Answer:
The velocity of the skier at the bottom of the ramp is approximately 26.288 meters per second.
Explanation:
We can determine the final velocity of the skier at the bottom of the ramp by Principle of Energy Conservation and Work-Energy Theorem, whose model is:
[tex]U_{g,1}+K_{1} = U_{g,2}+K_{2}+W_{disp}[/tex] (1)
Where:
[tex]U_{g,1}[/tex], [tex]U_{g,2}[/tex] - Initial and final gravitational potential energy, measured in joules.
[tex]K_{1}[/tex], [tex]K_{2}[/tex] - Initial and final translational kinetic energy, measured in joules.
[tex]W_{disp}[/tex] - Work dissipated by friction, measured in joules.
By definitions of gravitational potential and translational kinetic energy and work, we expand and simplify the model:
[tex]m\cdot g \cdot (z_{1}-z_{2})+\frac{1}{2}\cdot m \cdot (v_{1}^{2}-v_{2}^{2}) =\mu_{k}\cdot N\cdot \Delta s[/tex] (2)
Where:
[tex]m[/tex] - Mass, measured in kilograms.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]z_{1}[/tex], [tex]z_{2}[/tex] - Initial and final heights of the skier, measured in meters.
[tex]N[/tex] - Normal force from the incline on the skier, measured in newtons.
[tex]\Delta s[/tex] - Distance covered by the skier, measured in meters.
[tex]\mu_{k}[/tex] - Kinetic coefficient of friction, dimensionless.
The normal force exerted on the skier and the covered distance are, respectively:
[tex]N = m\cdot g\cdot \cos \theta[/tex] (3)
[tex]\Delta s = \frac{z_{1}-z_{2}}{\sin \theta}[/tex] (4)
Where [tex]\theta[/tex] is the angle of the incline above the horizontal, measured in sexagesimal degrees.
By applying (3) and (4) in (2), we get that:
[tex]m\cdot g \cdot (z_{1}-z_{2})+\frac{1}{2}\cdot m\cdot (v_{1}^{2}-v_{2}^{2}) = \mu_{k}\cdot m\cdot g \cdot \cos \theta \cdot \left(\frac{z_{1}-z_{2}}{\sin \theta} \right)[/tex]
[tex]g\cdot (z_{1}-z_{2}) +\frac{1}{2}\cdot (v_{1}^{2}-v_{2}^{2})= \mu_{k}\cdot g \cdot \left(\frac{z_{1}-z_{2}}{\tan \theta} \right)[/tex] (5)
Then, we clear the velocity of the skier at the bottom of the ramp is: ([tex]v_{1} = 0\,\frac{m}{s}[/tex], [tex]\mu_{k} = 0.1[/tex], [tex]\theta = 40^{\circ}[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]z_{1}-z_{2} = 40\,m[/tex])
[tex]\left[\frac{\mu_{k}}{\tan \theta}-1 \right]\cdot g\cdot (z_{1}-z_{2}) = \frac{1}{2}\cdot (v_{1}^{2}-v_{2}^{2})[/tex]
[tex]2\cdot \left[\frac{\mu_{k}}{\tan \theta}-1 \right]\cdot g\cdot (z_{1}-z_{2}) = v_{1}^{2}-v_{2}^{2}[/tex]
[tex]v_{2} = \sqrt{v_{1}^{2}-2\cdot \left[\frac{\mu_{k}}{\tan \theta}-1 \right]\cdot g\cdot (z_{1}-z_{2})}[/tex] (6)
[tex]v_{2} = \sqrt{\left(0\,\frac{m}{s} \right)^{2}-2\cdot \left(\frac{0.1}{\tan 40^{\circ}} -1\right)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (40\,m)}[/tex]
[tex]v_{2} \approx 26.288\,\frac{m}{s}[/tex]
The velocity of the skier at the bottom of the ramp is approximately 26.288 meters per second.
A car is accelerated at a constant rate from 15 m/s to 25 m/s. It takes the car 6 s to reach its final speed. What is the car’s acceleration?
Answer:
1.67 m/s²Explanation:
The car’s acceleration can be found by using the formula
[tex]a = \frac{v - u}{t} \\ [/tex]
where
v is the final velocity
u is the initial velocity
t is the time taken
a is the acceleration
From the question we have
[tex]a = \frac{25 - 15}{6} = \frac{10}{6} = \frac{5}{3} \\ = 1.666666...[/tex]
We have the final answer as
1.67 m/s²Hope this helps you
A small block of mass m1 = 0.4 kg is placed on a long slab of mass m2 = 2.8 kg. Initially, the slab is stationary and the block moves at a speed of vo = 3 m/s. The coefficient of kinetic friction between the block and the slab is 0.15 and there is no friction between the slab and the surface on which it moves.
Determine the speed v1.
Determine the distance traveled by the slab before it reaches the speed v1.
Answer:
v₁ = 0.375 m / s , x = 0.335 m
Explanation:
Let's analyze this interesting exercise, the block moves and has a friction force with the tile, we assume that the speed of the block is constant, so the friction force opposes the block movement. For the only force that acts (action and reaction) this friction force exerted by the block that is in the direction of movement of the tile.
We can also see that the isolated system formed by the block and the tile will reach a stable speed where friction cannot give the system more energy, this speed can be found by treating the system with the conservation of linear momentum.
initial moment. Right at the start of the movement
p₀ = m v₀ + 0
final moment. Just when it comes to equilibrium
[tex]p_{f}[/tex] = (m + M) v₁
how the forces are internal
p₀ =p_{f}
m v₀ = (m + M) v₁
v₁ = m /m+M v₀
let's calculate
v₁ = 0.4 /(0.4 + 2.8) 3
v₁ = 0.375 m / s
Let's apply Newton's second law to the Block, to find the friction force
Y axis
N - W = 0
N = W
N = m g
where m is the mass of the block
the friction force has the formula
fr = μ N
fr = μ m g
We apply Newton's second law to slab
X axis
fr = M a
where M is the mass of the slab
μ m g = M a
a = μ g m / M
let's calculate
a = 0.15 9.8 0.4 / 2.8
a = 0.21 m / s²
With kinematics we can find the position
v²= v₀²+2 a x
as the slab is initially at rest, its initial velocity is zero
v² = 2 a x
x = v2 / 2a
let's calculate
x = 0.375²/2 0.21
x = 0.335 m
7. It is the art of drawing solid objects on two-dimensional surfaces.
Explanation:
Perspective- the art of drawing solid objects on a two-dimensional surface so as to give the right impression of their height, width, depth, and position in relation to each other when viewed from a particular point.
A projector lens projects an image from a 6.35 cm wide LCD screen onto a
screen 3.25 m wide. If the focal length of the projector lens is 13.8 cm, the screen
must be how far from the projector
Answer:
For any given projector, the width of the image (W) relative to the throw distance (D) is know as the throw ratio D/W or distance over width. So for example, the most common projector throw ratio is 2.0. This means that for each foot of image width, the projector needs to be 2 feet away or D/W = 2/1 = 2.0.
Is a parked car potential or kinetic ?
Answer:
Potential energy is the energy that is stored in an object. ... When you park your car at the top of a hill, your car has potential energy because the gravity is pulling your car to move downward; if your car's parking brake fails, your vehicle may roll down the hill because of the force of gravity.
A cheetah can maintain a maximum constant velocity of 34.2 m/s for 8.70 s. What is
the displacement the cheetah covered at that velocity?
Answer:
297.54mExplanation:
step one:
given data
velocity v=34.2m/s
time t= 8.7s
Step two
Required is the distance the cheetah has covered on the condition
we know that speed= distance/time
make distance subject of formula we have
distance= velocity *time
distance= 34.2*8.7
distance = 297.54m
Therefore the displacement the cheetah covered at that velocity
is 297.54m
You serve a tennis ball of mass 60g at a speed of 50
m/s, what is the impulse exerted on the ball? ( ball starts from rest )
Answer:
[tex]J = 3~Kg.m/s[/tex]
Explanation:
Impulse and Momentum
The impulse-momentum theorem states that the change in momentum of an object equals the impulse applied to it.
The equation can be written as follows:
[tex]J =\Delta p = p_2-p_1[/tex]
Where:
J = Impulse
p2 = Final Momentum
p2 = Initial Momentum
The momentum can be calculated as:
p = m.v
Where m is the mass of the object and v is the velocity.
The tennis ball with mass m=60 g = 0.06 Kg was served from rest (v1=0) to v2=50 m/s. The change in momentum is:
[tex]\Delta p = 0.06Kg~50~m/s-0[/tex]
[tex]\Delta p = 3~Kg.m/s[/tex]
Thus the impulse is:
[tex]\marhbf{J = 3~Kg.m/s}[/tex]
If you start at a speed of 4m/s and slow down to 2m/s in 4s what is your
acceleration?
Answer:
penis
Explanation:
student measures the weight of a bag of bananas with a spring balance.
Describe what is inside a spring balance and explain how it works.
A spring balance measures the weight of an object by opposing the force of gravity acting with force of an extending spring. May be used to determine mass as well as weight by recalibrating the scale. Some spring balances are available in gram or kilogram markings and are used to measure the mass of an object. Spring balances consist of a cylindrical tube with a spring inside. One end (at the top) is fixed to an adjuster which can be used to calibrate the device. The other end is attached to a hook on which you can hang masses etc.
A student kicks a soccer ball upward at a 30º angle with an initial speed of 20 m∕s. What expression should the student use to calculate the magnitude of the ball’s initial velocity in the horizontal direction?
Answer:
[tex]\displaystyle x=10\sqrt{3}\ m/s[/tex]
[tex]y=10\ m/s[/tex]
Explanation:
Rectangular coordinates of vectors in 2D
Given a vector with a magnitude v and angle θ with respect to the positive horizontal direction, the x and y components of the vector are given by:
[tex]x=v\cos\theta[/tex]
[tex]y=v\sin\theta[/tex]
The soccer ball is kicked upward at an angle θ = 30° and at a speed v=20 m/s.
The rectangular components of the vector are:
[tex]x=20\cos 30^\circ[/tex]
[tex]\displaystyle x=20\cdot \frac{\sqrt{3}}{2}[/tex]
Operating:
[tex]\mathbf{\displaystyle x=10\sqrt{3}\ m/s}[/tex]
[tex]y=20\sin 30^\circ[/tex]
[tex]\displaystyle y=20\cdot \frac{1}{2}[/tex]
Operating:
[tex]\mathbf{y=10\ m/s}[/tex]
Ball 1 (1.5 kg) moves to the right at 2 m/s and ball 2
(2.5 kg) moves to the left at 1.5 m/s. The balls stick together after collision. What is the speed and direction of ball 2 after the collision?
Answer:
0.1875 m/s leftward
Explanation:
Taking rightwards as positive
We are given:
Ball 1:
Mass (m1) = 1.5 kg
velocity (u1) = 2 m/s
Ball 2:
Mass (m2) = 2.5 kg
velocity (u2) = -1.5 m/s [negative because it is in the opposite direction]
Speed and Direction of Ball 2:
We are told that the balls stick together after the collision
We can say that the balls have the same velocity since they are sticking together
So, Final velocity of Ball 1 (v1) = Final velocity of Ball 2 (v2) = V m/s
According to the law of conservation of momentum
m1u1 + m2u2 = m1v1 + m2v2
replacing the variables
1.5(2) + (2.5)(-1.5) = V (1.5 + 2.5) [v1 = v2 = V]
3 + (-3.75) = 4V
-0.75 = 4V
V = -0.75/4 [dividing both sides by 4]
V = -0.1875 m/s
Hence, the balls will move at a velocity of 0.1875 m/s in the Leftward direction
A freshly caught catfish is placed on a spring scale, and it oscillates up and down with a period of 0.19 s. If the spring constant of the scale is 2330 N/m, what is the mass of the catfish?
Answer:
The mass of the catfish is 2.13 kg
Explanation:
Period of oscillation, T = 0.19 s
spring constant, k = 2330 N/m
The period of oscillation of the spring is given by;
[tex]T = 2\pi \sqrt{\frac{m}{k} }\\\\\frac{T}{2\pi} = \sqrt{\frac{m}{k} }\\\\\frac{T^2}{4\pi^2} = \frac{m}{k}\\\\m = \frac{kT^2}{4\pi^2}[/tex]
where;
m is mass of the catfish
substitute the given values and solve for m;
[tex]m = \frac{kT^2}{4\pi^2} \\\\m = \frac{(2330)(0.19)^2}{4\pi^2} \\\\m = 2.13 \ kg[/tex]
Therefore, the mass of the catfish is 2.13 kg
Aluminum wire with a diameter of 0.8650 mm is wound onto a spool. The wire is insulated, but you have access to both ends. The resistivity of aluminum at 20.0 °C is 2.65 x 10^-8 Ω-m. You measure the resistance of the wire at that temperature, and it is 2.48 Ω. What is the length of the wire?
a. 8.10 x 10^4 m
b. 22.0 m
c. 5.68 m
d. 0.111 m
e. 55.0 m
Answer:
e. 55.0 m
Explanation:
Given;
diameter of the aluminum wire, d = 0.865 mm
radius of the wire, r = d/2 = 0.4325 mm = 0.4325 x 10⁻³ m
resistivity of the wire, ρ = 2.65 x 10⁻⁸ Ω-m
resistance of the wire, R = 2.48 Ω
The resistance of a wire is given by;
[tex]R = \frac{\rho \ L}{A} \\\\[/tex]
where;
L is length of the wire
A is area of the wire = πr² = π(0.4325 x 10⁻³ )² = 5.877 x 10⁻⁷ m²
Substitute the givens and solve for L,
[tex]L = \frac{RA}{\rho} \\\\L = \frac{(2.48)(5.877*10^{-7})}{2.65*10^{-8}}\\\\L = 55.0 \ m[/tex]
Therefore, the length of the wire is 55.0 m
An explanation of the relationships among particular phenomena.
Answer:
Theory
Explanation:
Theory is a term that is used often in academic work or scientific research to explain certain things or conditions established on universal principles or laws.
It is used to describe the "why and how" or the reason behind the occurrence of a situation.
Hence, it is correct to conclude that THEORY is "an explanation of the relationships among particular phenomena."
Answer:
E) Theory
Explanation:
Edge 2020
Brainliest?
A circular conducting loop with a radius of 1.00 m and a small gap filled with a 10.0 Ω resistor is oriented in the xy-plane. If a magnetic field of 2.0 T, making an angle of 30º with the z-axis, increases to 11.0 T, in 2.5 s, what is the magnitude of the current that will be caused to flow in the conductor?
Answer:
ill get back to this question once i find the answer to it
What is the voltage drop across the 10.0 2 resistor?
10.00
120.0 V
20.00
30.00
Answer: 20.0V
Answer:
20.0V
SANA MAKATULONG
The voltage drop across the 10.0-ohm resistor would be 20.0 volt.
What is resistance?Resistance is the obstruction of electrons in an electrically conducting material.
The mathematical relation for resistance can be understood with the help of the empirical relation provided by Ohm's law.
V=IR
For calculating equivalent resistance in series combination.
Re = R1 + R2 + R3
For the given problem the total resistance of the circuit would be as all the three resistors are connected in the series combination.
Re= 10+20+30
Re=60 ohm
As given in the figure all the resistance are connected in the series combination therefore the current flowing through them would be the same.
For the given problem we have to design and construct a circuit that has two resistors connected in series.
By using Ohms law
V=IR
120 = 60×I
I = 2 ampere
Given that we have a 120 V battery, that will produce a current of 2 Ampere
By using Ohm's law we can calculate the voltage drop across a 10-ohm resistor
V=IR
=2×10
= 20 volt
Thus, the voltage drop across the 10.0-ohm resistor comes out to be 20.0 volt.
Learn more about resistance from here
brainly.com/question/14547003
#SPJ2
SI unit differ from one country to another . true or false
Answer:
false ..........false
Answer:
FALSE
Explanation:
A ball is thrown vertically upward with an initial velocity of 23 m/s. What are its position and velocity after 2 s?
Answer:
The position of the ball after 2 s is 26.4 mThe velocity of the ball after 2 s is 3.4 m/sExplanation:
Given;
initial velocity of the ball, u = 23 m/s
time of motion, t = 2 s
The position of the ball after 2 s is given by;
h = ut - ¹/₂gt²
h = (23 x 2) - ¹/₂ x 9.8 x 2²
h = 46 - 19.6
h = 26.4 m
The velocity of the ball after 2 s is given by;
v² = u² + 2(-g)h
v² = u² - 2gh
v² = 23² - (2 x 9.8 x 26.4)
v² = 529 - 517.44
v² = 11.56
v = √11.56
v = 3.4 m/s
A 1kg cannon ball.is fired horizontally with an initial velocity of 5m/s. If the cannon was atop a wall 20m above the ground, what is the total
change in KE?
Answer:
Ek = 196.2 [J]
Explanation:
The question concerns the KE kinetic energy.
That is, we must find the kinetic energy at the moment the cannon is fired and the kinetic energy of when the ball hits the ground after having fallen 20 meters.
At the moment when the ball is fired it is 20 meters above ground level. If the ground level is taken as the reference level of potential energy, where it is equal to zero, in this way when the ball is at the highest (20 meters) you have the maximum potential energy.
In this way, the energy in the initial state is equal to the sum of the kinetic energy plus the potential energy. As the energy is conserved this same energy will be present when the ball hits the ground, where the potential energy is zero and will have only kinetic energy.
[tex]E_{1}=E_{2}\\E_{k1}+E_{p1}=E_{k2}\\\frac{1}{2} *m*v^{2} +m*g*h=E_{k2}\\E_{k2}=0.5*1*(5)^{2} +1*9.81*20\\E_{k2}=208.7[J][/tex]
The kinetic energy in the initial state can be easily calculated by means of the following equation.
[tex]E_{k1}=\frac{1}{2} *m*v^{2}\\E_{k1}=0.5*1*(5)^{2}\\E_{k1}=12.5 [J][/tex]
Therefore the change in KE
[tex]E_{k} = 208.7 - 12.5\\E_{k} = 196.2 [J][/tex]
An object in FREE-FALL on the MOON would experience which of the following
FORCES?
O a. Weight
O b. Normal
O c. Air Resistance
d. a and c
O e. None of these
Answer:
e. none of these
Explanation:
An object in FREE-FALL on the MOON would experience only acceleration
A force of 150 N is applied on an object at 60 degrees above the positive x-axis. Determine its
horizontal and vertical components.
Answer:
horizontal component=fcostita
=150cos60
use calculator to evaluate it
for vertical=fsintita
=150sin60
If 10 calories of energy are added to 2 grams of ice at -30° C, calculate the final temperature of the ice. (Notice that the specific heat of ice is different from that of water.) Assume the specific heat of ice is 0.5
-30° C
40° C
-20° C
30° C
Answer:
-20°C
Explanation:
The specific heat capacity of ice using the cgs system is 0.5cal/g°C
The enthalpy change is calculated as follows
ΔH=MC∅ where M represents mass C represents specific heat and ∅ represents the temperature change.
10cal = 2g×0.5cal/g°C×∅
∅=10cal/(2g×0.5cal/g°C)
∅=10°C
Final temperature= -30°C+ 10°C= -20°C
Answer:
-20 degrees Celsius
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An ideal gas expands quasi-statically and isothermally from a state with pressurepand volumeVto a state with volume 4V. How much heat is added to the expanding gas?
Answer:
Q = PV(In 4)
Explanation:
We are told that the volume expands from V to a state with volume 4V.
Thus, initial volume is V and Final volume is 4V.
We want to find How much heat is added to the expanding gas.
For an isothermal process, the work done is calculated from;
W = nRT(In(V_f/V_i))
Where;
V_f is final volume
V_i is initial volume
Thus;
W = nRT(In(4V/V))
W = nRT(In 4)
Now, from ideal gas equation, we know that;
PV = nRT
Thus;
W = PV(In 4)
Now from first law of thermodynamics, we know that internal energy is zero and thus; Q = W
Where Q is quantity of heat
Thus;
Q = PV(In 4)
You are hanging on to the edge of a merry-go-round, and must exert a force of 100 N to hang on. If the speed of the merry-go-round doubles, how much force will you need to exert to hang on?
Answer:
If the speed of the merry-go-round doubles, the force you will need to exert to hang on is 400 N.
Explanation:
Given;
initial force exerted to hang on, F₁ = 100 N
The force exerted on the merry-go-round in order to hang on must be an inward force known as centripetal force.
Centripetal force is given by;
[tex]F_c = \frac{mv^2}{r} \\\\keeping \ "m" \ and \ "r" \ constant, we \ will \ have \ the \ following \ equation;\\\\\frac{F_c_1}{v_1^2} = \frac{F_c_2}{v_2^2} \\\\F_c_2 = \frac{F_c_1*v_2^2}{v_1^2}\\\\when \ the \ speed\ doubles \ i.e, v_2 = 2v_1\\\\ F_c_2 = \frac{F_c_1*(2v_1)^2}{v_1^2}\\\\ F_c_2 = \frac{F_c_1*4v_1^2}{v_1^2}\\\\F_c_2 = F_c_1 *4\\\\F_c_2 = 4(F_c_1)\\\\F_c_2 = 4 (100 \ N)\\\\F_c_2 = 400 \ N[/tex]
Therefore, If the speed of the merry-go-round doubles, the force you will need to exert to hang on is 400 N.
The CEO, ellen misk, left her martian office but accidentally left a cylindricall can of coke (3.1 inches in diameter, 5.42 inches in height) on her desk. If the can exerts a pressure of 510 Pascals, what is the specific gravity of the can?
Answer:
Specific Gravity = 0.378
Explanation:
First, we will find the force exerted by the can on the table. This force will be equal to the weight of the can:
Pressure = Force/Area = Weight/Area
Weight = Pressure*Area
where,
Area = πdiameter²/4 = π[(3.1 in)(0.0254 m/1 in)]²/4 = 4.8 x 10⁻³ m²
Weight = (510 N/m²)(4.8 x 10⁻³ m²)
Weight = 2.48 N
Now, the weight is given as:
Weight = mg
2.48 N = m(9.8 m/s²)
m = (2.48 N)/(9.8 m/s²)
m = 0.25 kg
Now, we calculate volume of can:
Volume = (Area)(Height) = (4.8 x 10⁻³ m²)(5.42 in)(0.0254 m/1 in)
Volume = 6.6 x 10⁻⁴ m³
Hence, the density of can will be:
Density of Can = m/Volume = 0.25 kg/6.6 x 10⁻⁴ m³
Density of Can = 378.32 kg/m³
So, the specific gravity of Can will be:
Specific Gravity = Density of Can/Density of Water
Specific Gravity = (378.32 kg/m³)/(1000 kg/m³)
Specific Gravity = 0.378
Convert 451 milliliters to fluid
ounces. Round your answer to 2
decimal places. **There are 29.57
milliliters in 1 fluid ounce***
Answer:
451 milliliters equals 15.25 fluid ounces
Explanation:
The rule of three or is a way of solving problems of proportionality between three known values and an unknown value, establishing a relationship of proportionality between all of them. That is, what is intended with it is to find the fourth term of a proportion knowing the other three.
To solve a direct rule of three, the following formula must be followed:
a ⇒ b
c ⇒ x
So: [tex]x=\frac{c*b}{a}[/tex]
The direct rule of three is the rule applied in this case where there is a change of units.
In this case, the rule of three can be applied in the following way: if there are 29.57 milliliters in 1 fluid ounce, in 451 milliliters how many fluid ounces are there?
[tex]fluid ounces=\frac{451 mL*1 fluid ounce}{29.57 mL}[/tex]
fluid ounces= 15.25
451 milliliters equals 15.25 fluid ounces
The x and y coordinates of a particle at any time t are x = 5t - 3t2 and y = 5t respectively, where x and y are in meter and t in second. The speed of the particle at t = 1 second is
Answer:
[tex]v=\sqrt{26}~m/s[/tex]
Explanation:
Parametric Equation of the Velocity
Given the position of the particle at any time t as
[tex]r(t) = (x(t),y(t))[/tex]
The instantaneous velocity is the first derivative of the position:
[tex]v(t)=(v_x(t),v_y(t))=(x'(t),y'(t))[/tex]
The speed can be calculated as the magnitude of the velocity:
[tex]v=\sqrt{v_x^2+v_y^2}[/tex]
We are given the coordinates of the position of a particle as:
[tex]x=5t-3t^2[/tex]
[tex]y=5t[/tex]
The coordinates of the velocity are:
[tex]v_x(t)=(5t-3t^2)'=5-6t[/tex]
[tex]v_y(t)=(5t)'=5[/tex]
Evaluating at t=1 s:
[tex]v_x(1)=5-6(1)=-1[/tex]
[tex]v_y(1)=5[/tex]
The velocity is:
[tex]v=\sqrt{(-1)^2+5^2}[/tex]
[tex]v=\sqrt{1+25}[/tex]
[tex]\mathbf{v=\sqrt{26}~m/s}[/tex]
The earth's radius is 6.37×106m; it rotates once every 24 hours.What is the speed of a point on the earth's surface located at 3/4 of the length of the arc between the equator and the pole, measured from equator? (Hint: what is the radius of the circle in which the point moves?) Express your answer to two significant figures and include the appropriate units.
Answer:
v = 120 m/s
Explanation:
We are given;
earth's radius; r = 6.37 × 10^(6) m
Angular speed; ω = 2π/(24 × 3600) = 7.27 × 10^(-5) rad/s
Now, we want to find the speed of a point on the earth's surface located at 3/4 of the length of the arc between the equator and the pole, measured from equator.
The angle will be;
θ = ¾ × 90
θ = 67.5
¾ is multiplied by 90° because the angular distance from the pole is 90 degrees.
The speed of a point on the earth's surface located at 3/4 of the length of the arc between the equator and the pole, measured from equator will be:
v = r(cos θ) × ω
v = 6.37 × 10^(6) × cos 67.5 × 7.27 × 10^(-5)
v = 117.22 m/s
Approximation to 2 sig. figures gives;
v = 120 m/s
5.List the four goals of Psychology. Give your own example for each one using a behavior
Answer:
describe, explain, predict, and change/control behavior.
Explanation:
describe: What are they doing? -Pavlov noticed that dogs were salivating when they would see his lab assistant before food was presented to them. This observation acted as a description of what was happening to them.
explain: Why are they doing that?- Pavlov started to look into why they were doing it. There was a stimulus, the assistant giving them food in the past to where they started to salivate at the sight of the lab assistant
predict: What would happen if I responded in this way?- Pavlov predicted that he could get the same reaction if he used a bell as a stimulus. Using this he was able to condition dogs to salivate at the ring of the bell.
change/control: What can I do to get them to stop doing that? Because of this discovery we can use conditioning today. For example, in the classroom teachers can use conditioning with their students to make it easier, parents to teach their children right from wrong and to have good behavior. (you do this bad thing you get time out, do a good thing and I will praise you, etc) It can be used when training employees and many other places.
Two blocks with different masses are dropped, hitting the ground with the same velocity. Which of the following is true?
They have same change in velocity but different changes in kinetic energy
The lighter object started at a smaller height.
The heavier object started at a smaller height
They started at the same height
They have same change in kinetic energy but different changes in velocity
Answer: • They have same change in velocity but different changes in kinetic energy
•They started at the same height.
Explanation:
First and foremost, we need to note that both balls have thesame acceleration due to gravity and due to this, even though they've different masses, they'll fall at same speed.
Also, since kinetic energy that's, the energy relating to motion of a mass, us dependent on mass and speed, their kinetic energy will be different.
Therefore, based in the explanation, the correct options are:
• They have same change in velocity but different changes in kinetic energy
•They started at the same height.