Answer:
5 blocks east. His total distance is 15 feet.
Explanation:
I hope that this helps! Have a good day!
While making some observations at the top of the 66 m tall Astronomy tower, Ron
accidently knocks a 0.5 kg stone over the edge. How long will a student at the bottom
have to get out of the way before being hit?
Analysing the question:
Since the stone was dropped, there was no initial velocity applied on it and hence it's initial velocity of the stone is 0 m/s
We are given:
height of the tower (h) = 66 m
mass of the stone (m) = 0.5 kg
initial velocity of the stone (u) = 0 m/s
time taken by the stone to reach the ground (t) = t seconds
acceleration due to gravity = 10 m/s²
** Neglecting air resistance**
Finding the time taken by the stone to reach the ground:
from the second equation of motion
h = ut + 1/2at²
replacing the variables
66 = (0)(t) + 1/2 (10)(t)²
66 = 5t²
t² = 13.2
t = 3.6 seconds
I initially wanted to subtract the height of the student from the height of the tower since the time i calculated is the time taken by the stone to reach the ground and that means that the stone has already hit the student before 3.6 seconds
but since we were NOT given the height of a student, the person who posed this question wants the time taken by the stone to reach the ground and that is what we solved
I am a cell. I am long and thin. I reach all the way from the brain
to the tip of a finger. I have a special coat of fat that helps me do
my job. My job is to send electrical signals from one part of the
body to another.
Answer:
Neurons
Explanation:
We humans have a nervous system that coordinates our behavior and transmits signals between different parts of our body.
Now, this nervous system contains a lot of nerve cells which we call Neurons. These Neurons have a cell like body and their job is to transmit signals from one part of our body to another.
Thus, the cell is called Neurons.
A microwave oven operates at 2.50 GHzGHz . What is the wavelength of the radiation produced by this appliance? Express the wavelength numerically in nanometers.
Answer:
The wavelength is [tex]\lambda = 1.2 * 10^8 nm[/tex]
Explanation:
From the question we are told that
The frequency of operation of the microwave is [tex]f = 2.50 GHz = 2.50 *10^{9} \ Hz[/tex]
Generally the wavelength is mathematically represented as
[tex]\lambda = \frac{c}{f}[/tex]
Here c is the speed of light with value [tex]c = 3.0 *10^{8} \ m/s[/tex]
So
[tex]\lambda = \frac{3.0 *10^{8}}{ 2.50 *10^{9}}[/tex]
=> [tex]\lambda = 0.12 \ m [/tex]
converting to nanometer
[tex]\lambda = 1.2 * 10^8 nm[/tex]
QUESTION 10
An archer fires an arrow towards a tree with initial speed 65 m/s and angle 25 degrees above the horizontal. If the arrow takes 0.85
seconds to hit the tree, calculate the horizontal distance between the archer and the tree.
QUESTION 11
A monkey throws a banana from a tree into a nearby river. The banana has initial speed 7.6 m/s, is angled 40 degrees below the
horizontal, and takes 0.75 seconds to land in the river. Calculate the speed of the banana when it hits the water.
Answer:
10) The distance between the archer and the tree is 50.074 meters.
11) The speed of the banana when it hits the water is approximately 13.554 meters per second.
Explanation:
10) The arrow experiments a parabolic motion, which is the combination of horizontal motion at constant velocity and vertical uniform accelerated motion. In this case we need to find the horizontal distance between the archer and the tree, calculated by the following kinematic equation:
[tex]x = x_{o} +v_{o}\cdot t \cdot \cos \theta[/tex] (Eq. 1)
Where:
[tex]x_{o}[/tex] - Initial position of the arrow, measured in meters.
[tex]x[/tex] - Final position of the arrow, measured in meters.
[tex]v_{o}[/tex] - Initial speed of the arrow, measured in meters per second.
[tex]t[/tex] - Time, measured in seconds.
[tex]\theta[/tex] - Launch angle, measured in sexagesimal degrees.
If we know that [tex]x_{o} = 0\,m[/tex], [tex]v_{o} = 65\,\frac{m}{s}[/tex], [tex]t = 0.85\,s[/tex] and [tex]\theta = 25^{\circ}[/tex], the horizontal distance between the archer and the tree is:
[tex]x = 0\,m + \left(65\,\frac{m}{s}\right)\cdot (0.85\,s)\cdot \cos 25^{\circ}[/tex]
[tex]x = 50.074\,m[/tex]
The distance between the archer and the tree is 50.074 meters.
11) The final speed of the banana ([tex]v[/tex]), measured in meters per second, just before hitting the water is determined by the Pythagorean Theorem:
[tex]v = \sqrt{v_{x}^{2}+v_{y}^{2}}[/tex] (Eq. 2)
Where:
[tex]v_{x}[/tex] - Horizontal speed of the banana, measured in meters per second.
[tex]v_{y}[/tex] - Vertical speed of the banana, measured in meters per second.
Each component of the speed are obtained by using these kinematic equations:
[tex]v_{x} = v_{o}\cdot \cos \theta[/tex] (Eq. 3)
[tex]v_{y} = v_{o}\cdot \sin \theta +g\cdot t[/tex] (Eq. 4)
Where [tex]g[/tex] is the gravitational acceleration, measured in meters per square second.
If we know that [tex]v_{o} = 7.6\,\frac{m}{s}[/tex], [tex]\theta = -40^{\circ}[/tex], [tex]g = -9.807\,\frac{m}{s^{2}}[/tex] and [tex]t = 0.75\,s[/tex], the components of final speed are, respectively:
[tex]v_{x} = \left(7.6\,\frac{m}{s} \right)\cdot \cos (-40^{\circ})[/tex]
[tex]v_{x} = 5.822\,\frac{m}{s}[/tex]
[tex]v_{y} = \left(7.6\,\frac{m}{s}\right)\cdot \sin (-40^{\circ})+\left(-9.807\,\frac{m}{s^{2}} \right) \cdot (0.75\,s)[/tex]
[tex]v_{y} = -12.240\,\frac{m}{s}[/tex]
And the speed of the banana right before hitting the water is:
[tex]v = \sqrt{\left(5.822\,\frac{m}{s} \right)^{2}+\left(-12.240\,\frac{m}{s} \right)^{2}}[/tex]
[tex]v \approx 13.554\,\frac{m}{s}[/tex]
The speed of the banana when it hits the water is approximately 13.554 meters per second.
Supply the missing force necessary to achieve equilibrium. Show your work.
Analysing the Question:
We know that equilibrium is the state of a body when it has equal and opposite forces being applied on it
In this case, a net downward force of 496N is being applied and a net upward force of (106 + 106 + 142 + x) N
Finding the missing force:
Since we have to achieve equilibrium, the net upward forces have to be equal to the net downward forces
So, (106 + 106 + 142 + x) = 496
354 + x = 496
x = 496 - 354
x = 142 N
Therefore, the missing force is 142 N
I WILL MARK YOU AS BRAINLIEST IF RIGHT
A 75 kg skier travels downhill 1200 meters in 56 seconds. What is the velocity of the skier?
Radio station KBOB broadcasts at a frequency of 85.7 MHz on your dial using radio waves that travel at 3.00 × 108 m/s. Since most of the station's audience is due south of the transmitter, the managers of KBOB don't want to waste any energy broadcasting to the east and west. They decide to build two towers, transmitting in phase at exactly the same frequency, aligned on an east-west axis. For engineering reasons, the two towers must be AT LEAST 10.0 m apart. What is the shortest distance between the towers that will eliminate all broadcast power to the east and west?
Answer:
12.5 m
Explanation:
The first thing we would do is to calculate the wavelength. To do this, we use the formula
v = fλ, where
v = wave speed
f = frequency
λ = wavelength
If we make wavelength the formula, we have
wavelength = speed / frequency
Now, we substitute the values we had been given and we have
wavelength = (3 * 10^8 m/s) / (85.7 * 10^6 Hz) wavelength = 3.50 m
half of this said wavelength will be
= 3.50 / 2
= 1.75 m
As a result of the engineering constraints with the towers being more than 10 m apart, the distance can't be 1.75 m and as such, it has to be a multiple of 1.75m. So we say,
(10 / 1.75) = 5.7
So the separation will have to be 7 half wavelengths
= (7 * 1.75) = 12.5 m
Consider a river flowing toward a lake at an average speed of 3 m/s at a rate of 550 m3/s at a location 58 m above the lake surface. Determine the total mechanical energy of the river water per unit mass (in kJ/kg) and the power generation potential of the entire river at that location (in MW). The density of water is 1000 kg/m3, and the acceleration due to gravity is 9.81 m/s2. The total mechanical energy of the river per unit mass is kJ/kg. The power generation potential of the entire river at that location is MW..
Answer:
1. 0.574 kJ/kg
2. 315.7 MW
Explanation:
1. The mechanical energy per unit mass of the river is given by:
[tex] E_{m} = E_{k} + E_{p} [/tex]
[tex] E_{m} = \frac{1}{2}v^{2} + gh [/tex]
Where:
Ek is the kinetic energy
Ep is the potential energy
v is the speed of the river = 3 m/s
g is the gravity = 9.81 m/s²
h is the height = 58 m
[tex] E_{m} = \frac{1}{2}(3 m/s)^{2} + 9.81 m/s^{2}*58 m = 0.574 kJ/Kg [/tex]
Hence, the total mechanical energy of the river is 0.574 kJ/kg.
2. The power generation potential on the river is:
[tex] P = m(t)E_{m} = \rho*V(t)*E_{m} = 1000 kg/m^{3}*550 m^{3}/s*0.574 kJ/kg = 315.7 MW [/tex]
Therefore, the power generation potential of the entire river is 315.7 MW.
I hope it helps you!
help pls i don’t know what to dooooooo
Explanation:
distance from ground
mass
amount of compression
Matter is anything that takes up space and has mass. O A. True O B. False
Answer:True
Explanation:Matter is everything around you. Atoms and compounds are all made of very small parts of matter. Those atoms go on to build the things you see and touch every day. Matter is defined as anything that has mass and takes up space (it has volume).
what is the meaning of relative as a noun?
Answer:
noun. a person who is connected with another or others by blood or marriage. something having, or standing in, some relation or connection to something else. something dependent upon external conditions for its specific nature, size, etc. (opposed to absolute).
A plane flying horizontally at a speed of 40.0 m/s and at an elevation of 160 m drops a package. Two seconds later it drops a second package. How far apart will the two packages land on the ground?
Answer:
Package 1 will land at 228.0 m, package 2 will land at 308.0 m, and the distance between them is 80.0 m.
Explanation:
To find the distance at which the first package will land we need to calculate the time:
[tex] Y_{f} = Y_{0} + V_{0y}t - \frac{1}{2}gt^{2} [/tex]
Where:
Y(f) is the final position = 0
Y(0) is the initial position = 160 m
V(0y) is initial speed in "y" direction = 0
g is the gravity = 9.81 m/s²
t is the time=?
[tex] 0 = 160 m + 0t - \frac{1}{2}9.81 m/s^{2}t^{2} [/tex]
[tex] t = \sqrt{\frac{2*160 m}{9.81 m/s^{2}}} = 5.7 s [/tex]
Now we can find the distance of the first package:
[tex] X_{1} = V_{0x}*t = 40.0 m/s*5.7 s = 228.0 m [/tex]
Then, after 2 seconds the distance traveled by plane is (from the initial position):
[tex] X_{p} = V_{0x}*t = 40.0 m/s*2 s = 80.0 m [/tex]
Now, the distance of the second package is:
[tex] X _{2} = X_{1} + X_{p} = 228.0 m + 80.0 m = 308.0 m [/tex]
The distance between the packages is:
[tex] X = X_{2} - X_{1} = 308.0 - 228.0 m = 80.0 m [/tex]
Therefore, package 1 will land at 228.0 m, package 2 will land at 308.0 m and the distance between them is 80.0 m.
I hope it helps you!
If we throw a body upwards at time t=0 with an initial speed s=25 m/s, what would the body's speed, position and velocity?
Answer:
The speed is v = 25 - 9.8 t
the velocity is v = (25 - 9.8 t) j^
the position is y = 25 t - 4.9 t²
Explanation:
This is a vertical throwing exercise.
The first thing we must do is set a coordinate system, in this case we will make the upward direction positive.
Let's write the kinematics equations
v = v₀ - g t
v² = v₀² - g (y-y₀)
y = y₀ + v₀ t - ½ g t²
in our case the initial velocity is
v₀ = 25 m / s
and we zero the system at the launch point
y₀ = 0
the equations remain
v = 25 - 9.8 t
v² = 25² - 9.8 y
y = 0 + 25 t - ½ 9.8 t²
The speed is v = 25 - 9.8 t
the velocity is v = (25 - 9.8 t) j^
the position is y = 25 t - 4.9 t²
A mass (m = 30 g) falls onto a spring (k = 7.3 N/m) from a height (h = 25 cm). The spring compresses an additional amount x before temporarily coming to a stop. What is the value of x?
Answer:
x₁ = 0.1878 m
Explanation:
For this exercise we will use conservation of energy
starting point. Highest point
Em₀ = U = m g h
final point. Lowest point with fully compressed spring
Em_f = K_e + U
Em_f = ½ K x² + m g x
energy is conserved
Em₀ = Em_f
m g h = ½ K x² + m g x
½ K x² + mg (x- h) = 0
let's substitute
½ 7.3 x² + 0.030 9.8 (x- 0.25) = 0
3.65 x² + 0.294 (x- 0.25) = 0
x² + 0.080548 (x- 0.25) = 0
x² - 0.020137 + 0.080548 x = 0
x² + 0.080548 x - 0.020137 = 0
let's solve the quadratic equation
x = [0.080548 ±√ (0.080548² + 4 0.020137)] / 2
x = [0.080548 ± 0.29502] / 2
x₁ = 0.1878 m
x₂ = -0.1072 m
These are the compression and extension displacement of the spring
A block initially at rest is allowed to slide down a frictionless ramp and attains a speed v at the bottom. To achieve a speed 2v at the bottom, how many times as high must a new ramp be?
Answer:
new height of ramp must be 4 times that of the first ramp.
Explanation:
From conservation of energy, we know that;
Potential energy at the top of ramp = kinetic energy at the bottom of ramp.
Thus;
mgh_t = ½mv²
m will cancel out to give;
gh_t = ½v²
Thus means that the height of the first ramp is directly proportional to the square of the speed.
Thus;
h_t ∝ v²
Now, for the new ramp, we are told that we want to achieve a speed of 2v at the bottom.
Thus;
h'_t ∝ (2v)²
h'_t ∝ 4v²
From earlier we saw that;
h_t ∝ v²
Thus;
New height of ramp is;
h'_t ∝ 4h_t
Thus, new height of ramp must be 4 times that of the first ramp.
To increase the speed to 2v at the bottom of the ramp, the height will be 4 times higher.
The given parameters;
velocity at the bottom ramp, = vlet the height of the ramp = hApply the principle of conservation of energy;
P.E = K.E
[tex]mgh = \frac{1}{2} mv^2\\\\gh = \frac{1}{2} v^2\\\\h = \frac{v^2}{2g}[/tex]
To increase the speed 2v at the bottom of the ramp, the height will be;
[tex]H = \frac{(2v)^2}{2g} \\\\H = \frac{4v^2}{2g} \\\\H = 4(\frac{v^2}{2g} )\\\\H = 4(h)[/tex]
Thus, to increase the speed to 2v at the bottom of the ramp, the height will be 4 times higher.
Learn more here:https://brainly.com/question/20626677
Suppose the angle of incidence of a light ray is 42°.What is the angle of reflection?
Answer:
angle of reflection will be also 42°Explanation:
we know that ------------- angle of incidence=angle of reflection41
Adam is pushing his box of baseball
equipment with a force of 10 N and
the box is pushing back towards
Adam with a force of 6 N. What is the
total net force? What will happen to the motion of
the box? Explain.
The Magnolia loh
deneaker notes
Answer:
16
Explanation:
6+10=16
the box will go forward but it will be a little harder.
Which state of matter is most similar to solids
Answer:
liquids
Explanation
The magnetic field at the center of a 0.60-cm-diameter loop is 2.7 mT . What is the current in the loop
Answer:
12.898A
Explanation:
The formula for calculating the magnetic field in the loop bus expressed as;
B = Iμ0/2r
Given
Diameter d = 0.60cm
Radius r = d/2 = 0.30cm
r = 0.0030m
Permittivity of free space μ0 = 4π×10^-7
Magnetic field strength B = 2.7×10^-3T
Substitute into the formula and get I
2.7×10^-3 = 4π×10^-7I/2(0.0030)
2.7×10^-3 = 4π×10^-7I/0.0060
Cross multiply
2.7×10^-3 × 6.0×10^-3 = 4π×10^-7I
16.2×10^-6 = 4π×10^-7I
I = 16.2×10^-6/4π×10^-7
I = 16.2×10^-6/4(3.14)×10^-7
I = 16.2×10^-6/12.56×10^-7
I = 1.2898×10^{-6+7}
I = 1.2898×10
I = 12.898A
Hence the current in the loop is 12.898A
A Navy Seal of mass 80 kg parachuted directly down into an enemy harbor. At one point while she was falling, the resistive force that air exerted on her was 520 N upward. What can you determine about her motion at this point in time
Answer:
The Navy Seal is accelerating downwards at the rate of 3.3 m/s²
Explanation:
Given;
mass of the Navy Seal, m = 80 kg
the upward resistive force on her, F = 520 N
Her net downward force is given by;
[tex]F_{net} = F_{down} - F_{up}\\\\F_{net} = (80*9.8) - 520\\\\F_{net} = 264 \ N[/tex]
Her downward acceleration at this time is given by;
F = ma
a = F / m
a = 264 / 80
a = 3.3 m/s²
Therefore, the Navy Seal is accelerating downwards at the rate of 3.3 m/s²
A teacher pushes on a file cabinet sitting on a
level floor. Which force must the teacher
overcome if they want to slide the cabinet across
the floor: Gravity, Normal, or Friction?
Marking brainliest
A projectile is launched at an angle of 15 degrees above the horizontal and lands down range. For the same speed, what projection angle would produce the greatest downrange distance?
Answer:
45°
Explanation:
Range is the horizontal distance travelled by an object undergoing projectile motion.
Range is given by the formula:
[tex]R=\frac{u^2sin(2\theta)}{g}[/tex]
where u = velocity, g = acceleration due to gravity and θ = angle above the horizontal.
For angle of 15°:
[tex]R=\frac{u^2sin(2*15)}{g}=\frac{0.5u^2}{g}[/tex]
We get a maximum range when sin(2θ) = 1
sin(2θ) = 1
2θ = sin⁻¹(1)
2θ = 90⁰
θ = 90°/2
θ = 45⁰
For angle of 45°:
[tex]R=\frac{u^2sin(2*45)}{g}=\frac{u^2}{g}[/tex]
Write a conclusion to this activity in which you completely and intelligently describe the characteristics of an object that is traveling in uniform circular motion. Give attention to the quantities speed, velocity, acceleration and net force.
Answer:
This question is incomplete
Explanation:
This question is incomplete but there are some characteristics that are peculiar to an object traveling in uniform circular motion.
An object moving in a uniform circular motion generally has a constant speed with which it uses to move in a circle (the object moves round/tangent to the circle). This then means the object will move in different direction although with the same speed. This change in direction means the object will accelerate (inwards) at different velocities (since velocity is a vector quantity that measures both magnitude and direction). Because the object moves with different velocities; this makes the object an accelerating object.
From the descriptions above, it can be conceived/visualized that the net force acting on an object in a uniform circular motion is a centripetal force. This is because the net force acting on the object is directed towards the center of the circle the object in rotating/moving in. Without this net force, the object would have moved in a straight line and thus not changing it's direction. The formula used to calculate this centripetal force is
Fc = mv²/r
where Fc is the centripetal force
m is the mass of the object
v is the velocity of the object
r is the radius of the curvature/curved path
The net force on an object moving in a circular path is directed inwards and it is known as centripetal force. The centripetal force increases with increase in speed and acceleration of the object.
The acceleration of an object travelling in a circular path is directed inwards and the magnitude of the acceleration is given as;
[tex]a_c = \frac{v^2}{r} = \omega^2 r[/tex]
where;
v is the linear speed of the objectω is the angular speed of the objectr is the radius of the circle[tex]a_c[/tex] is the centripetal acceleration of the objectThe net force acting on the object is given as follows;
[tex]F_c = ma_c[/tex]
The net force on the object is directed inwards and it is known as centripetal force.
Learn more about circular motion here: https://brainly.com/question/14672628
A particle moves along a path described by y=Ax^3 and x = Bt, where tt is time. What are the units of A and B?
Answer:
In a nutshell, units of A and B are [tex]\frac{1}{[l]^{2}}[/tex] and [tex]\frac{[l]}{[t]}[/tex], respectively.
Explanation:
From Dimensional Analysis we understand that [tex]x[/tex] and [tex]y[/tex] have length units ([tex][l][/tex]) and [tex]t[/tex] have time units ([tex][t][/tex]). Then, we get that:
[tex][l] = A\cdot [l]^{3}[/tex] (Eq. 1)
[tex][l] = B\cdot [t][/tex] (Eq. 2)
Now we finally clear each constant:
[tex]A = \frac{[l]}{[l]^{3}}[/tex]
[tex]A = \frac{1}{[l]^{2}}[/tex]
[tex]B = \frac{[l]}{[t]}[/tex]
In a nutshell, units of A and B are [tex]\frac{1}{[l]^{2}}[/tex] and [tex]\frac{[l]}{[t]}[/tex], respectively.
Consider the force field and circle defined below. F(x, y) = x2 i + xy j x2 + y2 = 121 (a) Find the work done by the force field on a particle that moves once around the circle oriented in the clockwise direction.
Answer: the work done by the force is 0
Explanation:
F (x², xy)
121 = 11²
so R = x² + y² = 11²
p = x². Q = xy
Δp/Δy = 0, ΔQ/Δx
using Green's theorem
woek = c_∫F.Δr = R_∫∫ ΔQ/Δx - Δp/Δy) ΔA
= (x² + y² = 121)_∫∫ yΔA
now let x = rcosФ, y = rsinФ
ΔA = rΔrΔФ
so r from 0 to 11
and Ф from 0 to 2π
= 0_∫^2π 0_∫^11 rsinФ × rΔrΔФ
= 0_∫^2π SinФΔФ 0_∫^11 r²Δr
= [ -cosФ]^2π_0 [r³/3]₀¹¹ = ( -cos2π + cos0) (11³/3) = 0
therefore the work done by the force is 0
If 1.8 1016 electrons enter a light bulb in 3 milliseconds, what is the magnitude of the electron current at that point in the circuit?
Answer:
I = 0.96 A
Explanation:
No of electrons, [tex]n=1.8\times 10^{16}[/tex]
Time, t = 3 ms = [tex]3\times 10^{-3}\ s[/tex]
We need to find the electric current. We know that electric charge per unit time is equal to the electric current.
[tex]I=\dfrac{q}{t}[/tex]
q = ne (Quantization of electric charge)
[tex]I=\dfrac{ne}{t}\\\\I=\dfrac{1.8\times 10^{16}\times 1.6\times 10^{-19}}{3\times 10^{-3}}\\\\I=0.96\ A[/tex]
So, the electric current is 0.96 A.
Question 1 of 15
All digits shown on the measuring device, plus one estimated digit, are
considered
Answer here
SUBMIT
Answer:
significant
Explanation:
The digits in a measurement that are considered significant are all of those digits that represent marked calibrations on the measuring device plus one additional digit to represent the estimated digit (tenths of the smallest calibration).
A coin rests on a record 0.15 m from its center. The record turns on a turntable that rotates at variable speed. The coefficient of static friction between the coin and the record is 0.30.
Required:
What is the maximum coin speed at which it does not slip?
Answer:
0.66m/sExplanation:
We are expected to solve for the velocity with no slip condition
we know that the expression that relate coefficient of friction and velocity is given as
μs = v^2/rg
Given
coefficient of friction μs = 0.3
radius r= 0.15
assume g=9.81m/s^2
substituting into the expression we have
0.3= v^2/0.15*9.81
v^2=0.3*0.15*9.81
v^2=0.44145
v=√0.44145
v=0.66
therefore the velocity is 0.66m/s
Magnetic attraction is one of the chemical properties of matter *
True
False
Average velocity of Mike Phelps swimming 100 m race in the 50 m long pool (2 laps) is approximately equal to *
A. 0 m/s
B. 1 m/s
C. 2 m/s
D. 4 m/s
Answer:
2
Explanation:
