Answer:
Explanation:
When the balls meets, the time should be the same, let say t.
For the first ball (released from a loom):
[tex]\bar{h}=\frac{1}{2}gt^{2}[/tex] (Free fall motion) ([tex]\bar{h}[/tex] = the distance travelled by the first ball, measured from the loom)
So, we can find that [tex]\bar{h}=\frac{1}{2}(10t^{2})=5t^{2}[/tex]
For the second ball (which is thrown directly upwards):
[tex]y=v_{o}t-\frac{1}{2}gt^{2} = 25t-5t^{2}[/tex]
Because the height of the loom is h :
[tex]h=\bar{h}+y[/tex]
[tex]h=25t[/tex]
So, the time when two balls meets is : [tex]t=\frac{h}{5}[/tex] (In this question, the height of the loom h is not declared). When the height of the loom is known, you can calculate the time and the value of [tex]\bar{h}[/tex] or y
Two blocks are connected to identical ideal springs and are oscillating on a horizontal frictionless surface. Block A has mass m, and its motion is represented by the graph of position as a function of time shown above on the left. Block B’s motion is represented above on the right. Which of the following statements comparing
block B to block A is correct?
(A) Because block B covers more distance per cycle than block A, block B takes more time to complete each
cycle.
(B) Because the spring attached to block B is initially stretched a greater distance, the spring constant is smaller
and therefore block B has a slower average speed than block A does.
(C) Because block B has more mass, it has a slower average speed than block A does.
(D) Because block B has more mass, its acceleration is smaller than that of block A at any given displacement
from the equilibrium position.
Hi there!
We can begin by identifying key characteristics of both graphs.
Graph A.
Looking at the graph, we can see that the maximum distance (amplitude) is 10 cm (0.1 m). Additionally, its period (T) is 2 seconds (one full cycle).
We also know that:
[tex]T = 2\pi \sqrt{\frac{m}{k}}[/tex]
We can use this equation to compare with the other graph. Notice how the period does NOT depend on how far the spring is stretched. We can eliminate choice A for this reason.
Graph B.
The amplitude is 20 cm (0.2 m), and each period is 4 seconds.
We can now eliminate choice B because the springs are identical, so their spring constants are equal. Distance stretched has no impact on the spring constant.
For the other choices, we must look at forces and work.
Recall that:
Spring potential energy = [tex]\frac{1}{2}kx^2[/tex]
Kinetic energy = [tex]\frac{1}{2}mv^2[/tex]
Using the work-energy theorem:
[tex]\frac{1}{2}kx^2 = \frac{1}{2}mv^2[/tex]
Even if the mass of Block B is greater, its displacement is larger than that of Block A. Since displacement is squared in this equation, it would have a greater effect on the speed. Thus, Choice C is incorrect.
Using Hooke's Law:
[tex]\Sigma F = -kx\\\\ma = -kx\\[/tex]
[tex]a = \frac{-kx}{m}[/tex]
If the mass is greater, the acceleration will be smaller. Choice D is correct.
why did the ball move the same distance when it was on the flatbed truck that accelerated 0.2 m/s?
The relative velocity of the ball with respect to the truck is same as that of the truck with respect to the ball at equal time.
What is relative velocity?The relative velocity of an object is the velocity of the object observed with respect to rest frame of another object.
Distance traveled by the objectThe distance traveled by each object is determined from the prouduct of velocity and time of motion.
d = vt
where;
v is velocityt is timeThus, if the ball moved the same distance when it was on the flat bed truck, then the relative velocity of the ball with respect to the truck is same as that of the truck with respect to the ball at equal time.
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A 3 Hz wave has a wavelength of 0.4 m. Calculate the speed of the wave (in m/s).
The speed of the wave with the given frequency and wavelength is 1.2m/s.
Given the data in the question;
Wavelength of the wave; [tex]\lambda = 0.4m[/tex]Frequency; [tex]f = 3Hz = 3s^{-1}[/tex]Speed of the wave; [tex]v = \ ?[/tex]WavelengthWavelength is the distance over which the shapes of waves are repeated. It is the spatial period of a periodic wave.
It is expressed as;
[tex]\lambda = \frac{v}{f}[/tex]
Where v is velocity/speed and f is frequency.
Now, we can easily get the speed of the wave by substituting our given values into the expression above.
[tex]\lambda = \frac{v}{f} \\\\0.4m = \frac{v}{3s^{-1}}\\ \\v = 0.4m * 3s^{-1}\\\\v = 1.2ms^{-1}\\\\v = 1.2m/s[/tex]
Therefore, the speed of the wave with the given frequency and wavelength is 1.2m/s.
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A wave has a frequency of 5hz and a wave speed of 25m/s what is the wavelength?
25 m
5 m
25 m/s
125 m
Calculate the speed of sound in a string that has a tension of 100 N and a linear mass density of 0.0001 kg/m
4. The 50-kg crate shown in Fig. rests on a horizontal surface for which the coefficient of
kinetic friction is u - 0.3. If the crate is subjected to a 400-N towing force as shown,
determine the acceleration of the crate.
your
P = 400 N
490.5 N
400 N
too
30
30°
+
F-0.3 Nc
Nc
ΣΕ = max
Σ F, = may
A. a = 4. 185 m/s2
B. a = 5. 185 m/s2
C. a = 6. 185 m/s2
D. The entire above are false
Answer:
5.057 m/s^2
Explanation:
Force of kinetic friction = .3 = F /normal force
.3 = F /(50* 9.81) F of friction = 147.15
Net force = 400 - 147.5 = 252.85 N
F = m * a
252.85 = 50 * a a = 5.057 m/s^2
find the half life of a radioactive natural which decays to 6.25% of it's original no. of nuclei in 50 seconds.
Answer:
After 1 decay it has .9375 of its value left
.9375^n = 1/2
What is the value of n (number of decays)
n log .9375 = -.693
n =.693 / .0645 = 10.7
After 10.7 periods of 50 sec it will have 1/2 of its original nuclei left
1/2 life = 10.7 * 50 sec = 537 sec = 8.95 min
What should you do with electrical cells and batteries that no longer work
Answer:
Yes, single-use batteries are now made of common metals deemed non-hazardous by the federal government and can be disposed of in your regular trash in all states except California, where it is illegal to throw away all types of batteries.
What is maxwell’s theory about atoms?
Answer:
it states that the atom will rest confortably in the ground state and will not collapse for lack of energy
A dragster's top acceleration is 60 m/s.
If it accelerates for 3 seconds from the
starting line, how fast will it be going?
Answer:
very fast
Explanation:
list three things that spectra can tell us about objects in space
Answer:
The temperature, the density, and the element of stars.
Explanation:
The spectral line also can tell us about any magnetic field of the star. The width of the line can tell us how fast the material is moving. We can learn about winds in stars from this.
Which of the following examples of the streets Newton second law of motion
Explanation:
Newton’s second law is a quantitative description of the changes that a force can produce on the motion of a body. It states that the time rate of change of the momentum of a body is equal in both magnitude and direction to the force imposed on it.
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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explain the fleming left-hand rule with the diagram and what will the direction of the induced current in the figure if the magnet is going towards the coil.
A point charge q = -6.0nC is located at the origin. The electric field (in N/C) vector at the point x = -8.0m, y= +1.5m is?
Answer:
Explanation:
The position between point charge q to x = -8.0 m and y = 1.5 m is:
[tex]r=\sqrt{x^{2}+y^{2}}=\sqrt{(-8)^{2}+(1.5)^{2}}=\sqrt{66.25}\approx 8.13[/tex] meter
Then the magnitude of the electric field E is:
[tex]E=\frac{1}{4\pi\epsilon_{o}} \frac{q}{r^{2}}=(9\times 10^{9}) \frac{6.0\times 10^{-9}}{66.25}\approx 0.81 N/C[/tex]
For the vector of E:
[tex]\tan\theta=\frac{1.5}{-8}=-0.1875 \rightarrow \theta = -10.61^{0}[/tex]
[tex]E_{x}=E\cos (-10.16)=(0.81)(0.984)=0.79704 N/C[/tex]
[tex]E_{y}=E\sin\theta = -0.81(0.184)=-0.14904 N/C[/tex]
[tex]\vec{E}=0.79704\hat{i}-0.14904\hat{j}[/tex] in N/C
.. 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
7. A steel band exerts a horizontal force of 80.0 N on a tooth
at point B in Figure 7.3. What is the torque on the root of
the tooth about point A?
Gum
48.0°
A
F
1.20 cm
The torque experienced by the root of the tooth about point A when the steel band exerts the given force is 0.713 Nm.
Torque about point AThe torque experienced by the root of the tooth about point A is calculated as follows;
τ = rFsinθ
where;
F is applied forcer is the perpendicular distanceτ = (0.012) x (80) x (sin 48)
τ = 0.713 Nm
Thus, the torque experienced by the root of the tooth about point A when the steel band exerts the given force is 0.713 Nm.
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A box weighing 8 newtons requires a force of 2 newtons to push it along the floor. The coefficient or friction Berween the box and the floor is_?
I believe the answer is 16.
Have a blessed day!
By what percent is the torque of a motor decreased if its permanent magnets lose 6.5 % of their strength
A body is piloted at a point. A force of 10 N is applied at a distance of 30 cm from the pivot. Find the amount of force about the pivot .
Question :-
A Body is Pivoted at a Point. A Force of 10 N is Applied at a Distance of 30 cm from the Pivot. Find the Amount of Force about the Pivot.Answer :-
Amount of Force is 3 Nm .Explanation :-
As per the provided information in the given question, The Force is given as 10 Newton . The Distance is given as 30 cm [ 0.3 m ] . And, we have been asked to calculate the Amount of Force .
For calculating the Force , we will use the Formula :-
[tex] \bigstar \: \: \: \boxed{ \: \sf {Moment \: of \: Force \: = \: Force \: \times \: Distance} \: } [/tex]
Therefore , by Substituting the given values in the above Formula :-
[tex] \dag \: \: \: \sf {Moment \: of \: Force \: = \: Force \: \times \: Distance} [/tex]
[tex] \longmapsto \: \: \: \sf {Moment \: of \: Force \: = \: 10 \: \times \: 0.3} [/tex]
[tex] \longmapsto \: \: \: \textbf {\textsf {Moment \: of \: Force \: = \: 3}} [/tex]
Hence :-
Amount of Force = 0.3 Nm .[tex] \underline {\rule {185pt}{4pt}} [/tex]
Two pendula are shown in the figure. Each consists of a solid ball with uniform density and has a mass M. They are each suspended from the ceiling with massless rod as shown in the figure. The ball on the left pendulum is very small. The ball of the right pendulum has radius 1/2 L.
Find the period T of the right pendulum for small displacements in s.
The period (T) of the right pendulum for small displacements in s is equal to 1.49 seconds.
How to calculate the period (T).Mathematically, the time taken by this pendulum to undergo a small displacement is given by this formula:
[tex]T=2\pi \sqrt{\frac{L}{g} }[/tex]
Where:
L is the length.g is the acceleration due to gravity.Scientific data:
Acceleration due to gravity (g) = 9.8 m/s².
Assuming a randomized variable for the length of the rope to be 1.2 meter. Also, we know that the right pendulum has a radius of 1/2L.
1/2L = L/2
L/2 = 1.1/2
L/2 = 0.55 meter.
Substituting the parameters into the formula, we have;
[tex]T=2\times 3.142 \sqrt{\frac{0.55}{9.8} }\\\\T=6.284 \times \sqrt{0.0561} \\\\T=6.284 \times 0.2369[/tex]
T = 1.49 seconds.
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In a circuit, we are using conducting wires made from Manganese.
(i) If we assume there are 3 free electrons per an atom of manganese, what is its
electron density?
(ii) How much current flows through a cylindrical manganese wire of volume 27 cm3,
length 3cm if the circuit is switched on for 5 seconds?
i. The electron density of maganese is 2.44 × 10²⁸ electrons/m³
ii. The current that flows through a cylindrical manganese wire of volume 27 cm3 is 21.11 kA
To solve the question, we need to know what electron density is.
What is electron density?This is the number of free electrons per unit volume of material.
To find electron density, we need to find the atom density which is the number of atoms per unit volume, n.
Atom density of manganeseSo, the atom density of manganese is n = Nρ/A where
N = avogadro's number = 6.022 × 10²³ atoms/mol, ρ = density of manganese = 7430 kg/m³ and A = atomic mass of manganese = 0.0594 kg/molSo, n = Nρ/A
n = 6.022 × 10²³ atoms/mol × 7430 kg/m³/0.05494 kg/mol
n = 44743.46 × 10²³ atoms/mol × kg/m³/0.05494 kg/mol
n = 81440.59 × 10²³ atoms/m³
n = 8.144059 × 10²⁷ atoms/m³
n ≅ 8.14 × 10²⁷ atoms/m³
i. Electron density of manganese
The electron density of maganese is 2.44 × 10²⁸ electrons/m³
So, the electron density of manganese n' = n" × n where
n" = number of free electrons per atom = 3 and n = atom density = 8.14 × 10²⁷ atoms/m³So, n' = n" × n
= 3 electrons per atom × 8.14 × 10²⁷ atoms/m³
= 24.42 × 10²⁷ electrons/m³
= 2.442 × 10²⁸ electrons/m³
≅ 2.44 × 10²⁸ electrons/m³
So, the electron density of maganese is 2.44 × 10²⁸ electrons/m³
ii. How much current flows through the cylindrical maganese wire?
The current that flows through a cylindrical manganese wire of volume 27 cm3 is 21.11 kA
The current flowing through the wire is given by i = n'eV/t where
n = electron density = 2.44 × 10²⁸ electrons/m³, e = electron charge = 1.602 × 10⁻¹⁹ C, V = volume of wire = 27 cm³ = 27 × 10⁻⁶ m³ and t = time current flows = 5 sSo, substituting the values of the variables into the equation, we have
i = n'eV/t
i = 2.44 × 10²⁸ electrons/m³ × 1.602 × 10⁻¹⁹ C × 27 × 10⁻⁶ m³/5 s
i = 105.54 × 10³ C/5 s
i = 21.11 × 10³ C/s
i = 21.11 × 10³ A
i = 21.11 kA
So, the current that flows through a cylindrical manganese wire of volume 27 cm3 is 21.11 kA
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What is the gravitational potential energy of a 2100 kg car at the top of an 18 m parking garage?
[tex]\qquad\qquad\huge\underline{{\sf Answer}}♨[/tex]
Gravitational Potential of an object with mass m, and height of h metre is :
[tex]\qquad \sf \dashrightarrow \:mgh[/tex]
Now, if the mass of object is 2100 kg, height of 18 m is placed in there, then it's gravitational potential energy is ~
[tex]\qquad \sf \dashrightarrow \:2100 \times 10 \times 18[/tex]
[tex]\qquad \sf \dashrightarrow \:378000 \: \: joules[/tex]
[tex]\qquad \sf \dashrightarrow \:3.78 \times 10 {}^{5} \: \: joules[/tex]
You ride your bike north for 100 m at a constant speed of 5 m/Your acceleration is…
Answer:
our acceleration is 0 because there is a constant speed..
a = final speed - initial speed / time
final speed = initial speed ( since speed is constant)
therefore acceleration is 0
A fish uses its tail fin to push against the water. The water is pushed backward, in the fish is pushed for it. Is it Newton's first law of motion, Newton's second law of motion, or Newton's third law of motion?
Answer:
it is Newton's third law of motion?
Hope This Helps
A atom in its ground state contains 18 electrons. how many of these electrons are in orbital l=0 values?
An output gear has 10 teeth and an input gear has 40 teeth. What is the mechanical advantage of this gear combination?
Answer:
4
Explanation:
We know that :
Mechanical Advantage = Input / OutputSolving
MA = Input / OutputMA = 40 / 10MA = 4A Rich has been hooked and is puilling on the line as shown in the figure below.
RO
18
If the pole makes an angle of 25° with the horizontal, the distance between the point where the line is attached to the pole and the anglers hand is L. 2.45 m, and the fish
pulls on the line such that the tension in the line is T-166 Nat an angle 35.0 below the horizontal, determine the magnitude (in N.m) of the torque exerted by the fish
about an axis perpendicular to the page and passing through the angler's hand.
Nm
For the magnitude in N.m of the torque exerted by the fish about an axis perpendicular to the page and passing through the angler's hand is mathematically given as
Z = 360.33Nm
What is the magnitude in N.m of the torque exerted by the fish about an axis perpendicular to the page and passing through the angler's hand?
Generally, the equation for the Torque is mathematically given as
Z = force x perpendicular distance
Z = Tcos30° x L
Z = 166(0.886) x 2.45
Z = 360.33Nm
In conclusion s the magnitude in N.m of the torque exerted by the fish about an axis perpendicular to the page and passing through the angler's hand
Z = 360.33Nm
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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:
The sound intensity at a distance of 19 m from a noisy generator is measured to be 0.24 W/m2. What is the sound intensity at a distance of 27 m from the generator
Answer:
0.1111 W/m²
Explanation:
If all other parameters are constant, sound intensity is inversely proportional to the square of the distance of the sound. That is,
I ∝ (1/r²)
I = k/r²
Since k can be the constant of proportionality. k = Ir²
We can write this relation as
I₁ × r₁² = I₂ × r₂²
I₁ = 0.25 W/m²
r₁ = 16 m
I₂ = ?
r₂ = 24 m
0.25 × 16² = I₂ × 24²
I₂ = (0.25 × 16²)/24²
I₂ = 0.1111 W/m²