Answer:
(a) The net force is 80.394 N
The acceleration of the crate is 0.804 m/s²
(b) the final velocity of the crate is 5.02 m/s
Explanation:
Given;
mass of the crate, m = 100 kg
applied force, F = 250 N
angle of inclination, θ = 45°
coefficient of friction, μ = 0.12
Applied force in y-direction, [tex]F_y = Fsin \theta = 250sin45 = 176.78 \ N[/tex]
Applied force in x-direction, [tex]F_x = Fcos \theta = 250cos45 = 176.78 \ N[/tex]
The normal force is calculated as;
N + Fy -W = 0
N = W - Fy
N = (100 x 9.8) - 176.78
N = 980 - 176.78 = 803.22 N
The frictional force is given by;
Fk = μN
Fk = 0.12 x 803.22
Fk = 96.386 N
(a) The net force is given by;
[tex]F_{net} = F_x - F_k\\\\F_{net} = 176.78-96.386\\\\F_{net} = 80.394 \ N[/tex]
Apply Newton's second law of motion;
F = ma
[tex]a = \frac{F_{net}}{m}\\\\ a = \frac{80.394}{100}\\\\ a = 0.804 \ m/s^2[/tex]
(b) the velocity of the crate after 5.0 s
[tex]F = ma= \frac{m(v-u)}{t} \\\\Ft =m(v-u)\\\\v-u = \frac{Ft}{m}\\\\ v = \frac{Ft}{m} + u\\\\v = \frac{F_{net}*t}{m} + u\\\\v = \frac{80.394*5}{100} + 1\\\\v = 5.02 \ m/s[/tex]
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?
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
Which is the simplest?
Organ
Organism
Tissue
Cell
Which is most complex?
Tissue
Organ
Organ system
Whole organism
Answer:
1. Cell2. OrganHopes it helps you....Thank you ☺️
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.
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
Watching the World Series (only as an example of physics in action), you wonder about the ability of the catcher to throw out a base runner trying to steal second. Suppose a catcher is crouched down behind the plate when he observes the runner breaking for second. After he gets the ball from the pitcher, he throws as hard as necessary to second base without standing up. If the catcher throws the ball at an angle of 30 degrees from the horizontal so that it is caught at second base at about the same height as that catcher threw it, how much time does it take for the ball to travel the 120 feet from the catcher to second base
Answer:
The time is [tex]t_t = 3.7583 \ s [/tex]
Explanation:
From the question we are told that
The angle is [tex]\theta = 30^o[/tex]
The horizontal distance is [tex]d = 120 \ ft[/tex]
Generally when the ball is at maximum height before descending the velocity is zero and this velocity can be mathematically represented as
[tex]v = u + at[/tex]
here a = -g the negative sign is because the direction of motion is against gravity
So
[tex]v = v_i + at[/tex]
Here[tex] v_i [/tex] is the vertical component of the initial velocity of the ball which is mathematically
represented as
[tex]v_i = usin(\theta )[/tex]
So
=> [tex]0 = usin(\theta ) -9.8t[/tex]
Generally the total time taken to travel the 120 ft is mathematically represented as
[tex]t_t = \frac{120}{v_h}[/tex]
Here [tex]v_h[/tex] is the horizontal component of the initial velocity which is mathematically represented as
[tex]v_h = u cos(\theta )[/tex]
So
[tex]t_t = \frac{120}{ u cos(\theta )}[/tex]
Generally the time taken to reach the maximum height is
[tex]t = \frac{t_t}{2}[/tex]
=> [tex]t = \frac{120}{ u cos(\theta )} * \frac{1}{2} [/tex]
=> [tex]t = \frac{60}{ u cos(\theta )} [/tex]
So
[tex]0 = usin(\theta ) -9.8* [\frac{60}{ u cos(\theta )}][/tex]
[tex] usin(\theta ) = 9.8* [\frac{60}{ u cos(\theta )}][/tex]
[tex] usin(\theta ) * u cos(\theta) =60* 9.8 [/tex]
[tex] u^2 sin(\theta ) cos(\theta) =60* 9.8 [/tex]
[tex] u^2 sin(30 ) cos(30) =60* 9.8 [/tex]
[tex] u^2 * \frac{1}{2}* \frac{\sqrt{3}}{2} =588.6 [/tex]
[tex] u^2 *\sqrt{3} =2354.4 [/tex]
[tex] u^2 = 1359.31 [/tex]
[tex] u = 36.87 \ ft/s [/tex]
Substituting this value into the equation for total time
[tex]t_t = \frac{120}{36.87 cos(30 )}[/tex]
[tex]t_t = 3.7583 \ s [/tex]
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
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.
500m=?m² va rooooooog
Answer:
250000 [m²]
Explanation:
A unit analysis has to be performed, where the unit of length is the meter. And the unit for the area is the square meter (m²)
L = 500 [m]
Therefore if we want to convert this length to meters we must square the length.
A = 500² = 500*500 = 250000 [m²]
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 2-m3 rigid insulated tank initially containing saturated water vapor at 1 MPa is connected through a valve to a supply line that carries steam at 400°C. Now the valve is opened, and steam is allowed to flow slowly into the tank until the pressure in the tank rises to 2 MPa. At this instant the tank temperature is measured to be 300°C. Determine the mass of the steam that has e
Answer:
5.6449
9 mpa
Explanation:
we are to determine mass of steam that has entered and also the pressure of steam.
After solving
Mass of steam = m2 - m1
= 15.925-10.2901
= 5.6449kg
Then the enthalpy of steam was calculated to be 3109.26
Using steam table, tl = 400⁰c
Hl = 3109.26
Supply line pressure = 9mpa
Please refer to attachment for all calculations
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
Listening to the radio, you can hear two stations at once. Describe this wave interaction.
A)
diffraction
B)
interference. C)
reflection
D)
refraction
Answer:
Answer is C or B
Explanation:
Consider a system of a cliff diver and the Earth. The gravitational potential energy of the system decreases by 28,000 J as the diver drops to the water from a height of 40.0 m. Determine her weight in newtons.
Answer:
700 N
Explanation:
The gravitational potential energy of the system decreases by 28,000 joules
The diver jumps from a height of 40 meters
Therefore the weight in Newton can be calculated as follows
= 28,000/40
= 700 newtons
_______ are pictures of relationships.
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.
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 reflectionA 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
. A cathode ray tube (CRT) is a device that produces a focused beam of electrons in a vacuum. The electrons strike a phosphor-coated glass screen at the end of the tube, which produces a bright spot of light. The position of the bright spot of light on the screen can be adjusted by deflecting the electrons with electrical fields, magnetic fields, or both. Although the CRT tube was once commonly found in televisions, computer displays, and oscilloscopes, newer appliances use a liquid crystal display (LCD) or plasma screen. You still may come across a CRT in your study of science. Consider a CRT with an electron beam average current of 25.00μA25.00μA . How many electrons strike the screen every minute?
Answer:
The value is [tex]n= 9.375 *10^{15} \ electrons [/tex]
Explanation:
From the question we are told that
The average current is [tex]I = 25.0 \mu A = 25.0 *10^{-6} \ A[/tex]
Generally the quantity of charge (electron ) is mathematically represented as
[tex]Q = ne[/tex]
Here e is the charge on a single electron with value [tex]e = 1.60 *10^{-19} \ C[/tex]
Generally current is mathematically represented as
[tex]I = \frac{Q}{t}[/tex]
=> [tex]I = \frac{ne}{t}[/tex]
Here t is time which is given as 1 minutes = 60 seconds
and n is the number of electrons
So
[tex]25.0 *10^{-6} = \frac{ n* 1.60 *10^{-19}}{60}[/tex]
=> [tex] 60 * 25.0 *10^{-6} = n* 1.60 *10^{-19} [/tex]
=> [tex]n= \frac{60 * 25.0 *10^{-6} }{ 1.60 *10^{-19} }[/tex]
=> [tex]n= 9.375 *10^{15} \ electrons [/tex]
The number of electrons that strike the screen every minute is; n = 9.375 × 10¹⁵ electrons
What is the number of electrons?
We are given;
Average Current; I = 25 μA = 25 × 10⁻⁶ A
Formula for the current is;
I = Q/t = ne/t
where;
n is number of electrons
e is electron charge = 1.6 * 10⁻¹⁹ C
t is time = 1 minute = 60 seconds
Thus making n the subject gives;
n = It/e
n = (25 × 10⁻⁶ * 60)/(1.6 * 10⁻¹⁹)
n = 9.375 × 10¹⁵ electrons
Read more about number of electrons at; https://brainly.com/question/11406294
1 Which of the following is an example of a physical change?
A. Water freezing into ice.
B. A piece of wood burning.
C. A toy car rusting.
D. Zinc producing hydrogen gas when mixed with water.
Answer:
it's A trust me ok
Explanation:
i can't explain
Magnetic attraction is one of the chemical properties of matter *
True
False
4. According to Newton’s law of cooling, if an object at temperature T is immersed in a medium having the constant temperature M, then the rate of change of T is proportional to the difference of temperature M − T. This gives the differential equation dT dt = k(M − T). (a) Solve the differential equation for T. (b) A thermometer reading 100◦F is placed in a medium having a constant temperature of 70◦F. After 6 min, the thermometer reads 80◦F. What is the reading after 20 min?
Answer:
a) The solution of the differential equation is [tex]T(t) = M + (T_{o}-M) \cdot e^{-\frac{t}{\tau} }[/tex].
b) The reading after 20 minutes is approximately 70.770 ºF.
Explanation:
a) Newton's law of cooling is represented by the following ordinary differential equation:
[tex]\frac{dT}{dt} = -\frac{T-M}{\tau}[/tex] (Eq. 1)
Where:
[tex]\frac{dT}{dt}[/tex] - Rate of change of temperature of the object in time, measured in Fahrenheit per minute.
[tex]\tau[/tex] - Time constant, measured in minutes.
[tex]T[/tex] - Temperature of the object, measured in Fahrenheit.
[tex]M[/tex] - Medium temperature, measured in Fahrenheit.
Now we proceed to solve the differential equation:
[tex]\frac{dT}{T-M} = -\frac{t}{\tau}[/tex]
[tex]\int {\frac{dT}{T-M} } = -\frac{1}{\tau} \int \, dt[/tex]
[tex]\ln (T-M) = -\frac{t}{\tau} + C[/tex]
[tex]T(t) -M = (T_{o}-M)\cdot e^{-\frac{t}{\tau} }[/tex]
[tex]T(t) = M + (T_{o}-M) \cdot e^{-\frac{t}{\tau} }[/tex] (Eq. 2)
Where:
[tex]t[/tex] -Time, measured in minutes.
[tex]T_{o}[/tex] - Initial temperature of the object, measured in Fahrenheit.
b) From (Eq. 2) we obtain the time constant of the cooling equation for the object: ([tex]M = 70\,^{\circ}F[/tex], [tex]T_{o} = 100\,^{\circ}F[/tex], [tex]t = 6\,min[/tex], [tex]T(t) = 80\,^{\circ}F[/tex])
[tex]80\,^{\circ}F = 70\,^{\circ}F + (100\,^{\circ}F-70\,^{\circ}F)\cdot e^{-\frac{6\,min}{\tau} }[/tex]
[tex]e^{-\frac{6\,min}{\tau} } = \frac{80\,^{\circ}F-70\,^{\circ}F}{100\,^{\circ}F-70\,^{\circ}F}[/tex]
[tex]e^{-\frac{6\,min}{\tau} } = \frac{1}{3}[/tex]
[tex]-\frac{6\,min}{\tau} = \ln \frac{1}{3}[/tex]
[tex]\tau = -\frac{6\,min}{\ln \frac{1}{3} }[/tex]
[tex]\tau = 5.461\,min[/tex]
The cooling equation of the object is [tex]T(t) = 70 +30\cdot e^{-\frac{t}{5.461} }[/tex] and the temperature of the object after 20 minutes is:
[tex]T(20) = 70+30\cdot e^{-\frac{20}{5.461} }[/tex]
[tex]T(20) \approx 70.770\,^{\circ}F[/tex]
The reading after 20 minutes is approximately 70.770 ºF.
A thin, horizontal copper rod is 1.43 m long and has a mass of 236 g. What is the minimum current in the rod that can cause it to float in a horizontal magnetic field of 0.546 T
Answer:
The minimum current in the rod that can cause it to float 2.96 A.
Explanation:
Given;
length of the copper rod, L = 1.43 m
mass of the rod, m = 236 g = 0.236 kg
magnetic field, B = 0.546 T
The magnetic of the rod is given by;
Fm = BIL
The downward force on the rod is given by;
F = mg
For the rod to float, the difference in the two force will be zero;
BIL - mg = 0
BIL = mg
I = mg / BL
I = (0.236 x 9.8) / (0.546 x 1.43)
I = 2.96 A
Therefore, the minimum current in the rod that can cause it to float 2.96 A.
plz help me it is improtant
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!
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).
Is it true or false that the displacement always equals the product of the average velocity and the time interval?
Answer:
True.
Explanation:
Applying the definition of average velocity, we know that we can always write the following expression:[tex]v_{avg} = \frac{\Delta x}{\Delta t}[/tex] (1)
By definition, Δx is just the displacement, and Δt is the time interval.So, just rearranging terms in (1), we get:[tex]\Delta x} = v_{avg}* {\Delta t}[/tex]
A pole-vaulter just clears the bar at 5.53 m and falls back to the ground. The change in the vaulter's potential energy during the fall is -3200 J. What is his weight?
Answer:
578.66 N
Explanation:
The first step is to calculate the mass
mgh= 3200J
3200/9.8×5.53
3200/54.194
m = 59.047 kg
Therefore the weight can be calculated as follows
Weight = m × g
= 59.047 × 9.8
= 578.66 N
Suppose a NASCAR race car rounds one end of the Martinsville Speedway. This end of the track is a turn with a radius of approximately 57.0 m57.0 m . If the track is completely flat and the race car is traveling at a constant 24.5 m/s24.5 m/s (about 55 mph55 mph ) around the turn, what is the race car's centripetal (radial) acceleration
Answer:
10.53m/s²
Explanation:
Centripetal acceleration is the acceleration of an object about a circle. The formula for calculating centripetal acceleration is expressed by:
[tex]a = \frac{v^2}{r}[/tex]
v is the velocity of the car = 24.5m/s
r is the radius of the track = 57.0m
Substitute the given values into the formula:
[tex]a = \frac{24.5^2}{57} \\\\a = \frac{600.25}{57}\\ \\a = 10.53m/s^{2}[/tex]
Hence the centripetal acceleration of the race car is 10.53m/s²
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!
