Grant jumps 1.10 m straight up into the air to slam-dunk a basketball into the net, the speed from which he would have left the floor would be 4.64 m / s .
What are the three equations of motion?There are three equations of motion given by Newton
v = u + at
S = ut + 1/2 × a × t²
v² - u² = 2 × a × s
As given in the problem grant jumps 1.10 m straight up into the air to slam-dunk a basketball into the net.
By using the third equation of the motion,
v² - u² = 2 × a × s
0 - u² = 2 × -9.81 × 1.10
u = 4.64 m / s
Thus, the speed from which he would have left the floor would be 4.64 m / s .
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A circular loop of wire with a diameter of 13.478 cm is in the horizontal plane and carries a current of 1.607 A counterclockwise, as viewed from above. What is the magnetic field, in microTeslas, at the center of the loop?
Given:
The number of the loops, n = 1
The diameter of the loop is d = 2r = 13.478 cm
The current in the loop is I = 1.607 A
To find the magnetic field in micro Tesla
Explanation:
The magnetic field can be calculated by the formula
[tex]B\text{ =}\frac{n\mu_0I}{2r}[/tex]Here, the value of the constant is
[tex]\mu_0=\text{ 12.57}\times10^{-7}\text{ H/m}[/tex]On substituting the values, the magnetic field will be
[tex]\begin{gathered} B=\frac{1\times12.57\times10^{-7}\times1.607}{13.478\times10^{-2}} \\ =1.499\text{ }\times10^{-5}\text{ T} \\ =14.99\times10^{-6}\text{ T} \\ =14.99\text{ }\mu T \end{gathered}[/tex]The magnetic field is 14.99 micro Tesla
What is the net force on an object with an applied force of 800N (right) and friction resisting at 750 N (left)?1 1550 N left2 1550 N right3 50 N left4 50 N right
Given,
The applied force, F=800 N
The friction, f=750 N
Friction is a force that opposes the motion of an object. Thus the net force will be equal to the difference between the applied force and the friction. As the applied force is greater than the frictional force, the net force will be in the same direction as the applied force, that is to the right.
Thus the net force is given by,
[tex]F_n=F-f[/tex]On substituting the known values,
[tex]\begin{gathered} F_n=800-750 \\ =50\text{ N} \end{gathered}[/tex]Therefore the net force on the object is 50 N to the right. Thus, the correct answer is option 4.
Two cars in opposite directions were going at 32 mph before a collision. They had a head on inelastic collision, i.e. the two cars stuck together afterward. The common speed of the combined piece right after the collision is 20 mph. The mass of Car 1 was 2,000 lb. Car 2 was heavier. The mass of Car 2 was ____ lb.
The mass of Car 2 was 3000 lb.
We need to apply the concept of conservation of momentum.
The velocity of both cars= 32mph
Combined velocity = 20mph
Mass of Car 1= 2000 lb
According to the conservation of momentum
M1V1+ M2V2= (M1+M2)
2000x32- (-M2x 32)=20(2000+M2)
64000+32M2=40000 +20M2
24000= 8M2
M2= 3000lb
Therefore the mass of Car 2 is 3000lb.
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A car is driving around a turn with a radius of 20.7 m. If the coefficient of friction between the tires and the road is 2.07, what is the maximum speed the car can maintain around the turn without slipping?
Answer:
A car completes a turn that has a radius of 20 meters. The coefficient of friction between the tires and road is 0.50. What maximum speed can the car safely maintain in order to complete the turn without skidding? (A) 5 m/s (B) 10 m/s (C) 15 m/s (D) 20 m/s (E) 25 m/s
The figure shows a person whose weight is W = 607 N doing push-ups. Find the
normal force exerted by the floor on (a) each hand and (b) each foot, assuming that
the person holds this position.
The normal force exerted by the floor on each hand is 203.95N and the normal force exerted on each foot is 99.55 N.
Torque is defined as force times the distance from the line of force that is perpendicular to it.
(a) Each hand's typical response is, let's say, N1. For hands, the overall typical response is 2N1.
Balance the torque around the foot now.
torque is applied in a counterclockwise direction
Now ,
2N1 x 1.25 = W x 0.84.
2N1 × 1.25 = 607 × 0.84
N1 = (607 x 0.84)/(2x1.25)
N1 = 203.95N.
(a) The normal response for each foot is, let's say, N2, and the total normal response for feet is 2N2.
Now, distribute torque among the hands.
Anticlockwise torque is equal to clockwise torque.
Now,
2N2 x 1.25 = W x 0.41
2N2 × 1.25 = 607 × 0.41
N2 = (607× 0.41 )/2× 1.25
N2 = 99.55N
Hence,the normal force exerted by the floor on each hand is 203.95N and the normal force exerted on each foot is 99.55 N.
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Which of the following water molecules have the greatest kinetic energy?Select one:a. Cool water.b. Warm water.c. Boiling water.d. They all have the same kinetic energy.
Since all the molecules in the boiling water will have more energy introduced by heat, then the molecules with the greatest kinetic energy are the ones from the boiling water.
Assuming the jet slows with constant acceleration, find the magnitude and direction of its acceleration.
We are given that a jet is traveling with a speed of 78.6 m/s and travels a distance of 919m. We are asked to determine the constant acceleration when the jet stops. To do that we will use the following formula:
[tex]v^2_f=v^2_0+2ax[/tex]Where:
[tex]\begin{gathered} v_f=\text{ final speed} \\ v_0=\text{ initial speed} \\ a=\text{ acceleration} \\ x=\text{ distance traveled} \end{gathered}[/tex]Since the jet stops, this means that the final speed is zero. We will solve for the acceleration "a" in the formula. First, we will eliminate the term for the final speed since it is zero:
[tex]0=v^2_0+2ax[/tex]Now we will subtract the initial speed squared from both sides:
[tex]-v^2_0=2ax[/tex]Now we will divide by "2x" from both sides:
[tex]\frac{-v^2_0}{2x}=a[/tex]Now we replace the known values:
[tex]\frac{-(78.6\frac{m}{s})^2}{2(919m)}=a[/tex]Solving the operations:
[tex]-3.36\frac{m}{s^2}=a[/tex]Therefore, the magnitude of the acceleration is 3.36. Since the jet is deaccelerating in the direction due south, the direction of the acceleration is due north.
An airplane traveling at 1008 meters above the ocean at 135 km/h is going to drop a box of supplies to shipwrecked victims below. How many seconds before the plane is directly overhead should the box be dropped?
The horizontal velocity of the airplane is,
[tex]v=135\text{ km/h}[/tex]The height of the airplane is,
[tex]h=1008\text{ m}[/tex]The vertical initial velocity of the box is zero as the airplane is moving in the horizontal direction.
let the time to reach the victim is t.
we can write,
[tex]\begin{gathered} h=\frac{1}{2}gt^2 \\ t=\sqrt[]{\frac{2h}{g}} \end{gathered}[/tex]Substituting the values we get,
[tex]\begin{gathered} t=\sqrt[]{\frac{2\times1008}{9.8}} \\ =14.3\text{ s} \end{gathered}[/tex]Hence the required time is 14.3 s
What are the answers for a, b and c in MJ?
Given:
The orbital height of the satellite, h=94 km=94000 m
The mass of the satellite, m=1045 kg
The new altitude of the satellite, d=207 km=207000 m
To find:
a) The energy needed.
b) The change in the kinetic energy.
c) The change in the potential energy.
Explanation:
The radius of the earth, R=6.37×10⁶ m
The mass of the earth, M=6×10²⁴ kg
a) The orbital velocity is given by,
[tex]v=\sqrt{\frac{GM}{r}}[/tex]Where G is the gravitational constant and r is the radius of the satellite from the center of the earth.
Thus the initial orbital velocity of the earth,
[tex]\begin{gathered} v_1=\sqrt{\frac{6.67\times10^{-11}\times6\times10^{24}}{(6.37×10^6+94000)}} \\ =7868.43\text{ m/s} \end{gathered}[/tex]The orbital velocity after changing the altitude is,
[tex]\begin{gathered} v_2=\sqrt{\frac{6.67\times10^{-11}\times6\times10^{24}}{(6.37\times10^6+207000)}} \\ =7800.5\text{ m/s} \end{gathered}[/tex]Thus the total energy needed is given by,
[tex]E=(\frac{1}{2}mv_2^2-\frac{GMm}{(R+d)})-(\frac{1}{2}mv_1^2-\frac{GMm}{(R+h)})[/tex]On substituting the known values,
[tex]\begin{gathered} E=1045[(\frac{1}{2}\times7868.43^2-\frac{6.67\times10^{-11}\times6\times10^{24}}{(6.37\times10^6+207000)})-(\frac{1}{2}\times7800.5^2-\frac{6.67\times10^{-11}\times6\times10^{24}}{(6.37×10^6+94000)})] \\ =623\text{ MJ} \end{gathered}[/tex]b)
The change in the kinetic energy is given by,
[tex]\begin{gathered} KE=\frac{1}{2}mv_2^2-\frac{1}{2}mv_1^2 \\ =\frac{1}{2}m(v_2^2-v_1^2) \end{gathered}[/tex]On substituting the known values,
part 2 of 2 ASSUME BOTH snowballs are thrown with the same initial speed 39.9 m/s. the first snowball is thrown at an angle of 51 degrees above the horizontal. At what angle should you throw the second snowball to make it hit the same point as the first? how many seconds after the first snowball should you throw the second so that they arrive on target at the same time?
Explanation
Step 1
Let
a) for ball 1
[tex]\begin{gathered} \text{ Initial sp}eed=v_0=33.9\text{ }\frac{m}{s} \\ \text{ Angle=51 \degree} \end{gathered}[/tex]the formula for the distance is given by:
[tex]x=\frac{v^2_0\sin(2\theta)}{g}[/tex][tex]\begin{gathered} \text{hence, let v}_0=39.9,\text{ angle= 51 \degree , g=9.8 } \\ \text{replace to solve for x } \\ x=\frac{(39.9)^2\sin(2\cdot51)}{9.8} \\ x=158.9\text{ m} \\ \end{gathered}[/tex]hence, the horizontal distance reached by the ball 1 is 158.9 meters
Step 2
as the ball started from the same point at the same initial speed, the only way to make the second ball hits the same point as the first is thworing the second ball at the same angle, it is 51 °
A small object of mass 0.500 kg is attached by a 0.440 m-long cord to a pin set into the surface of a frictionless table top. The object moves in a circle on the horizontal surface with a speed of 5.34 m/s.What is the magnitude of the radial acceleration of the object? What is the tension in the cord?
Given data:
* The mass of the object attached is m = 0.5 kg.
* The radius of the circle is r = 0.44 m.
* The speed of the object moving in circular motion is v = 5.34 m/s.
Solution:
(a). The radial acceleration of the object is also known as the centripetal acceleration of the object.
The value of centripetal acceleration in terms of the velocity of the object is,
[tex]a_c=\frac{v^2}{r}[/tex]Substituting the known values,
[tex]\begin{gathered} a_c=\frac{5.34^2}{0.44} \\ a_c=64.8ms^{-2} \end{gathered}[/tex]Thus, the radial acceleration of the object is 64.8 meters per second squared.
(b). The tension in the chord is equivalent to the centripetal force acting on the object which helps it to move in the circular motion.
Thus, the tension acting on the chord is,
[tex]F=ma_c[/tex]Substituting the known values,
[tex]\begin{gathered} F=0.5\times64.8 \\ F=32.4\text{ N} \end{gathered}[/tex]Thus, the tension acting in the chord is 32.4 N.
An arrangement of two pulleys, as shown in the figure, is used to lift a 64.3-kg crate a distance of 4.36 m above the starting point. Assume the pulleys and rope are ideal and that all rope sections are essentially vertical.What is the change in the potential energy of the crate when it is lifted a distance of 4.36 m? (kJ)How much work must be done to lift the crate a distance of 4.36 m? (kJ)What length of rope must be pulled to lift the crate 4.36 m? (m)
Given data:
* The mass of the crate is m = 64.3 kg.
* The height of the crate is h = 4.36 m.
Solution:
(a). The potential energy of the crate at the initial state is zero (as the height of the crate is zero at the initial state), thus, the change in the potential energy of the crate is,
[tex]\begin{gathered} dU=\text{mgh}-0 \\ dU=mgh \end{gathered}[/tex]where g is the acceleration due to gravity,
Substituting the known values,
[tex]\begin{gathered} dU=64.3\times9.8\times4.36 \\ dU=2747.4\text{ J} \\ dU=2.75\times10^3\text{ J} \\ dU=2.75\text{ kJ} \end{gathered}[/tex]Thus, the change in the potential energy is 2.75 kJ.
(b). The work done to lift the crate is equal to the change in the potential energy of the crate.
Thus, the work done on the crate is 2.75 kJ.
(c). As the single is pulling the two ropes to increase the height of the crate, thus, the length of the rope pulled in terms of the height of the crate is,
[tex]l=2h[/tex]Substituting the known values,
[tex]\begin{gathered} l=2\times4.36 \\ l=8.72\text{ m} \end{gathered}[/tex]Thus, the length of the rope pulled to lift the crate is 8.72 meters.
In terms of area, about how much more pizza is given if the diameter is 12 inches compared to one with a diameter of 8 inches?
B. 2.3 times more
Explanation:The pizza is circular in shape
The diameter of the large-sized pizza, d₁ = 12 inches
Tha area of the large sized pizza is calculated as:
[tex]\begin{gathered} A_1=\frac{\pi{d^2_1}}{4} \\ A_1=\frac{\pi{12^2}}{4} \\ A_1=\frac{\pi{144^{}}}{4} \\ A_1=36\pi\text{ in}^{2} \end{gathered}[/tex]The diameter of the small-sized pizza, d₂ = 8 inches
The area of the small-sized pizza is calculated as:
[tex]\begin{gathered} A_2=\frac{\pi{d^2_2}}{4} \\ A_2=\frac{\pi{8^2}}{4} \\ A_2=\frac{64\pi{}}{4} \\ A_2=16\pi\text{ in}^{2} \end{gathered}[/tex]Ratio of A₁ to A₂
[tex]\begin{gathered} \frac{A_1}{A_2}=\frac{36\pi{}}{16\pi} \\ \frac{A_1}{A_2}=2.25 \\ \frac{A_1}{A_2}=2.3(to\text{ the nearest 1 dp)} \end{gathered}[/tex]The 12 inches pizza is 2.3 times more than the 8 inches pizza
What is held in orbit by the gravitational pull of earth
The international space station.
The moon.
All TV satellites.
All weather satellites.
All GPS satellites.
More than 4000 other artificial satellites.
Thousands of pieces of "space junk"
Answer:
The Moon.
Explanation:
The earths gravity holds the moon in place.
The block of a mass 10.2 kg is sliding at an initial velocity of 3.40 m/s in the positive x-direction. The surface has a coefficient of kinetic friction of 0.153.
m = mass = 10.2 kg
vo = initial velocity = 3.40 m/s
u = coefficient of kinetic friction = 0.153
g= gravity = 9.8m/s^2
a)
Fr = force of kinetic friction = u m g
Fr = 0.153 x 10.2 x 9.8 = -15.30N
b) Block's acceleration
Newton's second law of motion:
F = m*a
a = F/m = -15.30 / 10.2 = -1.5 m/s^2
c) USe the third equation of motion:
2as = vf^2 - vo^2
Where:
Vf= final velocity= 0 m/s
s = displacement
2 * -1.5 * s = 0^2 - 3.40^2
-3s = -11.56
s= -11.56/-3
s= 3.85 m
A 1.4N friction force slows a block to a stop after sliding 7m. How much work was done by the friction force
Answer:
9.8J
Explanation:
The work done by the friction force can be calculated as
W = Fd
Where F is the friction force, and d is the distance that the block slide.
So, replacing F = 1.4 N and d = 7 m
W = (1.4N)(7 m)
W = 9.8 J
Therefore, the work done by the friction was 9.8J against the movement of the block
You can hear sound from another room through a door that is slightly open because...the sound refracts as it goes from one room to another.the sound wave reflects off the air in the doorway.the sound diffracts as it goes through the opening.the sound is polarized as it goes through the narrow opening.
when sound wave go from small passages like the edge of a wall or from the slit of open door the sound wave diffracts.
So the 3rd option is correct option.
If an 800 kg roller coaster is at the top of its 50 m high track, it will have a potential energy 392,000 and a kinetic energy of 0J. This means the total mechanical energy is 392,000J. If the cart drops down to a new height of 10m, how much energy does the cart have now?
ANSWER:
313600 J
STEP-BY-STEP EXPLANATION:
We have that the gravitational potential energy is given by the following equation:
[tex]E_p=m\cdot g\cdot h[/tex]We substitute and calculate the potential energy, knowing that g is the acceleration of gravity and is equal to 9.8 m/s^2:
[tex]\begin{gathered} E_p=800\cdot9.8\cdot10 \\ E_p=78400\text{ J} \end{gathered}[/tex]We know that the total energy is 392,000 joules, so the energy it now carries would be the total minus the calculated potential energy:
[tex]\begin{gathered} E_k=392000-78400 \\ E_k=313600\text{ J} \end{gathered}[/tex]The energy carried by the cart is 313600 J
A raindrop has a mass of 7.7 × 10-7 kg and is falling near the surface of the earth. Calculate the magnitude of the gravitational force exerted (a) on the raindrop by the earth and (b) on the earth by the raindrop.(a)Fraindrop= _________________ units ________(b)Fearth= _________________ units_____________
ANSWER:
a) Fraindrop
[tex]F=7.546\cdot10^{-6}N[/tex](b) Fearth
[tex]F=-7.546\cdot10^{-6}N[/tex]STEP-BY-STEP EXPLANATION:
(a)
We calculate the force, multiplying the value of the mass by gravity, just like this:
[tex]\begin{gathered} F=m\cdot a \\ F=7.7\cdot10^{-7}\cdot9.8 \\ F=7.546\cdot10^{-6}N \end{gathered}[/tex](b)
by newton's 3rd law they are are equal and opposite so:
[tex]F=-7.546\cdot10^{-6}N[/tex]choose the 200 kg refrigerator. set the applied force to 400 n (to the right). be sure friction is turned off. what is the net force acting on the refrigerator?
Answer:
Explanation:
Ginen:
m₁ = 200 kg
F₂ = 400 N
g ≈ 10 m/s²
__________
R - ?
F₁ = m₁·g = 200· 10 = 2000 N
R = √ (F₁² + F₂²) = √ ( 2000² + 400²) ≈ 2040 N
The net force acting on the refrigerator having a mass of 200 kg and the applied force to 400 n (to the right) is 2040 Newtons.
What is force?Force is the influence of either pull or pushes in the body. Basically, gravitation forces, nuclear forces, and friction forces are the types of forces. For e.g. when the wall is hit by a hand then a force is exerted by the hand on the wall as well as the wall also exerts a force on the hand. There are different laws given to Newton to understand force.
Newton is a unit of force used by physicists that is part of the International System (SI). The force required to move a body weighing one kilogram one meter per second is known as a newton.
Given:
The mass of the refrigerator, m = 200 kg,
The force, F = 400 N,
Calculate the net force by the formula given below,
F = m × g
here, g is the gravitational acceleration.
Substitute the values,
F= 200 × 10 = 2000 N
[tex]R = \sqrt{F_1^2 +F_2^2}[/tex]
where R is the net force
[tex]R = \sqrt{2000^2 +400^2}[/tex]
R = 2040 Newton
Therefore, the net force acting on the refrigerator having a mass of 200 kg and the applied force to 400 n (to the right) is 2040 Newtons.
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S=3-2t+3t^2
What is the instantaneous velocity and it’s acceleration at t=3s
At what time is the particle at rest
Answer:
Explanation:
Given:
X(t) = 3 - 2*t + 3*t²
t = 3 s
_______________
V(t) - ?
a(t) - ?
Speed is the first derivative of the coordinate, acceleration is the second.
1)
V(t) = X' = (3 - 2*t + 3*t²)' = 0 - 2 +6*t = 6*t - 2
V(3) = 6*3 - 2 = 16 m/s
2)
a(t) = X'' = V' = (6*t - 2)' = 6 m/s²
a(3) = 6 m/s²
3)
The body will stop (V = 0 ) in (t) seconds:
V(t) = 6*t - 2
0 = 6*t - 2
6*t = 2
t = 2/6 = 1/3 ≈ 0,33 s
A jet engine applies a force of 300 N horizontally to the right against a 50 kg rocket. The force of air friction 200 N. If the rocket is thrust horizontally 2.0 m, the kinetic energy gained by the rocket is
We will have the following:
[tex](300N)(2m)+(-200N)(2m)=200J[/tex]Then, we from the work-kinetic force theorem we will have that the total kinetic energy gained by the rocket was 200 Joules.
this is a 2 part question2) Two drivers traveling side-by-side at the same speed suddenly see a deer in the road ahead of them and begin braking. Driver 1 stops by locking up his brakes and screeching to a halt; driver 2 stops byapplying her brakes just to the verge of locking, so that the wheels continue to turn until her car comes to a complete stop. (a) All other factors being equal, is the stopping distance of driver 1 greater than,less than, or equal to the stopping distance of driver 2? (b) Choose the best explanation from among thefollowing: 1. Locking up the brakes gives the greatest possible braking force.2. The same tires on thesame road result in the same force of friction.3. Locked-up brakes lead to sliding (kinetic) friction, which is less than rolling (static) friction.
The maximum static friction between two surfaces is greater than the kinetic friction between them.
If the wheels of a car get locked, the surface of the wheel slides through the floor and kinetic friction acts to stop the car.
If the wheels of the car don't get locked, they may turn fast enough to prevent the surface of the wheel from sliding through the floor and static friction acts on the car.
Since the force acting on the car with its wheel locked is less than the force acting on the car with the turning wheels, then, the stopping distance is greater for driver 1 than for diver 2.
Therefore, the answers are:
a) The stopping distance of driver 1 is greater than the stopping distance of driver 2.
b) The best explanation is:
3. Locked-up brakes lead to sliding (kinetic) friction, which is less than rolling (static) friction.
Tall pacific coast redwood trees can reach heights of about 100 m. If air drag is negligibly small, how fast is a sequoia come moving when it reaches the ground if it dropped from the top of a 100 m tree?
Given data:
Height of the tree;
[tex]h=100\text{ m}[/tex]Initial velocity;
[tex]u=0\text{ m/s}[/tex]The velocity of sequoia when it reaches the ground is given as,
[tex]v=\sqrt[]{u^2+2gh}[/tex]Here, g is the acceleration due to gravity.
Substituting all known values,
[tex]\begin{gathered} v=\sqrt[]{(0\text{ m/s})^2+2\times(9.8\text{ m/s}^2)\times(100\text{ m})} \\ \approx44.27\text{ m/s} \end{gathered}[/tex]Therefore, sequoia will reach the ground with a velocity of 44.27 m/s.
7/21/22, 7:37 AMProblem Set ThreeNotes: Use 9.8 m/s 2 for the acceleration due to gravity. Formust be expressed in m/sLaw calculations, mass must be expressed in kg and velocity.A steady 45 N horizontal force is applied to a 15kg object on a table. The object slides against a friction force of 30 N. Calculate the acceleration of the object in m/s.
Given:
The mass of the object is
[tex]m=15\text{ kg}[/tex]The applied force on the object is
[tex]F=45\text{ N}[/tex]The frictional force on the object is
[tex]f=30\text{ N}[/tex]To find:
The acceleration of the object
Explanation:
The net force on the object is
[tex]\begin{gathered} F_{net}=F-f \\ =45-30 \\ =15\text{ N} \end{gathered}[/tex]The acceleration of the object is,
[tex]\begin{gathered} a=\frac{F_{net}}{m} \\ =\frac{15}{15} \\ =1\text{ m/s}^2 \end{gathered}[/tex]Hence, the acceleration is
[tex]1\text{ m/s}^2[/tex]You push a 2.1 kg object on a table with 42N of force. The box then slowly skids to a stop over 2.2 m. What is the force of friction?
If you push a 2.1 kg object on a table with 42N of force. The box then slowly skids to a stop over 2.2 meters, then the force of the friction would be 42 Newtons as per the concept of limiting friction.
What is friction?Friction is a type of force that resists or prevents the relative motion of two physical objects when their surfaces come in contact.
As given in the problem if push a 2.1 kg object on a table with 42N of force. The box then slowly skids to a stop over 2.2 m, then we have to find the force of the friction.
The force of the friction = The limiting friction force
If you apply 42N of force to a 2.1-kilogram item on a table. According to the theory of limiting friction, when the box gently slides to a halt over a distance of 2.2 meters, the force of friction would be 42 Newtons.
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A cylinder of gas at room temperature has a pressure . To p_{1} what temperature in degrees Celsius would the temperature have to be increased for the pressure to be 1.5p_{1} ,
In order to calculate the temperature, we need to know that temperature and pressure are directly proportional, that is, if the pressure increases, the temperature (in Kelvin) also increases in the same proportion.
So, first let's convert the temperature from Celsius to Kelvin, by adding 273 units:
[tex]\begin{gathered} K=C+273 \\ K=20+273 \\ K=293 \end{gathered}[/tex]Then, let's calculate the proportion:
[tex]\begin{gathered} \frac{P_1}{T_1}=\frac{P_2}{T_2} \\ \frac{p_1}{293}=\frac{1.5p_1}{T_2} \\ \frac{1}{293}=\frac{1.5}{T_2} \\ T_2=1.5\cdot293 \\ T_2=439.5\text{ K} \end{gathered}[/tex]Now, converting back to Celsius, we have:
[tex]\begin{gathered} C=K-273 \\ C=439.5-273 \\ C=166.5\text{ \degree{}C} \end{gathered}[/tex]So the temperature would be 166.5 °C.
Given the wave described by y(x,t)=5cos[π(4x-3t)], in meters. Find the following. Giveexact answers with units.
Answer:
a) 5 m
b) 0.667 s
c) 0.5 m
d) 0.75 m/s
e) -5 m
Explanation:
In an equation of the form
y(x, t) = Acos(kx - ωt)
A is the amplitude, ω = 2π/T where T is the period, and k = 2π/λ where λ is the wavelength. In this case, the equation os
y(x,t) = 5cos(π(4x - 3t)
y(x,t) = 5cos(4πx - 3πt)
So, A = 5, k = 4π, and ω = 3π. Then, we can find each part as follows
a) Amplitude
The amplitude is A, so it is 5 m.
b) the period
Using the equation ω = 2π/T and solving for T, we get:
[tex]T=\frac{2\pi}{\omega}=\frac{2\pi}{3\pi}=\frac{2}{3}=0.667\text{ s}[/tex]So, the period is 0.667 s
c) the wavelength.
using the equation k = 2π/λ and solving for λ, we get:
[tex]\lambda=\frac{2\pi}{k}=\frac{2\pi}{4\pi}=0.5\text{ m}[/tex]So, the wavelength is 0.5 m
d) The wave speed
The wave speed can be calculated as the division of the wavelength by the period, so
[tex]v=\frac{\lambda}{T}=\frac{0.5\text{ m}}{0.667\text{ s}}=0.75\text{ m/s}[/tex]e) The height of the wave at (2, 1)
To find the height, we need to replace (x, t) = (2, 1) on the initial equation, so
[tex]\begin{gathered} y(x,t)=5\cos(\pi(4x-3t)) \\ y(2,1)=5\cos(\pi(4\cdot2-3\cdot1)) \\ y(2,1)=5\cos(\pi(8-3)) \\ y(2,1)=5\cos(\pi(5)) \\ y(2,1)=5\cos(5\pi) \\ y(2,1)=5(-1) \\ y(2,1)=-5 \end{gathered}[/tex]Then, the height of the wave is -5 m.
Therefore, the answers are
a) 5 m
b) 0.667 s
c) 0.5 m
d) 0.75 m/s
e) -5 m
a cheetah can accelerate from rest to a speed of 30.0 m/s in 7.00s. wht is its acceleration?
The given value of the speed of cheetah is,
[tex]v=30ms^{-1}[/tex]The time during the speed v is,
[tex]t=7\text{ s}[/tex]The relation between the acceleration, speed and time is,
[tex]a=\frac{v}{t}[/tex]Substituting the known values,
[tex]\begin{gathered} a=\frac{30}{7} \\ a=4.3ms^{-2} \end{gathered}[/tex]Thus, the value of the acceleration is 4.3 meter per second squared.
Answer:
30×7=210m it is easyokok
why free-fall acceleration can be regarded as a constant for objects falling within a few hundred miles of Earth’s surface.