Answer:
V = 168 km³
Explanation:
...............
Definition of distance in physics
Answer:
Distance is a scalar quantity that refers to "how much ground an object has covered" during its motion. Displacement is a vector quantity that refers to "how far out of place an object is"; it is the object's overall change in position.
Explanation:
The position of a particle is given by ~r(t) = (3.0 t2 ˆi + 5.0 ˆj j 6.0 t kˆ) m
Answer:
[tex]v=(6ti+6k)\ m/s[/tex]
Explanation:
Given that,
The position of a particle is given by :
[tex]r(t) = (3.0 t^2 i + 5.0j+ 6.0 tk) m[/tex]
Let us assume we need to find its velocity.
We know that,
[tex]v=\dfrac{dr}{dt}\\\\=\dfrac{d}{dt}(3.0 t^2 i + 5.0j+ 6.0 tk) \\\\=(6ti+6k)\ m/s[/tex]
So, the velocity of the particle is [tex](6ti+6k)\ m/s[/tex].
A motorcyclist start from rest to reaches 6m/s with uniform acceleration for 3s what his acceleration?
Answer:
[tex]\boxed {\boxed {\sf 2 \ m/s^2}}[/tex]
Explanation:
Acceleration is the rate of change in velocity with respect to time. It is calculated by dividing the change in velocity by the change in time. The formula is:
[tex]a= \frac{ \Delta v}{\Delta t}[/tex] or [tex]a= \frac{v_f-v_i}{\Delta t}[/tex]
The change in velocity is the difference between the initial velocity and the final velocity. The motorcycle starts at rest, or 0 meters per second and reaches 6 meters per second. The change in time is 3 seconds.
[tex]\bullet \ v_f= 6 \ m/s\\\bullet \ v_i= 0 \ m/s \\\bullet \ \Delta t = 3 \ s[/tex]
Substitute the values into the formula
[tex]a= \frac { 6 \ m/s - 0 m/s}{3 \ s}[/tex]
Solve the numerator.
[tex]a= \frac{6 \ m/s}{3 \ s}[/tex]
[tex]a= 2 \ m/s^2[/tex]
The motorcyclist's acceleration is 2 meters per second squared.
Cho lực F ⃗=6x^3 i ⃗-4yj ⃗ tác dụng lên vật làm vật chuyển động từ A(-2,5) đến B(4,7). Vậy công của lực là:
The work done by [tex]\vec F[/tex] along the given path C from A to B is given by the line integral,
[tex]\displaystyle \int_C \mathbf F\cdot\mathrm d\mathbf r[/tex]
I assume the path itself is a line segment, which can be parameterized by
[tex]\vec r(t) = (1-t)(-2\,\vec\imath + 5\,\vec\jmath) + t(4\,\vec\imath+7\,\vec\jmath) \\\\ \vec r(t) = (6t-2)\,\vec\imath+(2t+5)\,\vec\jmath \\\\ \vec r(t) = x(t)\,\vec\imath + y(t)\,\vec\jmath[/tex]
with 0 ≤ t ≤ 1. Then the work performed by F along C is
[tex]\displaystyle \int_0^1 \left(6x(t)^3\,\vec\imath-4y(t)\,\vec\jmath\right)\cdot\frac{\mathrm d}{\mathrm dt}\left[x(t)\,\vec\imath + y(t)\,\vec\jmath\right]\,\mathrm dt \\\\ = \int_0^1 (288(3t-1)^3-8(2t+5)) \,\mathrm dt = \boxed{312}[/tex]
A student wants to start a small business in school. Write down six items that
he/she can sell in school at a profit.
Answer:
packets of pen
packets of pencil
copies
books
bottles
mask
Six items that a student can sell in school at a profit:
- Homemade baked goods
- School supplies
-Drinks
- Healthy snacks
- Personalized accessories
- Stickers
What is a profit?Profit is the difference between the revenue earned by a business or individual and the costs incurred to produce the goods or services sold.
It is an important measure of financial success for companies and is often used to determine the value of a business.
We have,
Here are six items that a student can sell in school at a profit:
Homemade baked goods - cupcakes, cookies, brownies, and other treats can be sold individually or as a pack.
School supplies - items such as pens, pencils, erasers, rulers, notebooks, and binders are always in demand.
Drinks - bottled water, juices, and sodas are popular beverages that students may purchase during the school day.
Healthy snacks - fresh fruit, granola bars, and trail mix are nutritious snacks that many students are interested in buying.
Personalized accessories - items like keychains, bracelets, and bookmarks with unique designs or student names can be popular among peers.
Stickers - fun and colorful stickers can be sold individually or in packs and are often a favorite of younger students.
Thus,
Six items that a student can sell in school at a profit:
- Homemade baked goods
- School supplies
-Drinks
- Healthy snacks
- Personalized accessories
- Stickers
Learn more about profit here:
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A standard bathroom scale is placed on an elevator. A 28 kg boy enters the elevator on the first floor and steps on the scale. What will the scale read (in newtons) when the elevator begins to accelerate upward at 0.5 m/s2
Answer:
Explanation:
Newton's Second Law is pretty much the standard for all motion that involves a force. It applies to gravitational force and torque and friction and weight on an elevator. The main formula for force is
F = ma. We have to adjust that to take into account that when the elevator is moving up, that "surge" of acceleration weighs down a bit on the scale, causing it to read higher than the actual weight until the acceleration evens out and there is no acceleration at all (no acceleration simply means that the velocity is constant; acceleration by definition is a change in velocity, and if there is no change in velocity, there is 0 acceleration). The force equation then becomes
[tex]F_n-w=ma[/tex] where [tex]F_n[/tex] is normal force. This is what the scale will read, which is what we are looking for in this problem (our unknown). Since we are looking for [tex]F_n[/tex], that is what we will solve this literal equation for:
[tex]F_n=ma+w[/tex] . m is the mass of the boy, a is the acceleration of the elevator (which is going up so we will call that acceleration positive), and w is weight. We have everything but the unknown and the weight of the boy. We find the weight:
w = mg so
w = 28(9.8) and
w = 274.4 N BUT rounding to the correct number of significance we have that the weight is actually
w = 270 N.
Filling in the elevator equation:
[tex]F_n=28(.50)+270[/tex] and according to the rules of significant digits, we have to multiply the 28(.50) {notice that I did add a 0 there for greater significance; if not that added 0 we are only looking at 1 significant digit which is pretty much useless}, round that to 2 sig fig's, and then add to 170:
[tex]F_n=14+270[/tex] and adding, by the rules, requires that we round to the tens place to get, finally:
[tex]F_n=280N[/tex] So you see that the surge in acceleration did in fact add a tiny bit to the weight read by the scale; conversely, if he were to have moved down at that same rate, the scale would have read a bit less than his actual weight). Isn't physics like the coolest thing ever!?
Why are hydraulic brakes used?
Answer:
Hydraulic brake systems are used as the main braking system on almost all passenger vehicles and light trucks. Hydraulic brakes use brake fluid to transmit force when the brakes are applied.
Explanation:
Consider a 200-ft-high, 1200-ft-wide dam filled to capacity. Determine (a) the hydrostatic force on the dam and (b) the force per unit area of the dam near the top and near the bottom. Note: we will see that the resultant hydrostatic force will be
Answer:
a) [tex]F_g=1.5*10^9Ibf[/tex]
b) [tex]F_t=12490Ibf/ft^2[/tex]
[tex]F_b=0[/tex]
Explanation:
From the question we are told that:
Height [tex]h=200ft[/tex]
Width [tex]w=1200ft[/tex]
a)
Generally the equation for Dam's Hydro static force is mathematically given by
[tex]F_g=\rho*g*\frac{h}{2}(w*h)[/tex]
Where
[tex]\rho=Density\ of\ water[/tex]
[tex]\rho=62.4Ibm/ft^3[/tex]
Therefore
[tex]F_g=62.4*32.2*\frac{200}{2}(1200*200)[/tex]
[tex]F_g=1.5*10^9Ibf[/tex]
b)
Generally the equation for Dam's Force per unit area is mathematically given by
[tex]F=\rho*g*h[/tex]
For Top
[tex]F_t=\rho*g*h[/tex]
[tex]F_t=62.4*32.2*200[/tex]
[tex]F_t=12490Ibf/ft^2[/tex]
For bottom
[tex]Here \\H=0 zero[/tex]
Therefore
[tex]F_b=0[/tex]
The hydrostatic force on the dam is [tex]2.995 \times 10^9 \ lbF[/tex].
The force per unit area near the top is 86.74 psi.
The force per unit area near the bottom is zero.
Hydrostatic force
The hydrostatic force on the dam is the force exerted on the dam by the column of the water.
[tex]F = PA\\\\F = (\rho gh) \times (wh)\\\\F = (62.4 \times 32.17 \times 200) \times (1200 \times 200)\\\\F = 9.636 \times 10^{10} \ lb-ft/s^2\\\\1 \ lbF = 32.17\ lb-ft/s^2\\\\F = 2.995 \times 10^9 \ lbF[/tex]
Force per unit area near the topThe force per unit area is the pressure exerted near the top of the dam.
[tex]P = \rho gh\\\\P = 0.052 \times \rho h[/tex]
where;
P is pressure in PSI
ρ is density of water in lb/gal
h is the vertical height in ft
[tex]P = 0.052 \times 8.34 \times 200\\\\P = 86.74 \ Psi[/tex]
The pressure near the bottom is zero, become the vertical height is zero.
Learn more about hydrostatic pressure here: https://brainly.com/question/11681616
find the rate of energy radiated by a man by assuming the surface area of his body 1.7m²and emissivity of his body 0.4
The rate of energy radiated by the man is 3.86 x [tex]10^{-8}[/tex] J/s. [tex]m^{2}[/tex].
The amount of energy radiated by an object majorly depends on the area of its surface and its temperature. The is well explained in the Stefan-Boltzmann's law which states that:
Q(t) = Aeσ[tex]T^{4}[/tex]
where: Q is the quantity of heat radiated, A is the surface area of the object, e is the emmisivity of the object, σ is the Stefan-Boltzmann constant and T is the temperature of the object.
To determine the rate of energy radiated by the man in the given question;
[tex]\frac{Q(t)}{T^{4} }[/tex] = Aeσ
But A = 1.7 m², e = 0.4 and σ = 5.67 x [tex]10^{-8}[/tex] J/s.
So that;
[tex]\frac{Q(t)}{T^{4} }[/tex] = 1.7 * 0.4 * 5.67 x [tex]10^{-8}[/tex]
= 3.8556 x [tex]10^{-8}[/tex]
= 3.86 x [tex]10^{-8}[/tex] J/s. [tex]m^{2}[/tex]
Thus, the rate of energy radiated by the man is 3.86 x [tex]10^{-8}[/tex] J/s. [tex]m^{2}[/tex].
Learn more on energy radiation of objects by visiting: https://brainly.com/question/12550129
An observer on Earth sees rocket 1 leave Earth and travel toward Planet X at 0.3c. The observer on Earth also sees that Planet X is stationary. An observer on Planet X sees rocket 2 travel toward Earth at 0.4c. What is the speed of rocket 1 according to an observer on rocket 2?
Answer:
0.625 c
Explanation:
Relative speed of a body may be defined as the speed of one body with respect to some other or the speed of one body in comparison to the speed of second body.
In the context,
The relative speed of body 2 with respect to body 1 can be expressed as :
[tex]$u'=\frac{u-v}{1-\frac{uv}{c^2}}$[/tex]
Speed of rocket 1 with respect to rocket 2 :
[tex]$u' = \frac{0.4 c- (-0.3 c)}{1-\frac{(0.4 c)(-0.3 c)}{c^2}}$[/tex]
[tex]$u' = \frac{0.7 c}{1.12}$[/tex]
[tex]u'=0.625 c[/tex]
Therefore, the speed of rocket 1 according to an observer on rocket 2 is 0.625 c
How much amount of water can be decomposed
through electrolysis by passing 2 F charge?
Answer:
So, with 2 Faraday of electricity, we can decompose (2/4 × 2) = 1 mole of water. So 18 grams of water is decomposed.
Air contained in a rigid, insulated tank fitted with a paddle wheel, initially at 300 K, 2 bar, and a volume of 2 m3, is stirred until its temperature is 500 K. Assuming the ideal gas model, for the air, and ignoring kinetic and potential energy, determine
Answer:
The final pressure in bar will be "[tex]\frac{10}{3} \ Bar[/tex]".
Explanation:
As we know,
PV = nRT
[tex]\frac{P_1}{T_1} =\frac{P_2}{T_2} =CONST[/tex]
then,
⇒ [tex]\frac{2 \ bar}{300 \ K} = \frac{P_2}{500 \ K}[/tex]
⇒ [tex]P_2=(\frac{2}{300}\times 500 )Bar[/tex]
[tex]=\frac{10}{3} \ Bar[/tex]
Thus the above is the correct answer.
Solve numerical problem. Please give me step - step explanation Help me out plz
Answer:
You should multiply 60 kg*9.8 and answer will come.
Hope this will help you.
Answer:
yes she is right you should multiple 60*9.8
have a great day God bless you
A crude approximation is that the Earth travels in a circular orbit about the Sun at constant speed, at a distance of 150,000,000 km from the Sun. Which of the following is the closest for the acceleration of the Earth in this orbit?
A. exactly 0 m/s2.
B. 0.006 m/s2.
C. 0.6 m/s2.
D. 6 m/s2.
E. 10 m/s2.
Answer:
The answer is "Option B".
Explanation:
[tex]r=15\times 10^{7}\ km\ = 15\times 10^{10}\ m\\\\w=\frac{2\pi}{1\ year}\\\\=\frac{2\pi}{1\times 365.24 \times 24 \times 60 \times 60\ sec}\\\\a=w^2r\\\\=(\frac{2\pi}{1\times 365.24 \times 24 \times 60 \times 60\ sec})^2 \times 15 \times 10^{10}\ \frac{m}{s^2}\\\\[/tex]
[tex]=5.940 \times 10^{-3} \ \frac{m}{s^2}\\\\=6 \times 10^{-3} \ \frac{m}{s^2}\\\\=0.006\ \frac{m}{s^2}\\\\[/tex]
Twin skaters approach each other with identical speeds. Then, the skaters lock hands and spin. Calculate their final angular velocity, given each had an initial speed of 2.50 m/s relative to the ice. Each has a mass of 70.0 kg, and each has a center of mass located 0.800 m from their locked hands. You may approximate
Answer:
[tex]\omega=3.135rad/s[/tex]
Explanation:
From the question we are told that:
initial Speed [tex]V_1=2.50[/tex]
Mass [tex]m=70.0kg[/tex]
Center of mass [tex]d=0.0.800m\[/tex]
Generally the equation for angular velocity is is mathematically given by
[tex]\omega=\frac{v}{r}\\\\\omega=\frac{2.50}{0.0800}[/tex]
[tex]\omega=3.135rad/s[/tex]
In a double-slit experiment, the slit separation is 1.75 mm, and two coherent wavelengths of light, 425 nm and 510 nm, illuminate the slits. At what angle from the centerline on either side of the central maximum will a bright fringe from one pattern first coincide with a bright fringe from the other pattern
Answer:
the required angle is 0.0834879⁰
Explanation:
Given the data in the question;
slit separation; d = 1.75 mm = 1.75 × 10⁻³ m
wavelength λ₁ = 425 nm = 425 × 10⁻⁹ m
wavelength λ₂ 510 nm = 510 × 10⁻⁹ m
Now, we know that, the angle at which a particular bright fringe occurs on either side of the central bright fringe will be;
tanθ = [tex]y_m[/tex] / D = mλ/d
since they both coincides;
tanθ₁ = tanθ₂
m₁λ₁/d = m₂λ₂/d
multiply both sides by d
so,
m₁/m₂ = λ₂/λ₁
we substitute
m₁/m₂ = 510 nm / 425 nm
m₁/m₂ = 510 nm / 425 nm
divide through by 85
m₁/m₂ = 6 / 5
hence m₁ and m₂ are 6 and 5
so, from the previous formula
tanθ₂ = m₂λ₂/d
we substitute
tanθ₂ = [ 5 × ( 510 × 10⁻⁹ m ) ] / 1.75 × 10⁻³ m
tanθ₂ = 255 × 10⁻⁸ m / 1.75 × 10⁻³ m
tanθ₂ = 255 × 10⁻⁸ m / 1.75 × 10⁻³ m
tanθ₂ = 0.00145714
θ₂ = tan⁻¹( 0.00145714 )
θ₂ = 0.0834879⁰
Therefore, the required angle is 0.0834879⁰
given A=4i-10j and B= 7i+5j find b such that A+bB is a vector pointing along the x-axis (i.e has no y component)
Answer:
-4/7
Explanation:
Given the following
A=4i-10j and B= 7i+5j
A+ bB = 4i-10j + (7i+5j)b
A+ bB = 4i-10j + 7ib+5jb
A+ bB =
The vector along the x-axis is expressed as i + 0j
If the vector A+ bB is pointing in the direction of the x-axis then;
[tex]A+ bB * \frac{i+0j}{|i+0j|} = 0 \\ (4+7b)i-(10-5b)j* \frac{i+0j}{\sqrt{1^2+0^2} } = 0\\(4+7b)i-(10-5b)j *(i+0j) = 0\\4+7b-0 =0\\7b=-4\\b = -4/7[/tex]
Hence the value of b is -4/7
The value of [tex]\beta[/tex] such that [tex]\vec C = \vec A + \beta \cdot \vec B = c\,\hat{i}[/tex] is 2.
According to the statement, we have following system of vectorial equations:
[tex]\vec A = 4\,\hat {i} - 10\,\hat{j}[/tex] (1)
[tex]\vec {B} = 7\,\hat{i} + 5\,\hat{j}[/tex] (2)
[tex]\vec C = \vec A + \beta \cdot \vec B = c\,\hat{i}[/tex] (3)
By applying (1) and (2) in (3):
[tex](4\,\hat{i}-10\,\hat{j}) + \beta\cdot (7\,\hat{i}+5\,\hat{j}) = c\,\hat{i}[/tex]
[tex](4+7\cdot \beta)\,\hat{i} +(-10+5\cdot \beta)\,\hat{j} = c\,\hat{i}[/tex]
And we get two scalar equations after analyzing each component:
[tex]4+7\cdot \beta = c[/tex] (4)
[tex]-10+5\cdot \beta = 0[/tex] (5)
We solve for [tex]\beta[/tex] in (5):
[tex]\beta = 2[/tex]
And for [tex]c[/tex] in (4):
[tex]c = 4+7\cdot (2)[/tex]
[tex]c = 18[/tex]
The value of [tex]\beta[/tex] such that [tex]\vec C = \vec A + \beta \cdot \vec B = c\,\hat{i}[/tex] is 2.
Please see this question related to Sum of Vectors for further details: https://brainly.com/question/11881720
From the given picture What's the force? And where did it happen? (at least 2 forces)
Answer:
the force happens on the wall and couch
Explanation:
she is using her arm strength to lift and hold
a vector starts at the point (0.0) and ends at (2,-7) what is the magnitude of the displacement
Answer:
|x| = √53
Explanation:
We are told that the vector starts at the point (0.0) and ends at (2,-7) .
Thus, magnitude of displacement is;
|x| = √(((-7) - 0)² + (2 - 0)²)
|x| = √(49 + 4)
|x| = √53
What Are the type's of Tidal turbines?
Answer:
Types of tidal turbines
Axial turbines.
Crossflow turbines.
Flow augmented turbines.
Oscillating devices.
Venturi effect.
Tidal kite turbines.
Turbine power.
Resource assessment.
Answer:
Axial turbines
Crossflow turbines
flow augmented turbines
what is the dimensional formula of young modulas
Answer:
The dimensional formula of Young's modulus is [ML^-1T^-2]
Answer:
G.oogle : The dimensional formula for Young’s modulus is:
A. [ML−1T−2]A. [ML−1T−2]
B. [M0LT−2]B. [M0LT−2]
C. [MLT−2]C. [MLT−2]
D. [ML2T−2]
Rachel has good distant vision but has a touch of presbyopia. Her near point is 0.60 m. Part A When she wears 2.0 D reading glasses, what is her near point
Answer:
The right answer is "0.273 m".
Explanation:
Given:
Power (P),
[tex]\frac{1}{f} = 2D[/tex]
Near point,
u = 0.6 m
As we know,
⇒ [tex]\frac{1}{v} -\frac{1}{u}=\frac{1}{f} = 2[/tex]
By substituting the values, we get
⇒ [tex]\frac{1}{v} -\frac{1}{0.6} =2[/tex]
[tex]\frac{1}{v}=2+\frac{1}{0.6}[/tex]
[tex]\frac{1}{v} =\frac{1.2+1}{0.6}[/tex]
[tex]\frac{1}{v}=\frac{2.2}{0.6}[/tex]
By applying cross-multiplication, we get
[tex]0.6=2.2 \ v[/tex]
[tex]v = \frac{0.6}{2.2}[/tex]
[tex]S_{near} = 0.273 \ m[/tex]
herical piece of candy is suspended in flowing water. The candy has a density of 1950 kg/m3 and has a 1.0 cm diameter. The water velocity is 1.0 m/s, the water density is assumed to be 1000.0 kg/m3, and the water viscosity is 1.010-3 kg/m/s. The diffusion coefficient of the candy solute in water is 2.010-9 m2/s, and the solubility of the candy solute in water is 2.0 kg/m3. Calculate the mass tran
Answer: Below is the complete question
A spherical piece of candy is suspended in flowing water. The candy has a density of 1950 kg/m3 and has a 1.0 cm diameter. The water velocity is 1.0 m/s, the water density is assumed to be 1000.0 kg/m3, and the water viscosity is 1.0x10-3 kg/m/s. The diffusion coefficient of the candy solute in water is 2.0x10-9 m2/s, and the solubility of the candy solute in water is 2.0 kg/m3. Calculate the mass transfer coefficient (m/s)
answer:
mass transfer coefficient = 9.56 * 10^-5 m/s
Explanation:
Candy density = 1950 kg/m^3
Candy diameter = 1 cm
Velocity of water = 1 m/s
water density = 1000 kg/m^3
Viscosity of water = 1 * 10^-3 kg/m/s
diffusion coefficient of candy in water = 2 * 10^-9 m^2/s
solubility of candy = 2 kg/m^3
Determine the mass transfer coefficient ( m/s )
( Sh) mass transfer coefficient ( flow across sphere ) = 2 + 0.6Re^1/2 * SC^1/3
where : Re = vdp / μ , Sh = KLd / Deff
attached below is the remaining solution .
mass transfer coefficient = 9.56 * 10^-5 m/s
why do atom absorb photon since it makes it unstable??
[tex]\textsf{When an electron is hit by a }[/tex] [tex]\textsf{photon of light, it absorbs the quanta}[/tex] [tex]\textsf{of energy the photon was carrying}[/tex] [tex]\textsf{and moves to a higher energya}[/tex] [tex]\textsf{ state. Electrons therefore have to }[/tex] [tex]\textsf{jump around within the atom as }[/tex] [tex]\textsf{they either gain or lose energy. }[/tex]
When an electron is hit by a photon of light, it absorbs the quanta of energy the photon was carrying and moves to a higher energy state. Electrons therefore have to jump around within the atom as they either gain or lose energy.
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The area around a charged object that can exert a force on other charged objects is an electric ___
A rugby player passes the ball 7.00 m across the field, where it is caught at the same height as it left his hand.
(a) At what angle was the ball thrown if its initial speed was 12.0 m/ s, assuming that the smaller of the two possible angles was used?
(b) What other angle gives the same range, and why would it not be used?
(c) How long did this pass take?
Answer:
a) θ = 14.23º, b) θ₂ = 75.77, c) t = 0.6019 s
Explanation:
This is a missile throwing exercise.
a) the reach of the ball is the distance traveled for the same departure height
R = [tex]\frac{v_o^2 \ sin 2 \theta }{g}[/tex]
sin 2θ = [tex]\frac{Rg}{v_o^2}[/tex]
sin 2θ = 7.00 9.8 / 12.0²
2θ = sin⁻¹ (0.476389) = 28.45º
θ = 14.23º
the complementary angle that gives the same range is the angle after 45 that the same value is missing to reach 90º
θ ’= 90 -14.23
θ’= 75.77º
b) the two angles that give the same range are
θ₁ = 14.23
θ₂ = 75.77
the greater angle has a much greater height so the time of the movement is greater and has a greater chance of being intercepted by the other team.
C) the time of the pass can be calculated with the expression
x = v₀ₓ t
t = x / v₀ₓ
t = 7 / 11.63
t = 0.6019 s
A 2.5 kg block slides along a frictionless surface at 1.5 m/s.A second block, sliding at a faster 4.1 m/s , collides with the first from behind and sticks to it. The final velocity of the combined blocks is 2.5 m/s. What was the mass of the second block?
Answer:
1.5kg
Explanation:
Given data
mass m1= 2.5kg
mass m2=??
velocity of mass one v1= 1.5m/s
velocity of mass two v2= 4.1m/s
common velocity after impact v= 2.5m/s
Let us apply the formula for the conservation of linear momentum for inelastic collision
The expression is given as
m1v1+ m2v2= v(m1+m2)
substitute
2.5*1.5+ m2*4.1= 2.5(2.5+m2)
3.75+4.1m2= 6.25+2.5m2
collect like terms
3.75-6.25= 2.5m2-4.1m2
-2.5= -1.6m2
divide both sides by -1.6
m2= -2.5/-1.6
m2= 1.5 kg
Hence the second mass is 1.5kg
If at a particular instant and at a certain point in space the electric field is in the x-direction and has a magnitude of 3.70 V/m, what is the magnitude of the magnetic field of the wave at this same point in space and instant in time
Answer:
the magnitude of the magnetic field is 1.23 x 10⁸ T.
Explanation:
Given;
magnitude of the electric field, E = 3.7 V/m
The magnitude of the magnetic field is calculated as;
E = cB
where;
B is the magnitude of the magnetic field
c is the speed of light = 3 x 10⁸ m/s
From the above equation, the magnetic field, B, is calculated as;
[tex]B = \frac{E}{c} \\\\B = \frac{3.7 }{3\times 10^8 } \\\\B = 1.23 \times 10^{-8 } \ T[/tex]
Therefore, the magnitude of the magnetic field is 1.23 x 10⁸ T.
What is the mass of the diver in (Figure 1) if she exerts a torque of 2200 N⋅m on the board, relative to the left (A) support post?
A-->B = 1.0m
B--> end of board = 3.0m
Answer:
56.1 kg
Explanation:
Given
[tex]T = 2200Nm[/tex] -- torque
[tex]d_1 = 1.0m[/tex]
[tex]d_2 = 3.0m[/tex]
Required
The mass of the diver
From the question, we understand that the diver is at the extreme of the board.
So, we make use of the following torque equation
[tex]T = F * (d_1 + d_2)[/tex]
Where:
[tex]F \to Force[/tex]
So, we have:
[tex]2200 = F * (1.0 + 3.0)[/tex]
[tex]2200 = F * 4.0[/tex]
Divide both sides by 4.0
[tex]550 = F[/tex]
[tex]F = 550 N[/tex] --- This is the force exerted by the diver (in other words, the weight).
To calculate the mass, we use:
[tex]F = mg[/tex]
Make m the subject
[tex]m = \frac{F}{g}[/tex]
This gives:
[tex]m = \frac{550}{9.8}[/tex]
[tex]m = 56.1kg[/tex]
A 15kg mass suspended from a ceiling is pulled aside with a horizontal force, F. Calculate the value of the tension.
Answer:
147 Newtons
Explanation:
To find tension, you can use the formula Tension = (mass)(gravity)
*Gravity's acceleration = 9.8 m/s^2 because of Newton's law of universal gravitation*
T = (15kg)(9.8m/s^2)
= 147 Newtons
Hope this helps! Best of luck <3
