
[tex]67.5 = (2.50+33 ) v_f[/tex]
[tex]v_f = 1.9014 \ m/s[/tex]
The change in kinetic energy is mathematically represented as
[tex]\Delta KE = KE_f -KE_i[/tex]
Where [tex]KE_f[/tex] is the final kinetic energy of the two-body system which is mathematically represented as
[tex]KE_f = \frac{1}{2} (m_1 +m_2) * v_f^2[/tex]
substituting values
[tex]KE_f = \frac{1}{2} (2.50 +33) * (1.9014)^2[/tex]
[tex]KE_f =64.17 J[/tex]
While [tex]KE_i[/tex] is the initial kinetic energy of the two-body system
[tex]KE_i = \frac{1}{2} * m_1 * u_1^2[/tex]
substituting values
[tex]KE_i = \frac{1}{2} * 2.5 * 27^2[/tex]
[tex]KE_i = 911.25 \ J[/tex]
So
[tex]\Delta KE = 64.17 -911.25[/tex]
[tex]\Delta KE =- 847.08 \ J[/tex]
man stands on a platform that is rotating (without friction) with an angular speed of 1.2 rev/s; his arms are outstretched and he holds a brick in each hand.The rotational inertia of the system consisting of the man, bricks, and platform about the central vertical axis of the platform is 6.0 k g times m squared. If by moving the bricks the man decreases the rotational inertia of the system to 2.0 k g times m squared, what is the resulting angular speed of the platform in rad/s? Express to 3 sig figs.
Answer:
w₂ = 22.6 rad/s
Explanation:
This exercise the system is formed by platform, man and bricks; For this system, when the bricks are released, the forces are internal, so the kinetic moment is conserved.
Let's write the moment two moments
initial instant. Before releasing bricks
L₀ = I₁ w₁
final moment. After releasing the bricks
[tex]L_{f}[/tex] = I₂W₂
L₀ = L_{f}
I₁ w₁ = I₂ w₂
w₂ = I₁ / I₂ w₁
let's reduce the data to the SI system
w₁ = 1.2 rev / s (2π rad / 1rev) = 7.54 rad / s
let's calculate
w₂ = 6.0/2.0 7.54
w₂ = 22.6 rad/s
Calculate the maximum deceleration (in m/s2) of a car that is heading down a 14° slope (one that makes an angle of 14° with the horizontal) under the following road conditions. You may assume that the weight of the car is evenly distributed on all four tires and that the static coefficient of friction is involved—that is, the tires are not allowed to slip during the deceleration.
The question is incomplete. Here is the complete question.
Calculate the maximum deceleration of a car that is heading down a 14° slope (one that makes an anlge of 14° with the horizontal) under the following road conditions. You may assum that the weight of the car is evenlydistributed on all four tires and that the sttic coefficient of friction is involved - that is, the tires are not allowed to slip during the deceleration. (Ignore rolling) Calculate for a car: (a) On a dry concrete. (b) On a wet concrete. (c) On ice, assuming that μs = 0.100, the same as for shoes on ice.
Answer: (a) a = - 11.05 m/s²; (b) a = - 10.64 m/s²; (c) a = - 9.84m/s²
Explanation: The image in the attachment describe the forces acting on the car. Observing that, we know that:
[tex]F_{net}[/tex] = - [tex]W_x[/tex] - [tex]f_s[/tex]
The [tex]W_x[/tex] is a x-component of force due to gravity (W) and, in this case, is given by: [tex]W_x[/tex] = W.sin(14)
W is described as: W = m.g
Force due to friction ([tex]f_s[/tex]) is given by: [tex]f_s[/tex] = μs.N
N is the normal force and, in the system, is equivalent of [tex]W_y[/tex], so:
[tex]W_y[/tex] = m.g.cos(14)
Therefore, the formula will be:
[tex]F_{net}[/tex] = - [tex]W_x[/tex] - [tex]f_s[/tex]
m.a = - (m.g.sin14) - (μs.mg.cos14)
a = - g (sin14 + μscos 14)
a) For dry concrete, μs = 1:
a = - g (sin14 + μscos 14)
a = - 9.8 (sin14 + 1.cos14)
a = - 11.05 m/s²
b) For wet concrete, μs = 0.7:
a = - g (sin14 + μscos 14)
a = - 9.8 (sin 14 + 0.7.cos14)
a = - 10.64 m/s²
c) For ice, μs = 0.1:
a = - g (sin14 + μscos 14)
a = - 9.8 (sin14 + 0.1cos14)
a = - 9.84 m/s²
Some cats can be trained to jump from one location to another and perform other tricks. Kit the cat is going to jump through a hoop. He begins on a wicker cabinet at a height of 1.765 m above the floor and jumps through the center of a vertical hoop, reaching a peak height 3.130 m above the floor. (Assume the center of the hoop is at the peak height of the jump. Assume that +x axis is in the direction of the hoop from the cabinet and +y axis is up. Assume g = 9.81 m/s2.)
(a) With what initial velocity did Kit leave the cabinet if the hoop is at a horizontal distance of 1.560 m from the cabinet?
v_0 = m/s
(b) If Kit lands on a bed at a horizontal distance of 3.582 m from the cabinet, how high above the ground is the bed?
m
Answer:
a. the initial velocity of the cat is 5.95 m/s at 60.2° from the horizontal
b. 0.847 m
Explanation:
a. Using v² = u² + 2as, we find the initial vertical velocity of the cat. Now at the peak height, v = final velocity = 0, u = initial velocity and a = -g = 9.8 m/s², s vertical distance travelled by the cat from its position on the cabinet = Δy = 3.130 m - 1.765 m = 1.365 m.
Substituting these variables into the equation, we have
0² = u² + 2(-9.8m/s²) × 1.365 m
-u² = -26.754 m²/s²
u = √26.754 m²/s²
u = 5.17 m/s
To find its initial horizontal velocity, u₁ we first find the time t it takes to reach the peak height from
v = u + at. where the variables mean the same as above.
substituting the values, we have
0 = 5.17 m/s +(-9.8m/s²)t
-5.17 m/s = -9.8m/s²t
t = -5.17 m/s ÷ (-9.8m/s²)
= 0.53 s
Now, the horizontal distance d = u₁t = 1.560 m
u₁ = d/t = 1.560 m/0.53 s = 2.96 m/s
So, the initial velocity of the cat is V = √(u² + u₁²)
= √((5.17 m/s)² + (2.96 m/s)²)
= √(26.729(m/s)² + 8.762(m/s)²)
= √(35.491 (m/s)²)
= 5.95 m/s
its direction θ = tan⁻¹(5.17 m/s ÷ 2.96 m/s) = 60.2°
So, the initial velocity of the cat is 5.95 m/s at 60.2° from the horizontal
(b)
First, we find the time t' it takes the cat to land on the bed from d' = u₁t'
where d' = horizontal distance of cabinet from bed = 3.582 m
u₁ = horizontal velocity = 2.96 m/s
t' = d'/u₁
= 3.582 m/2.96 m/s
= 1.21 s
The vertical between the bed and cabinet which is the vertical distance moved by the cat is gotten from Δy = ut' +1/2at'²
substituting u = initial vertical velocity = 5.17 m/s, t' = 1.21 s and a = -g = -9.8 m/s² into Δy, we have
Δy = ut' +1/2at'² = 5.17 m/s × 1.21 s +1/2(- 9.8 m/s²) × (1.21 s)² = 6.256 - 7.174 = -0.918 m
Δy = y₂ - y₁
Since our initial position is the position of the cabinet above the ground = y₁ = 1.765 m
y₂ = position of bed above ground.
Δy = y₂ - y₁ = -0.918 m
y₂ - 1.765 m = -0.918 m
y₂ = 1.765 m - 0.918 m
= 0.847 m
The Gulf Stream off the east coast of the United States can flow at a rapid 3.8 m/s to the north. A ship in this current has a cruising speed of 8.0 m/s . The captain would like to reach land at a point due west from the current position.
At this heading, what is the ship's speed with respect to land?
Answer:
61.6° west of South
Explanation:
The ship goes to the south at an equal rate just like water flows to the north. Thus, the velocities would balance making the ship move towards the west.
Since we're dealing with water, the ship goes 3.8 m / s to the South, but a lot still remains to the west. Finding this would require us drawing a triangle. 3.8 m/s point down side and the hypotenuse is 8
cos(θ) = [adjacent/hypotenuse]
Cos θ = 3.8/8
Cos θ = 0.475
θ = cos^-1 (0.475)
θ = 61.6°
Therefore the angle is 61.6° west of South.
On a brisk walk, a person burns about 331 Cal/h. If the brisk walk were done at 3.0 mi/h, how far would a person have to walk
to burn off 1 lb of body fat? (A pound of body fat stores an amount of chemical energy equivalent to 3,500 Cal.)
mi?
Answer:
32mi
Explanation:
If 1lb contains 3,500 Cal
It means the number of hours required to burn 3500cal would be;
3500/331 = 10.57hours
But a brisk walk is 3.0 mi/h,
It means a distance of 3.0 × 10.57 mi would be covered = 31.71 miles
32miles{ approximated to the nearest whole}
Note Distance = speed × time
A tuba may be treated like a tube closed at one end. If a tuba has a fundamental frequency of 40.4 Hz, determine the first three overtones. Use 343 m/s as the speed of sound in air.
If the speed of sound is 337 m/s, determine the length of an open tube (open at both ends) that has a fundamental frequency of 233 Hz and a first overtone frequency of 466 Hz.
Answer:
Explanation:
fundamental frequency at closed pipe = 40.4 Hz
overtones are odd harmonics in closed pipe
first three overtones are
3 x 40.4 , 5 x 40.4 , 7 x 40.4 Hz
= 121.2 Hz , 202 Hz , 282.8 Hz .
speed of sound given is 337 , fundamental frequency is 233 Hz
wavelength = velocity of sound / frequency
= 337 / 233
= 1.446 m
for fundamental note in open pipe
wavelength /2 = length of tube
length of tube = 1.446 / 2
= .723 m
= 72.30 cm .
first overtone will be two times the fundamental ie 466. In open pipe all the harmonics are found , ie both odd and even .
Richard is driving home to visit his parents. 150 mi of the trip are on the interstate highway where the speed limit is 65 mph . Normally Richard drives at the speed limit, but today he is running late and decides to take his chances by driving at 80 mph. How many minutes does he save?
Answer:
t = 25.5 min
Explanation:
To know how many minutes does Richard save, you first calculate the time that Richard takes with both velocities v1 = 65mph and v2 = 80mph.
[tex]t_1=\frac{x}{v_1}=\frac{150mi}{65mph}=2.30h\\\\t_2=\frac{x}{v_2}=\frac{150mi}{80mph}=1.875h[/tex]
Next, you calculate the difference between both times t1 and t2:
[tex]\Delta t=t_1-t_2=2.30h-1.875h=0.425h[/tex]
This is the time that Richard saves when he drives with a speed of 80mph. Finally, you convert the result to minutes:
[tex]0.425h*\frac{60min}{1h}=25.5min=25\ min\ \ 30 s[/tex]
hence, Richard saves 25.5 min (25 min and 30 s) when he drives with a speed of 80mph
The velocity of an object is given by the expression v (t) = 3.00 m / s + (2.00 m / s ^ 3) t ^ 2. Determine the position of the object as a function of time if it is located at x = 1.00 m at time t = 0.00 s.
Answer: [tex]x=\frac{2}{3}t^3+3t+1[/tex]
Explanation:
Given
velocity of object is given by
[tex]v(t)=3+2t^2[/tex]
and we know change of position w.r.t time is velocity
[tex]\Rightarrow \dfrac{dx}{dt}=v[/tex]
[tex]\Rightarrow \dfrac{dx}{dt}=3+2t^2[/tex]
[tex]\Rightarrow dx=(3+2t^2)dt[/tex]
Integrating both sides we get
[tex]\Rightarrow \int_{1}^{x}dx=\int_{0}^{t}(3+2t^2)dt[/tex]
[tex]\Rightarrow x\mid _{1}^{x}=(3t+\frac{2}{3}t^3)\mid _{0}^{t}[/tex]
[tex]\Rightarrow x-1=3(t-0)+\frac{2}{3}(t^3-0)[/tex]
[tex]\Rightarrow x=\frac{2}{3}t^3+3t+1[/tex]
A large box containing your new computer sits on the bed of your pickup truck. You are stopped at a red light. When the light turns green, you stomp on the gas and the truck accelerates. To your horror, the box starts to slide toward the back of the truck. Draw clearly labeled free-body diagrams for the truck and for the box. Indicate pairs of forces, if any, that are third-law action–reaction pairs. (The horizontal truck bed is not frictionless.)
Answer:
The description of that same situation has been listed throughout the explanation segment below.
Explanation:
When another huge box or container containing your new machine or device sits on someone's pick-up truck's bed, the third low portion of the operation response force. This same friction force of the box mostly on the truck bed as well as the friction force including its truck bed on either the box from either the immune response pair.So that the above seems to be the right answer.
What percent of our solar system's mass is in the sun?
Answer:
99.8
Explanation:
most massive the sun is at the center of the universe
The first antiparticle, the positron or antielectron, was discovered in 1932. It had been predicted by Paul Dirac in 1928, though the nature of the prediction was not fully understood until the experimental discovery. Today, it is well accepted that all fundamental particles have antiparticles.
Suppose that an electron and a positron collide head-on. Both have kinetic energy of 3.58 MeV and rest energy of 0.511 MeV. They produce two photons, which by conservation of momentum must have equal energy and move in opposite directions. What is the energy Eloton of one of these photons?
Answer:
4.09 MeV
Explanation:
Find the given attachment
What is the minimum frequency with which a 200-turn, flat coil of cross sectional area 300 cm2 can be rotated in a uniform 30-mT magnetic field if the maximum value of the induced emf is to equal 8.0 V
Answer:
The minimum frequency of the coil is 7.1 Hz
Explanation:
Given;
number of turns, N = 200 turns
cross sectional area, A = 300 cm² = 300 x 10⁻⁴ m²
magnitude of magnetic field strength, B = 30 x 10⁻³ T
maximum value of the induced emf, E = 8 V
Maximum induced emf is given as;
E = NBAω
where
ω is angular velocity (ω = 2πf)
E = NBA2πf
where;
f is the minimum frequency, measured in hertz (Hz)
f = E / (NBA2π)
f = 8 / (200 x 30 x 10⁻³ x 300 x 10⁻⁴ x 2 x 3.142)
f = 7.073 Hz
f = 7.1 Hz
Therefore, the minimum frequency of the coil is 7.1 Hz
The minimum frequency of the coil in the case when it should be rotated in a uniform 30-mT magnetic field is 7.1 Hz.
Calculation of the minimum frequency:Since
number of turns, N = 200 turns
cross-sectional area, A = 300 cm² = 300 x 10⁻⁴ m²
the magnitude of magnetic field strength, B = 30 x 10⁻³ T
the maximum value of the induced emf, E = 8 V
Now
Maximum induced emf should be
E = NBAω
here,
ω is angular velocity (ω = 2πf)
Now
E = NBA2πf
here,
f is the minimum frequency
So,
f = E / (NBA2π)
f = 8 / (200 x 30 x 10⁻³ x 300 x 10⁻⁴ x 2 x 3.142)
f = 7.073 Hz
f = 7.1 Hz
Therefore, the minimum frequency of the coil is 7.1 Hz.
Learn more about frequency here: https://brainly.com/question/24470698
