Given data
*The given moment of inertia is I = 3.45 kg.m^2
*The given braking torque is T = -9.40 N.m
*The angular distance traveled is
[tex]\theta=(1\times2\pi)rad_{}[/tex]*The final angular speed is
[tex]\omega=0\text{ rad/s}[/tex]The angular acceleration of the flywheel is calculated by using the torque and moment of inertia relation as
[tex]\begin{gathered} T=I\alpha \\ \alpha=\frac{T}{I} \\ =\frac{-9.4}{3.45} \\ =-2.72rad/s^2 \end{gathered}[/tex]The formula for the initial angular speed of the flywheel is given by the rotational equation of motion as
[tex]\omega^2-\omega^2_0=2a\theta[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} (0)^2-\omega^2_0=2\times(-2.72)(2\pi) \\ \omega_0=\sqrt[]{2\times2.72\times2\pi} \\ =5.88\text{ rad/s} \end{gathered}[/tex]Hence, the initial angular speed of the flywheel is 5.88 rad/s
What is the index of refraction for a medium where light has a velocity of 2.75 x10³ m/s?0.9176.751.090.250
n = c/v, where:
n: index of refraction
c: speed of light in a vacuum (3*10^8 m/s)
v: speed of light in medium
n = (3*10^8)/(2.75*10^8)
n = 3/2.75
n = 1.09
Bryan slid a glass of O] for Dalton across a counter with a speed of 2.1m/s. Unfortunately, Dalton missed it! If the countertop was 1.1m above the floor a) how long did it take for the glass to hit the floor? b) how far horizontally did the glass travel in the air?
A. The time taken for the glass to hit the floor is 0.5 s
B. The horizontal distance travelled by the glass in the air is 1.1 m
A. How to determine the time
The following data were obtained from the question:
Height (h) = 1.1 mAcceleration due to gravity (g) = 9.8 m/s²Time (t) = ?The time taken to hit the ground can be obtained as follow:
h = ½gt²
1.1 = ½ × 9.8 × t²
1.1 = 4.9 × t²
Divide both side by 4.9
t² = 1.1 / 4.9
Take the square root of both side
t = √(1.1 / 4.9)
t = 0.5 s
Thus, the time taken is 0.5 s
B. How to determine horizontal distance
The horizontal distance can be obatined as follow:
Horizontal speed (u) = 2.1 m/sTime (t) = 0.5 sHorizont distance (s) = ?s = ut
s = 2.1 × 0.5
s = 1.1 m
Thus, the horizontal distance is 1.1 m
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Describe the mathematical relationship between the distance (d) and the attractive force (F) between protons and electrons.
The attractive force and the distance are inversely proportional.
[tex]F\propto\frac{1}{r}[/tex]This relation means that the attractive force decreases as the distance increases, and the attractive force increases as the distance decrease.
Describe
protons.
Location:
Charge:
Mass:
The protons in an atom are classified according to their mass, charge, and location as follows:
Particle: Protons
Mass: 1.67262 × 10⁻²⁷ kg
Charge: Positive charge (+e or +1)
Location: Found in the nucleus of every atom
The proton is a stable subatomic particle with a rest mass of 1.67262 x 10⁻²⁷ kg, or 1,836 times the mass of an electron, with a positive charge that is equal to one electron's charge in magnitude.
All atomic nuclei, with the exception of the hydrogen nucleus, are composed of protons and neutrons, which are electrically neutral particles (that consist of a single proton). A given chemical element's nuclei all contain the same number of protons. This number establishes an element's atomic number and establishes the element's position in the periodic table. An atom is electrically neutral when the number of protons in its nucleus equals the number of electrons in its orbit.
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You have a concave mirror with a focal length of 100.0 cm, and you want an image that is upright and 10.0 times as tall as the object. Where should you place the object?Select one:a.9.09 cmb.12.09 cmc.7.12 cmd.5.8 cm
ANSWER:
a. 9.09cm
STEP-BY-STEP EXPLANATION:
Given:
Focal length = 100 cm
Distance image = -10 cm
We use the following formula to calculate the value of the distance of the object:
[tex]\frac{1}{d_o}+\frac{1}{d_i}=\frac{1}{f}[/tex]We substitute each value and calculate the distance of the object just like this:
[tex]\begin{gathered} \frac{1}{d_o}+\frac{1}{-10}=\frac{1}{100} \\ \\ \frac{1}{d_{o}}=\frac{1}{100}+\frac{1}{10} \\ \\ \frac{1}{d_o}=\frac{11}{100} \\ \\ d_o=\frac{100}{11} \\ \\ d_o=9.09\text{ cm} \end{gathered}[/tex]Therefore, the correct answer is option a. 9.09cm
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A pair of fuzzy dice is hanging by a string from your rearview mirror. While you are accelerating from a stop-light to 28 m/s in 6.0 s, what angle theta does the string make with the vertical?
For the pair of fuzzy dice hanging by a string from the rearview mirror, while accelerating from a stop-light to 28 m/s in 6.0 s, the string makes an angle of 25.5 degrees with the vertical.
What is an angle?An angle is the figure formed by two rays, called the sides of the angle, sharing a common endpoint, called the vertex of the angle.
Parameters given include:
velocity: 28,/s time= 6.0 seconds
a = v/t = 28m/s / 6.0s = 4.667 m/s²
Taking into account the forces acting on the dice...there is a force of gravity acting straight down with a magnitude of mg.
so we have that
Angle = tan-1(a/g)
Angle = tan-1(4.667 / 9.8) = 25.5 degrees.
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A fireman standing on a 14 m high ladderoperates a water hose with a round nozzle ofdiameter 2.65 inch. The lower end of the hose(14 m below the nozzle) is connected to thepump outlet of diameter 3.49 inch. The gaugepressure of the water at the pump isCalculate the speed of the water jet emerg-ing from the nozzle. Assume that water is anincompressible liquid of density 1000 kg/m3and negligible viscosity. The acceleration ofgravity is 9.8 m/s?Answer in units of m/s.
Given data,
The height, H = 14 m
The diameter, D = 2.65 inch
The gauge pressure, P = 317.84 kPa
We need to calculate the speed of the water jet emerging from the nozzle.
Using Bernoulli's equation,
[tex]\begin{gathered} \frac{1}{2}\rho(v^2_n-v^2_p)=P_{gauge\text{ }}-\rho gh \\ (v^2_n-v^2_p)=(\frac{2}{\rho})P_{gauge}-2gh \\ v^2_n-(\frac{A_n}{A_p})^2v^2_n=(\frac{2}{\rho})P_{gauge}-2gh \\ v^2_n-(\frac{r_n}{r_p_{}})^4v^2_n=(\frac{2}{\rho})P_{gauge}-2gh \end{gathered}[/tex]Further solved as,
[tex]\begin{gathered} v_n=\sqrt[]{\frac{(\frac{2}{\rho})P_{gauge}-2gh}{1-(\frac{r_n}{r_p})^4}} \\ v_n=\sqrt[]{\frac{(\frac{2}{1000})\times317.84\times10^3-2\times9.8\times14}{1-(\frac{1.325_{}}{1.745_{}})^4}} \\ v_n=\sqrt[]{\frac{635-274.4}{0.667}} \\ v_n=\sqrt[]{540.62} \end{gathered}[/tex]Thus, the speed of the water jet is
[tex]v=23.25\text{ m/s}[/tex]A marble was thrown vertically up the air and came back down to the samelevel in 1.6 s. (using g = 9.81 ms?).(a) What is it initial velocity?(b) What would be the highest height it can reach?(c) How long would it take to reach the highest point?
ANSWER:
(a) 15.7 m/s
(b) 12.56 m
(c) 0.8 sec
STEP-BY-STEP EXPLANATION:
We can calculate the initial velocity using the following formula:
(a)
[tex]\begin{gathered} v=u+a\cdot t \\ v=0 \\ \text{therefore:} \\ u=-a\cdot t \\ \text{ replacing} \\ u=-(-9.81\cdot1.6) \\ u=15.7\text{ m/s} \end{gathered}[/tex](b)
We calculate the height in this way:
[tex]\begin{gathered} h=\frac{(-u)^2}{2g} \\ \text{ replacing} \\ h=\frac{(-15.7)^2}{2\cdot9.81} \\ h=12.56\text{ m} \end{gathered}[/tex](c)The time to reach this height would be half the time it takes for the entire journey, that is, 0.8 seconds
In order to hear a sound, even though there is an obstacle between you and the source, the sound wave must:A.diffract.B.refract.C.shorten.D.reflect.
We will have the following:
The sound must diffract. [Option A]
Two cars travel in the same direction along a straight highway, one at a constant speed of 55 mi/h and the other at 60 mi/h.(a) Assuming they start at the same point, how much sooner does the faster car arrive at a destination 14 mi away?_____ min(b) How far must the faster car travel before it has a 15-min lead on the slower car?_____ mi
We will have the following:
a) We first determine the time it takes to travel the distance to both vehicles:
*
[tex]t_1=\frac{14mi\ast1h}{55}\Rightarrow t_1=\frac{14}{55}h[/tex]*
[tex]t_2=\frac{14mi\ast1h}{60mi}\Rightarrow t_2=\frac{11}{12}h[/tex]So, we determine now the difference in time:
[tex]\frac{11}{12}h-\frac{14}{55}h=\frac{437}{660}h\approx0.66h[/tex]So, the fastest car will arrive approximately 0.66 hours sooner.
b) We determmine the distance it must travel the fastest car to have a 15 minute lead on the other one as follows:
First, we determine the time difference required:
[tex]t=\frac{15min\ast1h}{60min}\Rightarrow t=0.25h[/tex]Then, since both vehicles will move relative to each other, we will have that:
[tex]d_{c1}=(60mi/h)(0.25h)\Rightarrow d_{c1}=15mi[/tex]So, the fastest car must be 15 miles ahead of the other car in order to have a 15 minute lead with respect to the second car.
An electron is in an infinite one dimension well that is 8.9 nm wide. What is the ground state energy of the electron?
The ground state energy of the electron is 2.23 x 10⁻¹⁷ J.
What is the ground state energy of the electron?The ground state energy of the electron is calculated by applying the formula for energy of photons.
E = hf
E = hc/λ
where;
h is Planck's constantc is speed of lightλ is the wavelengthE = (6.626 x 10⁻³⁴ x 3 x 10⁸) / (8.9 x 10⁻⁹)
E = 2.23 x 10⁻¹⁷ J
Thus, the ground state energy of the electron is determined by applying the principle or formula for energy of a single photon at the given wavelength of 8.9 nm.
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A wooden sphere of mass 4.0 kg is completely immersed in water. A pushing force of 20. N is
applied.
21°
20 N
4.2 ms²
19⁰
At the moment shown in the diagram, the sphere is stationary and it experiences an
acceleration upwards and to the right as shown.
Calculate the size of the upwards force due to the water (upthrust) acting on the sphere.
The size of the upwards force due to the water (upthrust) acting on the sphere is 12.64 N.
What is upthrust?
Buoyancy or upthrust, is an upward force exerted by a fluid on an object immersed in the fluid due to the weight of the object
Thus, upthrust is the upward force acting on an object immersed in a liquid.
Fu - Df = F(net_u)
where;
Fu is the upward forceDf is the downward force applied on the objectF(net_u) is the net upward forceFu - F x sin(21) = ma x sin(19)
where;
m is the mass of the wooden spherea is the upward acceleration of the wooden sphereFu - 20 x sin(21) = (4 x 4.2) x sin(19)
Fu - 7.17 = 5.47
Fu = 12.64 N
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Find the direction of this vector -22.2m, 12.6m
Answer:
Angle:
α = - 62°
Explanation:
Given:
Nₓ = 12.6 m
Ny = - 22.2 m
_________
α - ?
The direction of vector:
tg α = Ny / Nₓ
tg α = ( - 22.2 ) / 12.0 ≈ - 1.85
Angle:
α = arctg ( - 1.85) ≈ - 62°
The direction of the vector is [tex]\alpha=-62^o[/tex]. Given Components: Nₓ = 12.6 m and Ny = -22.2 m. The negative sign indicates that the angle is measured clockwise from the positive x-axis.
Finding the Tangent of Angle α:
- The tangent of an angle α can be calculated using the formula: tan(α) = opposite/adjacent. In this case, the opposite side is Ny, and the adjacent side is Nₓ. So, you have:
tan(α) = Ny / Nₓ
tan(α) = (-22.2) / 12.6 ≈ -1.85
Finding the Angle α:
- To find the angle α, you need to take the arctangent (inverse tangent) of the value you calculated above. This will give you the angle whose tangent is approximately -1.85:
α = arctan(-1.85)
α ≈ -62°
So, based on the given vector components Nₓ and Ny, the angle α is approximately -62 degrees.
The negative sign indicates that the angle is measured clockwise from the positive x-axis.
The angle obtained is in degrees. If you need the angle in radians, you could convert it using the relationship: radians = degrees × (π / 180).
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How much heat in kcal must be added to 0.68 kg of water at room temperature (20°C) to raise its temperature to 45°C?answer:____ kcal
Given:
Mass, m = 0.68 kg
Initial temperature, T1 = 20°C
Final temperature, T2 = 45°C
Let's find the amount of heat needed.
Apply the Specific Heat Capacity formula:
[tex]\begin{gathered} Q=mc\Delta T \\ \\ Q=mc(T_2-T_1) \end{gathered}[/tex]Where:
m is the mass = 0.68 kg
c is the specific heat of water = 4.187 J/kg °C−1
T1 = 20°C
T2 = 45°C
Plug in the values and solve for Q:
[tex]\begin{gathered} Q=0.68*4.187*(45-20) \\ \\ Q=2.84716(25) \\ \\ Q=71.179\text{ kJ} \end{gathered}[/tex]Where:
1 kJ = 0.239 kCal
Since the answer is to be in kCal, we have:
[tex]Q=71.179*0.239=17.01\text{ kCal}[/tex]Therefore, the amount if heat added is 17.01 kCal.
ANSWER:
17.01 kCal
A baseball player pitches a fastball toward home plate at a speed of 41.0 m/s. The batter swings, connects with the ball of mass 195 g, and hits it so that the ball leaves the bat with a speed of 37.0 m/s. Assume that the ball is moving horizontally just before and just after the collision with the bat.A. What is the impulse delivered to the ball by the bat? Enter a positive value if the impulse is in the direction the bat pushes the ball and enter a negative value if the impulse is in the opposite direction the bat pushes the ball. (kg m/s)B. If the bat and ball are in contact for 3.00 ms, what is the magnitude of the average force exerted on the ball by the bat? (kN)
Given:
Initial velocity, vi = 41.0 m/s
Mass of ball, m = 195 g = 0.195 kg
Final velocity, vf = 37.0 m/s
Assuming the ball is moving horizontally just before and after collision with the bat, let's solve for the following:
• (A). What is the impulse delivered to the ball by the bat?
To find the impulse, apply the change in momentum formula:
[tex]\Delta p=p_f-p_i[/tex]Where:
pi is the initial momentum = -mvi
pf is the final momentum = mvf
Thus, we have:
[tex]\begin{gathered} \Delta p=mv_f-(-mv_i) \\ \\ \Delta p=mv_f+mv_i \\ \\ \Delta p=m(v_f+v_i) \\ \\ \Delta p=0.195(37.0+41.0) \\ \\ \Delta p=15.21\text{ kg}\cdot\text{ m/s} \end{gathered}[/tex]Impulse can be said to equal change in momentum.
Therefore, the impulse delivered to the ball by the bat is 15.21 kg.m/s away from the bat.
• (B). If the bat and ball are in contact for 3.00 ms, what is the magnitude of the average force exerted on the ball by the bat?
Apply the formula:
[tex]\text{ Impulse = Force }\ast\text{ time}[/tex]Rewrite the formula for force:
[tex]\text{ Force=}\frac{impulse}{time}[/tex]Where:
time = 3.00 m/s
impulse = 15.21 kg.m/s
Hence, we have:
[tex]\begin{gathered} \text{ F=}\frac{15.21}{3} \\ \\ F\text{ = 5.07 kN} \end{gathered}[/tex]Therefore, the magnitude of the average force exerted on the ball by the bat is 5.07 kN away from the bat.
ANSWER:
(A). 15.21 kg.m/s away from the bat
(B). 5.07 kN.
A student makes the following claim, "Acceleration is when an object changes speed, so it can be discussed as a scalar quantity." Explain the error in the student's claim. Provide an example of each quantity to support your answer.
A student makes the following claim, "Acceleration is when an object changes speed, so it can be discussed as a scalar quantity." The error here is that acceleration is said to be done when either speed of that object changes or direction of that object changes. Hence , acceleration is not a scaler quantity.
Scalar quantities are quantities that are described only by a magnitude. They do not have a direction of action.
A vector quantity is defined as the physical quantity that has both directions as well as magnitude.
Acceleration is said to be occurred in two cases :
when the object changes its speed
or when the object changes its direction
since , acceleration depends upon both direction and magnitude ,hence it is a vector quantity not a scaler quantity.
for example : a stone attached to a string moving in a circular motion at a constant speed will be considered in accelerated motion because it is constantly changing its direction.
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According to Figure 2. the solar radiation Intensity 8.000 years ago was closest towhich of the following?490 watts/m20495 watts/m?O 500 watts/m2O 505 watts/m2
From the given figure, let's determine the solar radiation intensity 8000 years ago.
We can see the solar radiation intensity is represented on the left vertical (left side of the y-axis), while the number of years is represented on the x-axis.
Using the graph, at 8 thousand(8000) years ago, the radiation intensity was closest to 500 watts/m².
T
A 3 kg ball is dropped from a height of 100 m above the surface of Planet Z. If the ball reaches a velocity of 45 m/s in 7 s, what is the ball’s weight on Planet Z? What is the gravitational field strength on Planet Z?
We are given the following information
Mass of ball = 3 kg
Height = 100 m
Final velocity = 45 m/s
Time = 7 s
Recall from the equations of motion
[tex]s=u\cdot t+\frac{1}{2}\cdot g\cdot t^2[/tex]Where u is the initial velocity of the ball that is zero.
[tex]\begin{gathered} s=u\cdot t+\frac{1}{2}\cdot g\cdot t^2 \\ 100=0\cdot7+\frac{1}{2}\cdot g\cdot7^2 \\ 100=\frac{1}{2}\cdot g\cdot49 \\ g=\frac{2\cdot100}{49} \\ g=4.08\; \frac{m}{s^2} \end{gathered}[/tex]So, the gravitational acceleration of the planet Z is 4.08 m/s^2
The weight o the ball is given by
[tex]\begin{gathered} W=m\cdot g \\ W=3\cdot4.08 \\ W=12.24\; N \end{gathered}[/tex]Therefore, the weight of the ball is 12.24 N
Ball A with diameter d and ball B with diameter 2d are dropped from the same height. When the two balls have the same speed, what is the ratio of the drag force on ball A to the drag force on ball B?
Ball A with diameter d and ball B with diameter 2d are dropped from the same height. When the two balls have the same speed, the ratio of the drag force on ball A to the drag force on ball B will be F1 : F2 = 1 : 4
When objects travel through fluids (a gas or a liquid), they will undoubtedly encounter resistive forces called drag forces.
The drag force always acts in the opposite direction to fluid flow. If the body’s motion exists in the fluid-like air, it is called aerodynamic drag.
formula to calculate drag force is = F(d) = 1/2 * C * rho*A * [tex]v^{2}[/tex]
C = drag coefficient
A = area of object
rho = density in which object is moving
v = velocity of object
A = area of the object
F1 ( drag force on ball A ) = 1/2 * C * rho * area of ball A * [tex]v^{2}[/tex]
F2 (drag force on ball A ) = 1/2 * C * rho * area of ball B * [tex]v^{2}[/tex]
since , both the balls have same speed and falling in same environment hence , density and speeds are the same , the only difference is in area of both the balls
F1/F2 = area of ball A / area of ball B = 4 * pi * [tex]r1^{2}[/tex] / 4 * pi * [tex]r2^{2}[/tex]
= [tex]r1^{2}[/tex] / [tex]r2^{2}[/tex]
= [tex](\frac{d}{2} )^{2}[/tex]/ [tex](\frac{2d}{2}) ^{2}[/tex]
= 1/4
F1 : F2 = 1 : 4
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A sled of mass 26 kg has an 18 kg child on it. If big brother is pulling with a 30 N force to the right and 10 N up, and big sister is pushing with a 40 N force to the right and 16 N down, what is the normal force?
Given data:
* The mass of the sled is m_1 = 26 kg.
* The mass of the child is m_2 = 18 kg.
* The force in the upwards direction by the big brother is F_1 = 10 N.
* The force in the downwards direction by the big sister is F_2 = 16 N.
Solution:
The net mass on the sled along with the child is,
[tex]\begin{gathered} m=m_1+m_2 \\ m=26+18 \\ m=44\text{ kg} \end{gathered}[/tex]The net weight of the sled along with the child is,
[tex]\begin{gathered} w=mg \\ w=44\times9.8 \\ w=431.2\text{ N} \end{gathered}[/tex]The weight of the sled along the child is acting on the sled in the downwards direction.
Thus, the normal force acting on the sled (taking upward force as negative and downward force as positive) is,
[tex]\begin{gathered} N=w+F_2-F_1 \\ N=431.2+16-10 \\ N=437.2\text{ newton} \end{gathered}[/tex]Thus, the normal force acting on the sled is 437.2 N.
44) Find the x coordinate of the center of mass of the bricks shown.
We are asked to determine the x-coordinate of the center of mass of the given bricks. To do that, we will use the following formula:
[tex]\bar{x}=\frac{\Sigma x_im_i}{\Sigma m_i}[/tex]Where:
[tex]\begin{gathered} x_i=\text{ x-coordinate of the center of mass of each brick} \\ m_i=\text{ mass of each brick} \end{gathered}[/tex]Since we have three bricks, the formula expands to:
[tex]\bar{x}=\frac{x_1m_1+x_2m_2_{}+x_3m_3}{m_1+m_2+m_3}[/tex]Since we have three bricks with the same characteristics we will assume the three of them have the same mass:
[tex]\bar{x}=\frac{x_1m_{}+x_2m+x_3m_{}}{m_{}+m_{}+m_{}}[/tex]Taking "m" as a common factor and adding like terms in the denominator we get:
[tex]\bar{x}=\frac{m(x_1+x_2+x_3)}{3m}[/tex]Now we cancel out the "m":
[tex]\bar{x}=\frac{x_1+x_2+x_3}{3}[/tex]Now we determine the x-coordinates of each brick. Each brick is a parallelepiped, therefore, the x-coordinate is in the middle. Since each brick measures L, this means that the x-coordinate of the first brick is:
[tex]x_1=\frac{L}{2}[/tex]For the second brick, we have the L/2 of the separation from the first plus the L/2 of its length, therefore:
[tex]x_2=\frac{L}{2}+\frac{L}{2}=L[/tex]Now, for the third brick we have the L/4 of the separation from the second brick plus the L/2 of the separation of the second brick and the first brick and the L/2 of the length of the third brick, therefore:
[tex]x_3=\frac{L}{2}+\frac{L}{4}+\frac{L}{2}=\frac{5L}{4}[/tex]Now we substitute in the formula for the x-coordinate:
[tex]\bar{x}=\frac{(\frac{L}{2})+(L)+(\frac{5L}{4})}{3}[/tex]Adding like terms in the numerator:
[tex]\bar{x}=\frac{\frac{11L}{2}}{3}[/tex]Simplifying:
[tex]\bar{x}=\frac{11L}{6}[/tex]Therefore, the x-coordinate of the center of mass is located at 11L/6 from the origin.
A football player runs from his own goal line to the opposing team's goal line, returning to his twenty-yard line, all in 27.0 s. Calculate his average speed and the magnitude of his average velocity. (Enter your answers in yards/s.)HINTApply the definitions of average speed and average velocity.Click the hint button again to remove this hint.(a) Calculate his average speed. ____yards/s(b) Calculate the magnitude of his average velocity. ____yards/s
time = 27 s
d = 20 yard
a) speed = distance / time
d1 = 100 yard ( own goal to opposing team's goal line)
d2 = 80 yard ( returning to 20 yard line)
d= d1+d2
d= 100 + 80 = 180 yards
Speed = 180 y / 27s = 6.667 y/s
b) velocity = displacement / time
d = 100 - 80 = 20 y
Velocity = 20 / 27 = 0.74 y/s
A spring with a spring constant of 52N/m sits on a desk. The spring is 34cm long. A block of mass 0.12kg is placed on top of the spring. How high above the desk does the block rest?
Answer:
H = 31.7 cm
Explanation:
Given:
k = 52 N/m - Spring rate
L = 34 cm = 0.34 m - Spring length
m = 0.12 kg
g = 9.8 m/s²
____________
H - ?
Block weight:
F = m*g = 0.12·9.8 ≈ 1.18 N
According to Hooke's law:
F = k·ΔL
The spring is compressed by:
ΔL = F / k = 1.18 / 52 ≈ 0.023 м
Height of the block above the table:
H = L -ΔL = 0.34 - 0.023 = 0.317 m or H = 31.7 cm
A robotic arm lifts a barrel of radioactive waste, as shown in the figure.
If the maximum torque delivered by the arm about the axis O is 3.00 x 10° N•m and the distance r is 3.00 m, what is the maximum mass m of the barrel?
The maximum mass of the barrel, with maximum torque of 3.0×10° N.m delivered about the axis is 0.102 kg.
What is mass?Mass is the quantity of matter a body contains. The S.I unit of mass is kilogram (kg). Mass can also be defined as the ratio of the force to the acceleration of a body.
To calculate the maximum mass of the barrel, we use the formula below.
Formula:
m = τ/dg........... Equation 1Where:
m = Maximum Mass of the barrel τ = maximum Torque deliveredd = Distanceg = Acceleration due to gravity.From the question,
Given:
τ = 3.00×10° N.md = 3.00 mg = 9.8 m/s²Substitute the values above into equation 1
m = (3.00×10°/3×9.8)m = 1/9.8 kg.m = 0.102 kg.Hence, the maximum mass of the barrel is 0.102 kg.
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To BEST avoid any accidents when swimming, you should NEVER swim:
A. with a parent.
B. with a partner.
C. alone.
D. near a lifeguard.
Answer:
Explanation:
c. Alone
Answer:
C. alone.
Explanation:
Multiple Choice: A car with a mass of 825 kg moves along the roadway at aspeed of 15 m/s to the east. What impulse is required to decrease the speed of theboat to 10 m/s east?
Given:
The mass of the car is m = 825 kg
The intial speed of the car is
[tex]v_i\text{ = 15 m/s}[/tex]towards east.
The final speed of the car is
[tex]v_f\text{ = 10 m/s}[/tex]To find the impulse of the car.
Explanation:
The impulse can be calculated by the formula
[tex]Impulse\text{ = mv}_f-mv_i[/tex]On substituting the values, the impulse will be
[tex]\begin{gathered} Impulse\text{ = \lparen825}\times10\text{\rparen-\lparen825}\times15\text{\rparen} \\ =\text{ -4125 kg m/s} \end{gathered}[/tex]The impulse will be 4125 kg m/s due
A ski jumper competing for an Olympic gold metal wants to jump
a horizontal distance of 149 meters. The takeoff point of the ski
jump is at a height of 38.0 meters. With what horizontal velocity
must he leave the jump in order to travel 149 meters?
19.25 m/s is horizontal velocity must he leave the jump in order to travel 149 meters .
How fast is horizontal moving?Standard definitions of horizontal velocity include miles per hour and meters per second, which are horizontal displacement times time. The distance an object has traveled since its origin is simply referred to as displacement.
How can one calculate vertical velocity using horizontal velocity?V * cos() equals the horizontal velocity component Vx. V * sin() is equal to the vertical component of velocity, Vy.
Time before landing = sqrt ( 2 x height / gravity ), sqrt ( 2 x 38/ 9.81) = 7.75
distance / time = avg speed
149/ 7.75 ≅ 19.25 m/s
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Which picture below correctly identifies the effort length and lifting length of a lever?Select one:a. Ab. Bc. Cd. D
d.D
Explanation
A lever is a simple machine made of a rigid beam and a fulcrum.
where
the Load is the object which we are lifting,Fulcrum is the point at which the lever is pivoted. and Effort is he force applied to make the object move
Step 1
check for the graph that:
the lifting length goes from the fulcrum to the load
and
the effort length goes from the fulcrum to the applied force,
therefore,
the answer is
d.D
I hope this helps you
Which optical instrument produces a magnified, virtual, and inverted image of small objects?1) a refracting telescope2) a single lens reflex camera3) a microscope4) a pair of binoculars
Microscope
Explanations:Microscopes are known for magnifying tiny objects
The images produced by a microscope are:
virtual (formed behind the screen)
Inverted
Magnified or enlarged
Therefore, the optical instrument which produces a magnified, virtual, and inverted image is the microscope
A puck is moving on an air hockey table. Relative to an x,y coordinate system at time t =0s, the x
components of the puck's initial velocity and acceleration are Vix=1.0 m/s and ax=2.0 m/s². The y
components of the puck's initial velocity and acceleration are Viy=2.0 m/s and ay=2.0 m/s². Find the
magnitude and direction of the puck's velocity at a time of t=0.50 s. Specify the direction relative to
the x axis. HELPP!!!
The supplied puck is moving at a speed of v0x=+3.4m/s on an air hockey table at time t=0.
What is the meaning of velocity?The direction of the movement of the body or the object is defined by its velocity. Most of the time, speed is a scalar quantity. In its purest form, velocity is a vector quantity. It measures how quickly a distance changes. It is the rate at which displacement is changing.
What does the term "tangential velocity" refer to?Any object traveling in a circular motion has a linear speed known as tangential velocity. On a turntable, a point in the center moves less distance in a full rotation than a point near the outside edge.
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