Answer:
80m/s2
Explanation:
m=30kg. r = 5meter v = 20m/s2 a= ? centripetal acceleration = v*v/r
20*20/5=80m/s2
centripetal acceleration=80 m/s2
why do fennec fox have big ears? explain why in big detail!
Answer:
Explanation:
The fennec may be the world's smallest fox, but it sports outlandishly large ears! In fact, relative to body size, it has the largest ears of any member of the canid family. It uses those big ears to listen for sounds of prey in the sand. The ears also help dispel body heat to keep the fox cool.
You are pushing a 30 crate along a rough surface by applying a horizontal force. The coefficients of
frictions between the surface and the crate are 0.3 and 0.6. Sketch the situation and create a Free Body diagram
from it, then Calculate:
a. The normal force on the crate by the floor.
b. The force required to get the crate to move.
c. Determine the kinetic friction force that the floor applies on the object.
d. Determine the acceleration of the crate.
Answer:
ill say math too!
Explanation:
Which nucleus completes the following equation?
Answer:
I think that CL IS THE ONLY BEST ANSWER
Explanation:
PLEASE MARK ME BRAINLIEST IF MY ANSWER IS CORRECT PLEASE
Can a machine multiply both force and speed at same time??
Answer:
No
Explanation:
It is not possible for a machine to act as a force multiplier and speed multiplier simultaneously. This is because machines which are force multipliers cannot gain in speed and vice-versa.
It is not possible for a machine to act as a force multiplier and speed multiplier simultaneously. This is because machines which are force multipliers cannot gain in speed and vice-versa.
at 27 degrees C and 750 mmhg a sample of hydrogen occupies 5.0 L how much space will occupy at STP
Answer:
Explanation:
The number of mole of the hydrogen sample at 27 C and 750 mmHg pressure is:
[tex]n=\frac{PV}{RT}=\frac{(0.9898\times 10^{5})(5\times 10^{-3})}{(8.314)(27+273)}\approx 0.0198[/tex] mol
This number of mole will be the same although we change the pressure and temperature. For STP (Temperature = 0 C = 273 K, and Pressure = 760 mmHg = 76 cmHg = 1 Bar = 1.01 x [tex]10^{5}[/tex] Pa):
[tex]P'V'=nRT' \rightarrow V'=\frac{nRT'}{P'}=\frac{0.0198(8.314)(273)}{1.01\times 10^{5}}\approx 44.49 \times 10^{-5}[/tex] in [tex]m^{3}[/tex]. But in L, we find that V' = 0.4449 L
What does the atomic numbers equal? A number of protons in the atomic nucleus B number of protons and neutrons in the atomic nucleus C number of neutrons in the atomic nucleus
number of protons
explanation:the atomic number of an atom is equal to the number of protons
Find the magnitude of the sum of these two vectors:
Answer:
Explanation:
You can decompose those vectors into their components in x and y direction. For the first vector:
[tex]\vec{r}_{1}=r_{1}\cos 30\hat{i}+r_{1}\sin 30\hat{j}=3.14(\frac{1}{2}\sqrt{3}\hat{i}+\frac{1}{2}\hat{j})=1.57\sqrt{3}\hat{i}+1.57\hat{j}[/tex]
For the second vector:
[tex]\vec{r}_{2}=r_{2}\cos 60\hat{i}-r_{2}\sin60 \hat{j}=1.355\hat{i}-1.355\sqrt{3}\hat{j}[/tex]
The sum of two vectors will be:
[tex]\vec{r}=\vec{r}_{1}+\vec{r}_{2}=(1.57\sqrt{3}+1.355)\hat{i}+(1.57-1.355\sqrt{3})\hat{j}\approx 4.0711\hat{i}-0.77415\hat{j}[/tex]
The magnitude of the sum of two vectors is:
[tex]r=\sqrt{(4.0711)^{2}+(-0.77415)^{2}}\approx 4.14[/tex] meter
Answer:
4.14m
Explanation:
A 2.0-kg block slides on a rough horizontal surface. A force (magnitude P = 4.0 N) acting parallel to the surface is applied to the block. The magnitude of the block's acceleration is 1.2 m/s2. If P is increased to 5.0 N, determine the magnitude of the block's
When the applied force increases to 5 N, the magnitude of the block's acceleration is 1.7 m/s².
Frictional force between the block and the horizontal surfaceThe frictional force between the block and the horizontal surface is determined by applying Newton's law;
∑F = ma
F - Ff = ma
Ff = F - ma
Ff = 4 - 2(1.2)
Ff = 4 - 2.4
Ff = 1.6 N
When the applied force increases to 5 N, the magnitude of the block's acceleration is calculated as follows;
F - Ff = ma
5 - 1.6 = 2a
3.4 = 2a
a = 3.4/2
a = 1.7 m/s²
Thus, when the applied force increases to 5 N, the magnitude of the block's acceleration is 1.7 m/s².
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What is the average acceleration of a me
that reaches 24 m/s in 5.5 seconds?
F 0.20 m/s2
G 2.5 m/s
H 10 m/s2
J 208 m/s
Answer:
See below
Explanation:
Assuming starting from zero velocity
vfinal = at
24 = a (5.5)
a = 4.36 m/s^2 I think part of your question is missing....what was the initial velocity?
a 56-kg student runs at 6.2m/s, grabs a hanging 10.0m long rope, and swings out over a lake. He releases the rope when his velocity is zero.
a) what is the tension in the rope just before he releases it?
b) what is the maximum tension in the rope during the swing?
Answer:
0.52 8.9
Explanation:
calc.
The tension in the rope just before the student releases it is 576 N.
What is tension?Tension is defined in physics as the pulling force transmitted axially by a string, rope, chain, or similar object.
Just before the student releases the rope, his velocity is zero. At this point, all of his kinetic energy has been converted into potential energy. Therefore, we can write:
1/2 m [tex]v^2[/tex] = mgh
h = [tex]v^2[/tex] / (2g) = [tex](6.2 m/s)^2[/tex] / (2 x 9.8 [tex]m/s^2[/tex]) ≈ 1.95 m
m g = 56 kg x 9.8 [tex]m/s^2[/tex] ≈ 549 N
The tension required to keep the rope taut is equal to the horizontal component of the force due to the weight of the student, which is:
T = m g / cos(θ)
tan(θ) = h / L
tan(θ) = 1.95 m / 10.0 m ≈ 0.195
θ ≈ 10.9°
So, T = m g / cos(θ) ≈ 576 N
To find the maximum tension in the rope during the swing, we can use conservation of angular momentum.
L = I ω
I = (1/3) m [tex]L^2[/tex]
I = (1/3) x 10.0 kg x [tex](10.0 m)^2[/tex] ≈ 3333.3 kg[tex]m^2[/tex]
The angular velocity of the rope just after the student releases it is:
ω = v / L
ω = 6.2 m/s / 10.0 m ≈ 0.62 rad/s
L = I ω ≈ 2070 N m s
Thus, the angular momentum of the system just after the student releases the rope is 2070 N m s.
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A toy car, with mass of 6 Kg, is pushed with a force of 10 N. If the toy car is in the grass with a coefficient of friction of 0.1 then what is the acceleration?
How far does it get pushed in 10 s?
An object moving in a straight line has a velocity v in m/s that varies with time t in s according the following function. v= 8+2.5 12 to 11. The instantaneous acceleration of the object at t = 2 sis (A) 2 m/s (B) 4 m/s (0) 6 m/s2 D) 8 m/s2 (E) 10 m/s2 12. The displacement of the object between t = 0 and t = 6 s is (A) 120 m (B) 180 m (C) 228 m (D) 242 m (E) 260 m
The instant acceleration of the object is 10 m/s² and the displacement of the object is 228 m.
Instantaneous acceleration
The instant acceleration of the object is determined by taking the derivative of the velocity as follows;
v = 8 + 2.5t²
a = dv/dt
a = 5t
a(2) = 5(2)
a(2) = 10 m/s²
Displacement of objectThe displacement of the object is determined by taking the integral of velocity as follows;
v = 8 + 2.5t²
x = ∫v
x = 8t + 2.5t³/3
x(0) = 0
x(6) = 8(6) + 2.5(6)³/3
x(6) = 228 m
Thus, the displacement of the object is 228 m.
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How long does it take a photon to escape the sun?
Answer:
About 5000 years.
Explanation:
Answer:
Explanation:
it will take more than half a million years for the photon to escape the sun. If you think it's about a centimeter, then it will take about 5,000 years for the photon to get outside the su
Which statement is true about a wave that is created by causing vibrations in
particles of matter?
A. It can travel through empty space.
B. It is not an electromagnetic wave.
C. It does not need a medium.
D. It moves through a magnetic or electric field.
Answer:
A
Explanation:
A. it can travel through empty space
A 2 kg solid disk with a radius of 0.22 m has a tangential force of 300N applied to it.
a) What is the torque acting on the disk?
b) What is the moment of inertia of the disk?
c) What angular acceleration is produced by the torque?
d) If the disk starts from rest and the acceleration is constant for 3.0s, what is the angular velocity of the disk at the end of 3.0s?
e) Through what angle in radians has the disk rotated during this time?
(a) The torque acting on the disk is 66 Nm.
(b) The moment of inertia of the disk is 0.05 kgm².
(c) The angular acceleration is produced by the torque is 1,320 rad/s².
(d) The final angular velocity of the disk is 3,960 rad/s.
(e) The angle of rotation of the disk is 5,940 rad.
Torque acting on the diskThe torque acting on the disk is calculated as follows;
τ = Fr
τ = 300 x 0.22
τ = 66 Nm
Moment of inertiaThe moment of inertia of a solid disk is calculated as follows;
I = ¹/₂MR²
I = ¹/₂ x 2 x (0.22)²
I = 0.05 kgm²
Angular acceleration of the diskThe angular acceleration of the disk is calculated as follows;
τ = Iα
[tex]\alpha = \frac{\tau }{I} \\\\\alpha = \frac{66}{0.05} \\\\\alpha = 1,320 \ rad/s^2[/tex]
Angular velocity of the disk after 3 sωf = ωi + αt
ωf = 0 + (1320 x 3)
ωf = 3,960 rad/s
Angle of rotation of the diskωf² = ωi²+ 2αθ
(3,960)² = 0 + 2(1320)θ
θ = (3,960²) / (2 x 1320)
θ = 5,940 rad
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What is the best definition of
work?
A. net force equals zero
B. a force exerted over a distance
a
C. pair of forces where one is of greater strength
Answer:
Work is best defined as a force exerted on a body to cause it move over a certain distance.
work=force×displacement×cosθ
Explanation:
1. A drag racer accelerates from rest at 18ft/sec^2. How long does it take to acquire a speed of 60mph? What is required?
2. A contestant ran a 100-m dash in 10.6sec. What was his speed a) In feet per second and b) in miles per hr?
(1) The time taken for the drag racer to accelerate is 4.89 s
(2) The speed of the contestant (a) in feet per second is 30.89 ft/s (b) in miles per hr is 21.12 miles per hr.
(1) To calculate the time required to accelerate 18 ft/sec² from rest to a velocity of 60 mph, we use the formula below.
Formula:
t = (v-u)/a........... Equation 1Where:
t = timev = Final velocityu = initial velocitya = acceleration.From the question,
Given:
a = 18 ft/sec² = (18×0.3048) = 5.4864 m/s²v = 60 mph = (60×0.44704) = 26.82 m/su = 0 m/s ( from rest)Substitute these values into equation 1
t = (26.82-0)/5.4864t = 4.89 seconds(2) To calculate the speed of the contestant, we use the formula below
Formula:
s = d/t............ Equation 1Where:
s = speed of the contestantd = distancet = time.From the question,
Given:
d = 100 mt = 10.6 sSubstitute these values into equation 1
s = 100/10.6s = 9.43 m/s(a) In feet = (9.43/0.3048) = 30.94 ft/s
(b) in miles per hr = (9.43×2.24) = 21.12 miles per hr
Hence, (1) The time taken for the drag racer to accelerate is 4.89 s (2) The speed of the contestant (a) in feet per second is 30.89 ft/s (b) in miles per hr is 21.12 miles per hr.
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stion 3
Define the law of moments. It is applicable to a certain condition of the acting forces.
Name the condition.
Answer:
algebraic sum of moments of all the forces acting on the body about the axis of rotation is zero, the body is in equilibrium. According to the principle of moments, in equilibrium: Sum of anticlockwise Moments = Sum of clockwise moments.
Condition:
For an object to be in equilibrium, it must be experiencing no acceleration.
Hope it may helpful to you
In the figure below (Figure 1), the upper ball is released from rest, collides with the stationary lower ball, and sticks to it. The strings are both 50.0 cm long. The upper ball has a mass of 2.20 kg and it is initially 10.0 cm higher than the lower ball, which has a mass of 2.70 kg. Find the frequency of the motion after the collision. Find the maximum angular displacement of the motion after the collision.
(a) The frequency of the motion after the collision is 0.71 Hz.
(b) The maximum angular displacement of the motion after the collision is 16.3⁰.
Speed of the 2.2 kg ball when it collides with 2.7 kg ballThe speed of the 2.2 kg ball which was initially 10 cm higher that 2.7 kg ball is calculated as follows;
K.E = P.E
¹/₂mv² = mgh
v² = 2gh
v = √2gh
v = √(2 x 9.8 x 0.1)
v = 1.4 m/s
Final speed of both balls after collisionThe final speed of both balls after the collision is determined from the principle of conservation of linear momentum.
Pi = Pf
m₁v₁ + m₂v₂ = vf(m₁ + m₂)
2.2(1.4) + 2.7(0) = vf(2.2 + 2.7)
3.08 = 4.9vf
vf = 3.08/4.9
vf = 0.63 m/s
Maximum displacement of the balls after the collisionP.E = K.E
[tex]mgh_f = \frac{1}{2} mv_f^2\\\\h_f = \frac{v_f^2}{2g} \\\\h_f = \frac{(0.63)^2}{2(9.8)} \\\\h_f = 0.02 \ m[/tex]
Maximum angular displacementThe maximum angular displacement of the balls after the collision is calculated as follows;
[tex]cos \theta = \frac{L - h_f}{L} \\\\cos\theta = \frac{0.5 - 0.02}{0.5} \\\\cos\theta = 0.96\\\\\theta = cos^{-1}(0.96)\\\\\theta = 16.3 \ ^0[/tex]
Frequency of the motion[tex]f = \frac{1}{2\pi} \sqrt{\frac{g}{L} } \\\\f = \frac{1}{2\pi } \sqrt{\frac{9.8}{0.5} } \\\\f = 0.71 \ Hz[/tex]
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An aquarium heater transfer 1200kJ of the heat to 75000g of water. What is the increase in the waters temperature?
Answer:
Energy = mcΔt
where m the mass , c specific heat capacity
Δt is the change in temperature
1200x10^3=75x4184xΔt
Δt=3.82°
Explanation:
A simple circuit has a voltage of 8V and a current of 3A. Find the resistance (R) in the
circuit.
Answer:
2.67 ohm
Explanation:
V=IR
R=V/I
In which V is voltage and I is current and R is resistance
.. In the reaction 238 92U ---- X + 2 4 He, the particle represented by X is . answer choices . 234 90 Th. 234 92 U. 238 93 Np. 242 94Pu. ... Which conditions are required
My question
• Complete the following reaction:
238 92U = ______ + ________ + 4 2He
…Is this the same question as the answer
Answer:
234 90 +energy
Explanation:
because alpha decrease by 4 2helium
a current of 180 mini amphere passes through a conductor for 5minute calculate the quantity of electricity transported
Answer:
Explanation:
You can calculate the total electric charge that passes through the conductor as [tex]q=It=(180\times 10^{-3})(5\times 60)= 54 C[/tex]. It means that the number of electron that passes through the conductor is:
[tex]n=\frac{q}{e}=\frac{54}{1.6\times 10^{-19}}=33.75\times 10^{19}[/tex]
Females are theoretically more prone to anxiety disorders due to __________. A. their habit of overreacting to physiological changes B. the high levels of serotonin causing hormone imbalance C. increased levels of norepinephrine causing anxiety sensitivity D. stress levels, life experiences, and friends’ stresses Please select the best answer from the choices provided A B C D
Answer:
fluctuations in the levels of female reproductive hormones
Explanation:
Women are more prone to anxiety due to a variety of biological, psychological, and cultural factors. Although the exact cause is unknown, recent research suggests that fluctuations in the levels of female reproductive hormones and cycles play an important role in women's enhanced vulnerability to anxiety.
Answer: D. stress levels, life experiences, and friends’ stresses
Explanation:
• (-/1 Points) DETAILS OSCOLPHYS2016 9.5.WA.040.
MY NOTES ASK YOUR TEACHER PRACTICE ANOTHER
A supertanker uses a windlass (a type of winch) like the one shown in the figure below to hoist its 21,200-kg anchor. Determine the force that must be exerted on the
outside wheel to lift the anchor at constant speed, neglecting friction and assuming the anchor is out of the water.
IN
doorg
-0.45 m
- 1.5m
The force that must be exerted on the outside wheel to lift the anchor at constant speed is 6.925 x 10⁵ N.
Force exerted outside the wheelThe force exerted on the outside of the wheel can be determined by applying the principle of conservation of angular momentum as shown below.
∑τ = 0
Let the distance traveled by the load = 1.5 mLet the radius of the wheel or position of the force = 0.45 m∑τ = R(mg) - r(F)
rF = R(mg)
0.45F = 1.5(21,200 x 9.8)
F = 6.925 x 10⁵ N.
Thus, the force that must be exerted on the outside wheel to lift the anchor at constant speed is 6.925 x 10⁵ N.
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What is one benefit of smart meters?
Answer:
You get up-to-date readings
Explanation:
When a speeding truck hits a stationary car, the car is deformed and heat is generated. What can you say about the kinetic energy of the system after the collision?
nothing
Explanation:
we can say that it was certainly bad for the shopkeeper obviously and we should not be making questions about the physics behind that accident and should call the cops or 911
you move a 25 N object 5.0 meters. how much work did you do?
Answer:
125J
Explanation:
[tex]work \: = force \: \times distance \\ = 25 \times 5 \\ = 125joules[/tex]
Residents of a community currently pay a fee for recreational services in their neighborhood, such as tennis courts, running trails, a pool. To assess whether there is support for raising a fee for the upcoming summer, a random sample of 50 residents is contracted and asked this question: “Would you support a $10 increase in the recreational fee to support programs for the health and well-being of the children in the community?” Of the 50 residents contacted, 42 approved of the fee increase. Which statement describes a likely source of bias present in survey results?
A likely source of bias in the survey is undercoverage bias because some parts of the community may not be represented in the survey results.
Bias present in the surveyThe bias present in the survey may lead to error in the result obtained from the random sampling survey.
Source of bias in the surveyOne of the likely source of bias in the survey is undercoverage.
It is not possible to ascertain the accurate opinion of the entire community from the 50 residents chosen for the survey.
Thus, a likely source of bias present in survey is undercoverage bias because some parts of the community may not be represented in the survey results.
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Answer:
B. Question wording bias may be present because the question leads those responding toward a certain response.
Explanation:
The force between a pair of charges is 900 newtons. The distance between the charges is 0.01 meters. If one of the charges is 2e-10 C what is the strength of the other charge ?
Answer:
[tex] \fbox{strength \: of \: the \: other \: charge = - 0.0196 Ke \: Coulomb}[/tex]
Explanation:
Given:
Force between pair of charges= 900 newtons
The distance between the charges = 0.01 meters
Strength of Charge first q1 = 2e-10 Coulomb
To find:
Strength of Charge second q2 = ____ Coulomb?
Solution:
We know that,
Force between two charges separate by distance r is given by the equation,
[tex]|F| = K_e \frac{q1 \cdot \: q2}{ {r}^{2} } \\ 900 =K_e \frac{(2e - 10)\cdot \: q2}{ {0.01}^{2} } \\ 900 \times {10}^{ - 4} = K_e {(2e - 10)\cdot \: q2} \\ q2 = \frac{9 \times {10}^{ - 2} }{(2e - 10) K_e} \\ \\ \fbox{We \: know \: that \: e = 2.71 } \\ substituting \: the \: value \: \\ q2 = \frac{9 \times {10}^{ - 2} }{(2 \times 2.71 - 10)K_e} \\ q2 = \frac{0.09}{ - 4.58 K_e} \\ q2 = \frac{-0.0196}{K_e}\: coulomb[/tex]
[tex] \fbox{strength \: of \: the \: other \: charge = - 0.0196 Ke \: Coulomb}[/tex]
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