Answer:
Approximately [tex]325[/tex] (rounded down,) assuming that [tex]g = 9.81\; {\rm N \cdot kg^{-1}}[/tex].
The number of repetitions would increase if efficiency increases.
Explanation:
Ensure that all quantities involved are in standard units:
Energy from the cookie (should be in joules, [tex]{\rm J}[/tex]):
[tex]\begin{aligned} & 53\; {\rm kCal} \times \frac{1\; {\rm kJ}}{4.184\; {\rm kCal}} \times \frac{1000\; {\rm J}}{1\; {\rm kJ}} \approx 2.551 \times 10^{5}\; {\rm J} \end{aligned}[/tex].
Height of the weight (should be in meters, [tex]{\rm m}[/tex]):
[tex]\begin{aligned} h &= 2\; {\rm dm} \times \frac{1\; {\rm m}}{10\; {\rm dm}} = 0.2\; {\rm m}\end{aligned}[/tex].
Energy required to lift the weight by [tex]\Delta h = 0.2\; {\rm m}[/tex] without acceleration:
[tex]\begin{aligned} W &= m\, g\, \Delta h \\ &= 100\; {\rm kg} \times 9.81\; {\rm N \cdot kg^{-1}} \times 0.2\; {\rm m} \\ &= 196\; {\rm N \cdot m} \\ &= 196\; {\rm J} \end{aligned}[/tex].
At an efficiency of [tex]0.25[/tex], the actual amount of energy required to raise this weight to that height would be:
[tex]\begin{aligned} \text{Energy Input} &= \frac{\text{Useful Work Output}}{\text{Efficiency}} \\ &= \frac{196\; {\rm J}}{0.25} \\ &=784\; {\rm J}\end{aligned}[/tex].
Divide [tex]2.551 \times 10^{5}\; {\rm J}[/tex] by [tex]784\; {\rm J}[/tex] to find the number of times this weight could be lifted up within that energy budget:
[tex]\begin{aligned} \frac{2.551 \times 10^{5}\; {\rm J}}{784\; {\rm J}} &\approx 325 \end{aligned}[/tex].
Increasing the efficiency (the denominator) would reduce the amount of energy input required to achieve the same amount of useful work. Thus, the same energy budget would allow this weight to be lifted up for more times.
Water moves through a constricted pipe in steady, ideal flow. At the lower point shown in Figure 1, the pressure is and the pipe diameter is 6.0 cm. At another point higher, the pressure is and the pipe diameter is 3.0 cm. Find the speed of flow (a) in the lower section and (b) in the upper section. (c) Find the volume flow rate through the pipe
Answer:
full explanation and answer is on the picture
In a recent study of how mice negotiate turns, the mice ran around a circular 90∘ turn on a track with a radius of 0.10 m. The maximum speed measured for a mouse (mass = 18.5 g) running around this turn was 1.39 m/s. Part A What is the minimum coefficient of friction between the track and the mouse's feet that would allow a turn at this speed?
The minimum coefficient of friction between the track and the mouse's feet that would allow a turn at this speed without sleeping is 1.97.
Minimum coefficient of frictionThe minimum coefficient of friction is calculated as follows;
F = μmg = mac
μmg = mac
where;
ac is centripetal accelerationμg = ac
μg = v²/r
μ = v²/rg
where;
μ is the minimum coefficient of frictionv is velocityr is radius of the pathg is acceleration due to gravityμ = (1.39²)/(0.1 x 9.8)
μ = 1.97
Thus, the minimum coefficient of friction between the track and the mouse's feet that would allow a turn at this speed without sleeping is 1.97.
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As the pressure on a sample of gas increases, the volume of the sample
Decrease
Stays the same
Increase
Answer:
decreases
Explanation:
because all the particles are wanting to say together and hit each other in a small area
The half-life of Iodine-131 is 8.0252 days. If 14.2 grams of I-131 is released in Japan and takes 31.8 days to travel across the Pacific, how much I-131 will be left by the time it reaches California?
Answer:
Explanation:
Half-life problems are modeled as exponential equations. The half-life formula is [tex]P=P_o\left (\dfrac{1}{2} \right)^{\frac{t}{k}}[/tex] where [tex]P_o[/tex] is the initial amount, [tex]k[/tex] is the length of the half-life, [tex]t[/tex] is the amount of time that has elapsed since the initial measurement was taken, and [tex]P[/tex] is the amount that remains at time [tex]t[/tex].
[tex]P=14.2\left (\dfrac{1}{2} \right)^{\frac{t}{8.0252}}[/tex]
Deriving the half-life formula
If one forgets the half-life formula, one can derive an equivalent equation by recalling the basic an exponential equation, [tex]y=a b^{t}[/tex], where [tex]t[/tex] is still the amount of time, and [tex]y[/tex] is the amount remaining at time [tex]t[/tex]. The constants a and b can be solved for as follows:
Knowing that amount initially is 14.2g, we let this be time zero:
[tex]y=a b^{t}[/tex]
[tex](14.2)=ab^{(0)}[/tex]
[tex]14.2=a *1[/tex]
[tex]14.2=a[/tex]
So, [tex]a=14.2[/tex], which represents out initial amount of the substance, and our equation becomes: [tex]y=14.2 b^{t}[/tex]
Knowing that the "half-life" is 8.0252 days (note that the unit here is "days", so times for all future uses of this equation must be in "days"), we know that the amount remaining after that time will be one-half of what we started with:
[tex]\left(\frac{1}{2} *14.2 \right)=14.2 b^{(8.0252)}[/tex]
[tex]\dfrac{7.1}{14.2}=\dfrac{14.2 b^{8.0252}}{14.2}[/tex]
[tex]0.5=b^{8.0252}[/tex]
[tex]\sqrt[8.0252]{\frac{1}{2}}=\sqrt[8.0252]{b^{8.0252}}[/tex]
[tex]\sqrt[8.0252]{\frac{1}{2}}=b[/tex]
Recalling exponent properties, one could find that [tex]\left ( \frac{1}{2} \right )^{\frac{1}{8.0252}}=b[/tex], which will give the equation identical to the half-life formula. However, recalling this trivia about exponent properties is not necessary to solve this problem. One can just evaluate the radical in a calculator:
[tex]b=0.9172535661...[/tex]
Using this decimal approximation has advantages (don't have to remember the half-life formula & don't have to remember as many exponent properties), but one minor disadvantage (need to keep more decimal places to reduce rounding error).
So, our general equation derived from the basic exponential function is:
[tex]y=14.2* (0.9172535661)^t[/tex] or [tex]y=14.2*(0.5)^{\frac{t}{8.0252}}[/tex] where y represents the amount remaining at time t.
Solving for the amount remaining
With the equation set up, substitute the amount of time it takes to cross the Pacific to solve for the amount remaining:
[tex]y=14.2* (0.9172535661)^{(31.8)}[/tex] [tex]y=14.2*(0.5)^{\frac{(31.8)}{8.0252}}[/tex]
[tex]y=14.2* 0.0641450581[/tex] [tex]y=14.2*(0.5)^{3.962518068}[/tex]
[tex]y=0.9108598257[/tex] [tex]y=14.2* 0.0641450581[/tex]
[tex]y=0.9108598257[/tex]
Since both the initial amount of Iodine, and the amount of time were given to 3 significant figures, the amount remaining after 31.8days is 0.911g.
Imagine that you're doing an experiment to determine how the running speed of a hamster depends on the weight of the hamster when plotting your data what variable should be on the y-axis
The variable on the y-axis should be the running speed of hamsters since it is the dependent variable.
What is the dependent variable?The dependent variable is the variable whose value fluctuates depending on other variables in an experiment.
In this case, the running speed is the dependent variable because it depends on the weight of the hamster.
The dependent variable is generally graphed on the Y axes, whereas the independent variable is graphed on the X-axes.
In conclusion, the variable on the y-axis should be the running speed of hamsters since it is the dependent variable.
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Suppose a piano tuner stretches a steel piano wire 7.5 mm. The wire was originally 0.975 mm in diameter, 1.45 m long, and has a Young's modulus of 2.10 × 1011 N/m2. Calculate the force a piano tuner applies to stretch the steel piano wire in Newtons.
The forces a piano tuner applies to stretch the steel piano wire willl be 1.4 × 10¹¹ N.
What is force?Force is defined as the push or pulls applied to the body. Sometimes it is used to change the shape, size, and direction of the body. Force is defined as the product of mass and acceleration. Its unit is Newton.
The application of a force may be used to describe the steel as an elastic element within a certain range of applied force.
Given data;
Young modulus, E=2.10 × 10¹¹ N/m²
Cross-sectional area,A
Final length,[tex]\rm L_f = 1.45 m = 1450 \ mm[/tex]
Initial length,[tex]\rm L_i = 7.5 mm[/tex]
[tex]\rm F = \frac{(L_f-L_i)(E)(A)}{L_1} \\\\ \rm F = \frac{(1450 - 7.5)(2.0 \times 0^{11})(0.746)}{1450} \\\\ F = 1.4 \times 10^{11} \ N[/tex]
Hence, the force a piano tuner applies to stretch the steel piano wire willl be 1.4 × 10¹¹ N.
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Water flows with a volume flow rate of 1.44 m³/s in a pipe. Find the water speed where the pipe radius is 0.430 m.
Answer:
0.6192
Explanation:
1.44 x 0.430=0.6192
Explain why aircraft are carefully designed so that parts do not resonate.
Answer:
See the answer Explain why aircraft are carefully designed so that parts do not resonate. Expert Answer This virtually takes place, however maximum usually in small piston-engined airplanes, in particular dual-engined airplanes. The resonant frequency of the fuselage of a small plane goes to have numerous nodes, withinside the low loads of hertz.
What is matter ?
.......................
Answer:
Something which occupies space and have mass.
Grains (pounds)
Nonfat/1% milk (gallons)
Poultry (pounds)
Seafood (pounds)
12
Vegetables (pounds)
271
1
Yogurt (pounds)
Fats and oils (pounds)
55
24
52
Soft drinks (gallons)
123
150
Sugars (pounds)
Source: U.S. Department of Agriculture, Hope Health Letter, April 1999
135
2
34
More people are:
What assumption can be made about the eating habits over the years?
6
63
15
320
4
67
O
A. becoming vegetarians
OB. eating more beef than chicken
OC. consuming whole milk dairy products
D. trying to eat healthier foods
E. eliminating soft drinks from their diets
Answer:
D. trying to eat healthier foods
Explanation:
A skateboarder starts from rest and maintains a constant acceleration of 0.50 m/s² for 8.4 s. What is the rider's displacement during this time
meters
Answer:
4.2m/s
Explanation:
0.50x8.4=4.2m/s
A star near the visible edge of a galaxy travels in a uniform circular orbit. It is 41,200 ly (light-years) from the galactic center and has a speed of 275 km/s. Estimate the total mass of the galaxy based on the motion of the star.
Gravitational constant is 6.674×10−11 m3/(kg·s2) and mass of the Sun Ms=1.99 × 1030 kg.
*Answer in billion solar mass
The total mass of the galaxy is 443.4 Solar mass
Orbital velocity ([tex]v[/tex]) = [tex]\sqrt{\frac{MG}{R} }[/tex]
where M= weight of galaxy
G= gravitational constatnt = [tex]6.674*10^-^1^1[/tex] (given)
R = distance from centre = [tex]41200[/tex] Light years (given)= [tex]4.12*9.5*10^1^6[/tex] km (1 ly= [tex]9.5*10^3[/tex] billion km)
v= orbital velocity = [tex]275[/tex] [tex]km/s[/tex] (given)
∴ According to the formula
[tex](2.75*10^2)^2[/tex] = [tex]\frac{M*6.674*10^-^1^1}{4.12*9.5*10^1^6}[/tex]
⇒ [tex]7.56*10^4*4.12*9.5*10^1^6=M*6.674*10^-^1^1[/tex] (cross multiplying and expanding)
⇒ [tex]29.59*10^2^1=M*6.674*10^-^1^1[/tex]
⇒ [tex]\frac{29.59*10^2^1*10^1^1}{6.674}=M[/tex]
⇒ [tex]4.434*10^3^2=M[/tex]
1 solar mass = [tex]1.989*10^3^0 kg[/tex]
⇒ Mass in solar mass =443.4 Solar mass
⇒ M = 443.4 Solar mass
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A metallic spoon is placed in a hot cup of coffee. If the coffee gives away 190 calories to the spoon to cool down by 0.75°C, what is the mass of the coffee? (Assume that ccoffee = 1.0 cal/gC°.)
Taking into account the definition of calorimetry, the mass of the coffee is 253.33 g.
CalorimetryCalorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.
Sensible heat is defined as the amount of heat that a body absorbs or releases without any changes in its physical state (phase change).
So, the equation that allows to calculate heat exchanges is:
Q = c× m× ΔT
where:
Q is the heat exchanged by a body of mass m.c is the specific heat sudstance.ΔT is the temperature variation.Mass of coffeeIn this case, you know:
Q= 190 caloriesc= 1 [tex]\frac{cal}{gC}[/tex]m= ?ΔT= 0.75 CReplacing in the equation that allows to calculate heat exchanges is:
190 cal = 1 [tex]\frac{cal}{gC}[/tex]× m× 0.75 C
Solving:
m= 190 cal÷ (1 [tex]\frac{cal}{gC}[/tex]× 0.75 C)
m=253.33 g
Finally, the mass of the coffee is 253.33 g.
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What is the frequency of light with a wavelength of 2.10 × 10-7 m?
Answer:
V = ?
C = 3 × 10^8
wavelength = 2.10 × 10^-7
wavelength = C/V
V = C/wavelength
V = 3 × 10^8/2.10 × 10^-7
V = 1.43 × 10^ 15
A small mass m slides with negligible friction down an incline at an angle of 25.76° with respect to the horizontal. It then drops down to a horizontal surface and bounces elastically back up as shown.
The picture is to scale. It shows the position of the mass at equal time intervals starting from rest at T. The height of the mass at X is the same as at V. Click here to view the motion of the mass m.
(options are: less than, greater than, equal to)
• The speed change between T and S is (blank) between S and R.
• The speed of m at X is (blank) that at Q.
• The size of the total force on m at P is (blank) at U.
• The mechanical energy of m at P is (blank) that at V.
• The size of the total force on m at S is (blank) at P.
• The velocity of m at X is (blank) that at V.
• The speed change between T and S is greater between S and R.
• The speed of m at X is greater that at Q.
• The size of the total force on m at P is less at U.
• The mechanical energy of m at P is equal to that at V.
• The size of the total force on m at S is greater at P.
• The velocity of m at X is equal to that at V.
What is mechanical energy?The mechanical energy is the sum of kinetic energy and the potential energy of an object at any instant of time.
M.E = KE +PE
Given is a small mass m slides with negligible friction down an incline at an angle of 25.76° with respect to the horizontal. It then drops down to a horizontal surface and bounces elastically back up as shown.
The picture shows the position of the mass at equal time intervals starting from rest at T. The height of the mass at X is the same as at V.
Between T to S and S to R, the mass is under constant acceleration. Time taken to move from T to S is greater than S to R. Thus, the speed change between T and S is greater than between S and R.
At Q, there is only a horizontal velocity component, but at X. the speed will be greater and has both vertical and horizontal component. Thus, the speed of m at X is greater than that at Q.
Force is given as the rate of change of momentum with time. At U, change in momentum is large compared to P. Thus, the size of the total force on m at P is less at U.
There is no friction acting on the system. So the energy remains conserved. Mechanical energy at P = V.
The force on mass m at S is only the gravity force. The remaining forces are cancelled by the normal force. Thus, size of the total force on m at S is greater at P.
The energy is conserved at each point of motion of mass. If X and V are at same height, they have same potential energy and so their kinetic energy. Thus, velocity of m at X is equal to that at V.
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A 10 cm diameter pulley is used to lift a bucket of cement weighing 400 N. How much force must be applied to the rope to lift the bucket at the other end?
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[tex]\large \bold {ANSWER}[/tex]
Considering that the pulley is fixed, the force applied should be equal to the weight of the object - of 400N.
[tex]\large \bold {EXPLANATION}[/tex]
Pulleys or pulleys are mechanical tools used to assist in the movement of objects and bodies. There are two types of pulleys: fixed and movable. While the fixed pulley changes the direction of force, the moving pulley helps to decrease the force needed to move the object or body in question.
As the statement only tells us a pulley, we must consider that it is fixed, because generally when it is mobile, this information is highlighted in the question.
In this way, a fixed pulley only changes the direction of the applied force. Thus, the force must have the same magnitude as the weight of the object to be moved. If the bucket weighs 400N, the force applied to the pulley must be 400N.
Therefore, having a fixed pulley, the force applied must be equal to the weight of the object, and will be 400N.
An object weighs 200N, what is its mass?
Answer:
kg
Explanation:
the highr and jebad to kilogram mizans
Answer:
w=m×g
Explanation:
w is given and take g as 9.8m/s^2
Find the current flowing out of the battery in the circuit. I = [?] A 9.0 V 30 ww 40 Ω www 50 Ω 20Ω 10Ω
We need Net resistance
solve from right words
[tex]\\ \rm\Rrightarrow R_1=20+10=30\Omega[/tex]
[tex]\\ \rm\Rrightarrow \dfrac{1}{R_2}=\dfrac{1}{30}+\dfrac{1}{50}=\dfrac{5+3}{150}[/tex]
[tex]\\ \rm\Rrightarrow R_2=\dfrac{150}{8}=18.75\Omega[/tex]
[tex]\\ \rm\Rrightarrow \dfrac{1}{R_3}=\dfrac{1}{30}+\dfrac{1}{40}=\dfrac{4+3}{120}[/tex]
[tex]\\ \rm\Rrightarrow R_3=\dfrac{120}{7}=17.14\Omega[/tex]
[tex]\\ \rm\Rrightarrow R_{net}=17.14+18.75=35.89\Omega[/tex]
Use ohm's law
[tex]\\ \rm\Rrightarrow I=\dfrac{V}{R}[/tex]
[tex]\\ \rm\Rrightarrow I=\dfrac{9}{35.89}[/tex]
[tex]\\ \rm\Rrightarrow I\approx 0.25A[/tex]
It takes a minimum distance of 76.50 m to stop a car moving at 15.0 m/s by applying the brakes (without locking the wheels). Assume that the same frictional forces apply and find the minimum stopping distance when the car is moving at 32.0 m/s.
The minimum stopping distance when the car is moving at 32.0 m/s is 348.3 m.
Acceleration of the car
The acceleration of the car before stopping at the given distance is calculated as follows;
v² = u² + 2as
when the car stops, v = 0
0 = u² + 2as
0 = 15² + 2(76.5)a
0 = 225 + 153a
-a = 225/153
a = - 1.47 m/s²
Distance traveled when the speed is 32 m/sIf the same force is applied, then acceleration is constant.
v² = u² + 2as
0 = 32² + 2(-1.47)s
2.94s = 1024
s = 348.3 m
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Need help with this question!
Answer:(note that values will be on top in small)
cobalt :- 60Co
potassium :- 40K
neon :- 24Ne
lead :- 208Pb
Honey bees beat their wings, making a buzzing sound at a frequency of 2.3 × 102 hertz. What is the period of a bee's wing beat?
how do I calculate equilibrant and fx and fy. I don't understand what they are asking
(a) The equilibrant C for force of vector A and B is 3.43 N.
(b) The equilibrant C for fx of vector A and B is 2.1 N.
(c) The equilibrant C, for fy of vector A and B is 2.12 N.
What is equilibrant force?An equilibrant force is a single force that will bring other bodies into equilibrium.
From configuration 1:Vector A: mass = 0.2 kg, θ = 20⁰
Vector B: mass = 0.15 kg, θ = 80⁰
Fx = mg cosθ
Fy = mg sinθ
where;
m is mass g is acceleration due to gravityVector AForce of A due to its weight
F(A) = mg
F(A) = 0.2 x 9.8 = 1.96 N
Fx = (0.2 x 9.8) cos(20) = 1.84 N
Fy = (0.2 x 9.8) sin(20) = 0.67 N
Resultant forceR = √(0.67² + 1.84²)
R = 1.96 N
Vector BForce of B due to its weight
F(B) = mg
F(B) = 0.15 x 9.8
F(B) = 1.47 N
Fx = (0.15 x 9.8) cos(80) = 0.26 N
Fy = (0.15 x 9.8) sin(80) = 1.45 N
Resultant forceR = √(0.26² + 1.45²)
R= 1.47 N
Equilibrant C of vector A and BEquilibrant force:
Force, C = 1.96 N + 1.47 N
Force, C = 3.43 N
Equilibrant FX:
Fx, C = Fx(A) + Fx(B)
Fx, C = 1.84 N + 0.26 N = 2.1 N
Equilibrant FY:
Fy, C = Fy(A) + Fy(B)
Fy, C =0.67 N + 1.45 N = 2.12 N
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. A vertical electric field is set up in space to compensate for the gravitational force on a point charge. What is the required magnitude and direction of the field when the point charge is: (a) an electron? (b) a proton? Comment on the obtained values.
(a) The required magnitude of the electric field when the point charge is an electron is 5.57 x 10⁻¹¹ N/C.
(b) The required magnitude of the electric field when the point charge is an proton is 1.02 x 10⁻⁷ N/C.
Magnitude of electric fieldThe magnitude of electric field is given by the following equation.
F = qE
But F = mg
mg = qE
E = mg/q
where;
E is the electric fieldm is mass of the particleg is acceleration due to gravityq is charge of the particleFor an electronE = (9.11 x 10⁻³¹ x 9.8)/(1.602 x 10⁻¹⁹)
E = 5.57 x 10⁻¹¹ N/C
For protonE = (1.67 x 10⁻²⁷ x 9.8)/(1.602 x 10⁻¹⁹)
E = 1.02 x 10⁻⁷ N/C
Thus, the required vertical electric field is greater when the charge is proton.
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A cannonball is shot horizontally off a high castle wall at 47.4 m/s. What is the magnitude of the cannonball's velocity after 1.23 s? (Ignore direction)
The magnitude of the cannonball's velocity after 1.23 s will be 59.46 m/sec.
What is velocity?The change of distance with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.
The given data in the problem is
u is the initial velocity of canonball= 47.4 m/sec
g is the acceleration of free fall = 9.81 m/sec²
v is the velocity after 1.23 s
According to Newton's first equation of motion,
[tex]\rm v=u +gt \\\\ v= 47.4 \ m/sec + 9.81 \ m/sec^2 \times 1.23 sec \\\\\ v=59.46 \ m/sec[/tex]
Hence, the magnitude of the cannonball's velocity after 1.23 s will be 59.46 m/sec.
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state two factors that affect current carrying capacity of an accumulator
The two factors that affect current carrying capacity of an accumulator will be accumulator size and ambient temperature
What is the function of the accumulator?Utilizing the compressible and decompressible properties of nitrogen gas, an accumulation vessel is used to store hydraulic pressure.
These key deciding elements are:
1. Accumulator Size: The current capacity increases as the circumference of the conductor increases.
2. Ambient Temperature: The greater the ambient temperature, the less heat is needed to raise the insulation's maximum operating temperature.
3. Accumulator identification:
4. Conditions for Installation:
Hence, the two factors that affect current carrying capacity of an accumulator will be accumulator size and ambient temperature
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Water flows through a pipe of diameter 8.0 cm with a speed of 10.0 m/s. It then enters a smaller pipe of diameter 3.0 cm. What is the speed of the water as it flows through the smaller pipe? (4) Water flows through a pipe of diameter 8.0 cm with a speed of 10.0 m / s . It then enters a smaller pipe of diameter 3.0 cm . What is the speed of the water as it flows through the smaller pipe ? ( 4 )
Answer:
71m/s
Explanation:
when you convert it you get 158.822 miles per hour
which of the following is NOT a type of scientist?
A. Geologist
B. Meteorologist
C. Botanist
D. Cryptologist
A cryptologist is not a type of scientist. The correct answer is a cryptologist and the correct option is (D).
Experts in the study of methods for secure communication and data security through encryption and decryption are known as cryptologists.
Even though they are considered scientists in the broadest sense, they are not typically classed with the other natural scientists (geologists, meteorologists, and botanists) stated in the alternatives.
Due to its emphasis on the methods and mathematical principles used in encryption and decryption, cryptology is more intimately related to computer science and mathematics.
Therefore, The correct answer is a cryptologist and the correct option is (D).
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A force of 20N is applied to the end of a wire of length 5m to produce an extension of 0.20mm
calculate: the stress on the wire.
The stress on the wire is determined as 6.37 x 10⁶ N/m².
Stress on the wireThe stress on the wire is calculated as follows;
σ = F/A
where;
F is the force applied on the wireA is area of the wireLet the diameter of the wire = 2 mm
Area = πr²
Area = π(1 x 10⁻³)²
Area = 3.142 x 10⁻⁶ m²
Stress = 20/(3.142 x 10⁻⁶)
Stress = 6.37 x 10⁶ N/m²
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A certain spring has a spring constant k1 = 660 N/m as the spring is stretched from x = 0 to x1 = 35 cm. The spring constant then changes to k2 = 250 N/m as the spring is stretched to x2 = 65 cm. From x2 = 65 cm to x3 = 89 cm the spring force is constant at F3 = 105 N.
Write an equation for the work done in stretching the spring from x1 to x2.
Calculate the work done, in joules, in stretching the spring from x1 to x2.
Calculate the work, in joules, necessary to stretch the spring from x = 0 to x3.
(a) The equation for the work done in stretching the spring from x1 to x2 is ¹/₂K₂Δx².
(b) The work done, in stretching the spring from x1 to x2 is 11.25 J.
(c) The work, necessary to stretch the spring from x = 0 to x3 is 64.28 J.
Work done in the springThe work done in stretching the spring is calculated as follows;
W = ¹/₂kx²
W(1 to 2) = ¹/₂K₂Δx²
where;
k is spring constantΔx is compression of the springW(1 to 2) = ¹/₂(250)(0.65 - 0.35)²
W(1 to 2) = 11.25 J
W(0 to 3) = ¹/₂k₁x₁² + ¹/₂k₂x₂² + ¹/₂F₃x₃
where;
k₁ is first spring constantk₂ is second spring constantF₃ is third force applied to the springW(0 to 3) = ¹/₂(660)(0.35)² + ¹/₂(250)(0.65 - 0.35)² + ¹/₂(105)(0.89 - 0.65)
W(0 to 3) = 64.28 J
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The value of acceleration due to gravity (g) on Pluto is about 0.61 meters/second2. How much will an object that weighs 250 newtons on Earth weigh on Pluto? Note that the value of acceleration due to gravity on Earth is 9.8 meters/second2.
The weight of an object on pluto will be 15.56. Mass multiplied by the gravitational acceleration gives weight.
What is gravitation?Gravitation is a natural law by which all things with all matter are attracted towards one another. Gravity is responsible for large-scale structures present in the Universe.
By dividing the object's weight on Earth by 9.8 m/s², as illustrated below, one may calculate the object's mass.
m = 250 N / 9.8 m/s²
m = 25.51 kg
Multiply the acquired mass by Pluto's gravitational acceleration (g);
W = (25.51 kg) x (0.61 m/s²)
W = 15.56 N
Thus, the item will only be 15.56 N in weight.
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Answer: B.
15.6 newtons
Explanation: edmentum
