a) At the end of the braking period, the car is moving at a speed of 16.6m/s
b) The car has traveled a total distance of 135.4m
Explanations:The car starts from rest
The initial velocity, u = 0 m/s
Uniform acceleration, a = 2.4 m/s²
time, t = 9.0 seconds
Find the final velocity when the car accelerates at 2.4m/s² using the equation
v = u + at
v = 0 + 2.4(9)
v = 21.6 m/s
The distance covered when the car accelerates at 2.4m/s²
s = ut + 0.5at²
s = 0(9) + 0.5(2.4)(9²)
s = 97.2 m
The distance covered when the car accelerates at 2.4m/s² is 97.2 m
The driver then applied a brake for 2.0 s and accelerates at -2.5m/s²
The initial velocity now becomes 21.6 m/s
That is, u = 21.6 m/s
t = 2 seconds
a = -2.5m/s²
The final speed v is calculated as:
v = u + at
v = 21.6 + (-2.5)(2)
v = 21.6 - 5
v = 16.6m/s
At the end of the braking period, the car is moving at a speed of 16.6m/s
The distance covered during the braking period is calculated as:
[tex]\begin{gathered} s\text{ = (}\frac{u+v}{2})t \\ s\text{ = }\frac{21.6+16.6}{2}\times2 \\ s\text{ = }\frac{38.2}{2}\times2 \\ s\text{ = 38.2 m} \end{gathered}[/tex]The car traveled a distance of 38.2 m during the braking period
Total distance covered = 97.2m + 38.2m
Total distance covered = 135.4m
The car has gone a distance of 135.4m
Timmy walks 5 m North, 3m West, and finally 1 m South. What is his displacement from his starting point?
Timmy walks 5 m North, 3m West, and finally 1 m South then his displacement from the starting point would be 5 meters in the northwest direction.
What is displacement?Displacement describes this shift in location and it is calculated with the help of the initial and the final position of the object.
As given in the problem If Timmy walks 5 m North, 3m West, and finally 1 m South ,
The resultant displacement of the Timmy = √(4² + 3²)
= 5 meters
Thus, the resultant displacement of the Timmy would be 5 meters
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In lacrosse, a ball is thrown from a net on the end of a stick by rotating the stick and forearm about the elbow. If the angular velocity of the ball about the elbow joint is 31.3 rad/s and the ball is 1.45 m from the elbow joint, what is the velocity of the ball?
The linear velocity of the ball whose angular velocity is 31.3 rad/s about the elbow joint will be 45.385 m/s.
What is angular velocity?Angular velocity is the rate of change of angular displacement with respect to time. Mathematically -
ω = dθ/dt
from this we can write -
dθ = ω dt
∫dθ = ω ∫dt
θ₂ - θ₁ = ω(t₂ - t₁)
Δθ = ω Δt
ω = Δθ/Δt
Given is a ball thrown from a net on the end of a stick by rotating the stick and forearm about the elbow. The angular velocity of the ball about the elbow joint is 31.3 rad/s and the ball is 1.45 m from the elbow joint.
The relation between linear velocity and the angular velocity of a body undergoing circular motion is given by -
v = rω
From this we can write -
ω = 31.3 rad/s
r = 1.45 m
Substituting the values -
v = rω
v = 1.45 x 31.3
v = 45.385 m/s
Therefore, the linear velocity of the ball whose angular velocity is 31.3 rad/s about the elbow joint will be 45.385 m/s.
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An electric motor is used to do the 2.30 x104 J of work needed to lift an engine out of a car. If the motor draws a current of 3.2 A for 30 s, calculate the potential difference across the motor.
Given data
*The given energy is U = 2.30 x 10^4 J
*The given current is I = 3.2 A
*The given time is t = 30 s
The formula for the charge is given as
[tex]q=It[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} q=(3.2)(30) \\ =96\text{ C} \end{gathered}[/tex]The formula for the potential difference across the motor is given as
[tex]U=qV[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} 2.30\times10^4=(96)V \\ V=239.58\text{ V} \\ \approx240\text{ V} \end{gathered}[/tex]Hence, the potential difference across the motor is V = 240 V
Which of the following is true for an isolated system? I. Matter is able to freely enter or exit the system.II. Heat is able to freely enter or exit the system.III. Work is able to freely enter or exit the system.II onlyNone of the aboveI or II onlyI only
Note that, an isolated system does not allow:
• the exchange of energy
,• the exchange of matter
Therefore, in an isolated system, neither
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A 72.5 kg student sits at a desk 1.25 m away from a 80.0 kg student. What is the magnitude of the gravitational force between the two students?
Given:
The mass of one student is,
[tex]m_1=72.5\text{ kg}[/tex]The mass of the other student is,
[tex]m_2=80.0\text{ kg}[/tex]The distance between them is,
[tex]d=1.25\text{ m}[/tex]The gravitational force between them is,
[tex]F=G\frac{m_1m_2}{d^2}[/tex]Here the gravitational constant is,
[tex]G=6.6\times10^{-11}\text{ }\frac{N.m^2}{\operatorname{kg}}[/tex]Substituting the values we get,
[tex]\begin{gathered} F=\frac{(6.6\times10^{-11})\times72.5\times80.0}{(1.25)^2} \\ =2.5\times10^{-7}\text{N} \end{gathered}[/tex]Hence the second option is correct.
A 60.0 kg skier with an initial speed of 14 m/s coasts up a 2.50 m high rise as shown in the figure.
Find her final speed right at the top, in meters per second, given that the coefficient of friction between her skis and the snow is 0.38?
The final speed of the skier at the top mountain is determined as 9.27 m/s.
What is the change in the energy of the skier?
The change in the energy of the skier due to frictional force is calculated as follows;
ΔP.E = Pi + Ef
where;
Pi is the initial potential at the topEf is the energy lost to frictionThe distance of the plane travelled is calculated as;
sin35 = 2.5/L
L = 2.5 / sin35
L = 4.36 m
ΔP.E = mghi - μmgcosθ(L)
where;
m is the masshi is the initial heightg is acceleration due to gravityμ is coefficient of frictionΔP.E = (60 x 9.8 x 2.5) - (0.38)(60)(9.8) cos(35) x (4.36)
ΔP.E = 671.98 1 J
The final speed of the skier at the top of the plane;
P.E = K.E
P.E = ¹/₂mv²
v² = 2P.E /m
v = √(2P.E /m)
v = √(2 x 671.98) / 60)
v = 4.73 m/s
Total speed = -4.73 m/s + 14 m/s = 9.27 m/s
Thus, due to frictional force opposing the upward motion of the skier, the final speed at the top will be smaller than the initial speed.
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Convert each quantity to the indicated units.
a. 3.01 g to cg
b. 6200 m to km
c. 0.13 cal/g to kcal/g
Answer:
Explanation:
a) 301 cg
b) 6.2 km
c) 0.00013
What is absolute zero?0k0c273-100
The absolute zero is measured in Kelvin scale
Thus absolute zero is 0K.
order the colors of the discuses to show the size of the force applied to throw each discus
Since the acceleration will be the same for all, the order would be the following [From larger to smaller force]:
Blue
Green
Orange
Red
Purple
An object is dropped from a height of 65 m above ground level. A) determine the final speed in m/s, at which the object hits the ground c) determine the distance in meters, traveled during the last second of motion before hitting the ground.
Given:
height = 65 m
Given that the object is in free fall, let's solve for the following:
• (a). determine the final speed in m/s.
To find the final velocity, apply the kinematics equation:
[tex]v^2=u^2-2ax[/tex]Where:
v is the final velocity
u is the initial velocity = 0
a is the acceleration due to gravity = 9.8 m/s²
x is the displacement = 65 m
Thus, we have:
[tex]\begin{gathered} v^2=0^2-2(-9.8)(65) \\ \\ v^2=-(-1274) \\ \\ v^2=1274 \\ \\ \text{ Take the square root of both sides:} \\ \sqrt{v^2}=\sqrt{1274} \\ \\ v=35.69\text{ m/s} \end{gathered}[/tex]Therefore the final speed will be -35.69 m/s.
• (c). The distance traveled during the last second of motion before hitting the ground.
To find the distance, apply the formula:
[tex]H=ut+\frac{1}{2}at^2[/tex]Where:
H is the height.
u is the initial velocity = 0 m/s
t is the time
a is acceleration due to gravity.
Let's rewrite the formula to find the time traveled.
[tex]\begin{gathered} H=0t+\frac{1}{2}at^2 \\ \\ H=\frac{1}{2}at^2 \\ \\ t=\sqrt{\frac{2H}{a}} \end{gathered}[/tex]Thus, we have:
[tex]\begin{gathered} t=\sqrt{\frac{2*65}{9.8}} \\ \\ t=\sqrt{\frac{130}{9.8}} \\ \\ t=\sqrt{13.26} \\ \\ t=3.64\text{ s} \end{gathered}[/tex]Therefore, the time is 3.64 seconds.
Now, to find the distance traveled during the last second of motion, apply the formula:
[tex]s=\frac{1}{2}a(t_2^2-t_1^2)[/tex]Where:
t2 = 3.64 seconds
t1 = 3.64 seconds - 1 second = 2.64 seconds
Thus, we have:
[tex]\begin{gathered} s=\frac{1}{2}(9.8)((3.64)^2-(2.64)^2) \\ \\ s=4.9(13.2496-6.9696) \\ \\ s=4.9(6.28) \\ \\ s=30.77 \end{gathered}[/tex]Therefore, the distance in meters, traveled during the last second of motion before hitting the ground is 30.77 meters.
ANSWER:
(A). -35.69 m/s
(C). 30.77 m
How much potential energy due to gravity would a person have if they were standing on top of a building that is 36.2 m high? Assume that they have a mass of 79.2 kg.
Given:
The mass of thee person is
[tex]m=79.2\text{ kg}[/tex]The height at which person standing is
[tex]h=36.2\text{ m}[/tex]Required: calculate the potential energy of the person
Explanation:
when anybody of mass m is at a distance h from the earth's surface then it has potential energy that is given as
[tex]P.E=mgh[/tex]where g is the acceleration due to gravity whose value is
[tex]9.8\text{ m/s}^2[/tex]Plugging all the values in the above relation, we get;
[tex]\begin{gathered} P.E=79.2\text{ kg}\times9.8\text{ m/s}^2\times36.2\text{ m} \\ P.E=28096.992\text{ J} \end{gathered}[/tex]Thus, the potential energy is
[tex]28096.992[/tex]The radioactive isotope 14C has a half-life of approximately 5715 years. Now there are 50g of 14C.(1) How much of it remains after 1600 years? (Round your answer to three decimal places.)
We know that the amount of matter is given by:
[tex]N=N_0e^{-\lambda t}[/tex]where λ is the decay constant. The decay constant is related to the half-life of the element by the equation:
[tex]\lambda=\frac{\ln2}{t_{\frac{1}{2}}}[/tex]Then we can express our first equation as:
[tex]N=N_0e^{-\frac{\ln2}{t_{\frac{1}{2}}}t}[/tex]Plugging the initial amount, 50 g, the half-life of 5715 years and the time we want to know we have that:
[tex]\begin{gathered} N=50e^{-\frac{\ln2}{5715}(1600)} \\ N=41.181 \end{gathered}[/tex]Therefore, after 1600 years there are 41.181 g
n a slap shot, a hockey player accelerates the puck from a velocity of 5 m/s to 30 m/s in the same direction. If the puck moves over a distance of 10 m during this process, what was the acceleration?
Given data
*The given initial velocity of the puck is u = 5 m/s
*The given final velocity of the puck is v = 30 m/s
*The given distance is s = 10 m
The formula for the acceleration is given by the kinematic equation of motion as
[tex]\begin{gathered} v^2=u^2+2as \\ a=\frac{v^2-u^2}{2s} \end{gathered}[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} a=\frac{(30)^2-(5)^2}{2\times10} \\ =43.75m/s^2 \end{gathered}[/tex]Hence, the acceleration of the puck is a = 43.75 m/s^2
Part of a light ray striking an interface between air and water is refracted, and part is reflected, as shown. The index of refraction of air is 1.00 and the index of refraction of water is 1.33. The frequency of the light ray is 7.85 x 10^16 Hz.(a) If angle 1 measures 40°, find the value of angle 2.(b) If angle 1 measures 40°, find the value of angle 3.(c) Calculate the speed of the light ray in the water.(d) Calculate the wavelength of the light ray in the water.(e) What is the largest value of angle 1, that will result in a refracted ray?
We will use Snell's law, which states:
[tex]n_1\sin \theta_1=n_2\sin \theta_2[/tex]Where n1 and n2 are the refraction indexes and their respective angles are "theta1" and "theta2".
For part A we replace the values:
[tex]1\sin 40=1.33\sin \theta_2[/tex]Now we solve for "theta2" first by dividing both sides by 1.33:
[tex]\frac{\sin40}{1.33}=\sin \theta_2[/tex]Now we use the inverse function for sine:
[tex]\arcsin (\frac{\sin 40}{1.33})=\theta_2[/tex]Solving the operation:
[tex]28.9=\theta_2[/tex]For part B, since "theta1" and "theta3" are angles of reflection, according to the reflection law, these angles are equal, therefore:
[tex]\theta_3=\theta_1=40[/tex]For part C. The index of refraction is defined as:
[tex]n=\frac{c}{v}[/tex]Where "c" is the speed of light in a vacuum and "v" is the speed of light in the medium. Replacing the values:
[tex]1.33=\frac{3\times10^8\text{ m/s}}{v}[/tex]Now we solve for "v":
[tex]v=\frac{3\times10^8\text{ m/s}}{1.33}[/tex]Solving the operation:
[tex]v=2.26\times10^8\text{ m/s}[/tex]For part d. We will use the following formula:
[tex]\lambda=\frac{v}{f}[/tex]Where "v" is the speed and "f" is the frequency. Replacing we get:
[tex]\lambda=\frac{2.26\times10^8\text{ m/s}}{7.85\times10^{16}s^{-1}}[/tex]Solving the operations:
[tex]\lambda=0.288\times10^{-8}m[/tex]For part e. The largest value of the angle of incidence that will result in refraction is 90 degrees.
You observe waves on the beach and measure that a wave hits the beach every 5 seconds. What is the period and the frequency of the waves ?
Answer:
Period = 5 seconds
Frequency = 0.2 Hz
Explanation:
The period is the time per cycle. So, if a wave hits the beach every 5 seconds, the period will be 5 seconds.
Additionally, the frequency is the inverse of the period, so the frequency of the waves can be calculated as:
[tex]\text{frequency = }\frac{1}{Period}=\frac{1}{5}=0.2\text{ Hz}[/tex]So, the answers are:
Period = 5 seconds
Frequency = 0.2 Hz
carts, bricks, and bands
5. What acceleration results when 2 rubber bands stretched to 20 cm are used pull a cart with one brick?
a. About 0.25 m/s2
b. About 0.50 m/s2
c. About 0.75 m/s2
d. About 1.00 m/s2
B. The acceleration results when 2 rubber bands stretched to 20 cm are used pull a cart with one brick is 0.5 m/s².
What is the applied force on an object?The force applied on object is obtained by multiplying the mass and acceleration of the object.
According to Newton's second law of motion, the force exerted on an object is directly proportional to the product of mass and acceleration of the object.
Also, the applied force is directly proportional to the change in the momentum of the object.
Mathematically, the force acting on object is given as;
F = ma
a = F/m
where;
a is the acceleration of the objectm is the mass of the objectF is the applied forceFrom the trials, the acceleration results when 2 rubber bands stretched to 20 cm are used pull a cart with one brick is 0.5 m/s².
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How much power is created when you perform 55 Joule of work with a time 20 sec?
Answer:
2.75 watts
Explanation:
The power is equal to the work divided by time, so
P = W/t
Then, replacing W = 55 J and t = 20 sec, we get:
P = 55 J / 20 s = 2.75 Watts
Therefore, the power created is 2.75 watts
It’s an assessment Jane multiplied 825x22 and got 3,300. Flynn multiples the same numbers and got 18,150 Which student is correct?
In order to determine which student is correct, multiply the given numbers, as follow:
8 2 5
x 2 2
1 6 5 0
1 6 5 0
1 8 1 5 0
As you can notice, the result of the multiplcation is 18,150, hence, Flynn is right.
Jeff tosses a can of soda pop to Karen, who is standing on her 3rd floor balcony a distance of 8.5m above Jeff’s hand. Jeff gives the can an initial velocity of 16m/s, fast enough so that the can goes up past Karen, who catches the can on its way down. Calculate the velocity of the can the instant before Karen grabs the can. How long after Jeff tosses the can does Karen have to prepare to catch it?
ANSWER
9.51 m/s
EXPLANATION
We know that Jeff is 8.5m below Karen. He tosses the can up with initial velocity u = 16m/s and it passes where Karen is, so the maximum height of the can is 8.5m plus some more meters x. Then Karen catches the can in its way down, so when she does the can goes this distance x.
Let's find this distance. The height of an object thrown up with initial velocity u is:
[tex]y=ut-\frac{1}{2}gt^2[/tex]We know u = 16m/s but we don't know the time. This we can find from the final velocity of the can:
[tex]v=u-gt[/tex]At its maximum height the velocity is zero:
[tex]0=u-gt[/tex]Solving for t:
[tex]t=\frac{u}{g}[/tex]If we assume g = 9.8m/s²:
[tex]t=\frac{16m/s}{9.8m/s^2}=1.63s[/tex]We know that the can was in the air for 1.63 seconds until it reached its maximum height. The maximum height is:
[tex]y=16m/s\cdot1.63s-\frac{1}{2}\cdot9.8m/s^2\cdot1.63^2s^2[/tex][tex]y=26.08m-13.02m=13.06m[/tex]This is the maximum height of the can. The extra distance the can travelled above Karen is:
[tex]x=13.06m-8.5m=4.56m[/tex]In the can's way down, the initial velocity is 0, because it starts falling after stopping in its way up. The acceleration is still the acceleration of gravity and the height it falls is x. We can find the time it took to reach Karen's hand after it started falling:
[tex]y=\frac{1}{2}gt^2[/tex]Note that in this case we use the acceleration of gravity positive because it is in the same direction of the can's motion. Solving for t:
[tex]t=\sqrt[]{\frac{2y}{g}}[/tex][tex]t=\sqrt[]{\frac{2\cdot4.56m}{9.8m/s^2}}=\sqrt[]{0.93s^2}=0.97s[/tex]Knowing that the can was in the air for another 0.97 seconds after starting falling until it reached Karen's hand, we can find its velocity at that instant:
[tex]v=u+gt[/tex]Remember that in this case u = 0:
[tex]v=gt=9.8m/s^2\cdot0.97s=9.51m/s[/tex]The velocity of the can the instant before Karen grabs it is 9.51 m/s
A worker pushes horizontally on a large crate with a force of 245 N, and the crate is moved 3.5 m. How much work was done? answer in : ___J
The amount of work done by a force can be written as the following:
[tex]W=F.\Delta x[/tex]For our case, we can replace our values and we'll get:
[tex]W=245*3.5=857.5J[/tex]Thus, the amount of work done is 857.5J
carts, bricks, and bands
Which one of the following changes would increase the amount of mass?
a. Increase the number of bricks resting upon the cart.
b. Increase the number of bands that are used to pull the cart.
c. Decrease the number of bands that are used to pull the cart.
d. Use a cart that is identical in every way, except for the color that it is painted.
The change that would increase the amount of mass is Increase in the number of bricks resting upon the cart. That is option A.
What is mass?Mass is defined as the amount or quantity of matter that makes up an object which can be measured in grams and Kilograms.
The mass of an object is dependent on the change of its weight, that is, an increase in the weight of an object increases its mass.
Since the bricks had the same mass as the cart, adding one brick to the cart would double the mass of the object. Adding two bricks to the cart would triple the mass of the object.
That is to say, when more bricks are being laid on the cart, it's weight increase leading to an increase in its mass also.
Therefore, it can be concluded that increase the amount of mass is Increase the number of bricks resting upon the cart.
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Sean and Greg are on a job site standing on two beams 11.0 ft apart. they need to lift their crate of tools midway between them with ropes up 33.5 ft to where they are working. (a) What is the angle between the ropes when the crate is on the ground? (b) How much force do Sean and Greg need to exert on the ropes when lifting the 115-lb crate off the ground? (c) How much force do both Sean and Greg need to exert when the crate is 5.75 ft below them? (d) Explain why the force to lift the crate changes as it moves closer to them crate is 5.75 ft below them? (d) Explain why the force to lift the
crate changes as it moves closer to them.
a ) The angle between the ropes when the crate is on the ground = 18.4°
b ) Force exerted by Sean and Greg = 260.8 N
c ) Force exerted when the crate is 5.75 ft below them = 354.9 N
a ) The angle between the ropes,
Distance between crate and a person = 33.5 ft
Distance between Sean and Greg = 11 ft
Consider only one person side of the rope and the midway point between Sean and Greg. This forms a right angled triangle.
sin θ = 5.5 / 33.5
θ = [tex]sin^{-1}[/tex] ( 0.16 )
θ = 9.2°
Angle between the ropes = 2 θ
Angle between the ropes = 2 * 9.2
Angle between the ropes = 18.4°
b ) Force exerted,
Since both ropes pull the same amount of weight for the same amount of distance, their tensions will be equal,
T = T1 = T2
Resolving the tension into its horizontal and vertical component.
[tex]T_{y}[/tex] = T cos θ
[tex]T_{x}[/tex] = T sin θ
Since there is not time component mentioned assuming the crate is pulled at a constant velocity. Therefore acceleration will be zero and hence net force in vertical direction will be zero.
m = 115 lb = 52.16 kg
∑ [tex]F_{y}[/tex] = 0
T cos θ + T cos θ - m g = 0
2 T cos 9.2° = 52.16 * 9.8
T = 511.17 / 1.96
T = 260.8 N
c ) When the crate is 5.75 ft below them,
tan θ = 5.5 / 5.75
θ = [tex]tan^{-1}[/tex] ( 0.96 )
θ = 43.8°
∑ [tex]F_{y}[/tex] = 0
T cos θ + T cos θ - m g = 0
2 T cos 43.8° = 52.16 * 9.8
T = 511.17 / 1.44
T = 354.9 N
d ) The force to lift the crate changes as it moves closer to them is because in a shorter cable the horizontal force increases and the vertical force decreases. So it becomes harder to pull as it gets closer to the destination.
Therefore,
a ) The angle between the ropes when the crate is on the ground = 18.4°
b ) Force exerted by Sean and Greg on the ropes when lifting the 115-lb crate off the ground = 260.8 N
c ) Force exerted by Sean and Greg when the crate is 5.75 ft below them = 354.9 N
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How long does it take to stop a 1000 kg object moving at 20 m/s with a force 5000N? 2500 N, 1000 N, 500 N, 400 N, 200 N, 100 N
ANSWER
[tex]\begin{gathered} 5000N\Rightarrow4s \\ 2500N=8s \end{gathered}[/tex]EXPLANATION
To find the time taken to stop the object, we first have to find the acceleration of the object.
Since the force is working to stop the object (slow down the object), it means that the object is decelerating (slowing down).
To find the acceleration (deceleration), apply Newton's second law of motion:
[tex]F=ma[/tex]where F = force; m = mass; a = acceleration
Therefore, for a force of 5000N, we have that:
[tex]\begin{gathered} 5000=1000\cdot a \\ \Rightarrow\frac{5000}{1000}=a \\ \Rightarrow a=5m\/s^2 \end{gathered}[/tex]Now, we can apply Newton's equation of motion to find the time taken to stop the object:
[tex]v=u-at[/tex]where v = final velocity = 0 m/s; u = initial velocity = 20 m/s; t = time taken
Note: the negative sign indicates deceleration
Hence, the time taken for a force of 5000N to stop the object is:
[tex]\begin{gathered} 0=20-5\cdot t \\ \Rightarrow5t=20 \\ \Rightarrow t=\frac{20}{5} \\ t=4s \end{gathered}[/tex]For a force of 2500N, the deceleration is:
[tex]\begin{gathered} 2500=1000\cdot a \\ a=\frac{2500}{1000} \\ a=2.5m\/s^2 \end{gathered}[/tex]Hence, the time taken for a force of 2500N to stop the object is:
[tex]\begin{gathered} 0=20-2.5\cdot t \\ \Rightarrow2.5t=20 \\ t=\frac{20}{2.5} \\ t=8s \end{gathered}[/tex]Multiple part question Here are the needed details:Five rotations took 5.15 seconds 1 rotation took 1.07s Distance from shoulder to elbow is 29 cm distance from shoulder to middle of the hand is 57cm.Questions:2. A how far in degrees did the hand travel during the five rotations?B. How far in radians did the hand travel during the five rotations?C. How far in meters did the hand travel during the five rotations?3. A. What was the average angular speed (degrees/s and rad/s) of the hand?B. What was the average linear speed (m/s) of the hand?C. Are the answers to a and b the same or different? Explain.4. A. What was the average angular acceleration (degrees/s squared and rad/s squared) of the hand. How do you know?B. What was the average centripetal acceleration (m/s squared) of the hand?C. Are the answers to a and b the same or different. Explain.5. A. How far (degrees and rad) did the elbow travel during the five rotations?B. How far (m) did the elbow travel during the five rotations?C. How do these compare to the hand? Why are they the same and or/different?6. A. What was the average angular speed (degrees/s and rad/s) of the elbow?B. What was the average linear speed (m/s) of the elbow?C. How do these compare to the hand? Why are they the same and or/different?7. A. What was the average angular acceleration (degrees/s squared and rad/s squared) of the elbow?B. What was the average centripetal acceleration (m/s squared) of the elbow?C. How do these compare to the hand? Why are they the same and or/ different?
Given:
Time taken for 5 rotations = 5.15 seconds
Time for 1 rotation = 1.07 seconds
Distance from shoulder to elbow = 29 cm
Distance from shoulder to the middle of the hand = 57 cm
Let's use the information above to answer the following questions.
Question 2:
Let's determine how far in degrees the hand travelled during the five rotations.
In one full rotation, we have 360 degrees.
Thus, 5 full rotations = 5 * 360 = 1800 degrees
Therefore, in 5 full rotations, the hand travelled 1800 degrees.
B. In radians, we have:
180 degrees = π rad
[tex]1800\degree=\frac{\pi}{180}\ast1800=10\pi\text{ radians}[/tex]C. To find the distance in meters, we have:
Distance from elbow to shoulder = 29 cm = 0.29 meters
[tex]2\pi\ast5\ast0.29=9.11\text{meters}[/tex]Therefore, the hand travelled 9.11 meters during the five rotations.
Question 3:
A. To find the average angular speed, apply the formula:
[tex]\begin{gathered} w=\frac{10\pi}{t}\text{ (rad/s)} \\ \\ w=\frac{1800}{t}\text{ (degre}es\text{/s)} \end{gathered}[/tex]Where t = 5.15 seconds
Thus, we have:
[tex]\begin{gathered} w=\frac{10\pi}{5.15}=6.1\text{ rad/s} \\ \\ w=\frac{1800}{5.15}=349.5\text{ degre}es\text{/s} \end{gathered}[/tex]B. Average linear speed of the hand.
To find the average linear speed of the hand, we have:
[tex]v=\frac{10\pi r}{t}=\frac{10\pi}{5.15}\ast\frac{1}{2}=3.05\text{ m/s}[/tex]C. The average angular speed and average linear speed are the same
The mass of the Moon is about 1/80th of the mass of Earth. The force exerted by Earth on the Moon is about 80 times thatexerted by the Moon on Earth.Select one:O TrueO False
According to Newton's Third Law of Motion, the force that object A exerts to object B has the same magnitude as the force that object B to object A, but in the opposite direction:
[tex]\vec{F}_{AB}=-\vec{F}_{BA}[/tex]Then, the force exerted by Earth on the Moon has the same magnitude as the force exerted by the Moon on the Earth.
Therefore, the given statement is false.
A 2000 kg car is stopped by applying a braking force of 5000 newtons.
Determine the acceleration caused by this braking force.
A 2000 kg car is stopped by applying a braking force of 5000 newtons. The acceleration caused by this braking force is (a)=2.5 m/s²
What is acceleration?When a object start with a velocity and ends with different velocity, so the change in velocity in a given time is called the acceleration of the object. It is a vector quantity. It can be measured in m/s².
How can we calculate the acceleration?To calculate the acceleration we are using the formula here is ,
F=ma
Here we are given by the question is,
F= The amount of force applied on the car. = 5000N.
m= The mass of the car. = 2000kg
We have to calculate the change in acceleration = a m/s².
Now we put the values in above equation we get,
F=ma
Or, a= F/m
Or, a=5000/2000
Or, a=2.5 m/s².
Now from the above calculation we can say that, The acceleration caused by this braking force is (a)=2.5 m/s².
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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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A snowmobiler travels 65km [37° E of S]. How far east does she travel?
ANSWER:
39.1 km
STEP-BY-STEP EXPLANATION:
To better understand the problem, we make a sketch, like this:
Therefore, we can determine the distance east with the cosine function, like this:
[tex]\begin{gathered} \cos\theta=\frac{\text{ adjacent}}{\text{ hypotenuse}} \\ \\ \theta=53\degree \\ \\ \text{ adjacent =}E \\ \\ \text{ hypotenuse = 65} \\ \\ \text{ We replacing:} \\ \\ \cos53\degree=\frac{E}{65} \\ \\ E=65\cdot\cos53\degree \\ \\ E=39.1\text{ km} \end{gathered}[/tex]The distance to the east is 39.1 km
On which of the following does the speed of a falling object depend?a.) v ∝ mb.)v ∝ mc.)v ∝ Δh
Given:
The falling object
To find:
The dependence on the speed of the falling object
Explanation:
For an object falling freely, the total mechanical energy remains always constant. So, we can write, that the decrease in potential energy will be equal to the increase in the kinetic energy that is
[tex]\begin{gathered} \frac{1}{2}mv^2=mgh \\ v^2=2gh \end{gathered}[/tex]Hence, the speed of the falling object does not depend on the mass it depends on the height difference.
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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: