Answer:
138.46 ft
Explanation:
When the ball is dropped until the moment it hits the water, the ball moves in a uniform acceleration motion. Therefore, the equation that describes the movement of the ball is:
[tex]X = \frac{1}{2}*g*t^{2} + V_{0} *t + x_0[/tex]
Where X is the distance that the ball has fallen at a time t. [tex]V_0[/tex] is the initial velocity, which is 0 ft/s as the ball was simply dropped. [tex]x_0[/tex] is the initial position, we will say that this value is 0 in the position where the ball was dropped for simplicity, and it increases as the ball is falling. Now, we replace x with 16 feets and solves for t:
[tex]16 ft = \frac{1}{2} * 32.2 \frac{ft}{s^{2}} *t^{2} +0 \frac{ft}{s} * t + 0 ft[/tex]
[tex]t = \sqrt{2* \frac{16 ft}{32.2 \frac{ft}{s^2}}} = 1 s[/tex]
The velocity that the ball will have at the moment the ball that the ball hits the water will be:
[tex]V = V_o+g*t=0\frac{ft}{s}+32.2\frac{ft}{s^2}*1s =32.2\frac{ft}{s}[/tex]
The time that will take the ball to reach the bottom from the top of the lake will be t = 5.3s - 1s = 4.3s. And as the ball will travel with constant velocity equal to 32.2 ft/s^2, the depth of the lake will be:
[tex]d = v*t = 32.2\frac{ft}{s} * 4.3s = 138.46 ft[/tex]
The depth of the lake is approximately 153.6 feet.
Step 1: Understanding the Problem
We have a lead ball that is dropped from a height of 16.0 ft. After being dropped, it hits the water and sinks to the bottom with a constant velocity. We are told that it reaches the bottom of the lake 5.30 seconds after it is dropped. Our goal is to calculate the depth of the lake.
Step 2: Analyzing the Time
The total time from the moment the ball is dropped until it hits the bottom of the lake is 5.30 seconds. This time includes:
The time it takes to fall to the water's surface.The time it takes to sink to the bottom of the lake once it hits the water.Step 3: Calculate the Fall Time
We can first determine the time it takes for the ball to drop 16.0 ft. The formula for the distance fallen under gravity is:
[tex]d = \frac{1}{2} g t^2[/tex]
where:
[tex]d[/tex] is the distance (16.0 ft),[tex]g[/tex] is the acceleration due to gravity (approximately 32 ft/s²), and [tex]t[/tex] is the time in seconds.Rearranging the formula for time:
[tex]t = \sqrt{\frac{2d}{g}} = \sqrt{\frac{2 \times 16.0\text{ ft}}{32 \text{ ft/s}^2}} = \sqrt{1} = 1 \text{ second}[/tex]
Step 4: Calculate Time to Sink
If the total time from the drop to reaching the bottom is 5.30 seconds, the time taken to sink in the water is:
[tex]\text{Time to sink} = 5.30 \text{ s} - 1 \text{ s} = 4.30 \text{ s}[/tex]
Step 5: Calculate Depth of the Lake
Since the ball sinks at a constant velocity, we need to find the velocity of the ball once it is in the water. The velocity of the ball just before entering the water can be found using the formula:
[tex]v = g t = 32 \text{ ft/s}^2 \times 1 \text{ s} = 32 \text{ ft/s}[/tex]
This means the ball will sink at a velocity of 32 ft/s. Now, we can calculate the depth of the lake (d):
[tex]d = \text{velocity} \times \text{time} = 32 \text{ ft/s} \times 4.30 \text{ s} = 137.6 \text{ ft}[/tex]
Step 6: Add the Initial Drop
Finally, to find the total depth of the lake, we need to add the initial drop of 16.0 ft:
[tex]\text{Total depth} = 137.6 \text{ ft} + 16.0 \text{ ft} = 153.6 \text{ ft}[/tex]
Suppose two equal charges of 0.65 C each are separated by a distance of 2.5 km in air. What is the magnitude of the force acting between them, in newtons?
Answer:
Force between two equal charges will be 608.4 N
Explanation:
We have given charges [tex]q_1=0.65C\ and\ q_2=0.65C[/tex]
Distance between the charges = 2.5 km = 2500 m
According to coulombs law force between two charges is given by
[tex]F=\frac{1}{4\pi \varepsilon _0}\frac{q_1q_2}{r^2}=\frac{Kq_1q_2}{r^2}[/tex], here K is constant which value is [tex]9\times 10^9Nm^2/C^2[/tex]
So force [tex]F=\frac{9\times 10^9\times 0.65\times 0.65}{2500^2}=608.4N[/tex]
How many 1140 nm long molecules would you have to line up end to end to stretch a distance of 158 miles?
Answer:
221754385964.9123
Explanation:
Convert miles to nanometer
1 mile = 1.6 km
1 km = 1×10³×10³×10³×10³ nm
1 mile = 1.6×10¹² nm
So,
158 miles = 158×1.6×10¹² = 252.8×10¹² nm
Length of each molecule = 1140 nm
Number of molecules = Total length / Length of each molecule
[tex]\text{Number of molecules}=\frac{252.8\times 10^{12}}{1140}\\\Rightarrow \text{Number of molecules}=221754385964.9123[/tex]
There are 221754385964.9123 number of molecules in a stretch of 158 miles
A rock is thrown straight up and passes by a window. The window is 1.7m tall, and the rock takes 0.19 seconds to pass from the bottom of the window to the top. How far above the top of the window will the rock rise?
Answer:
The rock will rise 3.3 m above the top of the window.
Explanation:
The equations used to find the height and velocity of the rock at any given time are as follows:
y = y0 + v0 · t + 1/2 · g · t²
v = v0 + g · t
Where:
y = height of the rock at time t
y0 = initial height
v0 = initial velocity
t = time
g = acceleration due to gravity
v = velocity of the rock at time t
If we place the frame of reference at the bottom of the window, we can say that at time t = 0.19 s the height of the rock is 1.7 m. That will allow us to find the initial velocity needed to find the time at which the rock is at its maximum height.
y = y0 + v0 · t + 1/2 · g · t²
1.7 m = 0 m + v0 · 0.19 s - 1/2 · 9.8 m/s² · (0.19 s)²
1.7 m + 1/2 · 9.8 m/s² · (0.19 s)² = v0 · 0.19 s
(1.7 m + 1/2 · 9.8 m/s² · (0.19 s)²) / 0.19 s = v0
v0 = 9.9 m/s
With the initial velocity, we can find at which time the rock reaches its max- height. We know that at maximum height, the velocity of the rock is 0. Then, using the equation of velocity:
v = v0 + g · t
0 = 9.9 m/s - 9.8 m/s² · t
-9.9 m/s / -9.8 m/s² = t
t = 1.0 s
Now calculating the position at time t = 1.0 s, we will find the maximum heigth:
y = y0 + v0 · t + 1/2 · g · t²
y = 0 m + 9.9 m/s · 1.0 s - 1/2 · 9.8 m/s² · (1.0 s)²
y = 5.0 m (this is the max-height meassured from the bottom of the window)
Then, the rock will rise (5.0 m - 1.7 m) 3.3 m above the top of the window.
what is the approximate radius of the n = 1orbit of gold ( Z
=79 )?
Answer:
[tex]6.70\times 10^{-13}\ m[/tex]
Explanation:
Given:
[tex]n = n^{th}[/tex] orbit of gold = 1[tex]Z[/tex] = atomic number of gold = 79Assumptions:
[tex]h[/tex] = Planck's constant = [tex]6.62\times 10^{-34}\ m^2kg/s[/tex][tex]k[/tex] = Boltzmann constant = [tex]9\times 10^{9}\ Nm^2/C^2[/tex][tex]e[/tex] = magnitude of charge on an electron = [tex]1.6\times 10^{-19}\ C[/tex][tex]m[/tex] = mass of an electron = [tex]9.1\times 10^{-31}\ kg[/tex][tex]r[/tex] = radius of the [tex]n^{th}[/tex] orbit of the atomWE know that the radius of the [tex]n^{th}[/tex] orbit of an atom is given by:
[tex]r = \dfrac{n^2h^2}{4\pi^2kZe^2m}\\[/tex]
Let us find out the radius of the 1st orbit of the gold atom for which n = 1 and Z = 79.
[tex]r = \dfrac{n^2h^2}{4\pi^2kZe^2m}\\\Rightarrow r = \dfrac{(1)^2(6.62\times 10^{-34})^2}{4\pi^2\times 9\times 10^9\times 79\times (1.6\times 10^{-19})^2\times 9.1\times 10^{-31}}\\\Rightarrow r =6.70\times 10^{-13}\ m[/tex]
What is the time traveled by a pulse over a distance of lcm in air (n=1) and in 1cm of glass (n 1.5)? What is the difference in picoseconds?
Answer: in air 33.33 ps and in the glass 50 ps: so the difference 16.67 ps
Explanation: In order to calculate the time for a pulse travellin in air and in a glass we have to consider the expresion of the speed given by:
v= d/t v the speed in a medium is given by c/n where c and n are the speed of light and refractive index respectively.
so the time is:
t=d/v=d*n/c
in air
t=0.01 m*1/3*10^8 m/s= 33.33 ps
while for the glass
t=0.01 m*1.5* 3* 10^8 m/s= 50 ps
Finally the difference is (50-33.33)ps = 16.67 ps
An electron moves with a speed of 5.0 x 10^4m/s
perpendicularto a uniform magnetic field of .20T. What is the
magnitude ofthe magnetic force on the electron?
Final answer:
The magnitude of the magnetic force on an electron moving with a speed of 5.0 × [tex]10^4[/tex] m/s perpendicular to a magnetic field of 0.20 Tesla is calculated using the formula F = qvB, resulting in a force of 1.6 × [tex]10^{-15}[/tex] Newtons.
Explanation:
The magnetic force on an electron moving perpendicular to a magnetic field can be calculated using the formula F = qvB, where F is the magnetic force, q is the charge of the electron (-1.6 × [tex]10^{-19}[/tex] C), v is the velocity of the electron, and B is the magnetic field strength. Given that the electron moves with a speed of 5.0 × [tex]10^4[/tex] m/s perpendicular to a uniform magnetic field of 0.20 T, we use the formula to find the magnitude of the force:
F = (1.6 ×[tex]10^{-19}[/tex]C)( 5.0 × [tex]10^4[/tex] m/s)(0.20 T) =1.6 ×[tex]10^{-19}[/tex]C× 104 m/s × 2 × [tex]10^{-1}[/tex] T
F = 1.6 ×[tex]10^{-15}[/tex] N
The magnitude of the magnetic force on the electron is 1.6 ×[tex]10^{-15}[/tex] Newtons.
Two joggers are running with constant speed in opposite directions around a circular lake. One jogger runs at a speed of 2.15 m/s; The other runs at a speed of 2.55 m/s. The track around the lake is 300m long, and the two joggers pass each other at exactly 3:00 PM. How long is it before the next time the two joggers pass each other again?
Answer:
The two joggers will pass each other after 1 minute and 4 seconds at 3:01:04 PM.
Explanation:
The situation is analogous to two joggers running in opposite direction in a straight line where one jogger starts at the beginning of the line and the other starts at the other end, 300 m ahead.
The equation for the position of the joggers will be:
x = x0 + v · t
Where:
x = position of the jogger at time t
x0 = initial position
v = velocity
t = time
When the joggers pass each other, their position will be the same. Let´s find at which time both joggers pass each other:
x jogger 1 = x jogger 2
0 m + 2.15 m/s · t = 300 m - 2.55 m/s · t
(notice that the velocity of the joggers has to be of opposite sign because they are running in opposite directions).
2.15 m/s · t + 2.55 m/s · t = 300 m
4.70 m/s · t = 300 m
t = 300 m / 4.70 m/s = 63.8 s
The two joggers will pass each other after 1 minute and 4 seconds at 3:01:04 PM.
As a train accelerates away from a station, it reaches a speed of 4.6 m/s in 5.2 s. If the train's acceleration remains constant, what is its speed after an additional 7.0 s has elapsed? Express your answer using two significant figures.
Answer:
Vf = 10.76 m/s
Explanation:
Train kinematics
The train moves with uniformly accelerated movement
[tex]V_f = V_o + a*t[/tex] Formula (1)
Vf: Final speed (m/s)
V₀: Inital speed (m/s)
t: time in seconds (s)
a: acceleration (m/s²)
Movement from t = 0 to t = 5.2s
We replace in formula (1)
4.6 = 0 + a*5.2
a = 4.6/5.2 = 0.88 m/s²
Movement from t = 5.2s to t = 5.2s + 7s = 12.2s
We replace in formula (1)
[tex]V_f = 4.6 + 0.88*7[/tex]
Vf = 10.76 m/s
The wheel has a weight of 5.50 lb, a radius of r=13.0 in, and is rolling in such a way that the center hub, O, is moving to the right at a constant speed of v=17.0 ft/s. Assume all the mass is evenly distributed at the outer radius r of the wheel/tire assembly. What is the total kinetic energy of the bicycle wheel?
Answer:
[tex]E_{k}=1589.5ftlb[/tex]
Explanation:
[tex]E_{k}=E_{movement}+E_{rotational}\\[/tex]
[tex]E_{k}=\frac{1}{2}mv^{2}+\frac{1}{2}Iw^{2}[/tex] (1)
For this wheel:
[tex]w=\frac{v}{r}[/tex]
[tex]I=mr^{2}[/tex]: inertia of a ring
We replace (2) and (3) in (1):
[tex]E_{k}=\frac{1}{2}mv^{2}+\frac{1}{2}(mr^{2})(\frac{v}{r})^{2}=mv^{2}=5.5*17^{2}=1589.5ftlb[/tex]
Two particles, one with charge -6.29 × 10^-6 C and one with charge 5.23 × 10^-6 C, are 0.0359 meters apart. What is the magnitude of the force that one particle exerts on the other?
Answer:
Force, F = −229.72 N
Explanation:
Given that,
First charge particle, [tex]q_1=-6.29\times 10^{-6}\ C[/tex]
Second charged particle, [tex]q_2=5.23\times 10^{-6}\ C[/tex]
Distance between charges, d = 0.0359 m
The electric force between the two charged particles is given by :
[tex]F=k\dfrac{q_1q_2}{d^2}[/tex]
[tex]F=9\times 10^9\times \dfrac{-6.29\times 10^{-6}\times 5.23\times 10^{-6}}{(0.0359)^2}[/tex]
F = −229.72 N
So, the magnitude of force that one particle exerts on the other is 229.72 N. Hence, this is the required solution.
A red ball is thrown down with an initial speed of 1.1 m/s from a height of 28 meters above the ground. Then, 0.5 seconds after the red ball is thrown, a blue ball is thrown upward with an initial speed of 24.4 m/s, from a height of 0.9 meters above the ground. The force of gravity due to the earth results in the balls each having a constant downward acceleration of 9.81 m/s^2. How long after the red ball is thrown are the two balls in the air at the same height?
Answer:0.931 s
Explanation:
Given
initial speed=1.1 m/s
height(h)=28 m
after 0.5 sec blue ball is thrown upward
Velocity of blue ball is 24.4 m/s
height with which blue ball is launched is 0.9 m
Total distance between two balls is 28-0.9=27.1 m
Let in t time red ball travels a distance of x m
[tex]x=1.1t+\frac{gt^2}{2}[/tex] --------1
for blue ball
[tex]27.1-x=24.4t-\frac{g(t-0.5)^2}{2}[/tex] -----2
Add 1 & 2
we get
[tex]27.1=24.4t+1.1t+\frac{g(2t-0.5)(0.5)}{2}[/tex]
[tex]27.1=25.5t+g\frac{4t-1}{8}[/tex]
t=0.931 s
after 0.931 sec two ball will be at same height
What is the magnetic field at the center of a circular loop
ofwire of radius 4.0cm when a current of 2.0A flows in
thewire?
Answer:
The magnetic field at the center of a circular loop is [tex]3.14\times10^{-5}\ T[/tex].
Explanation:
Given that,
Radius = 4.0 cm
Current = 2.0 A
We need to calculate the magnetic field at the center of a circular loop
Using formula of magnetic field
[tex]B = \dfrac{I\mu_{0}}{2r}[/tex]
Where, I = current
r = radius
Put the value into the formula
[tex]B =\dfrac{2.0\times4\pi\times10^{-7}}{2\times4.0\times10^{-2}}[/tex]
[tex]B =0.00003141\ T[/tex]
[tex]B=3.14\times10^{-5}\ T[/tex]
Hence, The magnetic field at the center of a circular loop is [tex]3.14\times10^{-5}\ T[/tex].
On your wedding day, you leave for the church 25 minutes before the ceremony is to begin. The church is 8 miles away. On the way, you have to make an unanticipated stop for construction work. As a result, your average speed for the first 15 minutes is only 7 miles per hour. What average speed in miles per hour do you need for the rest of the trip to get to the church in time
Answer:
V=37.5miles/h
Explanation:
For convenience, let's convert the average speed to miles per minute:
V=7miles/h * 1h/60min = 0.1167 miles/min.
The distance traveled during the first 15 min was:
D = V*t = 1.75 miles So, the remaining distance is 6.25miles.
Since you only have 10min left:
Vr = Dr / tr = 6.25 / 10 = 0.625 miles / min. If we take that to miles per hour we get final answer:
Vr = 0.625 miles/min * 60min/1h = 37.5 miles/h
A type of transmission line for electromagnetic waves consists of two parallel conducting plates (assumed infinite in width) separated by a distance a. Each plate carries the same uniform surface current density of 16.0 A/m, but the currents run in opposite directions. What is the magnitude of the magnetic field between the plates at a point 1.00 mm from one of the plates if a = 0.800 cm? (μ0 = 4π × 10-7 T · m/A)
Answer:
[tex]B=2.01 \times 10^{-5}\ T[/tex]
Explanation:
Distance between plates = 0.8 cm
Distance from one plate = 1 mm
Current density (J)= 16 A/m
Currents are flowing in opposite direction.
[tex]\mu _o=4\pi \times 10^{-7}[/tex]
When current is flowing in opposite direction then magnetic field given as
[tex]B=\dfrac{\mu _oJ}{2}+\dfrac{\mu _oJ}{2}[/tex]
[tex]B=\mu _oJ[/tex]
Now by putting the values we get
[tex]B=4\pi \times 10^{-7}\times 16[/tex]
[tex]B=2.01 \times 10^{-5}\ T[/tex]
The magnitude of the magnetic field between the plates at the given point is 2.011 x 10⁻⁵ T.
What is magnetic field?The magnitude of magnetic field between the plates due to the current flowing in opposite directions is determined by using the following formula;
B = μ₀J/2 + μ₀J/2
B = μ₀J
where;
μ₀ is permeability of free space = 4π x 10⁻⁷ T.m/AJ is current densitySubstitute the given parameters and solve for the magnetic field as follows;
B = (4π x 10⁻⁷) x (16)
B = 2.011 x 10⁻⁵ T
Thus, the magnitude of the magnetic field between the plates at the given point is 2.011 x 10⁻⁵ T.
Learn more about magnetic field here: https://brainly.com/question/7802337
Serving at a speed of 164 km/h, a tennis player hits the ball at a height of 2.23 m and an angle θ below the horizontal. The service line is 11.6 m from the net, which is 0.99 m high. What is the angle θ in degrees such that the ball just crosses the net? Give a positive value for the angle.
Answer:
The angle θ is 6.1° below the horizontal.
Explanation:
Please, see the figure for a description of the situation.
The vector "r" gives the position of the ball and can be expressed as the sum of the vectors rx + ry (see figure).
We know the magnitude of these vectors:
magnitude rx = 11.6 m
magnitude ry = 2.23 m - 0.99 m = 1.24 m
Then:
rx = (11. 6 m, 0)
ry = (0, -1.24 m)
r = (11.6 m + 0 m, 0 m - 1.24 m) = (11.6 m, -1.24 m)
Using trigonometry of right triangles:
magnitude rx = r * cos θ = 11. 6 m
magnitude ry = r * sin θ = 2.23 m - 0.99 = 1.24 m
where r is the magnitude of the vector r
magnitude of vector r:
[tex]r = \sqrt{(11.6m)^{2} + (1.24m)^{2}} = 11.667m[/tex]
Then:
cos θ = 11.6 m / 11.667 m
θ = 6.1°
Using ry, we should obtain the same value of θ:
sin θ = 1.24 m/ 11.667 m
θ = 6.1°
( the exact value is obtained if we do not round the module of r)
A driver increases his velocity from 20 km/hr to 100 km/hr. BY what factor does he increase the kinetic energy of the car with this increase in speed? Kinetic energy is 4 times greater
Kinetic energy is 16 times greater
Kinetic energy is 25 times greater
Kinetic energy is 9 times greater
Kinetic energy is 2 times greater
Answer:
25 times greater
Explanation:
Let the mass of the car is m
Initial speed, u = 20 km/h = 5.56 m/s
Final speed, v = 100 km/h = 27.78 m/s
The formula for the kinetic energy is given by
[tex]K = \frac{1}{2}mv^{2}[/tex]
So, initial kinetic energy
[tex]K_{i} = \frac{1}{2}m(5.56)^{2}[/tex]
Ki = 15.466 m
final kinetic energy
[tex]K_{f} = \frac{1}{2}m(27.78)^{2}[/tex]
Kf = 385.86 m
Increase in kinetic energy is given by
= [tex]\left ( \frac{K_{f}}{K_{i}} \right )[/tex]
= 385.86 / 15.466 = 25
So, the kinetic energy is 25 times greater.
The kinetic energy of the car increases by a factor of 25.
Explanation:The increase in kinetic energy of the car can be determined by comparing the initial kinetic energy to the final kinetic energy. Kinetic energy is directly proportional to the square of velocity.
In this case, the velocity is increased from 20 km/hr to 100 km/hr. Let's calculate the ratio of the final kinetic energy to the initial kinetic energy.
The initial kinetic energy is given by 1/2 * (mass of the car) * (initial velocity)^2, and the final kinetic energy is given by 1/2 * (mass of the car) * (final velocity)^2.
Let's substitute the values and calculate the ratio:
Ratio = (1/2 * (mass) * (final velocity)^2) / (1/2 * (mass) * (initial velocity)^2) = (final velocity)^2 / (initial velocity)^2.
Substituting the numbers, Ratio = (100 km/hr)^2 / (20 km/hr)^2 = 10000 / 400 = 25.
Therefore, the factor by which the kinetic energy of the car increases is 25 times greater.
David is driving a steady 30 m/s when he passes Tina, who is sitting in her car at rest. Tina begins to accelerate at a steady 2.0 m/s2 at the instant when David passes. How far does Tina drive before passing David
Answer:
The distance traveled by Tina before passing David is 900 m
Given:
Initial speed of David, [tex]u_{D} = 30 m/s[/tex]
Acceleration of Tina, [tex]a_{T} = 2.0 m/s^{2}[/tex]
Solution:
Now, as per the question, we use 2nd eqn of motion for the position of David after time t:
[tex]s = u_{D}t + \frac{1}{2}at^{2}[/tex]
where
s = distance covered by David after time 't'
a = acceleration of David = 0
Thus
[tex]s = 30t[/tex]
Now, Tina's position, s' after time 't':
[tex]s' = u_{T}t + \frac{1}{2}a_{T}t^{2}[/tex]
where
[tex]u_{T} = 0[/tex], initially at rest
[tex]s' = 0.t + \frac{1}{2}\times 2t^{2}[/tex]
[tex]s' = t^{2}[/tex] (1)
At the instant, when Tina passes David, their distances are same, thus:
s = s'
[tex]30t = t^{2}[/tex]
[tex]t(t - 30) = 0[/tex]
t = 30 s
Now,
The distance covered by Tina before she passes David can be calculated by substituting the value t = 30 s in eqn (1):
[tex]s' = 30^{2}[/tex] = 900 m
The distance covered by Tina before passing David at an acceleration rate of 2 m/s² is 900 meters.
Given to us
Velocity of David, v = 30 m/s
Acceleration of Tina, a = 2 m/s²
Let the time taken by Tina pass David is t.
What is the Distance traveled by David before Tina pass him?According to the given information, the distance traveled by Tina will be the same as the distance traveled by David between Tina when she was at rest and when Tina passes her.
Distance traveled by Tina = Distance traveled by David
Distance traveled by David,
[tex]s = v \times t\\\\ = 30 \times t =30t[/tex]
What is the time taken by Tina to pass David?
Using the second equation of Motion
[tex]s= ut +\dfrac{1}{2}at^2[/tex]
Substitute,
[tex]30t= (0)t +\dfrac{1}{2}(2)t^2[/tex]
[tex]t = 30\rm\ sec[/tex]
Thus, the time taken by Tina to pass David is 30 seconds.
How far does Tina drive before passing David?We have already discussed,
Distance traveled by Tina = Distance traveled by David,
therefore,
[tex]s = v \times t\\\\ = 30 \times 30 =900\rm\ meters[/tex]
Hence, the distance covered by Tina before passing David at an acceleration rate of 2 m/s² is 900 meters.
Learn more about the Equation of motion:
https://brainly.com/question/5955789
A football quarterback runs 15.0 m straight down the playing field in 3.00 s. He is then hit and pushed 3.00 m straight backward in 1.74 s. He breaks the tackle and runs straight forward another 29.0 m in 5.20 s. (a) Calculate his average velocity (in m/s) for each of the three intervals. (Assume the quarterback's initial direction is positive. Indicate the direction with the sign of your answer.)
v1= m/s, v2= m/s, v3= m/s
(b) Calculate his average velocity (in m/s) for the entire motion. (Assume the quarterback's initial direction is positive. Indicate the direction with the sign of your answer.)
m/s
Answer:
a) [tex]v_{1}=14.29m/s\\v_{2}=9.25m/s\\v_{3}=6.36m/s[/tex]
b) [tex]v=+9.97m/s[/tex]
Explanation:
From the exercise we know that
[tex]x_{1} =15m, t_{1}=3s[/tex]
[tex]x_{2} =-3m, t_{1}=1.74s[/tex]
[tex]x_{3} =29m, t_{3}=5.20s[/tex]
From dynamics we know that the formula for average velocity is:
[tex]v=\frac{x_{2}-x_{1} }{t_{2}-x_{1} }[/tex]
a) For the three intervals:
[tex]v_{1}=\frac{x_{2}-x_{1} }{t_{2}-t_{1} }=\frac{(-3-15)m}{(1.74-3)s}=14.29m/s[/tex]
[tex]v_{2}=\frac{x_{3}-x_{2} }{t_{3}-t_{2} }=\frac{(29-(-3))m}{(5.20-1.74)s}=9.25m/s[/tex]
[tex]v_{3}=\frac{x_{3}-x_{1} }{t_{3}-t_{1} }=\frac{(29-15)m}{(5.20-3)s}=6.36m/s[/tex]
b) The average velocity for the entire motion can be calculate by the following formula:
[tex]v=\frac{v_{1}+v_{2}+v_{3} }{n} =\frac{(14.29+9.25+6.36)m/s}{3}=+9.97m/s[/tex]
A flat uniform circular disk (radius = 2.00 m, mass= 100
kg) is initially stationary. The disk is free to rotate inthe
horizontal plane about a frictionless axis perpendicular to
thecenter of the disk. A 40.0-kg person, standing 1.25 m from
theaxis, begins to run on the disk in a circular path and has
atangential speed of 2.00 m/s relative to the ground. Find
theresulting angular speed (in rad/s) of the disk.
Answer:
0.5 rad / s
Explanation:
Moment of inertia of the disk I₁ = 1/2 MR²
M is mass of the disc and R is radius
Putting the values in the formula
Moment of inertia of the disc I₁ = 1/2 x 100 x 2 x 2
= 200 kgm²
Moment of inertia of man about the axis of rotation of disc
mass x( distance from axis )²
I₂ = 40 x 1.25²
= 62.5 kgm²
Let ω₁ and ω₂ be the angular speed of disc and man about the axis
ω₂ = tangential speed / radius of circular path
= 2 /1.25 rad / s
= 1.6 rad /s
ω₁ = ?
Applying conservation of angular moment ( no external torque is acting on the disc )
I₁ω₁ = I₂ω₂
200 X ω₁ = 62.5 X 1.6
ω₁ = 0.5 rad / s
The force of magnitude F acts along the edge of the triangular plate. Determine the moment of F about point O. Find the general result and then evaluate your answer if F = 260 N, b = 580 mm, and h = 370 mm. The moment is positive if counterclockwise, negative if clockwise.
Answer:
The moment is 81.102 k N-m in clockwise.
Explanation:
Given that,
Force = 260 N
Side = 580 mm
Distance h = 370 mm
According to figure,
Position of each point
[tex]O=(0,0)[/tex]
[tex]A=(0,-b)[/tex]
[tex]B=(h,0)[/tex]
We need to calculate the position vector of AB
[tex]\bar{AB}=(h-0)i+(0-(-b))j[/tex]
[tex]\bar{AB}=hi+bj[/tex]
We need to calculate the unit vector along AB
[tex]u_{AB}=\dfrac{\bar{AB}}{|\bar{AB}|}[/tex]
[tex]u_{AB}=\dfrac{h\hat{i}+b\hat{j}}{\sqrt{h^2+b^2}}[/tex]
We need to calculate the force acting along the edge
[tex]\hat{F}=F(u_{AB})[/tex]
[tex]\hat{F}=F(\dfrac{h\hat{i}+b\hat{j}}{\sqrt{h^2+b^2}})[/tex]
We need to calculate the net moment
[tex]\hat{M}=\hat{F}\times OA[/tex]
Put the value into the formula
[tex]\hat{M}=F(\dfrac{h\hat{i}+b\hat{j}}{\sqrt{h^2+b^2}})\times(-b\hat{j})[/tex]
[tex]\hat{M}=\dfrac{F}{\sqrt{h^2+b^2}}((h\hat{i}+b\hat{j})\times(-b\hat{j}))[/tex]
[tex]\hat{M}=\dfrac{F}{\sqrt{h^2+b^2}}(-bh\hat{k})[/tex]
[tex]\hat{M}=-\dfrac{bhF}{\sqrt{h^2+b^2}}[/tex]
Put the value into the formula
[tex]\hat{M}=-\dfrac{580\times10^{-3}\times370\times10^{-3}\times260}{\sqrt{(370\times10^{-3})^2+(580\times10^{-3})^2}}[/tex]
[tex]\hat{M}=-81.102\ \hat{k}\ N-m[/tex]
Negative sign shows the moment is in clockwise.
Hence, The moment is 81.102 k N-m in clockwise.
The moment will be 81.102 k N-m in a clockwise direction. The moment is used to rotate or twist the object.
What is a moment?The moment is defined as the product of the force and the perpendicular distance from the pivot point. Its unit is KN-m.
The given data in the problem is;
F is the Force = 260 N
b is the Side = 580 mm
h is the distance = 370 mm
Position of the points is found by;
O(0,0)
A(0,-b)
B(h,0)
The position vector for the AB will be;
[tex]\vec AB = (h-0)+ (0-(-b))j \\\\ \vec AB =h \vec i + b \vec j[/tex]
The unit vector along with AB
[tex]\rm u_AB = \frac{\vec AB }{|\vec AB|} \\\\ \rm u_AB = \frac{h \vec i + b \vec j}{\sqrt{h^2+b^2} }[/tex]
The net moment is found by;
[tex]\hat M = \hat F \times OA \\\\ \hat M =F \frac{h \vec i + b \vec j}{\sqrt{h^2+b^2} }\times (b \vec j) \\\\ \hat M =\frac{F}{\sqrt{h^2+b^2}} \times (bh \hat k) \\\\ \hat M =- \ \frac{bhf}{\sqrt{h^2+b^2}}[/tex]
[tex]\hat M =- \ \frac{bhf}{\sqrt{h^2+b^2}} \\\\ \hat M =- \ \frac{580 \times 10^-3 \times 370 \times 10^-3 \times 260 }{\sqrt{(370\times 10^-3)^2+(580\times 10^-3)^2}} \\\\ \hat M =- 81.102 \ KNm[/tex]
-ve sign shows that moment is clockwise.
Hence the moment will be 81.102 k N-m in a clockwise direction.
To learn more about the moment refer to the link;
https://brainly.com/question/6278006
A loop of wire with cross-sectional area 1×10^−3 m^2 lays centered in the xy -plane. The wire carries a uniform current of 180A running counter-clockwise. What is the magnitude of the magnetic moment of the current loop?
Answer:
[tex]\mu=180\times 10^{-3}A-m^2[/tex]
Explanation:
Given that,
Area of the loop, [tex]A=10^{-3}\ m^2[/tex]
Current flowing in the wire, I = 180 A
We need to find the magnetic moment of the current loop. It is given by :
[tex]\mu=I\times A[/tex]
[tex]\mu=180\times 10^{-3}[/tex]
[tex]\mu=180\times 10^{-3}A-m^2[/tex]
So, the magnetic moment of the current loop is [tex]180\times 10^{-3}A-m^2[/tex]. Hence, this is the required solution.
Light with a wavelength of 494 nm in vacuo travels from vacuum to water. Find the wavelength of the light inside the water in nm (10-9 m). Note that wavelength and speed change when light transfers between media. Frequency does not change. Assume the index of refraction of water is 1.333.
Answer:
370.6 nm
Explanation:
wavelength in vacuum = 494 nm
refractive index of water with respect to air = 1.333
Let the wavelength of light in water is λ.
The frequency of the light remains same but the speed and the wavelength is changed as the light passes from one medium to another.
By using the definition of refractive index
[tex]n = \frac{wavelength in air}{wavelength in water}[/tex]
where, n be the refractive index of water with respect to air
By substituting the values, we get
[tex]1.333 = \frac{494}{\lambda }[/tex]
λ = 370.6 nm
Thus, the wavelength of light in water is 370.6 nm.
What magnitude charge creates a 1.70 N/C electric field at a point 4.60 m away? Express your answer with the appropriate units.
Answer:
[tex]Q=4.0*10^{-9}C[/tex]
Explanation:
Electric field of a charge:
[tex]E=k*\frac{Q}{R^{2}}[/tex]
[tex]Q=\frac{E*R^{2}}{K}=1.7*4.6^{2}/(9*10^{9})=4.0*10^{-9}C[/tex]
If the electric potential is zero at a particular point, must the electric field be zero at the point? Explain
Answer:
If the potential is zero , the electric field could be different to zero
Explanation:
The relation between the electric field and the potential is:
=−∇
∇: gradient operator
If the electric potential, , is zero at one point but changes in the neighbourhood of this point, then the Electric field, , at that point is different from zero.
If the stopping potential of a metal when illuminated with a radiation of wavelength 480 nm is 1.2 V, find (a) the work function of the metal, (b) the cutoff wavelength of the metal, and (c) the maximum energy of the ejected electrons
Answer:
Part a)
[tex]W = 1.38 eV[/tex]
Part b)
[tex]\lambda = 901.22 nm[/tex]
Part c)
[tex]KE = 1.2 eV[/tex]
Explanation:
As we know by Einstein's equation of energy that
incident energy of photons = work function of metal + kinetic energy of electrons
here we know that incident energy of photons is given as
[tex]E = \frac{hc}{\lambda}[/tex]
[tex]E = \frac{(6.6 \times 10^{-34})(3 \times 10^8)}{480 \times 10^{-9}}[/tex]
now we have
[tex]E = 4.125 \times 10^{-19} J[/tex]
[tex]E = 2.58 eV[/tex]
kinetic energy of ejected electrons = qV
so we have
[tex]KE = e(1.2 V) = 1.2 eV[/tex]
Part a)
now we have
[tex]E = KE + W[/tex]
[tex]2.58 = 1.2 + W[/tex]
[tex]W = 1.38 eV[/tex]
Part b)
in order to find cut off wavelength we know that
[tex]W = \frac{hc}{\lambda}[/tex]
[tex]1.38 eV = \frac{1242 eV-nm}{\lambda}[/tex]
[tex]\lambda = 901.22 nm[/tex]
Part c)
Maximum energy of ejected electrons is the kinetic energy that we are getting
the kinetic energy of electrons will be obtained from stopping potential
so it is given as
[tex]KE = 1.2 eV[/tex]
While skydiving, your parachute opens and you slow from 50.0 m/s to 8.0 m/s in 0.75 s . Determine the distance you fall while the parachute is opening.
Answer:
21.75 m
Explanation:
t = Time taken for the car to slow down = 0.75 s
u = Initial velocity = 50 m/s
v = Final velocity = 8 m/s
s = Displacement
a = Acceleration
Equation of motion
[tex]v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow a=\frac{8-50}{0.75}\\\Rightarrow a=-56\ m/s^2[/tex]
Acceleration is -56 m/s²
[tex]v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{8^2-50^2}{2\times -56}\\\Rightarrow s=21.75\ m[/tex]
The distance covered in the 0.75 seconds is 21.75 m
Estimate the mass of blood in your body Explain your reasoning (Note: It is not enough to provide a numeric answer. The main point of this problem is to assess your reasoning ability)
Answer: A little more that 5 Kg for a healthy person
Explanation: First, we know the following:
The regular adult has from 9 to 12 pints of blood. This is around 5 liters for a healthy male adult.
The human body is composed mostly on water, around 80%.
Blood is mostly composed on plasma, which makes blood thicker than water.
Knowing that, almost all the body is compose of water, it is safe to think that blood density should be near to that of water but higher.
The density on water is a know value. Which makes the following true:
1 Liter of Water weights 1 Kg
It could be said then, that the total mass of blood for a healthy person should be a little more that 5 kgs.
Find the volume of a sphere of radius 10 mm.
Answer:
Explanation: This is done using the equation:
[tex]\frac{4}{3} π R^{3}[/tex]
Because the Radius is a know value. We have the following.
[tex]\frac{4}{3} π (10mm)^{3}[/tex]
Which is:
4188.7902 mm
A ship leaves the island of Guam and sails a distance 255 km at an angle 49.0 o north of west. Part A: In which direction must it now head so that its resultant displacement will be 125 km directly east of Guam? (Express your answer as an angle measured south of east) Part B: How far must it sail so that its resultant displacement will be 125 km directly east of Guam?
Answer:
Explanation:
We shall represent displacement in vector form .Consider east as x axes and north as Y axes west as - ve x axes and south as - ve Y axes . 255 km can be represented by the following vector
D₁ = - 255 cos 49 i + 255 sin49 j
= - 167.29 i + 192.45 j
Let D₂ be the further displacement which lands him 125 km east . So the resultant displacement is
D = 125 i
So
D₁ + D₂ = D
- 167.29 i + 192.45 j + D₂ = 125 i
D₂ = 125 i + 167.29 i - 192.45 j
= 292.29 i - 192.45 j
Angle of D₂ with x axes θ
tan θ = -192.45 / 292.29
= - 0.658
θ = 33.33 south of east
Magnitude of D₂
D₂² = ( 192.45)² + ( 292.29)²
D₂ = 350 km approx
Tan
As part of calculations to solve an oblique plane triangle (ABC), the following data was available: b=50.071 horizontal distance, C=90.286° (decimal degrees), B=62.253° (decimal degrees). Calculate the distance of c to 3 decimal places (no alpha).
Answer:
The distance of c is 56.57
Explanation:
Given that,
Horizontal distance b = 50.071
Angle C = 90.286°
Angle B = 62.253°
We need to calculate the distance of c
Using sine rule
[tex]\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}[/tex]
[tex]\dfrac{b}{\sin B}=\dfrac{c}{\sin C}[/tex]
Put the value into the formula
[tex]\dfrac{50.071}{\sin62.253 }=\dfrac{c}{\sin90.286}[/tex]
[tex]c= \dfrac{50.071\times\sin90.286}{\sin62.253}[/tex]
[tex]c=56.575[/tex]
Hence, The distance of c is 56.575.