Answer:
Coefficient of kinetic friction = 0.146
Explanation:
Given:
Mass of sled (m) = 18 kg
Horizontal force (F) = 30 N
FInal speed (v) = 2 m/s
Distance (s) = 8.5 m
Find:
Coefficient of kinetic friction.
Computation:
Initial speed (u) = 0 m/s
v² - u² = 2as
2(8.5)a = 2² - 0²
a = 0.2352 m/s²
Nweton's law of :
F (net) = ma
30N - μf = 18 (0.2352)
30 - 4.2336 = μ(mg)
25.7664 = μ(18)(9.8)
μ = 0.146
Coefficient of kinetic friction = 0.146
HELPP MEE
Which image illustrates the desired interaction of a sound wave with
soundproofing material in a recording studio?
Soundproofing material is required for blocking sound during some works like recording voice in the studio. Image D represents the interaction of a sound wave with soundproofing material in a recording studio.
What is the basis of soundproofing?Soundproofing is done by absorbing the sound. A very much used material for this is a dense foam.
Foam and like materials absorbs sound and it travels directly into the soft surface resulting in soundproofing.
Thus, the correct option is C, as the D image is showing the absorption.
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Answer: C.D
Explanation:...
A sample of silver (with work function Φ=4.52 eV ) is exposed to an ultraviolet light source (????=200 nm), which results in the ejection of photoelectrons. What changes will be observed if:
1. The silver is replaced with copper (Φ= 5.10 eV)?
a. more energetic photoelectrons (on average)
b. no photoelectrons are emitted more photoelectrons ejected
c. less energetic photoelectrons (on average)
d. fewer photoelectrons ejected
2. A second (identical) light source also shines on the metal?
a. fewer photoelectrons ejected
b. no photoelectrons are emitted more
c. energetic photoelectrons (on average)
d. less energetic photoelectrons (on average)
e. more photoelectrons ejected
3. The ultraviolet source is replaced with an X-ray source that emits the same number of photons per unit time as the original ultraviolet source?
a. no photoelectrons are emitted
b. less energetic photoelectrons (on average)
c. fewer photoelectrons ejected
d. more energetic photoelectrons (on average)
e. more photoelectrons ejected
Answer:
1. c
2. e
3. d
Explanation:
1.
From Einstein's Photoelectric Equation, we know that:
Energy given up by photon = Work Function + K.E of Electron
hc/λ = φ + K.E
where,
h = Plank's Constant = 6.626 x 10⁻³⁴ J.s
c = speed of light = 3 x 10⁸ m/s
λ = wavelength of light source = 200 nm = 2 x 10⁻⁷ m
φ = (5.1 eV)(1.6 x 10⁻¹⁹ J/eV) = 8.16 x 10⁻¹⁹ J
Therefore,
(6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(2 x 10⁻⁷ m) - 8.16 x 10⁻¹⁹ = K.E
K.E = (9.939 - 8.16) x 10⁻¹⁹ J
K.E = 1.778 x 10⁻¹⁹ J
The positive answer shows that electrons will be emitted. Since it is clear from the equation the the K.E of electron decreases with the increase in work function. Therefore:
c. less energetic photo-electrons (on average)
2.
The increase in light sources means an increase in the intensity of light. The no. of photons are increased, due to increase of intensity. Thus, more photons hit the metal and they eject greater no. of electrons. Therefore,
e. more photo-electrons ejected
3.
X-rays have smaller wavelength and greater energy than ultraviolet rays. Thus, the photons with greater energy will strike the metal and as a result, electrons with higher energy will be ejected.
d. more energetic photo-electrons (on average)
Suppose the demand for air travel decreases (as illustrated in the graph below). A decrease in demand _____ the equilibrium price for air travel and _____ the equilibrium quantity for air travel. decreases, decreases increases, increases decreases, increases
Answer:
decreases, decreases
Explanation:
A decrease in the demand will create a fall in equilibrium prices and the quantity supplied will also decrease. As the equilibrium prices in the market are the price in which the quantity demanded equals to quantity supplied. If the demand for the air decreases then the quantity of the air travel will also decrease and thus when the supply and demand change so do the changes associated with the equilibrium prices.Assume the three blocks (m. = 1.0 kg, m = 20 kg and m = 40 ko) portrayed in the figure below move on a frictionless surface and a force F: 36w acts as shown on the 4.0 kg block.
a) Determine the acceleration given this system (in m/s2 to the right). m/s2 (to the right)
b) Determine the tension in the cord connecting the 4.0 kg and the 1.0 kg blocks in N). Determine the force exerted by the 1.0 kg block on the 2.0 kg block (in N). N (a) What If How would your answers to parts (a) and (b) of this problem change if the 2.0 kg block was now stacked on top of the 1.0 kg block? Assume that the 2.0 kg block sticks to and does not slide on the 1.0 kg block when the system is accelerated.
(Enter the acceleration in m/s2 to the right and the tension in N.) acceleration m/s (to the right) tension
Answer:
a) 5.143 m/s^2
b) T = 15.43 N
c) Fr = 10.29 N
d) 5.143 m/s^2 , T = 15.43 N
Explanation:
Given:-
- The mass of left most block, m1 = 1.0 kg
- The mass of center block, m2 = 2.0 kg
- The mass of right most block, m3 = 4.0 kg
- A force that acts on the right most block, F = 36 N
Solution:-
a)
- For the first part we will consider the three blocks with masses ( m1 , m2 , and m3 ) as one system on which a force of F = 36 N is acted upon. The masses m1 and m3 are connected with a string with tension ( T ) and the m1 and m2 are in contact.
- We apply the Newton's second law of motion to the system with acceleration ( a ) and the combined mass ( M ) of the three blocks as follows:
[tex]F = M*a\\\\36 = ( 1 + 2 + 4 )*a\\\\a = \frac{36}{7}\\\\a = 5.143 \frac{m}{s^2}[/tex]
Answer: The system moves in the direction of external force ( F ) i.e to the right with an acceleration of 5.143 m/s^2
b)
- The blocks with mass ( m1 and m3 ) are connected with a string with tension ( T ) with a combined acceleration of ( a ).
- We will isolate the massive block ( m3 ) and notice that two opposing forces ( F and T ) act on the block.
- We will again apply the Newton's 2nd law of motion for the block m3 as follows:
[tex]F_n_e_t = m_3 * a\\\\F - T = m_3 * a\\\\36 - T = 4*5.143\\\\T = 36 - 20.5714\\\\T = 15.43 N[/tex]
Answer:- A tension of T = 15.43 Newtons acts on both blocks ( m1 and m3 )
c)
- We will now isolate the left most block ( m1 ) and draw a free body diagram. This block experiences two forces that is due to tension ( T ) and a reaction force ( Fr ) exerted by block ( m2 ) onto ( m3 ).
- Again we will apply the the Newton's 2nd law of motion for the block m3 as follows:
[tex]F_n_e_t = m_1*a\\\\T - F_r = m_1*a\\\\15.43 - F_r = 1*5.143\\\\F_r = 15.43 - 5.143\\\\F_r = 10.29 N[/tex]
- The reaction force ( Fr ) is contact between masses ( m1 and m2 ) exists as a pair of equal magnitude and opposite direction acting on both the masses. ( Newton's Third Law of motion )
Answer: The block m2 experiences a contact force of ( Fr = 10.29 N ) to the right.
d)
- If we were to stack the block ( m2 ) on-top of block ( m1 ) such that block ( m2 ) does not slip we the initial system would remain the same and move with the same acceleration calculated in part a) i.e 5.143 m/s^2
- We will check to see if the tension ( T ) differs or not as the two block ( m1 and m2 ) both experience the same Tension force ( T ) as a sub-system. with a combined mass of ( m1 + m2 ).
- We apply the Newton's 2nd law of motion for the block m3 as follows:
[tex]T = ( m_1 + m_2 ) *a\\\\T = ( 1 + 2 ) * 5.143\\\\T = 15.43 N[/tex]
Answer: The acceleration of the whole system remains the same at a = 5.143 m/s^2 and the tension T = 15.43 N also remains the same.
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At what distance from a 70.0 Watt speaker is the intensity 0.0195 W/m^2
(Treat the speaker as point of the source)
(Unit=meters)
PLEASE HELP ME!
Answer:
Distance = 16.9 m
Explanation:
We are given;
Power; P = 70 W
Intensity; I = 0.0195 W/m²
Now, for a spherical sound wave, the intensity in the radial direction is expressed as a function of distance r from the center of the sphere and is given by the expression;
I = Power/Unit area = P/(4πr²)
where;
P is the sound power
r is the distance.
Thus;
Making r the subject, we have;
r² = P/4πI
r = √(P/4πI)
r = √(70/(4π*0.0195))
r = √285.6627
r = 16.9 m
Answer:
16.9 m
Explanation:
To practice Problem-Solving Strategy 6.1: Circular motion A highway curve with radius R = 274 m is to be banked so that a car traveling v = 25.0 m/s will not skid sideways even in the absence of friction. At what angle should the curve be banked?
Answer:
The curve should be banked at an angle of 13 degrees.
Explanation:
We have,
Radius of a highway curve is 274 m
Speed of car on this curve is 25 m/s
Let [tex]\theta[/tex] is the banking angle. On a banked curve, the angle of safe diving is given by following expression.
[tex]\tan\theta=\dfrac{v^2}{Rg}[/tex]
g = 10 m/s²
Plugging all the values in above formula,
[tex]\tan\theta=\dfrac{(25)^2}{274\times 9.8}\\\\\theta=\tan^{-1}\left(\dfrac{(25)^{2}}{274\times9.8}\right)\\\\\theta=13^{\circ}[/tex]
So, the curve should be banked at an angle of 13 degrees.
Mr. Patel is photocopying lab sheets for his first period class. A particle of toner carrying a charge of 4.0 * 10^9 C in the copying machine experiences an electric field of 1.2 * 10^6 N/C as it’s pulled toward the paper. What is the electric force acting on the toner particle?
Answer:
4.8 × 10^15 N
Explanation:
Electric Field is defined as Force per unit Charge.
This is expressed mathematically as;
E= F/Q
Where E- Electric Field
F- Force
Q- charge
From the expression above by change of subject of formula for F, we have;
F=E×Q
= 1.2 * 10^6 ×4.0 * 10^9
= 4.8 × 10^15 N
Parallel light rays with a wavelength of 610nm fall on a single slit. On a screen 3.10m away, the distance between the first dark fringes on either side of the central maximum is 4.00mm.
What is the width of the slit?
Answer:
The width of the slit will be ".946 mm".
Explanation:
The given values are:
Wavelength = 610 × 10⁻⁹
Length, L = 3 m
As we know,
⇒ [tex]\frac{y}{L} = \frac{m(wavelength)}{a}[/tex]
On putting the estimated values, we get
⇒ [tex]\frac{2\times 10^{-3}}{3.1} = \frac{(1)(610 X 10^{-9})}{a}[/tex]
On applying cross-multiplication, we get
⇒ [tex]a=9.46\times 10^{-4}[/tex]
⇒ [tex]a = .946 mm[/tex]
A rifle fires a 2.05 x 10-2-kg pellet straight upward, because the pellet rests on a compressed spring that is released when the trigger is pulled. The spring has a negligible mass and is compressed by 8.01 x 10-2 m from its unstrained length. The pellet rises to a maximum height of 4.46 m above its position on the compressed spring. Ignoring air resistance, determine the spring constant.
Answer:
Spring Constant = 279.58 N/m
Explanation:
We are given;
Mass; m = 2.05 x 10^(-2) kg = 0.0205 kg
Distance of compression; x = 8.01 × 10^(-2) m = 0.0801 m
Maximum height; h = 4.46 m
The formula for the energy in the spring is given by;
E = ½kx²
where:
k is the spring constant
x is the distance the spring is compressed.
Now, this energy of the spring will be equal to the energy of the pellet at its highest point. Energy of pallet = mgh So;
½kx² = mgh
Plugging in the relevant values, we have;
½ * k * 0.0801² = 0.0205 * 9.81 * 4.46
0.003208005k = 0.8969
k = 0.8969/0.003208005
k = 279.58 N/m
An object will sink in a liquid if the density of the object is greater than that of the liquid. The mass of a sphere is 0.723 g. If the volume of this sphere is less than ________ cm3, then the sphere will sink in liquid mercury (density
Answer:
= 0.0532 cm^3
Explanation:
The computation of volume of the sphere is shown below:-
[tex]Density = \frac{Mass}{Volume}[/tex]
Where,
Density = 13.6 g/cm^3
Mass of sphere = 0.723 g
now we will put the values into the above formula to reach volume of the sphere which is here below:-
[tex]Volume = \frac{0.723}{13.6}[/tex]
= 0.0532 cm^3
Therefore for computing the volume of the sphere we simply applied the above formula.
4) (7 pts.) A water molecule is centered at the origin of a coordinate system with its dipole moment vector aligned with the x axis. The magnitude of a water molecule dipole is 6.16 × 10−30 C·m. What is the magnitude of the electric field at x = 3.00 × 10−9 m?
Answer:
[tex]E=3.69*10^{-11}\frac{V}{m}[/tex]
Explanation:
To solve this problem you use the following formula, for the calculation of the electric field along the axis of the dipole.
[tex]E=\frac{p}{2\pi \epsilon_ox^3}[/tex] (1)
p: dipole moment = 6.16*10^-30 Cm
x: distance to the center of mass of the dipole = 3.00*10^-9m
eo: dielectric permittivity of vacuum = 8.85*10^-12 C^2/Nm^2
You replace the values of the variables in the equation (1):
[tex]E=\frac{6.16*10^{-30}Cm}{2\pi(8.85*10^{-12}C^2/Nm^2)(3.00*10^{-9}m)^3}\\\\E=3.69*10^{-11}\frac{V}{m}[/tex]
please help
Complete the first and second sentences, choosing the correct answer from the given ones.
1. A temperature of 100 K corresponds on a Celsius scale to 100 ° C / 0 ° C / 173 ° C / –173 ° C.
2. At 50 ° C, it corresponds to a Kelvin scale of 150 K / 323 K / 273 K / 223 K.
1) 100 ° C
2) 323 K
hope it helps youuuuuu
Two cars start moving from the same point. One travels south at 60 miyh and the other travels west at 25 miyh. At what rate is the distance between the cars increasing two hours later?
Answer:
65 m/h
Explanation:
Let the distance of the car moving south be y.
Let the distance of the car moving west be x.
Let the distance between the two cars be a.
These three distances can be represented as a right angled triangle. So we can say:
[tex]a^2 = x^2 + y ^2[/tex]
Let us differentiate with respect to time, since the distances are changing with respect to time:
[tex]2a\frac{da}{dt} = 2x\frac{dx}{dt} + 2y\frac{dy}{dt} \\\\=>a\frac{da}{dt} = x\frac{dx}{dt} + y\frac{dy}{dt}[/tex]__________(1)
da/dt = rate of change of distance between two cars
The speed of the car moving south (dy/dt) is 60 m/h and the speed of the car moving west (dx/dt) is 25 m/h.
Therefore:
dy/dt = 60 m/h and dx/dt = 25 m/h
After two hours, the distance of the two cars will be:
y = 2 * 60 = 120 miles
x = 2 * 25 = 50 miles
Therefore:
[tex]a^2 = 50^2 + 120^2\\\\a^2 = 2500 + 14400 = 16900\\\\a = \sqrt{16900}\\ \\a = 130 miles[/tex]
From (1):
130(da/dt) = 50(25) + 120(60)
130(da/dt) = 1250 + 7200 = 8450
da/dt = 8450/130 = 65 m/h
Therefore, after two hours, the distance between the two cars is changing at a rate of 65 m/h.
What is the velocity of a car that travels 556km northwest in 3.2 hours
Answer:
173.75 km/hr in the NW direction.
Explanation:
Velocity is the time rate of change in displacement of a body. Mathematically:
v = d / t
where d = displacement
t = time
Therefore, the velocity of the car is:
v = 556 / 3.2 = 173.75 km/hr
The velocity of the car is 173.75 km/hr in the NW direction.
The velocity of a car will be "173.75 km/hr".
Displacement and Velocity,The velocity of something like a car moving northward on something like a prominent motorway as well as the velocity of something like a rocket launching towards spacecraft both might be determined or monitored.
Displacement, d = 556 km
Time, t = 3.2 hours
We know the relation,
→ Velocity = [tex]\frac{Displacement}{Time}[/tex]
or,
→ V = [tex]\frac{d}{t}[/tex]
By substituting the values, we get
= [tex]\frac{556}{3.2}[/tex]
= [tex]173.75[/tex] km/hr
Thus the response above is right.
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Someone plzzz helpppppp with this last question
Answer:
I dont know someone deleted answers. But they were wrong. INERTIA IS CORRECT I DID THIS IN MY SCHOOL
C IS CORRECT
4. Mrs. Parker was married to her husband for
30 years. They lived together with their two
children,
(A) Single-parent family
(B) Nuclear family
(C) Blended family
(D) Extended family
I think it’sd
Explanation:
The answer is B because Nuclear family mean a family with two kids and Mrs. Parker have two kids
What type of device forms images by changing the speed at which light travels?
Answer:
A lens
Explanation:
A lens forms images when light passes Through it bending the rays of in the process.A phenomena called refraction and the speed of light changes in the process because it enters a medium since it's wavelength is reduced.
The type of device that forms images by changing the speed at which light travels is the lens.
What is refraction through the lens?
A lens bends a light beam at an aimed perspective and converges or diffuses bundles of rays by taking benefit of refraction taking vicinity while the mild travels from air into glass or plastic. For that purpose, the aspect geometry of a lens has a spherical parent, which can be kind of divided into sorts.
A lens bends a mild beam at an aimed perspective and converges or diffuses bundles of rays through taking gain of refraction taking area whilst the mild travels from air into glass or plastic. For that motive, the facet geometry of a lens has a round parent, which may be kind of divided into sorts.
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A subatomic particle X spontaneously decays into two particles, A and B, each of rest energy 1.40 × 10^2 MeV. The particles fly off in opposite directions, each with speed 0.827c relative to an inertial reference frame S. What is the total energy of particle A?
Answer:
E = 389 MeV
Explanation:
The total energy of particle A, will be equal to the sum of rest mass energy and relative energy of particle A. Therefore,
Total Energy of A = E = Rest Mass Energy + Relative Energy
Using Einstein's Equation: E = mc²
E = m₀c² + mc²
From Einstein's Special Theory of Relativity, we know that:
m = m₀/[√(1-v²/c²)]
Therefore,
E = m₀c² + m₀c²/[√(1-v²/c²)]
E = m₀c²[1 + 1/√(1-v²/c²)]
where,
m₀c² = rest mass energy = 140 MeV
v = relative speed = 0.827 c
Therefore,
E = (140 MeV)[1 + 1/√(1 - (0.827c)²/c²)]
E = (140 MeV)(2.78)
E = 389 MeV
How many significant figures does 0.09164500561 have?
Answer:
10 Sig Figs
Explanation:
Just start counting at the first non zero after the decimal so in this case the nine, and count all of the numbers including zeros after that.
A tank circuit consists of an inductor and a capacitor. Give a simple explanation for why the magnetic field in the induc- tor is strongest at the moment that the separated charge in the capacitor reaches zero.
Answer:
If you pull a permanent magnet rapidly away from a tank circuit, what is likely to happen in that circuit?
Charge will oscillate in the tank's capacitor and inductor.
Explanation:
A particle moving along the x-axis has a position given by m, where t is measured in s. What is the magnitude of the acceleration of the particle at the instant when its velocity is zero
Question:
A particle moving along the x-axis has a position given by x=(24t - 2.0t³)m, where t is measured in s. What is the magnitude of the acceleration of the particle at the instant when its velocity is zero
Answer:
24 m/s
Explanation:
Given:
x=(24t - 2.0t³)m
First find velocity function v(t):
v(t) = ẋ(t) = 24 - 2*3t²
v(t) = ẋ(t) = 24 - 6t²
Find the acceleration function a(t):
a(t) = Ẍ(t) = V(t) = -6*2t
a(t) = Ẍ(t) = V(t) = -12t
At acceleration = 0, take time as T in velocity function.
0 =v(T) = 24 - 6T²
Solve for T
[tex] T = \sqrt{\frac{-24}{6}} = \sqrt{-4} = -2 [/tex]
Substitute -2 for t in acceleration function:
a(t) = a(T) = a(-2) = -12(-2) = 24 m/s
Acceleration = 24m/s
Three sheets of plastic have unknown indices of refraction. Sheet 1 is placed on top of sheet 2, and a laser beam is directed onto the sheets from above so that it strikes the interface at an angle of 26.50 with the normal. The refracted beam in sheet 2 makes an angle of 31.70 with the normal. The experiment is repeated with sheet 3 on top of sheet 2, and with the same angle of incidence, the refracted beam makes an angle of 36.70 with the normal. If the experiment is repeated again with sheet 1 on top of sheet 3, determine the expected angle of refraction in sheet 3? Assume the same angle of incidence.
Answer:
The angle of refraction of sheet 3 when sheet 1 is on top of it is [tex]\theta_{r_s } = 23.1 ^o[/tex]
Explanation:
From the question we are told that
The angle of incidence is [tex]\theta _i = 26.50 ^o[/tex]
The angle of refraction angle for sheet 1 is [tex]\theta _{r_1}} = 31.70 ^o[/tex]
The angle of refraction for sheet 3 is [tex]\theta _{r_3}} = 36.70 ^o[/tex]
According to Snell's law
[tex]\frac{n_2}{n_1} = \frac{sin (\theta_1)}{sin (\theta_{r_1})}[/tex]
Where [tex]n_1 \ and \ n_2[/tex] are refractive index of sheet 1 and sheet 2
=> [tex]n_2 = n_1 \frac{sin(\theta_i)}{sin (\theta _{r_1})}[/tex]
Also when sheet 3 in on top of sheet 2
[tex]\frac{n_2}{n_3} = \frac{sin \theta_i}{sin \theta_{r_3}}[/tex]
substituting for [tex]n_2[/tex]
[tex]n_1 \frac{sin(\theta_i)}{sin (\theta _{r_1})} = n_3 \frac{sin \theta_i}{sin \theta_{r_3}}[/tex]
[tex]n_1 \frac{sin(\theta_i)}{sin (\theta _{r_1})} = n_3 \frac{sin \theta_i}{sin \theta_{r_3}}[/tex]
=> [tex]n_3 = n_1 * \frac{sin(\theta_{r_3})}{sin(\theta_{r_1})}[/tex]
when sheet 1 in on top of sheet 3
[tex]\frac{n_3}{n_1} = \frac{sin(\theta_i)}{\theta_{r_s}}[/tex]
where [tex]r_s[/tex] is the angle of refraction when sheet 1 is on top of sheet 3
substituting for [tex]n_3[/tex]
[tex]\frac{ n_1 * \frac{sin(\theta_{r_3})}{sin(\theta_{r_1})}}{n_1} = \frac{sin(\theta_i)}{\theta_{r_s}}[/tex]
=> [tex]sin (\theta _{r_s}) = n_1 * sin (\theta_i) * \frac{sin (\theta_{r_1})}{ n_1 * sin(\theta_{r_3})}[/tex]
substituting values
[tex]sin (\theta _{r_s}) = n_1 * sin (26.50) * \frac{sin (31.70)}{ n_1 * sin(36.70)}[/tex]
=> [tex]\theta_{r_s } = sin^{-1} (0.3923)[/tex]
=> [tex]\theta_{r_s } = 23.1 ^o[/tex]
If the outer conductor of a coaxial cable has radius 2.6 mm , what should be the radius of the inner conductor so that the inductance per unit length does not exceed 50 nH per meter? Express your answer using two significant figures.
Answer:
Inner radius = 2 mm
Explanation:
In a coaxial cable, series inductance per unit length is given by the formula;
L' = (µ/(2π))•ln(R/r)
Where R is outer radius and r is inner radius.
We are given;
L' = 50 nH/m = 50 × 10^(-9) H/m
R = 2.6mm = 2.6 × 10^(-3) m
Meanwhile µ is magnetic constant and has a value of µ = µ_o = 4π × 10^(−7) H/m
Plugging in the relevant values, we have;
50 × 10^(-9) = (4π × 10^(−7))/(2π)) × ln(2.6 × 10^(-3)/r)
Rearranging, we have;
(50 × 10^(-9))/(2 × 10^(−7)) = ln((2.6 × 10^(-3))/r)
0.25 = ln((2.6 × 10^(-3))/r)
So,
e^(0.25) = (2.6 × 10^(-3))/r)
1.284 = (2.6 × 10^(-3))/r)
Cross multiply to give;
r = (2.6 × 10^(-3))/1.284)
r = 0.002 m or 2 mm
1. (a) The battery on your car has a rating stated in ampere-minutes which permits you to
estimate the length of time a fully charged battery could deliver any particular current
before discharge. Approximately how much energy is stored by a 50 ampere-minute 12
volt battery?
Answer:
Energy is stored by a 50 ampere-minute 12
volt battery is approximately = 36,000 J = 36 kJ
Explanation:
Power in electrical circuits is given as
Power = IV
But power generally is defined as energy expended per unit time
Power = (Energy/time)
Energy = Power × Time
Energy = IV × Time
Energy = (I.t × V)
I.t = 50 Ampere-minute = 50 × 60 = 3000 Ampere-seconds
V = 12 V
Energy = 3,000 × 12 = 36,000 J = 36 kJ
Hope this Helps!!!
An 89.2-kg person with a density 1025 kg/m3 stands on a scale while completely submerged in water. What does the scale read?
Answer:
89.11kg
Explanation:
Note an object weighs less when in a fluid and the weight of the volume of the fluid displaced is known as the upthrust.
Now, the person is going to displace the volume 89/1025 =0.087m3 { from density D = mass(M)/volume(V)}
The weight of the fluid displaced is the density of the fluid × volume of fluid displaced.
The weight of the fluid=0.087m3× 1kg/me = 0.087kg
Now the weight of the fluid displaced is referred to as the upthrust.
Now the real weight - the apparent weight = the upthrust.
Hence the apparent weight = real weight - upthrust
Apparent weight = 89.2-0.087 = 89.11kg
¿Cuantos metros recorre una motocicleta en un segundo si circula a una velocidad de 90km/h?
Answer:
La motocicleta recorre 25 metros en 1 segundo si circula a una velocidad de 90 km/h
Explanation:
La velocidad es una magnitud que expresa el desplazamiento que realiza un objeto en una unidad determinada de tiempo, esto es, relaciona el cambio de posición (o desplazamiento) con el tiempo.
Siendo la velocidad es el espacio recorrido en un período de tiempo determinado, entonces 90 km/h indica que en 1 hora la motocicleta recorre 90 km. Entonces, siendo 1 h= 3600 segundos (1 h=60 minutos y 1 minuto=60 segundos) podes aplicar la siguiente regla de tres: si en 3600 segundos (1 hora) la motocicleta recorre 90 km, entonces en 1 segundo ¿cuánta distancia recorrerá?
[tex]distancia=\frac{1 segundo*90 km}{3600 segundos}[/tex]
distancia= 0.025 km
Por otro lado, aplicas la siguiente regla de tres: si 1 km es igual a 1,000 metros, ¿0.025 km cuántos metros son?
[tex]distancia=\frac{0.025 km*1,000 metros}{1 km}[/tex]
distancia= 25 metros
La motocicleta recorre 25 metros en 1 segundo si circula a una velocidad de 90 km/h
La cantidad de metros que recorre una motocicleta en un segundo si viaja a una velocidad de 90 km / h es de 25 m / s.
Para obtener la velocidad de la motocicleta en un segundo, necesitaremos convertir 90 km / h en metros por segundo
Usando la tasa de conversión;
1000 m = 1 km
1 hora = 3600 segundos
[tex]\frac{90km}{hr} = \frac{90km \times 1000 m}{1km \times 3600s} \\\\\frac{90km}{hr} = \frac{90,000 m}{3600s} =25m/s \\[/tex]
Esto muestra que la cantidad de metros que recorre una motocicleta en un segundo si viaja a una velocidad de 90 km / h es de 25 m / s.
Obtenga más información aquí: https://brainly.com/question/13136984
A high diver of mass 60.0 kg steps off a board 10.0 m above the water and falls vertical to the water, starting from rest. If her downward motion is stopped 2.10 s after her feet first touch the water, what average upward force did the water exert on her
Answer:
The average upward force exerted by the water is 988.2 N
Explanation:
Given;
mass of the diver, m = 60 kg
height of the board above the water, h = 10 m
time when her feet touched the water, t = 2.10 s
The final velocity of the diver, when she is under the influence of acceleration of free fall.
V² = U² + 2gh
where;
V is the final velocity
U is the initial velocity = 0
g is acceleration due gravity
h is the height of fall
V² = U² + 2gh
V² = 0 + 2 x 9.8 x 10
V² = 196
V = √196
V = 14 m/s
Acceleration of the diver during 2.10 s before her feet touched the water.
14 m/s is her initial velocity at this sage,
her final velocity at this stage is zero (0)
V = U + at
0 = 14 + 2.1(a)
2.1a = -14
a = -14 / 2.1
a = -6.67 m/s²
The average upward force exerted by the water;
[tex]F_{on\ diver} = mg - F_{ \ water}\\\\ma = mg - F_{ \ water}\\\\F_{ \ water} = mg - ma\\\\F_{ \ water} = m(g-a)\\\\F_{ \ water} = 60[9.8-(-6.67)]\\\\F_{ \ water} = 60 (9.8+6.67)\\\\F_{ \ water} = 60(16.47)\\\\F_{ \ water} = 988.2 \ N[/tex]
Therefore, the average upward force exerted by the water is 988.2 N
cellus
An object ends up at a position of
327 m after a displacement of -144 m.
What was its initial position?
(Unit = m)
Answer:
Its initial position was 471 m.
Explanation:
We have,
Final position of the object is 327 m
Displacement of the object is -144 m
It is required to find its initial position. The difference of final and initial position is equal to the displacement of the object. So,
[tex]d=\text{final position}-\text{initial position}\\\\-144=327-\text{initial position}\\\\\text{initial position}=327+144\\\\\text{initial position}=471\ m[/tex]
So, its initial position was 471 m.
g Tether ball is a game children play in which a ball hangs from a rope attached to the top of a tall pole. The children hit the ball, causing it to swing around the pole. What is the total initial acceleration of a tether ball on a 2.0 m rope whose angular velocity changes from 13 rad/s to 7.0 rad/s in 15 s
Answer:
a = -0.8 m/s²
Here, negative sign indicates that the acceleration has opposite direction to the direction of motion.
Explanation:
First we find the angular acceleration of the ball from the following formula:
α = (ωf - ωi)/t
where,
α = angular acceleration = ?
ωf = final angular velocity = 7 rad/s
ωi = initial angular velocity = 13 rad/s
t = Time taken = 15 s
Therefore,
α = (7 rad/s - 13 rad/s)/15 s
α = - 0.4 rad/s
negative sign shows that acceleration is in opposite direction to the direction of motion.
Now, for the linear acceleration, we use the formula:
a = rα
where,
a = linear acceleration = ?
r = radius of circular path = length of rope = 2 m
therefore,
a = (2 m)(- 0.4 rad/s²)
a = -0.8 m/s²
Here, negative sign indicates that the acceleration has opposite direction to the direction of motion.
wheel rotates from rest with constant angular acceleration. Part A If it rotates through 8.00 revolutions in the first 2.50 s, how many more revolutions will it rotate through in the next 5.00 s?
Answer:
x2 = 64 revolutions.
it rotate through 64 revolutions in the next 5.00 s
Explanation:
Given;
wheel rotates from rest with constant angular acceleration.
Initial angular speed v = 0
Time t = 2.50
Distance x = 8 rev
Applying equation of motion;
x = vt +0.5at^2 ........1
Since v = 0
x = 0.5at^2
making a the subject of formula;
a = x/0.5t^2 = 2x/t^2
a = angular acceleration
t = time taken
x = angular distance
Substituting the values;
a = 2(8)/2.5^2
a = 2.56 rev/s^2
velocity at t = 2.50
v1 = a×t = 2.56×2.50 = 6.4 rev/s
Through the next 5 second;
t2 = 5 seconds
a2 = 2.56 rev/s^2
v2 = 6.4 rev/s
From equation 1;
x = vt +0.5at^2
Substituting the values;
x2 = 6.4(5) + 0.5×2.56×5^2
x2 = 64 revolutions.
it rotate through 64 revolutions in the next 5.00 s
