Hi there!
We can begin by calculating the acceleration of the block and the wheel using the following equation:
d = vit + 1/2at², where initial velocity = 0 m/s
d = 1/2at²
2d/t² = a
2(1.5)/2² = 0.75 m/s²
Now, we can do a summation of torques:
∑τ = rT
Rewrite using Newton's 2nd Law for rotation:
Iα = rT
Convert α to a using the relationship α = a/r:
I(a/r) = rT
Ia = r²T
I = r²T/a
Plug in the values:
I = (0.40²)(20)/(0.75) = 4.267 kgm²
a student lifts a toy car from a bench and places the toy car at the top of a slope describe an energy transfer that occurs when the student lifts the toy car from the bench and places the toy car at the top of the slope.
Answer:
Assuming there are no energy losses due to friction or drag, the gravitational potential energy will change into kinetic potential energy as the car reaches the bottom of the slope.
G.P.E = m*g*h
K.E = (m*v^2)/2
where
m = mass of toy car (kg)
g = gravity (m/s^2)
h = heigh of your car from the bottom (m)
v = velocity of the toy car as it reaches the bottom (m/s)
Equate K.E to G.P.E
G.P.E = K.E
m*g*h = (m*v^2)/2
make v the subject of the formula
v = (2*g*h)^(1/2)
Substitute g = 9.81 m/s^2 and h = 2m into the equation to get v
v = (2*9.81*2)^(1/2)
v = 6.264 m/s
Does it appear that true average HAZ depth is larger for the high current condition than for the nonhigh current condition
Answer:
The data suggest that the true mean HAZ depth is larger when the current setting is higher.
To get the dimmest bulbs with two batteries and two bulbs you would connect the batteries in ____ and the bulbs in ____.
Answer:
batteries in parallel connection and bulbs in serial connection
To get the dimmest bulb with two batteries and two bulbs you would connect the batteries in parallel and the bulbs in series.
What is Parallel and series circuits?When two-terminal components and electrical networks that can be connected in series or parallel. This will result in two terminals in the electrical network, and may themselves participate in a series or parallel topology. When a two-terminal "object" is an electrical component or electrical network is a matter of perspective.
A circuit is said to be in series when the same current flows through all the components in the circuit where the current has only one path. A circuit is said to be parallel when there are multiple paths for the electric current to flow through it where the components which are part of the parallel circuit will have a constant voltage across all their ends.
Thus, to get the dimmest bulb with two batteries and two bulbs you would connect the batteries in parallel and the bulbs in series.
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Saturn's mass is 5.68 x 1024 kg and its radius is 6.03 x 107 m. A. Calculate the gravitational field strength at Saturn's surface. (2 marks) B. Calculate the force of gravity at Saturn's surface on an object with a mass of 50 kg.
Hi there!
A.
We can calculate the gravitational field strength using the following equation:
[tex]g = \frac{Gm_p}{r^2}[/tex]
G = Gravitational Constant
mp = mass of planet (kg)
r = radius (m)
Plug in the given values:
[tex]g = \frac{(6.67*10^{-11})*(5.68*10^{24})}{(6.03*10^7)^2} = \boxed{0.104 N/kg}[/tex]
B.
The force can be calculated using:
[tex]F_g = \frac{Gm_1m_2}{r^2}[/tex]
Plug in the values:
[tex]F_g = \frac{(6.67*10^{-11})(5.68*10^{24})(50)}{(6.04*10^7)^2} = \boxed{5.209N}[/tex]
Answer:
[tex]\boxed {\boxed {\sf g=0.104 \ N/kg \ and \ F_g= 5.2 \ N }}[/tex]
Explanation:
A. Gravitational Field Strength
The gravitational field strength can be calculated using the following formula:
[tex]g= \frac{Gm}{r^2}[/tex]
G, or the universal gravitational constant, is 6.67 × 10⁻¹¹ N*m²/kg². The mass of Saturn is 5.68 × 10²⁴ kilograms. The radius of Saturn is 6.03×10⁷ meters.
Substitute these values into the formula.
[tex]g= \frac{ (6.67 \times 10^{-11} \ N*m^2/kg^2) (5.68 \times 10^{24} \ kg)}{(6.03 \times 10^{7} \ m )^2}[/tex]
Multiply the numerator and square the denominator.
[tex]g= \frac{ 3.78856 \times 10^{14} \ N *m^2/kg }{3.63609 \times 10^{15} \ m^2}[/tex]
Divide.
[tex]g= 0.1041932405 \ N/kg[/tex]
The original measurements of mass and radius have 3 significant figures, so our answer must have the same. For the number we found, that is the thousandth place. The 1 in the ten-thousandth place tells us to leave the 4 in the thousandth place.
[tex]\boxed {g \approx 0.104 \ N/kg}[/tex]
B. Force of Gravity
The force of gravity is calculated using the following formula:
[tex]F_g= mg[/tex]
The mass of the object is 50 kilograms. We just calculated the gravitational field strength, which is 0.104 Newtons per kilogram. Substitute these values into the formula.
[tex]F_g= (50 \ kg)(0.104 \ N/kg)[/tex]
Multiply. The units of kilograms cancel.
[tex]\boxed {F_g=5.20 \ N}[/tex]
is a fuel cell a primary or secondary cell
Answer:
Enrol in our 50 studyscore masterclass. Click here! Fuel cells are a type of primary cell in that they are not recharged, However, they are unique because they never run out, if the reactants are constantly supplied.
Which region of electromagnetic spectrum will provide photons of the least energy
Answer:
Explanation:
Radio waves
Radio waves have photons with the lowest energies. Microwaves have a little more energy than radio waves. Infrared has still more, followed by visible, ultraviolet, X-rays and gamma rays.
The linear distance traveled by a wheel of radius 50cm after 99 complete revolutions is?
1)99m
2)210m
3)311
4)433
Answer:
3) 311 m
Explanation:
Circumference = 2πR = π m/rev
99 rev(π m/rev) = 99π m or about 311 meters
How does friction help us in walking.
Read the scenario below and answer the question that follows. Randall is hiring cooks for his restaurant. The first applicant is a handsome man with an average resumé and average job experience. The second applicant is a far less attractive man with a slightly above average resumé and above average job experience. Randall decides to hire the first applicant. Based on this information and on Randall’s decision, what might a psychologist conclude about Randall’s social perception? Randall has an unconscious assumption that attractive people are more competent. Randall has a unconscious assumption that unattractive people are bad cooks. Randall has a conscious assumption that attractive people make better cooks. Randall has a conscious assumption that unattractive people are more competent.
Randall has an unconscious assumption that attractive people are more competent.
What is meant by assumption ?The term assumption can be described as an unspoken premise that underlies the conclusion.
Here,
The capacity to accurately evaluate and draw conclusions about other individuals based on their overall physical appearance, verbal behaviour, and nonverbal attitudes is referred to as social perception.
Given that, for his restaurant, Randall is employing chefs. The first applicant is a dashing man with an average resume and career history. The second candidate is a far less appealing man with an average to slightly above average resume and work experience. Randall chooses to hire the first candidate.
This shows the social perception of Randall and his unconscious assumptions against unattractive people.
Hence,
Randall has an unconscious assumption that attractive people are more competent.
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if the Periodic time of an oscillating object Triples then its frequency will?
Answer:
it would decrease
Explanation:
f=1/T
please help me
please help me
please help me
Answer:
do it got a picture
on the edge
Explanation:
This question has two parts. First, answer part A. Then, answer part B. Part A: Which statement best summarizes the theme of the text? O A. You do not always recognize what is most valuable. O B. Keep your friends close, but your enemies closer. W O C. Fine possessions do not make a fine person. h 0 D. The best things come in small packages. Part B: Which evidence from the text best supports your answer in part A? O A. "The other animals of the forest bowed to him, and they often spoke of his antlers with admiration." B. "The Stag was so engrossed that he did not notice that a Lion had crept up alongside him." C. ". . . but the branches of some of the trees hung low, and vines curled around them." D. "The long legs that I hated would have saved me, but the antlers that I loved have led to my destruction!"
Answer: for part A Its A the ANSWER for part B its D
Explanation:
Answer:
a and b
Explanation:
A man is whirling a 0.25 kg ball on a 1.5 m long string at 3 m/s. Find the centripetal acceleration of this ball.
Question 2 options:
0.5 m/s2
13.5 m/s2
6 m/s2
2 m/s2
The centripetal acceleration of this ball is equal to 12 [tex]m/s^2[/tex]
Given the following data:
Diameter = 1.5 mSpeed, V = 3 m/s.Mass = 0.25 kgRadius = [tex]\frac{Diameter}{2} = \frac{1.5}{2} = 0.75 \;meters[/tex]
To find the centripetal acceleration of this ball:
The acceleration of an object along a circular track is referred to as centripetal acceleration.
Mathematically, the centripetal acceleration of an object is given by the formula:
[tex]A_c = \frac{V^2}{r}[/tex]
Where:
Ac is the centripetal acceleration.r is the radius of the circular track.V is the velocity of an object.
Substituting the given parameters into the formula, we have;
[tex]A_c = \frac{3^2}{0.75}\\\\A_c = \frac{9}{0.75}\\\\A_c = \frac{9}{0.75}[/tex]
Centripetal acceleration = 12 [tex]m/s^2[/tex]
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A baseball player notices the ball when it is 3.4 m above the
ground, traveling at 4.4 m/s. He wants to make the catch when
the ball is 1.5 m above the ground, how long does it take to reach
his glove?
Find the distance the ball travels:
3.4 meters - 1.5 meters = 1.9 meters
Now divide the distance the ball travels by the speed:
1.9 meters / 4.4 m/s = 0.43 seconds
Answer:
Explanation:
s = s₀ + v₀t + ½at²
There are an infinite number of solutions to this question as posed because we are not told the direction of the initial velocity.
Assuming ground is level and origin and UP the positive direction
The shortest amount of time possible is when the initial velocity is straight down
1.5 = 3.4 - 4.4t + ½(-9.8)t²
0 = -4.9t² - 4.4t + 1.9
t = (4.4 ±√(4.4² - 4(-4.9)(1.9))) / (2(-4.9))
positive answer is
t = 0.32 s
The longest amount of time possible is when the initial velocity is straight up.
1.5 = 3.4 + 4.4t + ½(-9.8)t²
0 = -4.9t² + 4.4t + 1.9
t = (-4.4 ±√(4.4² - 4(-4.9)(1.9))) / (2(-4.9))
positive answer
t = 1.22 s
If the initial velocity is horizontal, meaning no vertical velocity
1.5 = 3.4 + 0t + ½(-9.8)t²
-4.9t² = -1.9
t² = 0.38775...
t = 0.62 s
Any angle between UP and Down will have a different initial vertical velocity and result in a different time to catch height.
It appears from the comments on the other answer, that I have shown you how to arrive at three of the four possible solutions. The initial direction is very important.
What is discordant characteristic ?
[tex] \: \: \: \: [/tex]
being at variance; disagreeingor incongruous: discordant opinions. disagreeable to the ear; dissonant; harsh.hope it helps[tex] \: \: \: \: \: [/tex]
dissimilar with respect to one or more particular characters
the c component of vector a is 5.3 units, and it’s y component is -2.3 units. the angle that vector a makes with the +x axis is closest to
110
160
23
340
250
Answer:
340
Explanation:
Sorry I don't know how to do this one yet, I just found the answer in a textbook.
The angle that vector a makes with the +x axis is closest to 23.
What is direction of a vector?The direction of a vector is represented tangent of angle equal to the ratio of the y component and the x component of the vector quantity.
tangent of angle = y/x
angle = tan⁻¹ (-2.3/5.3)
angle = 23.46°
Thus, the angle that vector makes with +x is 23.
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help :”)
a skydiver jumps out of a plane and falls for 45 seconds before deploying his parachute. how far did he fall?
Answer:
200 feet
Explanation:
An object is travels 50 m in 4 s. It had no initial velocity and experiences constant acceleration. What is the magnitude of the acceleration?
Free-fall Acceleration is -10 m/s^2
I also need the Formula
Answer:
Explanation:
s = s₀ + v₀t + ½at²
50 = 0 + 0(4) + ½a(4²)
50 = 8a
a = 50/8 = 6.25 m/s²
73 ml of water is followed by 25 ml of juice. What is the percent strength of juice?
Total liquid: 73 + 25 = 98 ml
Percent juice = (25/98) x 100 = 25.5 %
A merry-go-round of radius R, shown in the figure, is rotating at constant angular speed. The friction in its bearings is so small that it can be ignored. A sandbag of mass m is dropped onto the merry-go-round, at a position designated by r. The sandbag does not slip or roll upon contact with the merry-go-round.
The figure shows a top view of a merry-go-round of radius capital R rotating counterclockwise. A sandbag is located on the merry-go-round at a distance lowercase r from the center.
Rank the following different combinations of m and r on the basis of the angular speed of the merry-go-round after the sandbag "sticks" to the merry-go-round.
The angular speed of the merry-go-round reduced more as the sandbag is
placed further from the axis than increasing the mass of the sandbag.
The rank from largest to smallest angular speed is presented as follows;
[m = 10 kg, r = 0.25·R]
[tex]{}[/tex] ⇩
[m = 20 kg, r = 0.25·R]
[tex]{}[/tex] ⇩
[m = 10 kg, r = 0.5·R]
[tex]{}[/tex] ⇩
[m = 10 kg, r = 0.5·R] = [m = 40 kg, r = 0.25·R]
[tex]{}[/tex] ⇩
[m = 10 kg, r = 1.0·R]
Reasons:
The given combination in the question as obtained from a similar question online are;
1: m = 20 kg, r = 0.25·R
2: m = 10 kg, r = 1.0·R
3: m = 10 kg, r = 0.25·R
4: m = 15 kg, r = 0.75·R
5: m = 10 kg, r = 0.5·R
6: m = 40 kg, r = 0.25·R
According to the principle of conservation of angular momentum, we have;
[tex]I_i \cdot \omega _i = I_f \cdot \omega _f[/tex]
The moment of inertia of the merry-go-round, [tex]I_m[/tex] = 0.5·M·R²
Moment of inertia of the sandbag = m·r²
Therefore;
0.5·M·R²·[tex]\omega _i[/tex] = (0.5·M·R² + m·r²)·[tex]\omega _f[/tex]
Given that 0.5·M·R²·[tex]\omega _i[/tex] is constant, as the value of m·r² increases, the value of [tex]\omega _f[/tex] decreases.
The values of m·r² for each combination are;
Combination 1: m = 20 kg, r = 0.25·R; m·r² = 1.25·R²
Combination 2: m = 10 kg, r = 1.0·R; m·r² = 10·R²
Combination 3: m = 10 kg, r = 0.25·R; m·r² = 0.625·R²
Combination 4: m = 15 kg, r = 0.75·R; m·r² = 8.4375·R²
Combination 5: m = 10 kg, r = 0.5·R; m·r² = 2.5·R²
Combination 6: m = 40 kg, r = 0.25·R; m·r² = 2.5·R²
Therefore, the rank from largest to smallest angular speed is as follows;
Combination 3 > Combination 1 > Combination 5 = Combination 6 >
Combination 2
Which gives;
[m = 10 kg, r = 0.25·R] > [m = 20 kg, r = 0.25·R] > [m = 10 kg, r = 0.5·R] > [m =
10 kg, r = 0.5·R] = [m = 40 kg, r = 0.25·R] > [m = 10 kg, r = 1.0·R].
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Why is the sky blue and why do we get a sunset
Answer:
Small particles of dust and pollution in the air can contribute to (and sometimes even enhance) these colors, but the primary cause of a blue sky and orange/red sunsets or sunrises is scattering by the gas molecules that make up our atmosphere. Large particles of pollution or dust scatter light in a way that changes much less for different colors.
Explanation:
1.25 is the closest to 1.04 or not I want to answer please. I think it's true, but I want to prove it scientifically, please.
Answer/Explanation:
It False, because if You Round Both of them..
1.25= 1.30
1.04= 1.00
it's like, 1 dollar and 4 cents; compared to 1 dollar and 25 cents. Obviously 25 cents is a lot more than 4 cents.
Suppose a ball is thrown vertically upward (positive direction) from an initial height LaTeX: h_0 with initial velocity LaTeX: v_0. Find the position function LaTeX: s(t) of the ball after LaTeX: t seconds assuming the gravitational acceleration LaTeX: g is a positive constant pointing downward (negative direction).
After time t, the position function of the ball is determined as [tex]y(t) = h_0\ +\ v_0t \ - \ \frac{1}{2} gt^2[/tex]
The given parameters;
initial velocity of the ball, = [tex]v_0[/tex]initial position of the ball, = [tex]h_0[/tex]acceleration due to gravity, = gThe position function of the ball after time t, is calculated as follows;
[tex]y(t) = h_0\ +\ v_0t \ - \ \frac{1}{2} gt^2[/tex]
The negative sign of acceleration of due to gravity is because the ball is moving upward against gravity.
Thus, after time t, the position function of the ball is determined as [tex]y(t) = h_0\ +\ v_0t \ - \ \frac{1}{2} gt^2[/tex]
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A small, free-to-rotate magnet is placed in a strong magnetic field. In what orientation will it come to rest
Answer:
South-North
Explanation:
A gold doubloon 6.1 cm in diameter and 2.0 mm thick is dropped over the side of a pirate sheep. When it comes to rest on the ocean floor at a depth of 770 m, how much has its volume changed?
The definition of volume modulus and the variation of pressure with depth allows to find the result for the variation of the volume of the coin is:
ΔV = 2.15 10⁻⁸ m³
The pressure with the depth is given by the relation
P = P₀ + ρ g h
Where P is the pressure, ρ is the density anf h depth.
The size of the bodies is determined by the distance of their atomic and molecular bonds, therefore the size of the bodies changes under external interations, in the case of hydrostatic pressure a constant called volumetric modulus is defined.
[tex]B = - \frac{\Delta P}{\frac{\Delta V}{Vo} } \\\Delta V = - \frac{\Delta P }{B} \ V_o[/tex]
Where ΔP is the pressure change, V₀ and V are the volume change and the initial volume of the body, the negative sign is introduced so that the volumetric modulus is a positive quantity.
They indicate the diameter and thickness of the coin (d = 6.1 cm and e =0.20 cm) on the sea surface and the depth to which it is submerged
h = 770 m
Let's look for the volume of the coin.
V₀ = π r² h = [tex]\pi \ \frac{d^2}{4} \ e[/tex]
V₀ = [tex]\pi \ \frac{0.061^2 }{4} \ 0.002[/tex]
V₀ = 5.84 10-6 m³
Let's find the pressure at the depth of y = 770 m, the density of sea water is ρ = 1025 kg / m³, the pressure at the surface is the atmospheric pressure P₀ = 1 10⁵ Pa, the volumetric modulus of water is B = 0.21 10¹⁰ Pa.
P = 1 10⁵ + 1025 9.8 770
P = 1 10⁵ + 7,735 10⁶
P = 7.84 10⁶ Pa
Let's calculate
ΔV =[tex]- \frac{1 \ 10^5 - 7.84 \ 10^6 }{0.21 \ 10^{10}} \ 5.845 \ 10^{-6}[/tex]
ΔV = 2.15 10-8 m³
In conclusion using the definition of volume modulus and the variation of pressure with depth we can find the result for the variation of the volume of the coin is:
ΔV = 2.15 10-8 m³
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An initially stationary object experiences an acceleration of 6 m/s2 for a time of 15 s. How far will it travel during that time?
Free-fall Acceleration is -10 m/s^2
Answer:
Explanation:
s = s₀ + v₀t + ½at²
s = 0 + 0(15) + ½(6)(15²)
s = 675 m
Not sure what the free fall acceleration is needed for, but if the object is dropped from a high enough point, it will travel in 15 seconds
s = ½10(15²) = 2250 m if air resistance is ignored
A race car traveling at 100 m/s enters an unbanked turn of 400 m radius. The coefficient of (static) friction between the tires and the track is 1.1. The track has both an inner and an outer wall. Which statement is correct
Answer:
The race car will crash into the outer wall
Explanation:
max fr = μsN = 1.1 mg = 11 m
mv2/R = m(100)2/(400) = 25 m > fr
what is the acceleration of the cart at t=8 seconds?
a) 0 m/s^2
b) 10 m/s^2
c) 20 m/s^3
d) -20m/s^2
What is the acceleration of the cart at t=8 seconds?
a) 0 m/s^2b) 10 m/s^2c) 20 m/s^3d) -20m/s^2Hence the answer us letter a) 0 m/s^2.
That's all I know, Hope it help :)
Carbon tetrachloride (CCl4) is diffusing through benzene (C6H6), as the drawing illustrates. The concentration of CCl4 at the left end of the tube is maintained at 1.71 x 10-2 kg/m3, and the diffusion constant is 21.9 x 10-10 m2/s. The CCl4 enters the tube at a mass rate of 5.86 x 10-13 kg/s. Using these data and those shown in the drawing, find (a) the mass of CCl4 per second that passes point A and (b) the concentration of CCl4 at point A.
We have that for the Question "find (a) the mass of CCl4 per second that passes point A and (b) the concentration of CCl4 at point A."
Answers:
Mass of CCI_4 per second = [tex]5.86*10^{-13} kg/s[/tex] Concentration of CCI_4 = [tex]12.6*10^{-3}kg/m^3[/tex]
From the question we are told
The concentration of [tex]CCl_4[/tex] at the left end of the tube is maintained at 1.71 x 10-2 kg/m3, and the diffusion constant is 21.9 x 10-10 m2/s. The CCl4 enters the tube at a mass rate of 5.86 x 10-13 kg/s
A) the mass flow rate of CCI_4 as it passes point A is the same as the mass flow rate at which CCI_4 enters the left end of the tube
Therefore, the mass flow rate of CCI_4 at point A
= [tex]5.86*10^{-13} kg/s[/tex]
B) From Fick's law
[tex]\deltaC = \frac{mL}{DAt}\\\\ Assume L = 5*10^{-3}, A = 3*10^{-4}\\\\\deltaC = \frac{5.86*10^{-13} * 5*10^{-3}}{21.9*10^{-10} * 3*10^{-4}}\\\\\deltaC = 4.46*10^{-3}kg/m^3[/tex]
Then,
[tex]Concentration = 1.71*10^{-2} - 4.46*10^{-3}\\\\= 12.6*10^{-3}kg/m^3[/tex]
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If the voltage across a 5-F capacitor is 2*e^-3
V find the current and the power