Answer:
A) r = 0.03 m
B) r = 0.0533 m
C) B_max = 0.00003 T
Explanation:
Formula for magnetic field inside the capacitor when it is parallel to the length element is;
B_in = (μ_o•I•r/(2πR²)
Formula for maximum magnetic field is;
B_max = (μ_o•I/(2πR)
Formula for magnetic field outside the capacitor is; B_out = (μ_o•I/(2πr)
A) Magnetic field inside the capacitor is gotten from our first equation above;
B_in = (μ_o•I•r/R²)
Since we want to find the radius at which the magnitude of the induced magnetic field equal to 75% or 0.75 of its maximum value.
Thus;
B_in = 0.75B_max
(μ_o•I•r/(2πR²) = 0.75((μ_o•I/(2πR))
μ_o•I and 2πR will cancel out to give;
r/R = 0.75
r = 0.75R
We are given R = 40 mm = 0.04 m
r = 0.75 × 0.04
r = 0.03 m
B) magnetic field outside the capacitor is; B_out = (μ_o•I/(2πr)
Thus for the magnitude of the induced magnetic field equal to 75% or 0.75 of its maximum value:
B_out = 0.75B_max
(μ_o•I/(2πr) = 0.75((μ_o•I/(2πR))
μ_o•I and 2π will cancel out to give;
1/r = 0.75/R
r = R/0.75
r = 0.04/0.75
r = 0.0533 m
C) B_max = μ_o•I/(2πR)
μ_o is a constant known as vacuum of permeability with a value of 4π × 10^(-7) T.m/A
Thus;
B_max = (4π × 10^(-7) × 6)/(2π × 0.04)
B_max = 0.00003 T
STATION 1
Jane moved a 800kg piano to the right across the
carpet with a coefficient of friction of 0.4. What is the
magnitude of the force of friction acting on the
piano?
If she moved it at a constant velocity what is the
applied force acting on the piano?
Imagine you are running PE class. You run the first 1000 meters in 3 minutes and then get tired and run the last 600 meters in 5 minutes. What was your *average* speed?
Answer:
200 meters per minute
Explanation:
You run the 1600 meters in a total of 8 minutes, so the average speed if 200 meters per minute.
At 20 oC the densities of fresh water and ethyl alcohol are, respectively, 998 and 789 kg/m3. Find the ratio of the adiabatic bulk modulus of fresh water to the adiabatic bulk modulus of ethyl alcohol at 20 oC.
Answer:
The ratio is [tex]\frac{B_1}{B_2} = 1.265[/tex]
Explanation:
From the question we are told that
The density of fresh water is [tex]\rho__{f}} = 998 \ kg/m^3[/tex]
The density of ethanol is [tex]\rho_{e} = 789 \ kg /m^3[/tex]
Generally speed of a wave in a substance is mathematically represented as
[tex]v = \sqrt{\frac{B}{\rho} }[/tex]
Here B is the adiabatic bulk modulus of the substance while [tex]\rho[/tex] is the density of the substance
So at constant wave speed
[tex]\sqrt{\frac{B_1}{\rho_1} } = \sqrt{\frac{B_2}{\rho_2} }[/tex]
=> [tex]\frac{B_1}{\rho_1} = \frac{B_2}{\rho_2}[/tex]
=> [tex]B_1 \rho_2 = B_2\rho_1[/tex]
=> [tex]\frac{B_1}{B_2} = \frac{\rho_1}{\rho_2}[/tex]
Here [tex]\rho_1 =\rho__{f}} = 998 \ kg/m^3[/tex] and [tex]\rho_2 = \rho_{e} = 789 \ kg /m^3[/tex]
So
=> [tex]\frac{B_1}{B_2} = \frac{998}{789}[/tex]
=> [tex]\frac{B_1}{B_2} = 1.265[/tex]
The propeller of an aircraft accelerates from rest with an angular acceleration α = 7t + 8, where α is in rad/s2 and t is in seconds. What is the angle in radians through which the propeller rotates from t = 1.00 s to t = 6.10 s?
Answer:
The value is [tex]\theta =407.3 \ radian[/tex]
Explanation:
From the question we are told that
The angular acceleration is [tex]\alpha = (7t + 8) \ rad/ s^2[/tex]
The first time is [tex]t_1 = 1.00 \ s[/tex]
The second time [tex]t_2 = 6.10 \ s[/tex]
Generally the angular velocity is mathematically represented as
[tex]w = \int\limits {\alpha } \, dt[/tex]
=> [tex]w = \int\limits {7t + 8 } \, dt[/tex]
=> [tex]w =\frac{ 7t^2}{2} + 8 t[/tex]
Generally the angular displacement is mathematically represented as
[tex]\theta = \int\limits^{t_2}_{t_1} { w} \, dt[/tex]
=> [tex]\theta = \int\limits^{t_2}_{t_1} { \frac{7t^2}{2} + 8t } \, dt[/tex]
=> [tex]\theta = { \frac{7t^3}{6} + \frac{8t^2}{2} } | \left \ t_2} \atop {t_1}} \right.[/tex]
=> [tex]\theta = { \frac{7t^3}{6} + 4t^2} } | \left \ 6.10} \atop {1}} \right.[/tex]
=> [tex]\theta =[ { \frac{7}{6}[6.10 ]^3 + 4[6.10]^2} } ] -[ { \frac{7}{6}[1 ]^3 + 4[1]^2} } ][/tex]
=> [tex]\theta =407.3 \ radian[/tex]
In a warehouse, the workers sometimes slide boxes along the floor to move them. Two boxes were sliding toward each other and crashed. The crash caused both boxes to change speed. Based on the information in the diagram, which statement is correct? In your answer, explain what the forces were like and why the boxes changed speed.
Box 1 has more mass than Box 2.
Box 1 and Box 2 are the same mass.
Box 1 has less mass than Box 2.
**YOU MUST BE DESCRIPTIVE! Any short answers not explaining it wont get brainliest!**
Answer:
box 1 has larger mass than box 2 (which agrees with the first answer option given.
Explanation:
We need to consider the linear momentum of the boxes immediately before and after they crash.
Recall that momentum is defined as mass times velocity.
So for before the collision, the linear momentum of the system of two boxes is:
m1 * 4km/h - m2 * 8km/h
with m1 representing mass "1" on the left, and m2 representing mass 2 on the right.
Notice the sign of the linear momentum (one positive (moving towards the right) and the other one negative (moving towards the left)
For after the collision, we have or the linear momentum of the system:
- m1 * 2km/h - m2 * 1km/h
Then, since the linear momentum is conserved in the collision, we make the initial momentum equal the final and study the mass relationship between m1 and m2:
4 m1 - 8 m2 = - 2 m1 - m2
combining like terms for each mas on one side and another of the equal sign, we get;
4 m1 + 2 m1 = 8 m2 - m2
6 m1 = 7 m2
therefore m1 = (7/6) m2
which (since 7/6 is a number larger than one) tells us that m1 is larger than m2 by a factor of 7/6
Therefore, the first answer option is the correct answer.
A rack of seven spherical bolwing balls (each 7.00 kg, radius of 9.50 cm) is positioned along a line located a distance =0.850 m from a point ,
Calculate the gravitational force
the bowling balls exert on a ping-pong ball of mass 2.70 g, centered at point .
Answer:
8.72*[tex]10^{-12}[/tex] N
Explanation:
force of attraction f = G m1m2/ r^2 = [tex]\frac{6.67*10^{-11}*7*5*2.70 }{.85*.85*1000}[/tex] = 8.72*[tex]10^{-12}[/tex] N
The gravitational force the bowling balls exert on a ping-pong is given as 8.72*[tex]10^{-12}[/tex] N.
What is force?A force is an effect that can alter an object's motion according to physics. An object with mass can change its velocity, or accelerate, as a result of a force. An obvious way to describe force is as a push or a pull. A force is a vector quantity since it has both magnitude and direction.
force of attraction
f = G [tex]m_1m_2[/tex]/r²
= [tex]6.67*10^{-11}[/tex]*7*5*2.70/.85²*1000
= 8.72*[tex]10^{-12}[/tex] N
The gravitational force the bowling balls exert on a ping-pong is given as 8.72*[tex]10^{-12}[/tex] N.
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What is the wavelength of light falling on double slits separated by 2.00 μm if the third-order maximum is at an angle of 60.0∘?
Answer:
λ = 5.773 x 10⁻⁷ m = 577.3 nm
Explanation:
In order to solve this problem we will use the grating equation:
mλ = d Sin θ
where,
m = order = 3
λ = wavelength of light = ?
d = slit separation = 2 μm = 2 x 10⁻⁶ m
θ = angle = 60°
Therefore,
(3)λ = (2 x 10⁻⁶ m)Sin 60°
λ = 1.732 x 10⁻⁶ m/3
λ = 5.773 x 10⁻⁷ m = 577.3 nm
Calculate the distance covered by a bus moving at a rate of 11.5km/h with a time of 2060seconds
Answer:
6.56km
Explanation:
Given parameters:
Speed = 11.5km/hr
Time = 2060s
Unknown:
Distance covered = ?
Solution:
Speed is distance divided by the time taken.
Speed = [tex]\frac{distance}{time}[/tex]
Distance = Speed x time
Let us convert the seconds to hours;
3600s = 1hr
2060s = [tex]\frac{2060}{3600}[/tex] = 0.57hr
Now
Distance = 11.5km/hr x 0.57hr = 6.56km
Use the above picture to fill in the blanks for the following statement.
One of the element carbon combines with one of the element oxygen to form one of the compound carbon dioxide.
Answer:
C + 2O ------ CO2
Explanation:
"One" element of Carbon combines with "Two" elements of Oxygen, to form "One" compound of Carbon dioxide.
I didn't really get what you meant but this is my guess of what you meant
When does a magnet induce an electric current in a wire coil?
O A. When the wire is connected to the coil
O B. When the magnet is near the coil
O C. When the magnet is moving back and forth in the coil
D. When the magnet is very strong
Answer:
B I believe
Explanation:
A 1200 N force acts on an object, resulting in an acceleration of 8.0 m/s2. What is the mass of the object?
Answer:
The answer is 150 kgExplanation:
The mass of the object can be found by using the formula
[tex]m = \frac{f}{a} \\ [/tex]
f is the force
a is the acceleration
From the question we have
[tex]m = \frac{1200}{8} \\ [/tex]
We have the final answer as
150 kgHope this helps you
Discuss two ways to determine your muscular strength explain the advantages
Answer:
Two ways to determine ones muscular strength is to either lift the heaviest weight possible or by doing one repetition. The advantages these two methods is that it is easy to determine and does not require calculations or estimating.
Explanation:
Answer:
Two ways to determine ones muscular strength is to either lift the heaviest weight possible or by doing one repetition. The advantages these two methods is that it is easy to determine and does not require calculations or estimating.
Explanation:
Your teacher placed a 3.5 kg block at the position marked with a “ + ” (horizontally, 0.5 m from the origin) on a large incline outlined on the graph below and let it slide, starting from rest. ***There are two images included!***
Answer:
x = 10.75 m
Explanation:
For this problem we will solve it in two parts, the first using energy and the second with kinematics
Let's use the energy work relationship to find the velocity of the block as it exits the ramp
W = [tex]Em_{f}[/tex] - Em₀
Starting point. Higher
Em₀ = U = m g h
the height from the edge of the ramp of the graph has a value
h = 9-3 = 6 m
Final point. At the bottom of the ramp
Em_{f} = K = ½ m v²
Friction force work
W = - fr d
The friction force has the formula
fr = μ N
On the ramp, we can use Newton's second law
N - W cos θ = 0
N = W cos θ
where the angle is obtained from the graph
tan θ = (9-3) / (0.5-4) = -6 / 3.5
θ = tan⁻¹ (-1,714)
θ = -59.7º
the distance d is
d = √ (Δx² + Δy²)
d = √ [(0.5-4)² + (9-3)²]
d = 6.95 m
for which the work is
W = - μ mg cos 59.7 d
we substitute
W = Em_{f} -Em₀
- μ mg cos 59.7 d = ½ m v² - m g h
In the graph o text the value of the friction coefficient is not observed, suppose that it is μvery = 0.2
- μ g cos 59.7 d = ½ v² - g h
v² = 2g (h - very d coss 59.7)
let's calculate
v² = 2 9.8 (6 - 0.2 6.95 cos 59.7)
v = √ 103.8546
v = 10.19 m / s
in the same direction as the ramp
in the second part we use projectile launch kinematics
let's look for the components of velocity
v₀ₓ = vo cos -59.7
[tex]v_{oy}[/tex] = vo sin (-59,7)
v₀ₓ = 10.19 cos (-59.7) = 5.14 m / s
v_{oy} = 10.19 if (-59.7) = -8.798 m / s
Let's find the time to get to the floor (y = o)
y = y₀ + v_{oy} t - ½ g t²
to de groph y₀=3 m
0 = 3 - 8.798 t - ½ 9.8 t²
t² - 1.796 t - 0.612 = 0
we solve the quadratic equation
t = [1.796 ±√(1.796² + 4 0.612)] / 2
t = [1,795 ± 2,382] / 2
t₁ = 2.09 s
t₂ = -0.29 s
since time must be a positive quantity the correct value is t = 2.09 s
we calculate the horizontal displacement
x = v₀ₓ t
x = 5.14 2.09
x = 10.75 m
The motion of the box, after it exits the incline is the motion and trajectory
of a projectile.
Horizontal distance from the right-hand edge of the incline to the point of
contact with the floor is approximately 1.24613 m.
Reasons:
Mass of the block, m = 3.5 kg
Coefficient of kinetic friction, μ = 1.2
Location of the = 0.5 m from the origin
Required:
Horizontal distance between the block's point of contact with the floor and
the bottom right-hand edge of the incline.
Solution:
Let θ represent the angle the incline make with the horizontal.
The normal reaction of the incline on the block, [tex]F_N[/tex] = m·g·cos(θ)
Work done on friction = [tex]F_N[/tex]×μ×Length of the incline, L
Rise of the incline = 10 - 3 = 7
Run of the incline = 4
L = √(6.125² + 3.5²) = [tex]\dfrac{7 \times \sqrt{65} }{8}[/tex]
Let ΔP.E.₁ represent the potential energy transferred to kinetic energy
and work along the incline, we have;
Energy of the block at the bottom of the incline, M.E.₂, is found as follows;
K.E.₂ = mgh - m·g·μ·cos(θ)·L
[tex]K.E. =\frac{1}{2} \times 3.5 \times v^2 = 3.5 \times 9.81 \times 6.125 - 3.5 \times 9.81 \times 1.2 \times \dfrac{4}{\sqrt{65} } \times \dfrac{7 \times \sqrt{65} }{8}[/tex]
v ≈ 6.1456 m/s
The vertical component of the velocity is therefore;
[tex]v_y = v \cdot sin(\theta)[/tex]
[tex]v_y = 6.1456 \times \dfrac{7}{\sqrt{65} } \approx 5.33588[/tex]
From the equation, h = u·t + 0.5·g·t² derived from Newton's Laws of motion, we have;
ΔP.E.₁ = 3.5×9.81×7
3 = 5.33588·t + 0.5×9.81·t²
Factorizing, the above quadratic equation, we get;
The time it takes the block to reach the floor, t ≈ 0.40869 seconds
Horizontal component of the velocity is [tex]v_x \approx 6.1456 \times \dfrac{4}{\sqrt{65} } \approx 3.04908[/tex]
The horizontal distance, x = vₓ × t
∴ x = 3.04908 × 0.40869 ≈ 1.08194
Horizontal distance from the right-hand edge of the incline to the point of
contact with the floor, x ≈ 1.24613 m.
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A twirlers baton is 0.76 m long and spins around its center. The end of the baton has a centripetal acceleration of 47.8 m/s2?
A 0.31 s
B 0.56 s
C 4.3s
D 70s
Answer:
C. 4.3 seconds
Explanation:
B 0.56 s is the time period of a twirlers baton.
What is Centripetal Acceleration?Centripetal acceleration is defined as the property of the motion of an object which traversing a circular path.
Any object that is moving in a circle and has an acceleration vector pointed towards the center of that circle is known as Centripetal acceleration.
The centripetal acceleration is given by:
a = 4π²R/T²
Given values are:
a = 47.8 m/s²
D = 0.76 m so , R = 0.76/2 = 0.38m
Using this formula,
47.8*T² = 4π² x0.38
T² = [tex]\frac{4*3.14^2*0.38}{47.8}[/tex]
T = 0.56 s
Therefore,
A twirlers baton is 0.76 m long and spins around its center. The end of the baton has a centripetal acceleration of 47.8 m/s2 which have time period of 0.56 s.
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.
During a circus performance, a 72-kg humancannonball is shot out of an 18-m-long cannon. If thehuman cannonball spends 0.95 s in the cannon,determine the average net force exerted on him in thebarrel of the cannon.
Answer:
2872.8 N
Explanation:
We have the following information
m =n72kg
Δy = 18m
t = 0.95s.
From here we use the equation
Δy=1/2at2 in order to solve for the acceleration.
So a
=( 2x 18m)/(0.95s²)
= 36/0.9025
= 39.9m/s2.
From there we use the equation
F = ma
F=(72kg) x (39.9)
= 2872.8N.
2872.8N is the average net force exerted on him in the barrel of the cannon.
Thank you!
When an aluminum bar is connected between a hot reservoir at 720 K and a cold reservoir at 358 K, 3.00 kJ of energy is transferred by heat from the hot reservoir to the cold reservoir. (a) In this irreversible process, calculate the change in entropy of the hot reservoir._______ J/K
(b) In this irreversible process, calculate the change in entropy of the cold reservoir.
_______ J/K
(c) In this irreversible process, calculate the change in entropy of the Universe, neglecting any change in entropy of the aluminum rod.
_______ J/K
(d) Mathematically, why did the result for the Universe in part (c) have to be positive?
Answer:
a. -4.166 J/K
b. 8.37 J/K
c. 4.21 J/K
d. entropy always increases.
Explanation:
Given :
Temperature at hot reservoir , [tex]$T_h$[/tex] = 720 K
Temperature at cold reservoir , [tex]$T_c$[/tex] = 358 K
Transfer of heat, dQ = 3.00 kJ = 3000 J
(a). In the hot reservoir, the change of entropy is given by:
[tex]$dS_h= -\frac{dQ}{t_h}$[/tex] (the negative sign shows the loss of heat)
[tex]$dS_h= -\frac{3000}{720}$[/tex]
= -4.166 J/K
(b) In the cold reservoir, the change of entropy is given by:
[tex]$dS_c= \frac{dQ}{t_c}$[/tex]
[tex]$dS_c= \frac{3000}{358}$[/tex]
= 8.37 J/K
(c). The entropy change in the universe is given by:
[tex]$dS=dS_h+dS_c$[/tex]
= -4.16+8.37
= 4.21 J/K
(d). According to the concept of entropy, the entropy of the universe is always increasing and never decreasing for an irreversible process. If the entropy of universe decreases, it violates the laws of thermodynamics. Hence, in part (c), the result have to be positive.
if you increase the frequency of a wave by 5x whats it’s period?
We know that Period of a wave is the inverse of its Frequency
So, Period = 1 / Frequency
From the above, we can say that Period is inversely proportional to Frequency and hence, any change in Frequency will be the inverse change in the period
Therefore, we can say that if the frequency is increased by 5 times, the period will increase by 1/5 times
what transition metal has 5 more protons than the halogen found in period 3?
Answer: Titanium (Ti)
Explanation:
First, each element has a unique atomic number Z, that is equal to the number of protons in the nucleus.
The halogen in period 3 is chlorine (Ch)
Chlorine's atomic number is Z = 17, this means that it has 17 protons.
Now we want to find a transition metal that has 5 more protons, then this transition metal has Z = 17 + 5 = 22
Now we can look at the periodic table and find the element with Z = 22, and if this is in the d-block, then this will be a transition metal.
The element with Z = 22 is titanium (Ti)
What statement is TRUE about all the substances listed in the data table? A) All the substances conduct electricity. B) All the substances are strong electrolytes. All the substances have high dissociation constants. D) All the substances contain an equal amount of ions in solution. Eliminate
Answer:
A) all substance conduct electricity
Which statement best describes energy and matter in a closed system? (2 po
O Energy and matter flow into and out of the system.
Energy can flow into and out of the system but matter cannot.
Energy and matter are contained within a closed system.
O There is no energy in a closed system; there is matter.
Answer:
Energy can flow into and out of the system but matter cannot.
Explanation:
In a closed system, energy can flow in and out of the system but matter cannot.
A closed system prevents double way flow of matter. A closed system conserves matter.For an isolated system, energy and matter cannot flow out of the system.
For open systems, energy and matter can flow out of the system.
Such systems are used for certain thermodynamics experiment.
What happens when the elevator is in free-fall, that is, what is the value for FN and what is the sensation experienced by the person in the elevator? Also, discuss the meaning of FN<0.
Answer:
Weightlessness
Explanation:
When the elevator is in free fall, this can only occur when the cord of the elevator breaks.
The acceleration of the elevator will be equal to the acceleration due to gravity. That is:
a = g
Then, the body will experience what we called WEIGHTLESSNESS
Where the normal reaction N of the person will tend to zero. That is
N = 0
The value of FN < 0 because the person inside the lift will experience weightlessness.
why does sound energy even exist
A parallel plate capacitor is made up of two metal squares with sides of length 8.8 cm, separated by a distance 5.0 mm. When a voltage 187 V is set up across the terminals of the capacitor, the charge stored on the positive plate is equal to __________ nC. g
Answer:
2.56 nC
Explanation:
By definition, the capacitance is expressed by the following relationship between the charge stored on one of the plates of the capacitor and the potential difference between them, as follows:[tex]C =\frac{Q}{V} (1)[/tex]
For a parallel-plate capacitor, assuming a uniform surface charge density σ, if the area of the plates is A, the charge on one of the plates can be written as follows:[tex]Q = \sigma * A (2)[/tex]
Assuming an uniform electric field E, the potential difference V can be expressed as follows:[tex]V = E*d (3)[/tex]
where d is the distance between plates.
Applying Gauss 'Law to a closed surface half within one plate, half outside it, we find that E can be written as follows:[tex]E =\frac{\sigma}{\epsilon_{0}} (4)[/tex]
Replacing (4) in (3), and (2) in (1), we can express the capacitance C as follows:[tex]C= \frac{\epsilon_{0}*A}{d} (5)[/tex]
Taking (1) and (5), as both left sides are equal each other, the right sides are also equal, so we can write the following equality:[tex]\frac{Q}{V} = \frac{\epsilon_{0}*A}{d} (6)[/tex]
Solving for Q, we get:[tex]Q = \frac{\epsilon_{0}*A*V}{d} = \frac{8.85e-12F/m*(0.088m)^{2}*187 V}{5.0e-3m} = 2.56 nC[/tex]
I need emergency help we only have 3 minutes left
Answer:
Option A
Explanation:
Ast the force is equal and the diayance is equal the beam is also balanced
A 4.00-kg particle moves along the x axis. Its position varies with time according to x= 5t +1 2.0t^3, where x is in meters and t is in seconds. Find:
a. the kinetic energy of the particle at any time t.
b. the acceleration of the particle and the force acting on it at time t.
c. the power being delivered to the particle at time t.
d. the work done on the particle in the interval t = 0 to t =5
Answer:
a) The kinetic energy of the particle at any time t is [tex]K = 50+120\cdot t^{2}+72\cdot t^{4}[/tex].
b) The acceleration of the particle at time t is [tex]a= 12\cdot t[/tex]. The force acting on the particle at time t is [tex]F = 48\cdot t[/tex].
c) The power being delivered to the particle at time t is [tex]\dot W = 240\cdot t +288\cdot t^{3}[/tex].
d) The work done on the particle in the interval t = 0 to t = 5 is 48000 joules.
Explanation:
a) The kinetic energy of the particle is entirely translational, whose formula is:
[tex]K = \frac{1}{2}\cdot m \cdot v^{2}[/tex] (1)
Where:
[tex]K[/tex] - Translational kinetic energy, measured in joules.
[tex]m[/tex] - Mass of the particle, measured in kilograms.
[tex]v[/tex] - Velocity of the particle, measured in meters per second.
The velocity of the particle is the rate of change of the position of the particle in time, that is:
[tex]v = 5+6\cdot t^{2}[/tex] (2)
Where [tex]t[/tex] is the time, measured in seconds.
By substituting on (1), we have the following expression: ([tex]m = 4\,kg[/tex])
[tex]K = 2\cdot (5+6\cdot t^{2})^{2}[/tex]
[tex]K = 2\cdot (25+60\cdot t^{2} +36\cdot t^{4})[/tex]
[tex]K = 50+120\cdot t^{2}+72\cdot t^{4}[/tex]
The kinetic energy of the particle at any time t is [tex]K = 50+120\cdot t^{2}+72\cdot t^{4}[/tex].
b) The acceleration of the particle is the rate of change of the velocity of the particle in time, that is:
[tex]a= 12\cdot t[/tex] (3)
Where [tex]a[/tex] is the acceleration of the particle, measured in meters per square second.
The acceleration of the particle at time t is [tex]a= 12\cdot t[/tex].
The force is obtained by multiplying (3) by the mass of the particle. That is to say: ([tex]m = 4\,kg[/tex])
[tex]F = m\cdot a[/tex] (4)
[tex]F = 48\cdot t[/tex]
The force acting on the particle at time t is [tex]F = 48\cdot t[/tex].
c) According to the Work-Energy Theorem, the change in kinetic energy of the particle equals the change in the net work done on the particle. In this case, the power is equal to the rate of change in kinetic energy.
[tex]\dot W = \dot K[/tex] (5)
[tex]\dot W = \frac{d}{dt}(50+120\cdot t^{2}+72\cdot t^{4})[/tex]
[tex]\dot W = 240\cdot t +288\cdot t^{3}[/tex]
The power being delivered to the particle at time t is [tex]\dot W = 240\cdot t +288\cdot t^{3}[/tex].
d) The work done on the particle ([tex]W[/tex]), measured in joules, is equal to the change of the kinetic energy of the particle:
[tex]W = K(5)-K(0)[/tex] (6)
[tex]W = [50+120\cdot (5)^{2}+72\cdot (5)^{4}]-[50+120\cdot (0)^{2}+72\cdot (0)^{4}][/tex]
[tex]W = 48000\,J[/tex]
The work done on the particle in the interval t = 0 to t = 5 is 48000 joules.
A 27-g steel-jacketed bullet is fired with a velocity of 640 m/s toward a steel plate and ricochets along path CD with a velocity 500 m/s. Knowing that the bullet leaves a 50-mm scratch on the surface of the plate and assuming that it has an average speed of 600 m/s while in contact with the plate, determine the magnitude and direction of the impulsive force exerted by the plate on the bullet.
Answer:
F = - 3.56*10⁵ N
Explanation:
To attempt this question, we use the formula for the relationship between momentum and the amount of movement.
I = F t = Δp
Next, we try to find the time that the average speed in the contact is constant (v = 600m / s), so we say
v = d / t
t = d / v
Given that
m = 26 g = 26 10⁻³ kg
d = 50 mm = 50 10⁻³ m
t = d/v
t = 50 10⁻³ / 600
t = 8.33 10⁻⁵ s
F t = m v - m v₀
This is so, because the bullet bounces the speed sign after the crash is negative
F = m (v-vo) / t
F = 26*10⁻³ (-500 - 640) / 8.33*10⁻⁵
F = - 3.56*10⁵ N
The negative sign is as a result of the force exerted against the bullet
1. Opposite charges
O repel
attract
Answer:
attract
Explanation:
that is the answer
Answer:
Attract.
Explanation:
I took the quiz.
Help me please!
On the earth, the gravitational field strength is 10 N/kg. On the Moon, the gravitational field strength is 1.6 N/kg.
If an object has a weight of 50 N on earth, what is its weight on the Moon?
A: 1.6 N
B: 5.0 N
C: 8.0 N
D: 80 N
Answer:
This is a way of measuring how much gravity there is. The formula is: weight/mass = gravitational field strength.
Gravitational field strength = Weight/mass unit is N/kg
Weight = mass x gravitational field strength unit is N
On Earth the gravitational field strength is 10 N/kg. Other planets have different gravitational field strengths. The Moon has a gravitational field strength of 1.6 N/kg. You might have seen films of astronauts leaping high on the moon.
Here on Earth, if I jump I am pulled back to ground by gravity. What is my weight? My mass is 80kg and if we multiply by gravitational field strength (10N/kg) - my weight is 800N. Now if I go to the moon, my mass will be the same, 80kg. We multiply that by the moon's gravitational field strength, which is 1.6 N/ kg. That means my weight on the moon is 128N. So I have different weights on the Earth and on the Moon. That's why astronauts can jump high into the air on the moon - they're lighter up there.
Jupiter is a very large planet with strong gravitational field strength of 25 N/ kg. My body is 80kg. If I go to Jupiter my weight is going to be 25 x 80 = 2,000 N. That means I wouldn't be able to get off the ground or stand up straight! I would probably be lying down all the time there. So weight varies depending on which planet you are on. You can find out more yourself by looking up tables of weight on different planets.
Mr. Jones starts from rest and begins to accelerate straight to the bathroom at a rate of 0.5 m/s for 10 seconds. What kind of motion is this
A. linear
b centripetal
c.free fall
d projectile
Answer:
A. linear
Explanation:
because they are going in a straight line
Orb spiders make silk with a typical diameter of 0.15 mm. a. A typical large orb spider has a mass of 0.50 g. If this spider suspends itself from a single 12-cm-long strand of silk, by how much will the silk stretch?b. What is the maximum weight that a single thread of this silk could support?
Answer:
(a) the change in length of the silk is 0.001585 cm
(b) the maximum weight that a single thread can support is 17.67 N
Explanation:
Given;
mass of the spider, m = 0.50 g = 0.5 x 10⁻³ kg
length of the silk, L = 12 mm = 0.012 m
diameter of the silk, d = 0.15 mm
radius of the silk, r = d / 2 = 0.075 mm = 0.075 x 10⁻³ m
The cross sectional area of the silk;
A = πr² = π(0.075 x 10⁻³)²
A = 1.767 x 10⁻⁸ m²
The Young's modulus of elasticity of spider-silk is given by;
2.1 Gpa = 2.1 x 10⁹ N/m²
(a)
Apply Young's modulus of elasticity equation to determine the change in length of the silk;
[tex]E = \frac{FL}{Ax} = \frac{F_gL}{Ax}\\\\x = \frac{F_gL}{AE}\\\\x = \frac{(0.5*10^{-3}*9.8)(0.12)}{(2.1*10^9)(1.767*10^{-8})}\\\\x = 1.585*10^{-5} \ m\\\\x = 0.01585 \ mm[/tex]
[tex]x = 0.001585 \ cm[/tex]
(b)
the maximum weight that a single thread can support is given by;
[tex]T_{tensile \ strength} = \frac{F_{max}}{A}[/tex]
The tensile strength of spider-silk is given by 1 Gpa = 1 x 10⁹ N/m²
[tex]F_{max} = T_{tensile \ strength}*A\\\\F_{max} = (1*10^9)(1.767*10^{-8})\\\\F_{max} = 17.67 \ N[/tex]