Answer:
10945.9 kg/m^3
Explanation:
mass of cylinder, m = 1 kg
Height of cylinder, h = 55 mm = 0.055 m
Diameter of cylinder = 46 mm
Radius of cylinder, r = 23 mm = 0.023 m
The formula of the volume of the cylinder is given by
[tex]V = \pi r^{2}h[/tex]
V = 3.14 x 0.023 x 0.023 x 0.055
V = 9.136 x 10^-5 m^3
Density is defined as mass per unit volume .
[tex]Density = \frac{1}{9.136 \times 10^{-5}}[/tex]
Density = 10945.9 kg/m^3
A solid conducting sphere of radius 2.00 cm has a charge of 6.88 μC. A conducting spherical shell of inner radius 4.00 cm and outer radius 5.00 cm is concentric with the solid sphere and has a charge of −2.96 μC. Find the electric field at the following radii from the center of this charge configuration. (a) r = 1.00 cm magnitude:_________ N/C
direction:____________
(b) r = 3.00 cm magnitude:____________ N/C
direction:_________________
(c) r = 4.50 cm magnitude:_____________ N/C direction:______________
(d) r = 7.00 cm magnitude:_____________ N/C
direction:______________
Explanation:
Given that,
Radius R= 2.00
Charge = 6.88 μC
Inner radius = 4.00 cm
Outer radius = 5.00 cm
Charge = -2.96 μC
We need to calculate the electric field
Using formula of electric field
[tex]E=\dfrac{kq}{r^2}[/tex]
(a). For, r = 1.00 cm
Here, r<R
So, E = 0
The electric field does not exist inside the sphere.
(b). For, r = 3.00 cm
Here, r >R
The electric field is
[tex]E=\dfrac{kq}{r^2}[/tex]
Put the value into the formula
[tex]E=\dfrac{9\times10^{9}\times6.88\times10^{-6}}{(3.00\times10^{-2})^2}[/tex]
[tex]E=6.88\times10^{7}\ N/C[/tex]
The electric field outside the solid conducting sphere and the direction is towards sphere.
(c). For, r = 4.50 cm
Here, r lies between R₁ and R₂.
So, E = 0
The electric field does not exist inside the conducting material
(d). For, r = 7.00 cm
The electric field is
[tex]E=\dfrac{kq}{r^2}[/tex]
Put the value into the formula
[tex]E=\dfrac{9\times10^{9}\times(-2.96\times10^{-6})}{(7.00\times10^{-2})^2}[/tex]
[tex]E=5.43\times10^{6}\ N/C[/tex]
The electric field outside the solid conducting sphere and direction is away of solid sphere.
Hence, This is the required solution.
You drive 6.0 km at 50 km/h and then another 6.0 km at 90 km/h. Your average speed over the 12 km drive will be ____
(A) a greater than 70 km/h.
(B) less than 70 km/h.
(C) exactly 38 km/h.
(D) It cannot be determined from the information given because we must also know directions traveled.
(E) equal to 70 km/h.
Answer:
The average speed is less than 70 km/h.
(B) is correct option.
Explanation:
Given that,
distance = 6.0 km
Speed = 50 km/h
Speed = 90 km/h
We need to calculate the time in 6.0 km distance
Using formula of time
[tex]t = \dfrac{d}{v}[/tex]
Put the value in to the formula
[tex]t=\dfrac{6.0}{50}[/tex]
[tex]t=0.12\ hr[/tex]
We need to calculate the time in another distance
Using formula of time
[tex]t = \dfrac{d}{v}[/tex]
Put the value in to the formula
[tex]t=\dfrac{6.0}{90}[/tex]
[tex]t=0.067\ hr[/tex]
We need to calculate the average speed
Using formula of average speed
[tex]v=\dfrac{D}{T}[/tex]
Where, D = total distance
T = total time
Put the value into the formula
[tex]v=\dfrac{12}{0.12+0.067}[/tex]
[tex]v=64.17\ km/hr[/tex]
Hence, The average speed is less than 70 km/h.
The average speed over the 12 km drive is less than 70 km/h since the total time for the trip is 0.187 hours, leading to an average speed of 64.2 km/h.
To calculate the average speed for the entire trip, you must divide the total distance by the total time taken. In the given scenario, you drive 6.0 km at 50 km/h and then another 6.0 km at 90 km/h. First, calculate the time taken for each portion of the trip.
For the first 6.0 km at 50 km/h: time = distance/speed = 6.0 km / 50 km/h = 0.12 hours.For the second 6.0 km at 90 km/h: time = distance/speed = 6.0 km / 90 km/h = 0.067 hours.Now, add the times together and divide the total distance by the total time to find the average speed.
Total time = 0.12 hours + 0.067 hours = 0.187 hours.Average speed = Total distance / Total time = 12 km / 0.187 hours = 64.2 km/h.Therefore, the average speed over the 12 km drive is less than 70 km/h, which corresponds to option (B).
The magnitude of the acceleration of gravity, in the International System of Measurements, in an object that falls vertically is: a. 32.2 ft / s^2
b. 9.81 m / s
c. 32.2 m / s^2
d. 9.81 m / s^2
Answer:
option D
Explanation:
the correct answer is option D
acceleration of a body under free fall due to the effect of gravity is the called acceleration due to gravity.
Acceleration due to gravity is a constant value.
According to the International System of Measurements the acceleration due to gravity is 9.81 m/s² when an object falls vertically.
acceleration due to gravity is denoted by 'g'.
An Earth satellite needs to have its orbit changed so the new orbit will be twice as far from the center of Earth as the original orbit. The new orbital period will be twice as long as the original period. O true O false
Answer:
False.
Explanation:
From Kepler's Third Law of plenetary motion, we know that:
"The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit."
Or, as expressed in mathematical terms:
[tex]\frac{a^3}{T^2}=constant[/tex], where a is the semi-major axis of the orbit (the distance from the center), and T is the orbital period of the satellite.
From this expression we can clearly see that if the orbit's semi-major axis is doubled, orbital period will be [tex]\sqrt{8}[/tex] times longer to compensate the variation.
A policeman, parked at arn intersection, is passed by a speeder traveling at 120 km/hr. Two seconds later, the policeman starts and uniformly accelerates at 4.0 m/s^2. How far will the policeman travel before he catches the speeder?
Answer:
682.32 m
Explanation:
Speed of the passing speeder = 120 km/hr = 120 × 0.2777 = 33.33 m/s
time after which police man starts = 2 seconds
Acceleration of the policeman = 4.0 m/s²
let the time taken to catch be 't' seconds
Now,
The total distance to be covered by the policeman
= Distance covered by the speeder in the 2 seconds + Distance further traveled by the speeder in time t
thus,
From Newton's equation of motion
[tex]s=ut+\frac{1}{2}at^2[/tex]
where,
s is the total distance traveled by the police man
u is the initial speed = 0
a is the acceleration
t is the time
thus,
[tex]0\times t+\frac{1}{2}\times4t^2[/tex]= 33.33 × 2 + 33.33 × t
or
2t² = 66.66 + 33.33t
or
2t² - 33.33t - 66.66 = 0
on solving the above equation, we get
t = 18.47 seconds (negative value is ignored as time cannot be negative)
therefore,
the total distance covered = 33.33 × 2 + 33.33 × 18.47 = 682.32 m
You are racing your dog next to a football field. You run from the 10 yard line to the 50 yard line, then turn around and run back to the 30 yard line. What is your distance? What is your displacement?
Answer:60 m,20 m
Explanation:
Given
You ran 10 yard line to 50 yard line therefore total distance traveled is 40 yard
Finally you turned around and came back to 30 yard line
Therefore total distance is 40+20=60 m(From 50 yard line to 30 yard line it is 20 yards)
Displacement=30-10=20 yards
While John is traveling along a straight interstate highway, he notices that the mile marker reads 245 km. John travels until he reaches the 135 km marker and then retraces his path to the 176 km marker. What is John’s resultant displacement from the 245 km marker? Answer in units of km.
Answer:Displacement =69 km
Explanation:
Given
First John notices 245 km marker i.e. he is 245 away from destination.
John travels until he reach 135 km marker i.e. he is 135 km away from destination.
After that he retraces his path to reach 176 km marker
so his resultant displacement is i.e. shortest distance between initial and final point is 245-176 =69 km
A particle starts from the origin at t = 0 with an initial velocity having an x component of 29.8 m/s and a y component of −18 m/s. The particle moves in the xy plane with an x component of acceleration only, given by 5.21 m/s 2 . Determine the x component of velocity after 7.22 s. Answer in units of m/s.
Answer:
Vx(7,22s)=67,42m/s
Explanation:
V(t)=Vo+a(t) [tex]Vx(7,22s)=Vxo+5,21\frac{m}{s^{2} } *7,22s
Vx(7,22s)=29,8m/s +37,62m/s=67,42m/s[/tex]
you just consider the x component of your velocity, then you apply the Velocity formula of uniform accelerated rectilinear motion, and you get to the result, by replacing t (time) by the value given 7,22s.
A transverse sinusoidal wave is moving along a string in the positive direction of an x axis with a speed of 86 m/s. At t = 0, the string particle at x = 0 has a transverse displacement of 4.0 cm from its equilibrium position and is not moving. The maximum transverse speed of the string particle at x = 0 is 18 m/s. (a) What is the frequency of the wave? (b) What is the wavelength of the wave? If the wave equation is of the form y(x, t) = ym sin(kx ± ωt + φ), what are (c) ym, (d) k, (e) ω, (f) φ, and (g) the correct choice of sign in front of ω?
Final answer:
The frequency of the wave is 1075 Hz, and the wavelength is 0.08 m. The wave equation of the form y(x, t) = ym sin(kx ± ωt + φ) represents a sinusoidal wave with specific parameters. The amplitude is 0.04 m, the wave number is 25π m⁻¹, the angular frequency is 6755π rad/s, and the phase angle can be either +π or -π. The choice of sign in front of ω determines the direction of wave propagation.
Explanation:
To find the frequency of the wave, we can use the formula f = v/λ, where f is the frequency, v is the speed of the wave, and λ is the wavelength. In this case, the speed of the wave is given as 86 m/s. Since the wave is sinusoidal, the wavelength is equal to twice the amplitude of the transverse displacement of the string particle at x = 0. So, the wavelength is equal to 2 times 4.0 cm, which is 8.0 cm or 0.08 m.
Substituting these values into the formula, we get f = 86 m/s / 0.08 m = 1075 Hz. Therefore, the frequency of the wave is 1075 Hz.
To find the wavelength of the wave, we can use the formula λ = v/f, where λ is the wavelength, v is the speed of the wave, and f is the frequency. Substituting the given values, we get λ = 86 m/s / 1075 Hz = 0.08 m.
Therefore, the wavelength of the wave is 0.08 m.
The wave equation of the form y(x, t) = ym sin(kx ± ωt + φ) represents a sinusoidal wave. In this equation, ym is the amplitude of the wave, k is the wave number, ω is the angular frequency, and φ is the phase angle. The choice of the sign in front of ω determines the direction of wave propagation. If it is positive, the wave propagates in the positive x-direction, and if it is negative, the wave propagates in the negative x-direction.
In this case, the amplitude of the wave is given as 4.0 cm or 0.04 m. The wave number can be calculated using the formula k = 2π/λ, where k is the wave number and λ is the wavelength. Substituting the given wavelength of 0.08 m, we get k = 2π/0.08 m = 25π m⁻¹.
The angular frequency can be calculated using the formula ω = 2πf, where ω is the angular frequency and f is the frequency. Substituting the given frequency of 1075 Hz, we get ω = 2π × 1075 Hz = 6755π rad/s.
The phase angle is given as ± π. The choice between +π and -π determines the phase of the wave at x = 0 and t = 0.
Therefore, (c) ym = 0.04 m, (d) k = 25π m⁻¹, (e) ω = 6755π rad/s, (f) φ = ±π, and (g) the correct choice of sign in front of ω depends on the desired direction of wave propagation.
Objects that are thrown upwards as they gain height their speed goes: a. Increasing
b. Staying the same
c. Decreasing
d. Duplicating the initial
Answer:
The correct option is 'c': Decreases.
Explanation:
As we know that when we throw a ball upwards we imoart a kinetic energy to the bsll. As the ball moves upwards the kinetic energy of the ball starts to decrease as the ball slows down as it moves upwards as work has to be done for movement against gravity. This work done against gravity is stored as potential energy of the ball.
Mathematically as we throw the ball upward's with velocity 'u' it's initial kinetic energy equals [tex]E_{initial}=\frac{1}{2}mu^{2}[/tex]
As the ball attain's a height 'h' it's total energy is sum of the potential and the remaining kinetic energy as follows
[tex]E(h)=mgh+\frac{1}{2}mv^{2}[/tex]
Equating both the energies we get
[tex]\frac{1}{2}mu^{2}=mgh+\frac{1}{2}mv^{2}\\\\\\therefore v=\sqrt{u^{2}-2gh}[/tex]
Hence we can see as the height increases the velocity decreases.
A toy rocket, launched from the ground, rises vertically with an acceleration of 18 m/s^2 for 18 s until its motor stops. Disregarding any air resistance, what max- imum height above the ground will the rocket achieve? The acceleration of gravity is 9.8 m/s^2. Answer in units of km.
Answer:
8271.92 m
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration
Equation of motion
[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow s=0\times t+\frac{1}{2}\times 18\times 18^2\\\Rightarrow s=2916\ m[/tex]
The height reached at 18 seconds is 2916 m
[tex]v=u+at\\\Rightarrow v=0+18\times 18\\\Rightarrow v=324\ m/s[/tex]
The velocity at 2916 m is 324 m/s
[tex]v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-324^2}{2\times -9.8}\\\Rightarrow s=5355.92\ m[/tex]
Maximum height reached by the toy rocket is 2916+5355.92 = 8271.92 m
The maximum height above the ground that the rocket will achieve is 8,271.92m.
To get the maximum height, we need to calculate the height reached by the rocket.
Using the equation:
[tex]S=ut+\frac{1}{2}at^2\\S =0(t) + \frac{1}{2}\times18\times 18^2\\S=9 \times 18^2\\S=2,916m[/tex]
Get the velocity also using the equation of motion:
[tex]v=u+at\\v=0+18(18)\\v=324m/s[/tex]
Get the maximum height above the ground the rocket will achieve:
[tex]v^2=u^2+2as\\324^2=0^2+2(9.8)s\\104,976=19.6s\\s=\frac{104,976}{19.6}\\s= 5,355.92m[/tex]
The total maximum height reached by rocket is 2,916 + 5,355.92 = 8,271.92m.
Learn more here: https://brainly.com/question/24018491
How fast does a rocket ship have to go for its length to be contracted to 99% of its rest length?
Answer:
4.2×10⁷ m/s
Explanation:
L = length observed by an observer
L₀ = Proper length
v = Velocity of the rocket ship
Length contraction formula
[tex]L=L_{0}{\sqrt{1-\frac{v^{2}}{c^{2}}}}\\\Rightarrow 0.99=1{\sqrt{1-\frac{v^{2}}{c^{2}}}}\\\Rightarrow \frac{0.99}{1}={\sqrt{1-\frac{v^{2}}{c^{2}}}}\\\Rightarrow 0.99^2=1-\frac{v^{2}}{c^{2}}\\\Rightarrow 0.99^2-1=-\frac{v^2}{c^2}\\\Rightarrow 0.0199\times c^2=v^2\\\Rightarrow v=\sqrt{0.0199}c=v\\\Rightarrow v =\sqrt{0.0199}\times 3\times 10^8\\\Rightarrow v=4.2\times 10^7\ m/s[/tex]
The speed of the ship would be 4.2×10⁷ m/s
you walk from the park to your friend's house, then back to your house. (a) What is the distance traveled? (b) What is your displacement? If you walk from your house to the library, then back to your house, repeat (a) and (b).
Explanation:
The total distance in a path is called distance.
The shortest distance between two points is called displacement.
a) Here, the distance travelled between the park to your friend's house and back is
Distance between park to friends house + Distance from friend's house to your house.
b) Displacement would be the shortest distance between the park and your house.
a) Distance walked between your house to library and back is
Distance between your house and library + Distance between your house and library
b) Displacement would be zero (0) as the distance between you initial point and final point is zero. Here, the initial and final points are the same
Final answer:
The distance traveled is the total length of the path taken, while displacement is the distance between the initial and final positions in a straight line. Displacement is a vector quantity because it has both magnitude and direction.
Explanation:
a. The distance traveled is the total length of the path taken. In this case, you traveled from the park to your friend's house, and then back to your own house. So, the distance traveled is the sum of these two distances.
b. Displacement is the distance between the initial and final positions, considering only the straight line between them. In this case, since you ended up back at your own house, the displacement is zero because you did not change your position.
c. Displacement is a vector quantity because it has both magnitude (the distance between the initial and final positions) and direction. In this case, the direction is the straight line between the initial and final positions.
Your FHE brother (mass 144.1 kg) is canoeing on Utah Lake with his sister (mass 81.6 kg). He had the only paddle but managed to drop it in the lake. It has drifted just out of his reach along the long axis of the canoe. His sister has taken Physics 121 and suggests that if they trade places she might be able to reach the paddle. Due to the high "slime" content of the lake and the slick surface of the canoe, the canoe moves in the water without any friction or drag. The passengers sit 2.03 meters apart and the canoe has a mass of 48.7 kg. After the passengers switch places, how much closer to the paddle (which does not move) is the end of the canoe? (Enter a positive number.)
Answer:
The end of the canoe is 46.27 cm closer to the paddle.Explanation:
This problem can be solved using the center of mass of the system.
As there are no external forces in the horizontal direction, and the external forces in the vertical direction, the gravitational force and the buoyancy, cancel each other, we know:
[tex]\vec{F}_{net_{external}} = M \frac{d\vec{P}}{dt} = 0[/tex]
where M is the total mass of the system, and [tex]\vec{P}[/tex] is the linear momentum of the system. This implies:
[tex]\frac{d\vec{P}}{dt} = 0[/tex]
so
[tex]\vec{P} = constant[/tex]
the linear momentum of the system is:
[tex]\vec{P} = M \vec{r}_{cm}[/tex]
where [tex] \vec{r}_{cm}[/tex] is the position of the center of mass. As the mass in this problem is constant, this implies that the position of the center of mass is conserved.
Finding the center of massFor a system formed by i particles, each of mass [tex]m_i[/tex] and position [tex]\vec{r}_i[/tex], the center of mass can be found with the equation:
[tex]\vec{r}_{cm} = \frac{\sum\limits_i m_i \vec{r}_i }{M}[/tex]
where M is the total mass of the system, found with the equation:
[tex]M = \sum\limits_i m_i [/tex]
For this problem, the total mass of the system is :
[tex]M = m_{brother} + m_{sister} + m_{canoe}[/tex]
[tex]M = 144.1 \ kg + 81.6 \ kg + 48.7 \ kg[/tex]
[tex]M = 274.4 \ kg[/tex]
To find the positions, we need an frame of reference. Putting the origin of coordinate at the position of the sister.
[tex]\vec{r}_{sister_i} = 0 \ \hat{i}[/tex]
The brother is 2.03 meters apart in the positive direction:
[tex]\vec{r}_{brother_i} = 2.03 m \ \hat{i}[/tex]
and the center of mass of the canoe must be at half this distance:
[tex]\vec{r}_{canoe_i} = 1.015 m \ \hat{i}[/tex]
taking all this, the center of mass of the system is at :
[tex]\vec{r}_{cm} = \frac{ 81.6 \ kg * 0 \ \hat{i} + 144.1 \ kg \ 2.03 m \ \hat{i} + 48.7 \ kg \ 1.015 m \ \hat{i} }{274.4 \ kg}[/tex]
[tex]\vec{r}_{cm} = 1.2462 \ m\ \hat{i} [/tex]
Finding the displacementAfter the switch, we found that the center of mass has to be displaced. If its displaced a distance d, we find that the new positions must be:
[tex]\vec{r}_{sister_f} = 2.03 m \ \hat{i} + d \ \hat{i}[/tex]
[tex]\vec{r}_{brother_f} = 0 \ \hat{i} + d \ \hat{i}[/tex]
[tex]\vec{r}_{canoe_f} = 1.015 m \ \hat{i} + d \ \hat{i}[/tex]
where the brother and the sister had switched the original position. Now, the center of mass will be located at:
[tex]\vec{r}_{cm} = 1.2462 \ m\ \hat{i} [/tex]
And the equation is :
[tex]\vec{r}_{cm} = \frac{ 81.6 \ kg * (2.03 m \ \hat{i} + d \ \hat{i} ) + 144.1 \ kg \ d\ \hat{i} + 48.7 \ kg \ ( 1.015 m \ \hat{i} + d \ \hat{i}) }{274.4 \ kg}[/tex]
[tex]\vec{r}_{cm} = 0.7838 \ m \ \hat{i} + \frac{ 81.6 \ kg d \ \hat{i} ) + 144.1 \ kg \ d\ \hat{i} + 48.7 \ kg \ d \ \hat{i}) }{274.4 \ kg}[/tex]
[tex]\vec{r}_{cm} = 0.7838 \ m \ \hat{i} + \frac{ 81.6 \ kg + 144.1 \ kg + 48.7 \ kg \ }{274.4 \ kg} \ d \ \hat{i}[/tex]
[tex]\vec{r}_{cm} = 0.7838 \ m \ \hat{i} + d \ \hat{i}[/tex]
[tex]1.2462 \ m\ \hat{i} = 0.7838 \ m \ \hat{i} + d \ \hat{i}[/tex]
[tex]d \hat{i} = 0.4627 \ m \ \hat{i} [/tex]
And this is how much closer the paddle is to the end of the canoe.
After the passengers switch places, the distance to the paddle (which does not move) is the end of the canoe is 46.27 m.
What is center of mass?The center of mass of a object or for a system of object is the point, where the center of distribution of mass in space.
FHE brother (mass 144.1 kg) is canoeing on Utah Lake with his sister (mass 81.6 kg) and canoe has a mass of 48.7 kg.
Let the position of the sister at the origin, his brother is at 2.03 meters. The position of the center of the mass of the canoe is at half distance (1.015 m). Thus, the center of mass of the system is,
[tex]r_{COM}=\dfrac{144.1(2.03)+81.6(0)+48.7(1.015)}{144.1+81.6+48.7}\\r_{COM}=1.2462[/tex]
For the displacement, the equation can be given as,
[tex]r_{COM}=\dfrac{144.1(d)+81.6(2.03+d)+48.7(1.015+d)}{144.1+81.6+48.7}\\1.2462=\dfrac{144.1(d)+81.6(2.03+d)+48.7(1.015+d)}{144.1+81.6+48.7}\\d=0.4627\rm \; m[/tex]
Thus, after the passengers switch places, the distance to the paddle (which does not move) is the end of the canoe is 46.27 m.
Learn more about the center of mass here;
https://brainly.com/question/13499822
The driver of a car slams on the brakes when he sees a tree blocking the road. The car slows uniformly with acceleration of −5.25 m/s2 for 4.15 s, making straight skid marks 62.5 m long, all the way to the tree. With what speed (in m/s) does the car then strike the tree? You may want to calculate the initial velocity of the car first. m/s (b) What If? If the car has the same initial velocity, and if the driver slams on the brakes at the same distance from the tree, then what would the acceleration need to be (in m/s2) so that the car narrowly avoids a collision? m/s2
Answer:
Part a)
Final speed of the car is
[tex]v_f = 4.17 m/s[/tex]
Part b)
Acceleration of the car is
[tex]a = -5.39 m/s^2[/tex]
Explanation:
As we know that car makes a skid of 62.5 m
here acceleration of the car is
[tex]a = - 5.25 m/s^2[/tex]
now we have
[tex]d = v_i t + \frac{1}{2}at^2[/tex]
[tex]62.5 = v_i (4.15) + \frac{1}{2}(-5.25)(4.15^2)[/tex]
[tex]v_i = 25.95 m/s[/tex]
Part a)
Speed of the car by which it will hit the tree
[tex]v_f = v_i + at[/tex]
[tex]v_f = 25.95 - (5.25)(4.15)[/tex]
[tex]v_f = 4.17 m/s[/tex]
Part b)
Now if car will stop after travelling same distance which same initial speed
Then we can use kinematics
[tex]v_f^2 - v_i^2 = 2 a d[/tex]
[tex]0 - 25.95^2 = 2(a)(62.5)[/tex]
[tex]a = -5.39 m/s^2[/tex]
Acceleration of a object is the rate of change of velocity of the object per unit time.
a) The speed does the car then strike the tree 4.17 m/s.b) The acceleration need to be, so that the car narrowly avoids a collision -5.39 m/s squared.What is the acceleration of a object?Acceleration of a object is the rate of change of velocity of the object per unit time.
Given information-
The car slows uniformly with acceleration of −5.25 m/s squared for 4.15 s.
The skid made by the car is 62.5 m long.
The initial velocity of the car can be find out using the distance formula of motion as,
[tex]d=ut+\dfrac{1}{2}at^2[/tex]
Put the values as,
[tex]62.5=u\times4.15+\dfrac{1}{2}\times(-5.25)\times(4.15)^2\\u=25.95[/tex]
Thus the value of initial velocity is 25.95 m/s.
(a) The speed does the car then strike the tree-The velocity formula using the equation of motion can be given as,
[tex]v=u+at\\v=25.95+(-5.25)\times4.15\\v=4.17\rm m/s[/tex]
Thus, the speed does the car then strike the tree 4.17 m/s.
(b) The acceleration need to be, so that the car narrowly avoids a collision-To stop the car the final velocity of it must be 0. Thus,
[tex]2ad=v^2-u^2\\2\times a\times62.5=0-25.95^2\\a=-5.93\rm m/s^2[/tex]
The acceleration need to be, so that the car narrowly avoids a collision -5.39 m/s squared.
Hence,
a) The speed does the car then strike the tree 4.17 m/s.b) The acceleration need to be, so that the car narrowly avoids a collision -5.39 m/s squared.Learn more about the acceleration here;
https://brainly.com/question/605631
A 75.0 kg stunt man jumps from a balcony and falls 24.0
mbefore colliding with a pile of mattresses. If the mattresses
arecompressed 1.15 m before he is brought to rest, what is the
averageforce exerted by the mattresses on
thestuntman?
Answer:
The force that acts on the man equals 15354.75 newtons.
Explanation:
After falling through a distance of 24.0 meters the speed of the stunt man upon hitting the mattress can be obtained using third equation of kinematics as
[tex]v^{2}=u^{2}+2gh\\\\\therefore v=\sqrt{2gh}\\\\v=\sqrt{2\times 9.81\times 24}\\\\v=21.7m/s[/tex] ( u=0 since the man falls from rest)
Now since the man is decelerated through a distance of 1.15 meters thus the acceleration produced can be obtained from third equation of kinematics as
[tex]v^{2}=u^{2}+2as\\\\0=u^{2}=2as\\\\a=\frac{-u^{2}}{2s}\\\\a=-\frac{(21.7^{2}}{2\times 1.15}=-204.73m/s^{2}[/tex]
Now by newton's second law the force that produced deceleration of the calculated magnitude is obtained as
[tex]F=mass\times acceleration\\\\F=75.0\times -204.73=-15354.75Newtons[/tex]
The negative sign indicates that the direction of force is opposite of motion.
X-rays are scattered from a target at an angle of 37.2° with the direction of the incident beam. Find the wavelength shift of the scattered x-rays.
Answer:
[tex]\Delta \lambda=4.94\times 10^{-13}\ m[/tex]
Explanation:
It is given that,
X-rays are scattered from a target at an angle of, [tex]\theta=37.2^{\circ}[/tex]
We need to find the wavelength shift of the scattered x-rays. The shift in wavelength is given by :
[tex]\Delta \lambda=\dfrac{h}{mc}(1-cos\ \theta)[/tex]
h is the Planck's constant
m is the mass of electron
c is the speed of light
[tex]\Delta \lambda=\dfrac{6.63\times 10^{-34}}{9.1\times 10^{-31}\times 3\times 10^8}(1-cos(37.2))[/tex]
[tex]\Delta \lambda=4.94\times 10^{-13}\ m[/tex]
So, the wavelength shift of the scattered x- rays is [tex]4.94\times 10^{-13}\ m[/tex]. Hence, this is the required solution.
A car and a truck are both traveling at the constant speed of 35 mi/h; the car is 40 ft behind the truck. The driver of the car wants to pass the truck, i.e., he wishes to place his car at B, 40 ft in front of the truck, and then resume the speed of 35 mi/h. The maximum acceleration of the car is 5 ft/s2 and the maximum deceleration obtained by applying the brakes is 20 ft/s2. What is the shortest time in which the driver of the car can complete the passing operation if he does not at any time exceed a speed of 50 mi/h
Final answer:
To find the shortest time for the passing maneuver, we must consider the car's acceleration and deceleration phases and the distance to be traveled. The task involves calculations using the kinematic equations for uniformly accelerated motion.
Explanation:
To calculate the shortest time in which the driver of the car can complete the passing operation, we must consider the car's acceleration, maximum speed, and the distance to be covered. The car must not only reach a point 40 feet in front of the truck but also get back to the speed of 35 mi/h after overtaking, all while not exceeding a speed of 50 mi/h. The entire maneuver consists of accelerating to the maximum speed, maintaining that speed for a portion of the pass, and then decelerating back to 35 mi/h.
First, convert speeds from miles per hour to feet per second:
35 mi/h = 51.33 ft/s (approximately),
50 mi/h = 73.33 ft/s (approximately).
The car needs to cover an initial 40 ft behind the truck and a further 40 ft to be ahead, for a total of 80 ft. We would calculate the time needed to accelerate to 50 mi/h (maximum passing speed), the time spent traveling at this top speed, and then the time to decelerate back to 35 mi/h, all while covering the required distance. The actual calculations involve using the kinematic equations for uniformly accelerated motion; however, without further details on the constraints such as road length, traffic laws, and exact acceleration and deceleration phases, a precise number cannot be produced.
A simple AC generator consists of a single loop of wire rotating 50 times per second within a magnetic field. The loop is a rectangle 5cm by 12 cm. The field strength is 0.21 T. What is the peak output emf?
Answer:
39.56 x 10⁻² V
Explanation:
Peak emf in a rotating coil
= n BAω
Where n is no of turns of coil , B is magnetic field , A is area of coil and ω is angular velocity of rotation of coil
Here n = 1
B = .21 T
A = 5 X 12 X 10⁻⁴ = 60 X 10⁻⁴
ω = 2π X 50 = 100π
Emf = 1 x .21 x 60 x 10⁻⁴ x 100 x 3.14
= 39.56 x 10⁻² V
A 1500kg space probe is moving at constant velocity of 4200m/sec. In a course change maneuver the rocket engines fire for 5 minutes increasing the probe's velocity to 5000m/sec. What is the probe's change in kinetic energy? A. 3.68x10^6 J. B. 5.52x10^9 J C. 1.1x10^10 J D. not enough info.
Answer:
B)5.52x10^9 J
Explanation:
The equation for calculating kinetic energy is as follows
E=[tex]\frac{1}{2} mv^{2}[/tex]
so what we have to do is calculate the kinetic energy of the space probe in each state and calculate the difference
E=[tex]\frac{1}{2} m.V2^{2}-\frac{1}{2} m.V1^{2}=E[/tex]
E=[tex]\frac{1}{2} m(V2^{2} -V1^{2})=E[/tex]
E=(1500)(5000^2-4200^2)/2
E=5.52x10^9 J
Suppose you take a 50gram ice cube from the freezer at an initial temperature of -20°C. How much energy would it take to completely vaporize the ice cube? (Hint: think of this process as four separate steps and calculate the energy needed for each one.)
Answer:
The amount of energy required is [tex]152.68\times 10^{3}Joules[/tex]
Explanation:
The energy required to convert the ice to steam is the sum of:
1) Energy required to raise the temperature of the ice from -20 to 0 degree Celsius.
2) Latent heat required to convert the ice into water.
3) Energy required to raise the temperature of water from 0 degrees to 100 degrees
4) Latent heat required to convert the water at 100 degrees to steam.
The amount of energy required in each process is as under
1) [tex]Q_1=mass\times S.heat_{ice}\times \Delta T\\\\Q_1=50\times 2.05\times 20=2050Joules[/tex]
where
[tex]'S.heat_{ice}[/tex]' is specific heat of ice =[tex]2.05J/^{o}C\cdot gm[/tex]
2) Amount of heat required in phase 2 equals
[tex]Q_2=L.heat\times mass\\\\\therefore Q_{2}=334\times 50=16700Joules[/tex]
3) The amount of heat required to raise the temperature of water from 0 to 100 degrees centigrade equals
[tex]Q_3=mass\times S.heat\times \Delta T\\\\Q_1=50\times 4.186\times 100=20930Joules[/tex]
where
[tex]'S.heat_{water}[/tex]' is specific heat of water=[tex]4.186J/^{o}C\cdot gm[/tex]
4) Amount of heat required in phase 4 equals
[tex]Q_4=L.heat\times mass\\\\\therefore Q_{4}=2260\times 50=113000Joules\\\\[/tex][tex]\\\\\\\\[/tex]Thus the total heat required equals [tex]Q=Q_{1}+Q_{2}+Q_{3}+Q_{4}\\\\Q=152.68\times 10^{3}Joules[/tex]
It would take 200 kJ of energy to completely vaporize the ice cube.
Explanation:The process of completely vaporizing an ice cube can be broken down into four separate steps, each requiring a certain amount of energy:
Heating the ice cube from -20°C to 0°C: This step requires 4.1 kJ of energy.Melting the ice at 0°C: This step requires 133.6 kJ of energy.Heating the resulting water from 0°C to 37°C: This step requires 61.9 kJ of energy.Vaporizing the water at 37°C: This step requires an additional 0 kJ of energy, since the water is already at its boiling point.In total, it would take 200 kJ of energy to completely vaporize the ice cube.
A rocket is fired straight upward, starting from rest with an acceleration of 25.0 m/s^2. It runs out of fuel at the end of 4.00 s and continues to coast upward, reaching a maximum height before falling back to earth. (a) Find the rocket height and velocity when it runs out of fuel. (b) Find the maximum height the rocket reaches. (c) Find the velocity the instant before the rocket crashes into the ground. (d) Find the total elapsed time from launch to ground impact.
Answer:
a) 200m, 100m/s
b) 710.20m
c) -117.98 m/s
d) 26.24 s
Explanation:
To solve this we have to use the formulas corresponding to a uniformly accelerated motion problem:
[tex]V=Vo+a*t[/tex] (1)
[tex]X=Xo+Vo*t+\frac{1}{2}*a*t^2\\[/tex] (2)
[tex]V^2=Vo^2+2*a*X[/tex] (3)
where:
Vo is initial velocity
Xo=intial position
V=final velocity
X=displacement
a)
[tex]X=0+0*4+\frac{1}{2}*25*4^2[/tex]
the intial position is zero because is lauched from the ground and the intial velocitiy is zero because it started from rest.
[tex]X=200m[/tex]
[tex]V=0+25*4[/tex]
[tex]V=100m/s[/tex]
b)
The intial velocity is 100m/s we know that because question (a) the acceleration is -9.8[tex]\frac{m}{s^2}[/tex] because it is going downward.
[tex]0=100^2+2*(-9.8)*X\\X=510.20\\totalheight=200+510.20=710.20m[/tex]
c)
In order to find the velocity when it crashes, we can use the formula (3).
the initial velocity is 0 because in that moment is starting to fall.
[tex]V^2=0^2+2*(-9.8)*(710.20)\\V=-117.98 m/s[/tex]
the minus sign means that the object is going down.
d)
We can find the total amount of time adding the first 4 second and the time it takes to going down.
to calculate the time we can use the formula (2) setting the reference at 200m:
[tex]-200=0+100*t+\frac{1}{2}*(-9.8)*t^2[/tex]
solving this we have: time taken= 22.24 seconds
total time is:
total=22.24+4=26.24 seconds.
Imagine that you are standing 5 feet from, and looking directly toward, a brass ball (a convex mirror) 1 foot in diameter. Describe the image you world see in the ball (a) Determine the image distance. (b) Is this a real or virtual image?
Answer:
virtual the image is virtual because image formed viruvirualconvex mirror
While you are driving down a country road at dusk with a velocity of 22 m/s, a deer suddenly runs across the road in front of you. You apply the brakes, reducing your velocity to 7 m/s in 5 s as you avoid hitting the deer. What is the acceleration of the car?
Answer:
[tex]a=\frac{v_{f}-v_{o} }{t_{f}-t_{o} } =\frac{(7-22)\frac{m}{s} }{5s-0s} =-3\frac{m}{s^{2} }[/tex]
Explanation:
We know that the initial velocity is 22 m/s
The final velocity after applying the brakes is 7 m/s, and the time that takes it to break is 5 seconds
We know that the concept of acceleration is the change of velocity in time
[tex]a=\frac{v_{f}-v_{o} }{t_{f}-t_{o} } =\frac{(7-22)\frac{m}{s} }{5s-0s} =-3\frac{m}{s^{2} }[/tex]
The answer is negative since the car is slowing down
In a game of basketball, a forward makes a bounce pass to the center. The ball is thrown with an initial speed of 4.4 m/s at an angle of 24° above the horizontal. It is released 0.70 m above the floor. Q1: What horizontal distance does the ball cover before bouncing? please show how to solve for t in the problem.
Answer:
The horizontal distance is 2.41 mts
Explanation:
For this problem, we will use the formulas of parabolic motion.
[tex]Y=Yo+Voy*t+\frac{1}{2}*a*t^2\\Vx=V*cos(\alpha)\\Vy=V*sin(\alpha)[/tex]
We need to find the time of the whole movement (t), for that we will use the first formula:
we need the initial velocity for that:
[tex]Vy=4.4*sin(24^o)\\Vy=1.79m/s[/tex]
so:
[tex]0=0.70+1.79*t+\frac{1}{2}*(-9.8)*t^2[/tex]
now we have a quadratic function, solving this we obtain two values of time:
t1=0.60sec
t2=-0.234sec
the obvious value is 0.60sec, we cannot use a negative time.
Now we are focusing on finding the horizontal distance.
the movement on X is a constant velocity motion, so:
[tex]x=Vx*t[/tex]
[tex]Vx=4.4*cos(24^o)=4.02m/s\\[/tex]
so:
[tex]x=4.02*(0.60)=2.41m[/tex]
At a certain instant, a particle is moving in the direction with momentum 18 kg·m/s. During the next 0.5 s, a constant force <-4, 12, 0> N acts on the particle. What is the momentum of the particle at the end of this 0.5 s interval?
Answer:
24.325 kg m/s
Explanation:
Initial momentum, pi = 18 kg m/s
F = < -4, 12, 0>
t = 0.5 s
Let the final momentum is pf.
The magnitude of force is
[tex]F=\sqrt{(-4)^{2}+12^{2}+0^{2}}=12.65 N[/tex]
According to the Newton's second law, the rate of change of momentum is equal to the force.
[tex]F = \frac{p_{f}-p_{1}}{t}[/tex]
[tex]12.65= \frac{p_{f}-18}}{0.5}[/tex]
pf - 18 = 6.325
pf = 24.325 kg m/s
Thus, the momentum of body after 0.5 s is 24.325 kg m/s.
What component of a longitudinal sound wave is analogous to a trough of a transverse wave?
Explanation:
There are two components of a longitudinal sound wave which are compression and rarefaction. Similarly, there are two components of the transverse wave, the crest, and trough.
The crest of a wave is defined as the part that has a maximum value of displacement while the trough is defined as the part which corresponds to minimum displacement.
While compression is that space where the particles are close together while the rarefaction is that space where the particles are far apart from each other.
So, the refraction or the rarefied part of a longitudinal sound wave is analogous to a trough of a transverse wave.
You are riding in an elevator on the way to the
eighteenthfloor of your dormitory. The elevator is accelerating
upwardwith a = 1.90 m/s2. Beside you is the
boxcontaining your new computer; box and contents have a total
mass28.0 kg. While the elevator is accelerating upward, you
pushhorizontally on the box to slide it at constant speed toward
thedoor. If the coefficient of kinetic friction between the boxand
elevator floor is mk = 0.32, what magnitudeof
force must you apply?
Answer:
F = 104.832 N
Explanation:
given,
upward acceleration of the lift = 1.90 m/s²
mass of box containing new computer = 28 kg.
coefficient of friction = 0.32
magnitude of force = ?
box is moving at constant speed hence acceleration will be zero.
Now force acting due to lift moving upward =
F = μ m ( g + a )
F = 0.32 × 28 × ( 9.8 + 1.9 )
F = 104.832 N
hence, the force applied should be equal to 104.832 N
The average American uses electrical energy at the rate of about 1.5 kW. Solar energy reaches Earth’s surface at an average rate of about 300 watts on every square meter (a value that accounts for nights and clouds). What fraction of the United State’s land area would have to be covered with 20% efficient solar cells to provide all of our electricity?
Answer: the fraction is [tex]0.88*10^{-6}[/tex]
Explanation:
Hi!
The land area per capita in USA is 28,310 square km. Then the solar power per capita is (rounding some numbers):
[tex]P_s = 300 \frac{W}{m^2} (28,310) km^2 = 3*10^2*2.83*10^4*10^6 W = 8.49*10^{12} W = 8.49*10^9 kW[/tex]
If we take 20% of this power, the fraction k to have 1.5 kW is:
[tex]1.5 kW = k(0.2*8.49 * 10^9 kW)[/tex]
[tex]k = \frac{1.5}{1.7}*10^{-6} = 0.88 *10^{-9} [/tex]
Final answer:
To answer the question, we need to calculate the total electrical energy consumption in the U.S., understand the solar energy per square meter, consider the 20% efficiency of solar cells, and determine the needed land area for solar panels to meet the country's electrical energy needs.
Explanation:
The question involves calculating what fraction of the United States' land area would need to be covered with 20% efficient solar cells to provide all of its electricity, given that the average American uses electricity at the rate of about 1.5 kW and solar energy reaches the Earth's surface at an average rate of about 300 watts per square meter. To find this fraction, we first need to consider the total electrical energy consumption of the United States and then assess how much of this demand can be met by the solar energy available on a given area of land, taking into consideration the efficiency of the solar cells.
The key steps include calculating the total power used by Americans in watts, understanding the energy provided by the sun per square meter, adjusting for the efficiency of solar conversion, and finally, determining the land area required. Specifically, a 20% conversion efficiency means that only a fifth of the solar energy striking the solar cells is converted into usable electrical energy. With these considerations, the solution entails a mix of energy requirement calculations and solar energy potential assessments to identify the required land area for solar panels.
A 3730-kg SUV can climb up a 600-m high hill in one minute (60 seconds). How much is its power output in horse-powers? 245 hp
490 hp
800 hp
373 hp
600 hp
Answer:
490hp
Explanation:
Power is energy per unit of time, an in this case the energy needed is the gravitational potential energy, so we have:
[tex]P=E/t=mgh/t=(3730kg)(9.8m/s)(600m)/60s=365540W[/tex]
Since all units are in S.I. we get our result in Watts (Joules/s). To convert to horse power (imperial), we need to know that:
[tex]745.7 W = 1 hp[/tex]
Which obviously means:
[tex]\frac{1 hp}{745.7 W} = 1[/tex]
So we can write:
[tex]P=365540W=365540W(\frac{1 hp_m}{745.7 W})=490hp[/tex]
There is also metric horse power ([tex]735.5 W = 1 hp[/tex]), and using this value we get [tex]P=497hp[/tex], although both results are close.
Final answer:
To calculate the SUV's power output, we find the work done against gravity and divide it by the time taken to get power in watts. This is then converted to horsepowers leading to the answer of 490 hp. The correct answer is b) 490 hp.
Explanation:
To find the power output in horsepowers of the 3730-kg SUV climbing a 600-meter hill in 60 seconds, first we calculate the work done against gravity, which is equal to the force due to gravity (weight of the SUV) multiplied by the height of the hill:
Work = Weight × Height = (mass × gravity) × height
Then, we convert the weight to newtons by multiplying the mass (kg) by the acceleration due to gravity (9.81 m/s2, approximately), and multiply that result by the height (in meters).
Power in watts is calculated by dividing the work done by the time taken in seconds:
Power (W) = Work done (Joules) / Time (seconds)
To convert watts to horsepowers, we use the conversion rate where 1 horsepower is equivalent to 746 watts:
Power (hp) = Power (W) / 746
After plugging in the values, the SUV's power output in horsepowers would be:
Power (W) = (3730 kg × 9.81 m/s2) × 600 m / 60 s = 365,130 W
Power (hp) = 365,130 W / 746 = 489.43 hp, which we round to 490 hp.
Thus, the correct answer from the given options is 490 hp.