Question: A) A car is driving too fast on a flat non-banked curve. The car cannot stay on the road. What path will the car take as it leaves the road? See Image. (a) (b) B) The driver wants to go faster around the same curve mentioned above and not run off the road. He adds sandbags to his car to make it weigh more so that the friction force between the tires and the road will be increased. Will this work? Explain your answer.
Explanation:
(a) The car will leave tangentially to the road, so it will take path d.
(b) Sum of forces on the car in the vertical direction:
∑F = ma
N − mg = 0
N = mg
Sum of forces on the car in the centripetal direction:
∑F = ma
Nμ = m v²/r
mgμ = m v²/r
gμ = v²/r
v = √(grμ)
The maximum velocity is independent of mass, so adding sandbags will not work.
Using the equation for for Newton's Second law, m=F/a solve the following problem. You have been given an object with a force of 10N and an acceleration of 2 m/s2, what is the mass?
Group of answer choices
1. 8g
2. 3g
3. 20g
4. 5g
Answer:
4. 5g
Explanation:
F=ma so, m=Fa. All you have to do is 10/2. Don't be confused by the units. Mass will normally be in grams.
ball is thrown vertically upward from an initial position 5m above the ground. At the same time, a cube is released from rest down aslippery incline, from an unknownheight. The two objects reach the ground at the same instant, and both have a final speed of 15m/s. What is the angle ofthe incline
Answer:
The value is [tex]\theta = 34.8 ^o[/tex]
Explanation:
Generally the time it takes the ball to move from maximum height to the ground is
[tex]t = \frac{v - u}{g}[/tex]
Here u= 0 m/s and v = 15 m/s
So
[tex]t = \frac{15 - 0}{9.8}[/tex]
=> [tex]t = 1.53 \ s [/tex]
Generally the maximum height traveled in time t is mathematically represented as
[tex]s = u * t + \frac{1}{2} * g * t^2[/tex]
=> [tex]s = 0 * 1.53 + \frac{1}{2} * 9.8 * 1.53^2[/tex]
=> [tex]s = 11.5 \ m [/tex]
The interval in distance between the maximum height and the initial position of the ball is
[tex]d = s - k[/tex]
Here k = 5 m (given)
[tex]d = 11.5 - 5[/tex]
[tex]d = 6.5 \ m [/tex]
Generally from the kinematic equation we have that
[tex]d = ut_1 + \frac{1}{2} g * t_1 ^2[/tex]
Here [tex] t_1 [/tex] is the time taken to travel through d
[tex]t_1 = \sqrt{ \frac{2 * d }{g} }[/tex]
=> [tex]t_1 = \sqrt{ \frac{2 * 6.5 }{9.8} }[/tex]
=> [tex]t_1 = 1.15 \ s [/tex]
Generally the total time for this fight is mathematically represented as
[tex]t_f = t + t_1[/tex]
=> [tex]t_f = 1.15 + 1.53[/tex]
=> [tex]t_f = 2.68 \ s [/tex]
Generally the vertical force exerted on the cube is mathematically represented as
[tex]F = m * g * sin (\theta )[/tex]
=> [tex] m * a = m * g * sin (\theta )[/tex]
=> [tex] a = g * sin (\theta )[/tex]
Now a is mathematically represented as
[tex]a = \frac{v - u }{t_f}[/tex]
=> [tex]a = \frac{15 - u }{2.68}[/tex]
=> [tex]a = 5.597 \ m/s^2[/tex]
So
[tex] 5.597 = 9.8 * sin (\theta )[/tex]
[tex]\theta = sin^{-1} [\frac{5.597}{9.8} ][/tex]
[tex]\theta = 34.8 ^o[/tex]
What were the physical activities in your childhood that you still do today? Do you spend more time now in doing these activities as compared before?
Understand that the acceleration vector is in the direction of the change of the velocity vector. In one dimensional (straight line) motion, acceleration is accompanied by a change in speed, and the acceleration is always parallel (or antiparallel) to the velocity. When motion can occur in two dimensions (e.g. is confined to a tabletop but can lie anywhere in the x-y plane), the definition of acceleration is
Answer:
a = √ (a_t² + a_c²)
a_t = dv / dt , a_c = v² / r
Explanation:
In a two-dimensional movement, the acceleration can have two components, one in each axis of the movement, so the acceleration can be written as the components of the acceleration in each axis.
a = aₓ i ^ + a_y j ^
Another very common way of expressing acceleration is by creating a reference system with a parallel axis and a perpendicular axis. The axis called parallel is in the radial direction and the perpendicular axis is perpendicular to the movement, therefore the acceleration remains
a = √ (a_t² + a_c²)
where the tangential acceleration is
a_t = dv / dt
the centripetal acceleration is
a_c = v² / r
2. The components of vector A are given as follows:
Ax = 5.6 Ay = -4.7
What is the angle between vector A and positive direction of x-axis?
Answer:
50 degree.
Explanation:
Given that the components of vector A are given as follows: Ax = 5.6 Ay = -4.7
The angle between vector A and B in the positive direction of x-axis will be achieved by using the formula:
Tan Ø = Ay/Ax
Substitute Ay and Ax into the formula above.
Tan Ø = -4.7 / 5.6
Tan Ø = -0.839
Ø = tan^-1(-0. 839)
Ø = - 40 degree
Therefore, the angle between vector A and B positive direction of x-axis will be
90 - 40 = 50 degree.
A column of soldiers, marching at 100 steps per minute, keep in step with the beat of a drummer at the head of the column. It is observed that the soldiers in the rear end of the column are striding forward with the left foot when the drummer is advancing with the right. What is the approximate length of the column? (Take the speed of sound to be 343 m/s.)
Answer:
The value is [tex]D = 205.8 \ m [/tex]
Explanation:
The time taken for the column to take a step mathematically represented as
100 steps => 1 minutes => 60 seconds
1 step => t
=> [tex]t = 0.6 \ s [/tex]
Generally the length of the column is mathematically represented as
[tex]D = v * t[/tex]
substituting 343 m/s for v we have
[tex]D = 343 * 0.6 [/tex]
=> [tex]D = 205.8 \ m [/tex]
Hitting a ball off a tee would be considered which of the following?
A.
A closed serial skill
B.
An open continuous skill
C.
An open discrete skill
D.
A closed discrete skill
How fast do you go if you move 80 meters in 0.20 seconds?
Answer: I HOPE IT HELPS
Use this Speed Calculator to calculate the speed you have travelled based ... (so add three 0s to your kilometres to get metres) and a mile is 1609 metres. ... You will have distance per time such as kilometres per hour or metres per second. ... If you want help with that, click on hyperlink for metres or for minutes and seconds.
Explanation:
An object is released from rest from a top of a building 90 meters high. Neglect air friction.
What is the volume of its acceleration?
Calculate the time it takes to reach the floor?
With what velocity does it reach the floor?
How fast is it moving when it is 50 meter above the floor?
Answer:
1. a = 9.8 m/s²
2. t = 4.28 s
3. Vf = 42 m/s
4. Vf = 28 m/s
Explanation:
1.
Since, the body is under free fall motion. Therefore, the value of its acceleration shall be equal to the acceleration due to gravity.
a = 9.8 m/s²
2.
The time taken by the ball to reach the ground can be calculated by using second equation of motion:
h = Vi t + (1/2)gt²
where,
h = height = 90 m
Vi = initial velocity = 0 m/s
t = time taken = ?
Therefore,
90 m = (0 m/s)t + (1/2)(9.8 m/s²)t²
t = √(18.36 s²)
t = 4.28 s
3.
In order to find final velocity we use first equation of motion:
Vf = Vi + gt
Vf = 0 m/s + (9.8 m/s²)(4.28 s)
Vf = 42 m/s
4.
when the ball is at height of 50 m, it means it has covered:
h = 90 m - 50 m = 40 m
we use third equation of motion at this point:
2gh = Vf² - Vi²
(2)(9.8 m/s²)(40 m) = Vf² - (0 m/s)²
Vf = √(784 m²/s²)
Vf = 28 m/s
An elevator held by a single cable is ascending but slowing down. Is the work done by tension positive, negative, or zero? What about the work done by gravity? Explain
Explanation:
1.) Work = Force*Distance
2.) The direction of motion lies in the same direction as the tension in the cable. So the work done by tension would be positive.
3.) The direction of the weight (due to gravity) would act opposite to the direction of motion of the elevator, so work done by gravity becomes negative.
A car starts at 80 m/s but sees a cop and hits the brakes slowing down to 50 m/s in 2 sec.
Answer:
-15m/s/s
Explanation:
Acceleration = change in speed/ change in time
The change in speed is calculated by subtracting the initial speed from the final speed, so the change in speed is: 50 - 80 = -30m/s. The change in time is 2 - 0 = 2.
So the car acceleration is -30/2 = -15m/s/s
It is negative because it is decelerating.
Hope this helped!
What are some reasons people may not prepare to be safe and comfortable when they participate in physical activity?
Answer: ummm arent you in my physical fundations class lol
Explanation:
are you or am i tripping? and sorry i dont have the answer :(
Common reasons people give for not being active include not having enough time, finding physical activity inconvenient, lacking self-motivation, etc.
What is physical activity?Any consensual bodily mobilization generated by skeletal muscles that necessitate caloric expenditure is characterized as physical activity.
Physical activity includes all activities of any frequency, at any moment of day or night. It incorporates both strength training and incidental activity into the daily routine.
Physical activity can improve your brain health, help you manage your weight, lower your risk of disease, strengthen your bones and muscles, and improve your ability to do everyday tasks.
Adults who sit less and engage in moderate-to-vigorous physical activity reap health benefits.
Internal barriers were classified into three types namely, a lack of energy, a lack of motivation, and a lack of self-efficacy.
External barriers were also classified into three categories namely a lack of resources, a lack of social support, and a lack of time.
Thus, these are some reasons that may not prepare to be safe and comfortable when they participate in physical activity
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A fisherman’s boat moves up and down periodically due to water waves. The boat travels from its highest to its lowest point, a vertical distance of 0.60 m, in 2.0 s. The horizontal distance between wave crests is 7.0 m. (a) What is the period of the wave motion? (b) What is the frequency of the wave motion? (c) W
Complete question is;
A fisherman’s boat moves up and down periodically due to water waves. The boat travels from its highest to its lowest point, a vertical distance of 0.60 m, in 2.0 s. The horizontal distance between wave crests is 7.0 m. (a) What is the period of the wave motion? (b) What is the frequency of the wave motion? (c) What is the amplitude of each wave
Answer:
A) Period = 4 s
B) Frequency = 0.25 Hz
C) Amplitude = 0.3 m
Explanation:
A) We are told that the distance it takes for the boat to travel from its highest to its lowest point is 2 s.
Thus, t = 2 s
Now, in waves, period (T) it the time between two successive waves.
This means that;
T = 2t
T = 2 × 2
T = 4 s
B) Also, the frequency(f) is given by the formula;
f = 1/T
f = 1/4
f = 0.25 Hz
C) We are given a vertical distance of 0.60 m from highest to lowest point of waves.
Now, amplitude is half of this distance.
Thus;
Amplitude(A) = 0.6/2
A = 0.3 m
A small block with mass 0.200 kg is released from rest at the top of a frictionless incline. The block travels a distance 0.394 m down the incline in 2.00 s. The 0.200 kg block is replaced by a 0.400 kg block. If the 0.400 kg block is released from rest at the top of the incline, how far down the incline does it travel in 2.00 s
Answer:
The distance traveled by the second block is 0.197 m
Explanation:
Given;
mass of the small block, m₁ = 0.2 kg
distance traveled by the block, d₁ =0.394 m
time of travel, t = 2 s
mass of the second block, m₂ = 0.4 kg
distance traveled by the second block, d₂ = ?
The work done per unit time on the inclined plane is given by;
[tex]\frac{f_1d_1}{t} = \frac{f_1d_1}{t}\\\\f_1d_1 = f_2d_2\\\\m_1gd_1 = m_2gd_2\\\\m_1d_1 = m_2d_2\\\\d_2 = \frac{m_1d_1}{m_2} \\\\d_2 = \frac{0.2*0.394}{0.4}\\\\d_2 = 0.197 \ m[/tex]
Therefore, the distance traveled by the second block is 0.197 m
3) An explorer walks 13 km due east, then 18 km north, and finally 3 km west.
a) What is the total distance walked?
b) What is the resulting displacement of the explorer from the starting point?
Answer: 34 km, 21 km 61 degrees north of east
Explanation: distance = 13 + 3 + 18 = 34
displacement = 13 - 3 = 10
10^2 + 18^2 = 424
find the square root of 424 ( 20.5 rounded to 21 )
The total distance walked is 34 km and the resulting displacement is 20.6 km.
a) The total distance is gotten by summing up all the distance.
Total distance = Distance moved east + distance moved north + distance moved west
Total distance = 13 km + 18 km + 3 km = 34 km
b) The displacement is the distance from the beginning point to end point.
Displacement² = 18² + (13 - 3)² = 18² + 10²
Displacement² = 424
Displacement = 20.6 km
Therefore the total distance walked is 34 km and the resulting displacement is 20.6 km.
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The position of a particle
Answer:
vector
Explanation:
The position of a particle is vector
Which of the following is the best thermal conductor?
A. Fiberglass.
B. Stainless.
C. Steel.
D. Wood.
E. Silver
Answer:
steel because being alloy in metal it has free electrons in metals
Consider the displacement vectors Ā=(i +6j)m, B = (3i– 7j)m,
If A-B-4D=0, what are the components of D? (Dx = ?, and Dy = ?)
Answer:
Dx = -0.5
Dy = -0.25
Explanation:
Two vectors are given in rectangular components form as follows:
A = i + 6j
B = 3i - 7j
It is also given that:
A - B - 4D = 0
so, we solve this to find D vector:
(i + 6j) - (3i - 7j) - 4D = 0
- 2i - j = 4D
D = - (2/4)i - (1/4)j
D = - (1/2)i - (1/4)j
D = - 0.5i - 0.25j
Therefore,
Dx = -0.5
Dy = -0.25
how many legs a cow has
Calculate the length of segment RS with midpoint, M , if RM = 5x and MS = x + 12.
A. 3
B. 25
C. 30
D. 36
E. 15
Answer:
the full segment is: 30 units long
which coincides with answer C in your list
Explanation:
If M is the midpoint, then it divides the segment in two equal parts. Then, we can say that:
RM = MS (equality among the two parts of the divided segment)
replacing RM with "5 x" and MS with "x + 12", we get:
5 x = x + 12
solving for x:
5 x - x = 12
4 x = 12
then x = 3
With this info, we can calculate the length of each half of the segment and consequently its full length:
RM = 5 x = 5 (3) = 15
then the full segment is: 30 units long
A sprinter practicing for the 200-m dash accelerates uniformly from rest at A and reaches a top speed of 41 km/h at the 61-m mark. He then maintains this speed for the next 83 meters before uniformly slowing to a final speed of 34 km/h at the finish line. Determine the maximum horizontal acceleration which the sprinter experiences during the run. Where does this maximum acceleration value occur
Answer:
2.13 m/s^2.
Find the remaining answer in the explanation
Explanation:
Given that a sprinter practicing for the 200-m dash accelerates uniformly from rest at A and reaches a top speed of 41 km/h at the 61-m mark. He then maintains this speed for the next 83 meters before uniformly slowing to a final speed of 34 km/h at the finish line.
1.) The maximum horizontal acceleration will be when she attained her maximum speed. That is,
Speed = 41 km/h
Convert km/h to m/s
(41×1000) / 3600
Speed = 11.39 m/s
Using speed formula to calculate time
Speed = distance/time
11.39 = 61/t
t = 61/11.39
t = 5.36s
Where the distance = 61 m
Maximum acceleration = velocity /time
Maximum acceleration = 11.39/5.36
Maximum acceleration = 2.13 m/s^2
The maximum acceleration value occurs when the sprinter is starting from rest and attaining the maximum speed.
The rain gauge has 2 inches of precipitation.
Qualitative or Quantitative
Answer:
Quantitative is the correct term
Which of these is caused by the interaction between the atmosphere and the hydrosphere?
Answer:
Explanation: All the spheres interact with other spheres. For example, rain (hydrosphere) falls from clouds in the atmosphere to the lithosphere and forms streams and rivers that provide drinking water for wildlife and humans as well as water for plant growth (biosphere).
What is the volume of a marble with a diameter of 3.0
Explanation:
You can solve for volume using radius or diameter.
Sphere Volume = 4/3 • π • r³ = ( π •d³)/6
We're given the diameter so let's use that.
Volume = PI * d^3 / 6
Volume = 3.14159 * 3.0^3 / 6
Volume = 3.14159 * 9 / 2
Volume = 14.137 cubic centimeters
A pendulum built from a steel sphere with radius r cm 5 and density stl kg m S 3 7800 is attached to an aluminum bar with length l m 1 thickness t cm 0 8. and width w cm 4 and density . al kg m S 3 2820 a. Calculate the mass moment of inertia of the pendulum about its center of mass, . cm I b. Calculate the mass moment of inertia of the pendulum about its pivot point, o I .
Answer:
a) I = 0.0198 kg m² , b) I = 21.85 kg m²
Explanation:
For this exercise we will use the definition of moment of inertia
I = ∫ r² dm
For body with high symmetry they are tabulated
sphere I = 2/5 m r²
bar with respect to center of mass I = 1/12 m L²
let's calculate the mass of each body
bar
ρ = m / V
m = ρ V
m = ρ l w h
where we are given the density of the bar rho = 32840 kg / m³ and its dimensions 1 m, 0.8 cm and 4 cm
m = 32820 1 0.008 0.04
m = 10.5 kg
Sphere
M = ρ V
V = 4/3 pi r³
M = rgo 4/3 π r³
give us the density 37800 kg / m³ and the radius of 5 cm
M = 37800 4/3 π 0.05³
M = 19.8 kg
a) asks us for the moment of inertia of the sphere with respect to its center of mass
I = 2/5 M r²
I = 2/5 19.8 0.05²
I = 0.0198 kg m²
b) the moment of inertia with respect to the turning point, for this we will use the theorem of parallel axes
I = I_cm + M d2
where d is the distance from the body to the point of interest
I_cm = 0.0198 kg m²
the distance to the pivot point is
l = length of the bar + radius of the sphere
l = 1 + 0.05 = 1.005 m
I = 0.0198 + 19.8 1.05²
I = 21.85 kg m²
- Over the weekend you go for an urban hike, traveling 3 miles north then 4 miles west before stopping to take a break. If this hike takes 1 hour, what is your average velocity during that time? - You leave your house to go for a drive to clear your head, driving 10 miles over the course of 30 minutes before arriving back home. What is your average velocity?
Answer:
A.) 5 mph
B.) 20 mph
Explanation:
Given that you are traveling 3 miles north then 4 miles west before stopping to take a break. If this hike takes 1 hour,
Let's first calculate the displacement be using pythagorean theorem
D = sqrt ( 3^2 + 4^2 )
D = sqrt ( 9 + 16 )
D = sqrt ( 25 )
D = 5
Average velocity = displacement/time
Average velocity = 5/1
Average velocity = 5 mph
Therefore, your average velocity during that time is 5 mph
- You leave your house to go for a drive to clear your head, driving 10 miles over the course of 30 minutes before arriving back home.
Average velocity = 10/ 0.5
Average velocity = 20 mph
sam exerts a force of 65 N on a lawn mower with a mass of 25 kg. which formula can be used to calculate the acceleration of a lawn mower
Answer:
F = ma - with this you will get 2.6 m/s^2
Explanation:
Use this formula to get the acceleration...
where F is force, M is mass and A is acceleration
by using this we get...
65 = 25 * a
so, a = 65/25
Therefore, the acceleration is 2.6 m/s^2
Hope that helped :)
The acceleration of a lawn mower will be 2.6 m/s². It is the ratio of force and the mass.
What is acceleration?The rate of change of velocity with respect to time is known as acceleration. According to Newton's second law, the eventual effect of all forces applied to a body is its acceleration.
The given data in the problem is;
The force act on a lawn mower,[tex]\rm F = 65 \ N[/tex]
Mass of lawn mower,[tex]\rm m = 25 kg[/tex]
The acceleration of a lawn mower is,[tex]\rm a = ?[/tex]
Acceleration, is found as the ratio of force and the mass.
[tex]\rm F= ma \\\\ a = \frac{F}{m} \\\\ a= \frac{65}{25} \\\\ a = 2.6 \ m/s^2[/tex]
Hence, the acceleration of a lawn mower will be 2.6 m/s².
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What the differences between static and kinetic friction?
Answer:
Static friction prevents a stationary object from moving while kinetic or dynamic friction slows down a moving object.
Explanation:
Static Friction is the maximum force that must be overcome before a stationary object begins to move, while kinetic or dynamic friction is the maximum force that must be overcome for an object in motion to continue moving at a uniform velocity.
Static friction keeps a stationary object at rest, once the Force of Static friction is overcome, the Force of Kinetic friction is what slows down the moving object.
A magnetic field is created by ____.
A moving electric charges
b elctromagnetic pulses
c a strong current
d a change in the current of a wire