A basketball player wishes to make a stunning last-second basket. With .8 seconds on the clock, she wants the ball to enter the hoop (height 3.05m) as time expires in the game. She shoots the ball with an initial speed of 7 m/s, from 2.1 meters above the ground.
1. At what angle (from the HORIZONTAL) does she shoot the ball?
2. Draw qualitatively accurate position, velocity, and acceleration graphs for both dimensions of the ball's motion while it is in flight.
Answers
Answer:
(1) [tex]46.86^\circ[/tex]
(2) Diagram has been attached in the solution.
Explanation:
This question is from projectile motion.
From the given question, we will discuss the motion of the basket ball only in the vertical direction from which we will be able to find out the angle of the initial velocity with the horizontal with which it should be shoot to enter the hoop.
Part (1):
Let us assume:
[tex]y_i[/tex] = initial position of the basket ball = 2.1 m[tex]y_f[/tex] = final position of the basket ball = 3.05 m[tex]a_y[/tex] = acceleration of the ball along the vertical = [tex]-9.8\ m/s^2[/tex][tex]t[/tex] = time taken to reach the goal = 0.8 s[tex]\theta[/tex] = angle of the initial velocity with the horizontal[tex]u[/tex] = initial speed of the ball = 7 m/s[tex]u_y[/tex] = initial vertical velocity of the ball = u\sin \thetaUsing the equation of motion for constant acceleration, we have
[tex]y_f-y_i=u_yt+\dfrac{1}{2}a_yt^2\\\Rightarrow 3.05-2.1=u\sin \theta (0.8) +\dfrac{1}{2}\times (-9.8)(0.8)^2\\\Rightarrow 0.95=7\times \sin \theta (0.8) -3.136\\\Rightarrow 0.95=5.6\sin \theta -3.136\\\Rightarrow 5.6\sin \theta= 0.95+3.136\\\Rightarrow 5.6\sin \theta= 4.086\\\Rightarrow \sin \theta= \dfrac{4.086}{5.6}\\\Rightarrow \sin \theta=0.729\\\Rightarrow \theta=\sin^{-1}(0.729)\\\Rightarrow \theta=46.86^\circ[/tex]
Hence, the angle of the shoot of the basket ball with the horizontal is [tex]46.86^\circ[/tex] such that it reaches the hoop on time.
Part (2):
For this part, a diagram has been attached.
Related Questions
Consider projectile thrown horizontally at 50 m/s from height of 19.6 meters. The projectile will take ______________ time to hit the ground as it would if it were dropped from the same height. a) More
b) Less
C)The Same
Answers
Answer:
C)The Same
Explanation:
Kinematics equation:
[tex]y=v_{oy}*t+1/2*g*t^2[/tex]
for both cases the initial velocity in the axis Y is the same, equal a zero.
So the relation between the height ant temps is the same for both cases (the horizontal velocity does not play a role)
C)The Same
At a certain elevation, the pilot of a balloon has a mass of 125 lb and a weight of 119 lbf. What is the local acceleration of gravity, in ft/s2, at that elevation? If the balloon drifts to another elevation where g = 32.05 ft/s2, what is her weight, in lbf, and mass, in lb?
Answers
Final answer:
The local acceleration of gravity at the pilot's initial elevation is 0.952 ft/s². At a different elevation with gravity at 32.05 ft/s², her weight would be 4006.25 lbf, but her mass remains the same at 125 lb.
Explanation:
To find the local acceleration of gravity, we use the formula weight = mass × gravity. The pilot's weight is 119 lbf, and her mass is 125 lb. We rearrange the formula to find gravity: gravity = weight / mass, which gives us 119 lbf / 125 lb.
The local acceleration of gravity at the pilot's elevation is therefore 0.952 ft/s². Now, if the pilot drifts to another elevation where gravity is 32.05 ft/s², her weight in pounds-force would be her mass times the new acceleration due to gravity, which is 125 lb × 32.05 ft/s². Hence, her new weight would be 4006.25 lbf. Her mass remains unchanged as mass is not dependent on gravity.
Final answer:
The local acceleration of gravity at the given elevation is 0.952 ft/s². When the balloon drifts to another elevation with an acceleration of gravity of 32.05 ft/s², the pilot's weight is 4006.25 lbf and the mass is 125 lb.
Explanation:
At a certain elevation, the pilot's weight is less than the mass due to the reduction in the acceleration of gravity. To find the local acceleration of gravity, we need to use the equation:
weight = mass * acceleration of gravity
For the given values, the pilot's weight is 119 lbf, and the mass is 125 lb. Rearranging the equation, we have:
acceleration of gravity = weight / mass
Substituting the values, we get:
acceleration of gravity = 119 lbf / 125 lb = 0.952 ft/s²
When the balloon drifts to another elevation where the local acceleration of gravity is 32.05 ft/s², we can use the same equation to find the new weight and mass. Rearranging the equation, we have:
weight = mass * acceleration of gravity
Substituting the new acceleration of gravity and the previous mass, we get:
weight = 125 lb * 32.05 ft/s² = 4006.25 lbf
Therefore, at the new elevation, the pilot's weight is 4006.25 lbf and the mass is 125 lb.
An electron is a negatively charged particle that has a charge of magnitude, e - 1.60 x 10-19 C. Which one of the following statements best describes the electric field at a distance r from the electron? The electric field is directed toward the electron and has a magnitude of ke/r. The electric field is directed away from the electron and has a magnitude of ke/2. The electric field is directed toward the electron and has a magnitude of ke/? The electric field is directed toward the electron and has a magnitude of ke?/r. The electric field is directed away from the electron and has a magnitude of ke/r.
Answers
Explanation:
The charge on the electron is, [tex]q=-1.6\times 10^{-19}\ C[/tex]
The electric field at a distance r from the electron is :
[tex]E=k\dfrac{q}{r^2}[/tex]
Where
k is the electrostatic constant, [tex]k=\dfrac{1}{4\pi \epsilon_o}[/tex]
We know that the electric field lines starts from positive charge and ends at the negative charge. Also, for a positive charge the field lines are outwards while for a negative charge the field lines are inwards.
So, the correct option is " the electric field is directed toward the electron and has a magnitude of [tex]k\dfrac{q}{r^2}[/tex]. Hence, this is the required solution.
When leaping at an angle of 47.7° above the horizontal, a froghopper reaches a maximum height of 42.4 cm above the level ground. What was the takeoff speed for such a leap?
What horizontal distance did the froghopper cover for this world-record leap?
Answers
Answer:
39 cm /s
77.25 cm approx
Explanation:
Angle of projection θ = 47.7°
Maximum height H = 42.4 cm
Initial velocity = u =?
we know that
maximum height
H = U² x sin²θ / 2g
U² = H x 2g /sin²θ
Putting the values
U² =( 42.4 X 2 X9.8 ) / (sin47.7)²
U = 39 cm /s
Horizontal Range R = U²sin2θ / 2g
= 39 x 39 x (sin95.4) / 2 x 9.8
R = 77.25 cm approx
Final answer:
In this question, we determine the takeoff speed and horizontal distance covered by a froghopper during a leap. Calculations involve kinematic equations and trigonometry to find these values accurately.
Explanation:
Takeoff speed: To find the takeoff speed, we can use the fact that the maximum height reached by the froghopper is related to its initial velocity. By using the kinematic equation for projectile motion, we can calculate the takeoff speed to be approximately 0.75 m/s.
Horizontal distance covered: The horizontal distance covered can be determined by analyzing the horizontal component of the motion. With the takeoff angle and the calculated initial speed, the horizontal distance traveled can be found using trigonometry to be around 0.91 m.
Zeta Anderson, futuristic super-spy for the Terran Confederation, has completed her objective of stealing intelligence from the Zorn collective. Stealthily, she slips into her space suit (with jet-pack), and slips from an airlock, headed for her stealth ship. Her jet-pack can supply her with a constant acceleration, and gravity can be neglected. When she turns on her jet-pack, how does her velocity change? Since she does not want to be going too fast (and either overshoot, or collide with her ship), how does her velocity change when she turns the jet-pack off?
Answers
Explanation:
First lets understand the 2nd law of motion by Sir Isaac Newton. According to this law,
[tex]Force = mass\times acceleration[/tex]
[tex]Velocity = acceleration\times time[/tex]
Since the spy is in space there is no medium and hence no friction to restrict the motion. Thus, when the spy turns on the jet pack, she will be accelerated and her velocity will increase. As there is no medium so no friction. So even when she turns her jet-pack off the velocity will not change. Although the acceleration will be zero but she will be moving with a constant velocity until an opposite force is applied. That can be done using reverse thrust.
Red and blue light enter together into a glass plate of 10 cm. What is the distance between red and blue light whenthe light emerges from the plate. n(blue) = 1.6, n(red) = 1.3
Answers
Answer:
The shift in the color's depends on the angle of incidence, for a special case when the angle of incidence is along the normal to the surface no shift will be observed.
Explanation:
When a ray of light is incident on a medium perpendicular to it it does not undergo any refraction thus no shift will be seen.
Answer:
The distance between the emergent red and blue light is 3 cm
Solution:
As per the question:
Thickness of the glass plate, s = 10 cm = 0.1 m
Refractive index of blue light, [tex]n_{blue} = 1.6[/tex]
Refractive index of blue light, [tex]n_{red} = 1.3[/tex]
Now, to calculate the distance between red and blue light as it emerges from the plate:
We know that refractive index is given as the ratio of speed of light in vacuum, c or air to that in medium, [tex]v_{m}[/tex].
[tex]n = \frac{c}{v_{m}}[/tex]
[tex]v_{m} = \frac{c}{n}[/tex] (1)
Since, c is constant, thus
n ∝ [tex]\frac{1}{v_{m}}[/tex]
Now, the refractive index of blue light is more than that of red light thus its speed in medium is lesser than red light.
Now, time taken, t by red and blue light to emerge out of the glass slab:
[tex]s = v_{m}\times t[/tex]
[tex]t = \frac{s}{v_{blue}} = \frac{sn_{blue}}{c}[/tex]
In the same time, red light also traveled through the glass covering some distance in air say x
[tex]t' = \frac{s}{v_{red}} = \frac{sn_{red}}{c}[/tex] (2)
Time taken by red light to cover 'x' distance in vacuum is t'':
[tex]t" = \frac{x}{c}[/tex]
Now,
t = t' + t" (3)
From eqn (1), (2) and (3):
[tex]\frac{sn_{blue}}{c} = \frac{sn_{red}}{c} + \frac{x}{c}[/tex]
Now, putting appropriate values in the above eqn:
[tex]\frac{0.1\times 1.6}{c} = \frac{0.1\times 1.3}{c} + \frac{x}{c}[/tex]
[tex]\frac{0.16}{c} - \frac{0.13}{c} = \frac{x}{c}[/tex]
x = 0.03 m = 3 cm
Explain the meaning of the term "potential divider" as applied to the circuit?
Answers
Answer and Explanation:
A potential divider, as the name itself gives a clear description of a simplified circuit which divides the potential across different circuit elements.
The potential divider circuit divides the potential depending upon the respective value of the elements of the circuit,
Given circuit diagram shows a potential divider circuit.
The working of this circuit is as :
[tex]V' = \frac{R'}{R + R'}V[/tex]
where
V = Input voltage
V' = output voltage
The average velocity for a trip has a positive value. Isit
possible for the instantaneous velocity at any point during thetrip
to have a negative value? Justify the answer.
Answers
Answer:
Yes, it's possible.
Explanation:
The average velocity is a mean value:
[tex]Vavg=\frac{displacement}{time taken}[/tex].
during that displacement, it may occur that the acceleration would negative at any time so at that moment if the velocity goes in the same direction with the acceleration, the velocity will be negative, it may take just a few moments and then go positive again. The velocity can also take negative values if for a moment the object was going backward (opposite direction). so the average velocity only means that the major of the velocity was positive.
A bird, accelerating from rest at a constant rate,
experiencesa displacement of 28 m in 11 s. What is the final
velocity after 11s?
Answers
Answer:
the speed of the bird is 2.8 m/s in after 11 seconds.
Explanation:
given,
displacement of bird = 28 m
time taken of the displacement of 28 m = 11 s
distance = speed × time
velocity = [tex]\dfrac{displacement}{time}[/tex]
=[tex]\dfrac{28}{10}[/tex]
= 2.8 m/s
hence, the speed of the bird is 2.8 m/s in after 11 seconds.
A particle of charge Q is fixed at the origin of an xy coordinate system. At t = 0 a particle (m = 0.727 g, q = 2.73 µC is located on the x axis at x = 19.3 cm, moving with a speed of 49.2 m/s in the positive y direction. For what value of Q will the moving particle execute circular motion? (Neglect the gravitational force on the particle.)
Answers
Answer:
Q = 12.466μC
Explanation:
For the particle to execute a circular motion, the electrostatic force must be equal to the centripetal force:
[tex]Fe = \frac{K*q*Q}{r^{2}} = \frac{m*V^{2}}{r}[/tex]
Solving for Q:
[tex]Q = \frac{m*V^{2}*r}{K*q}[/tex]
Taking special care of all units, we can calculate the value of the charge:
Q = 12.466μC
For a growing quantity to reach a value 32 times its initial value, how many doubling times are required? A) 4
B) 5
C) 16
D) 8
Answers
Answer:
B) n=5
Explanation:
We call the initial value Xo. We start to double this initial value
X1=2*Xo n=1
We double again:
X2=2*X1=2*(2Xo) n=2
X2=2*X1=2^2*Xo n=2
In general:
Xn=(2^n)*Xo
If we want to reach a value 32 times its initial value:
2^n=32
then: n=5
2^5=2*2*2*2*2=32
A squirrel is trying to locate some nuts he buried for the winter. He moves 4.3 m to the right of a stone and dogs unsuccessfully. Then he moves 1.1 m to the left of his hole, changes his mind, and moves 6.3 m to the right of that position and digs a second hole. No luck. Then he moves 8.0 m to the left and digs again. He finds a nut at last. What is the squirrel's total displacement from his starting point
Answers
Answer:
The total displacement from the starting point is 1.5 m.
Explanation:
You need to sum and substract, depending on the movement (to the right, sum; to the left, substract).
First, it moves 4.3 m right and return 1.1 m. So the new distance from the starting point is 3.2 m.
Second, it moves 6.3 m right, so the new distance is 9.5 m.
Finally it moves 8 m to the left, so 9.5 m - 8 m= 1.5 m.
Summarizing, at the end the squirrel is 1.5 m from its starting point.
A jet lands with a speed of 100 m/s and can accelerate uniformly at a rate of -5.0 m/s^2 as it comes to rest. What is the minimum length runway to land this plane? A. 1000 m
B. 20 m
C. 675 m
D. 500 m
Answers
Answer:
A ) 1000 m.
Explanation:
Here initial velocity u = 100 m /s
Final velocity v = 0
Acceleration a = -5 ms⁻²
Distance travelled = S
v² = u² + 2aS
0 = (100)² -2 x 5 S
S = 10000/ 10
=1000 m.
A crate is resting on a ramp that is inclined at an angleabove
the horizontal. As the angle is incrased the crateremains in place
until the angle reaches a value of 38degrees. Then the crate begins
to slide down the slope. (a) Determine the coefficient of static
friction between the crateand the ramp surface. (b) The coefficient
of kinetic frictionbetween the crate and the ramps surface is
0.600. Find theacceleration of the moving crate.
Answers
Answer:
1.4 m/s^2
Explanation:
The angle of repose or the angle of sliding is defined as the angle of inclined plane with the horizontal at which an object placed on the plane is just start to slide down.
The relation between the angle of friction and the coefficient of static friction is given by
[tex]\mu _{s}=tan\theta[/tex]
[tex]\mu _{s}=tan38=0.78[/tex]
Thus, the coefficient of static friction is 0.78.
(b) As the crate is moving down, then the friction force is kinetic friction.
The force acting along the plane downward = mg Sinθ
The normal reaction, N = mg cosθ
The friction force acting upward along the plane, f = μk N = μk mg Cosθ
The net force acting along the plane downward
Fnet = mg Sinθ - μk mg Cosθ
According to the newton's second law, Fnet = mass x acceleration
so, m x a = mg Sinθ - μk mg Cosθ
a = g Sinθ - μk g Cosθ
here, μk = 0.6 and θ = 38°, g = 9.8 m/s^2
By substituting the value, we get
a = 9.8 ( Sin38 - 0.6 x Cos 38)
a = 9.8 (0.6156 - 0.6 x 0.788)
a = 1.4 m/s^2
The coefficient of static friction between the crate and the ramp surface is 0.781. The acceleration of the moving crate is 5.88 m/s².
Explanation:To find the coefficient of static friction, we need to calculate the tangent of the angle at which the crate begins to slide. The tangent of 38 degrees is approximately 0.781. Since the crate remains in place until this angle, the coefficient of static friction must be equal to or greater than this value. Therefore, the coefficient of static friction between the crate and the ramp surface is 0.781.
To find the acceleration of the moving crate, we can use the formula a = μk * g, where a is the acceleration, μk is the coefficient of kinetic friction, and g is the acceleration due to gravity. Given that the coefficient of kinetic friction is 0.600, and the acceleration due to gravity is approximately 9.8 m/s², we can substitute these values into the formula to find the acceleration.
Substituting the values, we get a = (0.600) * (9.8) = 5.88 m/s². Therefore, the acceleration of the moving crate is 5.88 m/s².
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A blacksmith drops a 550 °F piece of iron into a vat of 75 °F water in order to cool it to 100 °F. How many kilograms of water are needed per kilogram of iron? Assume all the thermal energy from the iron is transferred to the water and none of the water evaporates. The specific heats of water and iron are 4186 J/kg×°C and 448 J/kg×°C, respectively.
Answers
Answer:
1.93 kg water/kg iron
Explanation:
All the thermal energy from the iron is transferred to the water. In equilibrium, the temperature of both the water and the iron will be the same. The heat that an object loses or gains after a change in value in its temperature is equal to:
[tex]Q = mc*(T_f-T_o)[/tex]
Then,
[tex]-Q_{iron} = Q_{water}[/tex]
Before solving the problem, let's convert the values of temperature to Celsius:
(550°F -32)*5/9 = 287.78 °C
(75°F - 32)*5/9 = 23.89 °C
(100°F -32)*5/9 = 37.78°C
Now, we can solve:
[tex]-Q_{iron} = Q_{water}\\m_{iron}*c_{iron}*(T_o_i-T_f_i) = m_{water}*c_{water}*(T_f_w-T_o_w)\\\frac{m_{water}}{m_{iron}} = \frac{c_{iron}(T_o_i-T_f_i)}{c_{water}(T_f_w-T_o_w)} =\frac{448J/kg^oC(287.78^oC - 37.78^oC)}{4186J/kg^oC(37.78^oC-23.89^oC)}= 1.93 kg_{water}/kg_{iron}[/tex]
A small silver (10.5 g/cm^3) cylinder of diameter 1.4 cm and a cylinder of lead (11.3 g/cm^3) balance each other when placed on a triple beam balance. If they have the same length, what must be the diameter of the lead cylinder?
Answers
Answer:
The diameter of the lead cylinder is 1.35 cm.
Explanation:
Given that,
Density of silver = 10.5 g/cm³
Density of lead = 11.3 g/cm³
Diameter = 1.4 cm
As mass of both is equal.
Let diameter of lead [tex]d_{l}[/tex]
We need to calculate the the diameter of the lead cylinder
Using balance equation of density
[tex]V\times \rho_{s}=V\times \rho_{l}[/tex]
[tex]\dfrac{\pi\times d_{s}^2\times h}{4}\times\rho_{s}=\dfrac{\pi\times d_{l}^2\times h}{4}\times\rho_{s}[/tex]
[tex]d_{l}^2=\dfrac{d_{s}^{2}\times\rho_{s}}{\rho_{l}}[/tex]
put the value into the formula
[tex]d_{l}^2=\dfrac{(1.4\times10^{-2})^2\times10.5}{11.3}[/tex]
[tex]d_{l}=\sqrt{0.00018212}[/tex]
[tex]d_{l}=0.0135\ m[/tex]
[tex]d_{l}=1.35\ cm[/tex]
Hence, The diameter of the lead cylinder is 1.35 cm.
Two cars drive on a straight highway. At time t=0, car 1 passes mile marker 0 traveling due east with a speed of 20.0 m/s . At the same time, car 2 is 0.80 km east of mile marker 0 traveling at 26.0 m/s due west. Car 1 is speeding up with an acceleration of magnitude 0.40 m/s^2 , and car 2 is slowing down with an acceleration of magnitude 0.80 m/s^2 . a.) Write x-versus-t equations of motion for both cars, taking east as the positive direction.
b.) At what time do the cars pass next to one another?
Answers
Answer:
Explanation:
a ) Equation for car 1
X = 20 t + 1/2 x 0.40 t² ( initial velocity is 20 and acceleration is 0.4 )
Equation for car 2
X = 800 - ( 26 t - 1/2 x0.80 t² ) [ when t = 0 , X = 800 m and with time t , X decreases ]
b ) Let after time t they meet , then X will be equal for both of them at t.
20 t + 1/2 x 0.40 t² = 800 - ( 26 t - 1/2 x0.80 t² )
0.20 t²- 46 t +800 = 0
t = 19 s and 211 s .
The equation of motion for car 1 is x = 20t + 0.2t^2. For car 2, it's x = 800 + 26t - 0.4t^2. To find when the cars meet each other, set these equal to each other and solve for t.
Explanation:The x-versus-t equations of motion would be determined by rate of movement and acceleration. For car 1, which is accelerating positively, the equation of motion is x = v1*t + 0.5*a1*t^2 (where v1 = initial velocity of car 1, a1 = acceleration of car 1, t is time). So, it would be x = 20t + 0.5 * 0.4 * t^2 = 20t + 0.2t^2.
For car 2, decelerating negatively and starting at 0.80 km east of the origin (which equates to 800m), the equation is x = x2 + v2*t - 0.5*a2*t^2. This works out to x = 800 + 26t - 0.5 * 0.8 * t^2 = 800 + 26t - 0.4t^2.
To find the time (t) when cars pass each other, we equate the two displacements: 20t + 0.2t^2 = 800 + 26t - 0.4t^2. Solving that equation gives us the time in seconds when the cars will meet next to each other.
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Joseph DeLoach of the United States set an Olympic record in 1988 for the 200-meter dash with a time of 19.75 seconds. What was his average speed? Give your answer in meters per second and miles per hour.
Answers
Answer:
The average speed was 10.12 m/s or 22.96 mi/h
Explanation:
The average speed is defined as:
[tex]v = \frac{d}{t}[/tex] (1)
Where d is the total distance traveled and t is the passed time.
For the special case of Joseph DeLoach, he traveled a total distance of 200 meters in 19.75 seconds. Those values can be introduced in equation 1:
[tex]v = \frac{200 m}{19.75 s}[/tex]
[tex]v = 10.12 m/s[/tex]
That means that Joseph DeLoach traveled a distance of 10.12 meters per second.
To represent the result in miles per hour, it is necessary to know that 1 mile is equivalent to 1609 meters.
[tex]200 m x \frac{1 mi}{1609 m}[/tex] ⇒ 0.124 mi
it is needed to express the given time in units of hour. 1 hour is equivalent to 3600 seconds.
[tex]19.75 s x \frac{1 h}{3600 s}[/tex] ⇒ 0.0054 h
Then, equation 1 is used with the new representation of the values.
[tex]v = \frac{0.124 mi}{0.0054 h}[/tex]
[tex]v = 22.96 mi/h[/tex]
From a crouched position an excellent human jumper launches herself upwards. The jumpers full height is 168 cm, but in the crouched position her height is half of that value. The jumper accelerates from the crouched position until she reaches her full height. Her motion then carries her upward until her center of mass is 90 cm above its position when she is standing. At what velocity must the jumper leave the ground to reach this height? What constant acceleration must be maintained to reach this initial velocity? Assuming a constant acceleration, how long in seconds does it take to accelerate from rest to this initial velocity?
Answers
Answer:
a) 4.2 m/s
b) 13.6 m/s^2
Explanation:
She is jumping, and when her feet no longer touch the ground she is in free fall, only affected by the acceleration of gravity.
The equation for position under constant acceleration is:
Y(t) = Y0 + Vy0 * t + 1/2 * a * t^2
We set up a reference system that has its origin at the point her center of mass is when she is standing and the positive Y axis points upwards, then:
Y0 = 0 m
a = -9.81 m/s^2
The equation for speed under constant acceleration is:
V(t) = Vy0 + a * t
We know that when she reaches her highes point her vertical speed will be zero because that is wehn her movement changes direction. We'll call this moment t1.
0 = Vy0 + a * t1
a * t1 = -Vy0
t1 = -Vy0/a
If we replace this value on the position equation we can find her initial speed:
Y(t1) = Y0 - Vy0 * Vy0/a + 1/2 * a * (-Vy0/a)^2
Y(t1) = - Vy0^2/a + 1/2 * Vy0^2/a
Y(t1) = -1/2 * Vy0^2 / a
Vy0^2 = -2 * a * Y(t1)
[tex]Vy0 = \sqrt{-2 * a * Y(t1)}[/tex]
[tex]Vy0 = \sqrt{-2 * (-9.81) * 0.9} = 4.2 m/s[/tex]
I assume her center of mass is at half her height, so when she is standing it would be at 93 cm of the grouind, and when she is crouching at 46.5 cm.
Therefore when she jumps her centr of mass moves 0.465 m before leaving the ground.
During that trajectory she moves with acceleration.
Y(t) = Y0 + Vy0 * t + 1/2 * a *t^2
In this case her initial position is
Y0 = -0.465
Her initial speed is
Vy0 = 0
At t=t0 her position will be zero
The equation for speed under constan acceleration is
Vy(t) = Vy0 + a * t
Her speed at t0 will be 4.2 m/s
4.2 = a * t0
t0 = 4.2 / a
0 = -0.465 - 1/2 * 9.81 * (4.2 / a)^2
0.465 = 4.9 * 17.6 / a^2
a^2 = 86.2 / 0.465
[tex]a = \sqrt{185.4} = 13.6 m/s^2[/tex]
I know that my heat recovery ventilator consumes 83.2 Watts of power to run its fans when supplying balanced fresh air into my house. If it runs 24 hrs a day, how many kWh's of energy would it consume in a year?
Answers
Answer: 730 kWh
Explanation: To solve this problem we have to use the following considerations:
If teh heat recovery ventilator consumes 83.2 W running 24 hs we have a consumed energy of:
Energy consumed= 83.2* 24= 2 kWh
so in a year we must multiply by 365 then
Energy consumed in a year= 2kWh*365= 730 kWh
A golf ball is hit into the air, but NOT Staight Up, and encounters no significant air resistance. Which statement accurately describes its motion while it is in air? On the way up, both its horizontal and vertical velocity components are decreasing, on the way down they are both increasing. Its acceleration is zero at the highest point. Its horizontal velocity doesn't change once it is in air, but its vertical velocity does change. On the way up its acceleration is 9.8m/s2 upward, on the way down its acceleration is 9.8m/s2 downward, Its velocity is zero at the highest point.
Answers
Answer:
Answered
Explanation:
A) False , horizontal velocity component is constant since there's no horizontal acceleration component
B) False, acceleration remains constant through out and only verticle acceleration is there horizontal acceleration is zero.
C) True , Its horizontal velocity doesn't change once it is in air, but its vertical velocity does change ( because of acceleration in verticle direction)
D)False, the acceleration is constant through the motion.
E)False, Vertical velocity is Zero but there will be Horizontal velocity.
You are jogging for one hour at a speed of 5 m/s and for the final 200 m you decide to increase your velocity to 11 m/s. Assuming you were able to do so instantaneously. Once you reach that speed how long it will take to cover 200m?
Answers
Answer:
[tex]t=18.18s[/tex]
Explanation:
From the exercise we know that the person change its initial velocity from 5m/s to 11 m/s for 200m
According to the formula:
[tex]v=\frac{x}{t}[/tex]
If we want to know how much time does it take to cover 200 m at 11 m/s we need to calculate the following formula:
[tex]t=\frac{x}{v}=\frac{200m}{11m/s}=18.18s[/tex]
What is the state of an object being acted upon by an unbalanced force? A. at rest B. zero speed C. in motion with a constant velocity D. accelerated E. any of these
Answers
Answer:
D. accelerated
Explanation:
According to Newton's first law of motion, an object always remains in its state of rest or in uniform motion until an external unbalanced force is acted upon the object. This means if an external unbalanced force is acted on the object, it will come to a non-uniform motion i.e., an accelerated motion.
Let me first get you clear that uniform motion is determined by a constant velocity and a state of rest is considered by a zero velocity or speed. But, here we have an unbalanced force acting on the object. This means the object will change its velocity and hence, it has an accelerated motion.
Final answer:
The state of an object being acted upon by an unbalanced force is accelerated. This is because an unbalanced force results in a change in the velocity of the object, causing it to accelerate as per Newton's First Law of Motion.
Explanation:
The state of an object being acted upon by an unbalanced force is accelerated. According to Newton's First Law of Motion, an object at rest stays at rest, and an object in motion continues in motion with the same speed and in the same direction unless acted upon by an unbalanced external force. This implies that the presence of an unbalanced force (net force not equal to zero) changes the velocity of an object, causing it to accelerate.
In other words, option D, accelerated, is the correct answer. When an unbalanced force acts on an object, it results in the object's acceleration, which is a change in its velocity over time. This scenario directly contradicts situations where an object is at rest, at zero speed, or moving with a constant velocity, where no net force or balanced forces are acting on the object.
The time base on the oscilloscope is set for 2 ms/cm and the vertical input has a frequency of 3000 Hz. How many wave cycles will appear within the 10 cm width of the screen?
Answers
Answer:
60 cycles
Explanation:
The first thing we must do to solve the problem is to find how many cycles are presented in 1cm by multiplying the frequency by the base time of the
K=time base=2ms/cm=2x10-3s/cm
f=frecuency=3000s^-1
N=fk
N=(3000)(2x10^-3)=6cycles/cm
Ntot=6x10=60cycles
Answer:
60 wave cycles
Explanation:
As the horizontal axis in a oscilloscope represents time, the time base is simply the scale, in other words, the amount of time that each division of oscilloscope represents. Therefore, multiplying the width of the screen times the time base will give us the total amount of time graphed on the screen.
The frequency is the amount of oscillations or waves cycles per second. So, in order to find the total amount of oscillations:
[tex]10cm * \frac{0.002 s}{cm} *\frac{3000cycles}{s} = 60 cycles[/tex]
An object (with charge -2.4 µC and mass 0.026 kg) hovers at rest above the ground, its weight being held up by a uniform electric field. (a) Find the size and direction of this electric field.
(b) If the electric charge on the object is quadrupled while its mass remains the same, find the size and direction of its acceleration.
Answers
Answer:
a ) 10.61 X 10⁴ N/C
b )29.37 ms⁻² .
Explanation:
Since the weight of the object is balanced by force by electric field , the force must be acting in upward direction . In an electric field , an electron experiences a force opposite to the direction of field , therefore ,direction of electric field must be in downward direction .
Let E be the electric field . Force on the charge
= electric field x charge = E x 2.4 x 10⁻⁶
Weight acting downwards = .026 x 9.8
For balancing
.026 x 9.8 = E x 2.4 x 10⁻⁶
E = .026 x 9.8 / 2.4 x 10⁻⁶
= 10.61 X 10⁴ N/C
b )
If the charge is quadrupled , charge becomes 4 x 2.4 x 10⁻⁶
Upward force = 4 x 2.4 x 10⁻⁶ x 10.61 X 10⁴ = 101.85 x 10⁻²
Down ward force =.026 x 9.8
Net force = 101.85 x 10⁻² - .026 x 9.8
= .7637 N
acceleration = .7637 / .026 = 29.37 ms⁻² .
A mail carrier parks her postal truck and delivers packages. To do so, she walks east at a speed of 0.80 m/s for 4.0 min, then north at a speed of 0.50 m/s for 5.5 min, and finally west at a speed of 1.1 m/s for 2.8 min. Define east as +x and north as +y. (a) Write unit-vector velocities for each of the legs of her journey. (b) Find unit-vector displacements for each of the legs of her journey. (c) Find her net displacement from the postal truck after her journey is complete.
Answers
Explanation:
Given that,
She walks in east,
Speed = 0.80 m/s
Time = 4.0 min
In north,
Speed = 0.50 m/s
Time = 5.5 min
In west,
Speed = 1.1 m/s
Time = 2.8 min
(a). We need to calculate the unit-vector velocities for each of the legs of her journey.
The velocity of her in east
[tex]\vec{v_{1}}=0.80\ \hat{x}\ m/s[/tex]
[tex]\vec{v_{2}}=0.50\ \hat{y}\ m/s[/tex]
[tex]\vec{v_{3}}=1.1\ \hat{-x}\ m/s[/tex]
(b). We need to calculate the unit-vector displacements for each of the legs of her journey
Using formula of displacement
[tex]\vec{d_{1}}=v_{1}\times t_{1}[/tex]
In east ,
[tex]\vec{d_{1}}=0.80\times4.0\times60[/tex]
[tex]\vec{d_{1}}=192\ \hat{x}\ m[/tex]
In north,
[tex]\vec{d_{2}}=0.50\times5.5\times60[/tex]
[tex]\vec{d_{2}}=165\ \hat{y}\ m[/tex]
In west,
[tex]\vec{d_{3}}=1.1\times2.8\times60[/tex]
[tex]\vec{d_{3}}=184.8\ \hat{-x}\ m[/tex]
(c). We need to calculate the net displacement from the postal truck after her journey is complete
[tex]\vec{d}=\vec{d_{1}}+\vec{d_{2}}+\vec{d_{3}}[/tex]
Put the value in the formula
[tex]\vec{d}=192\hat{x}+165\hat{y}+184.8\hat{-x}[/tex]
[tex]\vec{d}=7.2\hat{x}+165\hat{y}[/tex]
We need to calculate the magnitude of the displacement
[tex]d=\sqrt{(7.2)^2+(165)^2}[/tex]
[tex]d=165.16\ m[/tex]
The magnitude of the displacement is 165.16 m.
Hence, This is the required solution.
If your front lawn is 16.016.0 feet wide and 20.020.0 feet long, and each square foot of lawn accumulates 13501350 new snowflakes every minute, how much snow, in kilograms, accumulates on your lawn per hour? Assume an average snowflake has a mass of 2.20 mg.
Answers
Answer:
The mass of snow accumulated in the lawn in 1 hour equals 57.024 kilograms
Explanation:
Given
Width of lawn = 16 feet
Length of lawn = 20 feet
Thus total area of lawn equals [tex]16\times 20=320sq.feet[/tex]\
Now it is given that 1 square foot accumulates 1350 new snowflakes each minute thus number of snowflakes accumulated by 320 square feet in 1 minute equals
[tex]320\times 1350=432000[/tex]
Now it is given that average mass of each snowflake is [tex]2.20mg=2.20\times 10^{-3}g=2.20\times 10^{-6}kg[/tex]
Hence the mass accumulated per minute equals [tex]432000\times 2.20\times 10^{-6}=0.9504kg[/tex]
Now since there are 60 minutes in 1 hour thus the mass accumulated in 1 hour equals [tex]0.9504kg\times 60=57.024kg[/tex]
A charged wire of negligible thickness has length 2L units and has a linear charge density λ. Consider the electric field E-vector at the point P, a distance d above the midpoint of the wire. The field E-vector points along one of the primary axes, yWhat is the magnitude E of the electric field at point P? Throughout this part, express your answers in terms of the constant k, defined by k=1/(4πε)
Answers
Answer:
[tex]E=2K\lambda d\dfrac{L }{d^2\sqrt{L^2+d^2}}[/tex]
Explanation:
Given that
Length= 2L
Linear charge density=λ
Distance= d
K=1/(4πε)
The electric field at point P
[tex]E=2K\int_{0}^{L}\dfrac{\lambda }{r^2}dx\ sin\theta[/tex]
[tex]sin\theta =\dfrac{d}{\sqrt{d^2+x^2}}[/tex]
[tex]r^2=d^2+x^2[/tex]
So
[tex]E=2K\lambda d\int_{0}^{L}\dfrac{dx }{(x^2+d^2)^{\frac{3}{2}}}[/tex]
Now by integrating above equation
[tex]E=2K\lambda d\dfrac{L }{d^2\sqrt{L^2+d^2}}[/tex]
A bullet with momentum of 2.8 kg·m/s [E] is traveling at a speed of 187 m/s. The mass of the bullet is: a) 67 g b) 15 g c) 0.067 g d) 0.015 g
Answers
Answer:
The mass of the bullet is 15 g.
(b) is correct option.
Explanation:
Given that,
Momentum = 2.8 kg m/s
Speed = 187 m/s
We need to calculate the mass of the bullet
Using formula of momentum
[tex]P = mv[/tex]
[tex]m = \dfrac{P}{v}[/tex]
Where, P = momentum
v = speed
Put the value into the formula
[tex]m = \dfrac{2.8}{187}[/tex]
[tex]m = 0.015\ kg[/tex]
[tex]m = 15\ g[/tex]
Hence, The mass of the bullet is 15 g.
Two foreces act on a block of mass 4.5 kg resting on
africtionless, horizontal surface, as shown. The horizontal force
is3.7 N; The other force of 5.9 N acts at an angle of 43 degrees
fromthe horizontal. what is the magnitude of the acceleration of
theblock?
Answers
Answer:[tex]=\frac{8.014}{4.5}=1.78 m/s^2[/tex]
Explanation:
Given
mass of block(m)=4.5 kg
Horizontal force([tex] F_h[/tex])=3.7 N
another force F at angle of [tex]43^{\circ}[/tex]
if F is pulling Block then
Net Normal reaction=mg-Fsin43=40.12 N
Net Force in Horizontal direction =3.7+Fcos43
=3.7+4.31=8.014 N
thus Net acceleration is a[tex]=\frac{8.014}{4.5}=1.78 m/s^2[/tex]