We will have the following:
1)
[tex]I=\frac{V}{R}\Rightarrow I=\frac{15V}{120000\Omega}\Rightarrow I=1.25\cdot10^{-4}A[/tex][tex]\Rightarrow I=0.125mA[/tex]2)
First we determine the quantity of coulombs:
[tex]Q=(1.25\cdot10^{-4}A)(1s)\Rightarrow Q=1.25\cdot10^{-4}C[/tex]Now, we know that 1 coulomb represents 6.24*10^18 electrons, so:
[tex]1.25\cdot10^{-4}C\cdot\frac{6.24\cdot10^{18}\text{electrons}}{1C}=7.8\cdot10^{14}\text{electrons}[/tex]So, approximately 7.8*10^14 electrons pass each second.
Calculate the current needed to carry the 2 000 MW generated at alarge power station if the distribution voltage is kept at 22 kV, thegenerator output voltage.
Given data
*The given power is P = 2000 MW = 2000 × 10^6 W
*The given distribution voltage is V = 22 kV = 22 × 10^3 V
The formula for the current needed to carry the 2000 MW generated at a large power station is given as
[tex]I=\frac{P}{V}[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} I=\frac{2000\times10^6}{22\times10^3} \\ =90.90\times10^3\text{ A} \end{gathered}[/tex]Hence, the current needed to carry the 2000 MW generated at a large power station is I = 90.90 × 10^3 A
I would appreciate if you can help me solve this problem. I’m having trouble with it
The frequency of a wave is given by:
[tex]f=\frac{v}{\lambda}[/tex]Where v is the speed of the wave and lambda is the wavelenght. In this case the speed is 345 and the wavelenght is 0.85 meters; plugging the values given we have:
[tex]\begin{gathered} f=\frac{345}{0.8} \\ f=431.25 \end{gathered}[/tex]Therefore the frequency of the wave is 431.25 Hz
A submarine is stranded on the bottom of the ocean with its hatch 25 m below the surface. In this problem, assume the density of sea water is 1.03 x 10^3 kg/m^3. A) calculate the magnitude of the force in newtons pressing on the hatch from outside by the sea water given it is circular and 0.25m in diameter. The air pressure inside the submarine is 1.00atm (101,325 Pa). B) calculate the magnitude of the force, in newtons needed to open the hatch from the inside
Given:
The height (depth) of the hatch from the seawater surface is: h = 25 m
The density of seawater is: ρ = 1.03 × 10³ kg/m³
The diameter of the hatch is: d = 0.25 m
The pressure inside the submarine is: Ps = 1 Atm = 101325 Pa
To find:
The magnitude of the force pressing the hatch from the outside by the seawater and the magnitude of force to open the hatch from the inside.
Explanation:
The net pressure on the hatch area is given as:
[tex]P_{net}=\frac{F_{net}}{A}..........(1)[/tex]Here, Pnet is the net pressure on the hatch of area A, and Fnet is the net force acting on the hatch from the outside.
The net pressure Pnet is given as:
[tex]P_{net}=P_{atm}+P_w-P_s[/tex]Here, Patm is the atmospheric pressure above the seawater surface, Pw is the pressure of the water and Ps is the pressure inside the submarine.
The atmospheric pressure Patm of 1 Atm will contribute to the net pressure on the hatch from the outside. The atmospheric pressure Patm will have no effect as it will cancel out with the pressure Ps of equal magnitude as it is in an opposite direction from inside the submarine.
Substituting the values of the atmospheric pressure Patm and the pressure inside the submarine Ps in the above equation, we get:
[tex]\begin{gathered} P_{net}=1\text{ Atm}+P_w-1\text{ Atm} \\ \\ P_{net}=P_w..........(2) \end{gathered}[/tex]Thus the net pressure on the hatch will be the pressure of the seawater only.
Now, the net pressure Pnet can be calculated as:
[tex]P_{net}=h\rho g[/tex]Here, g is the acceleration due to the gravity having a value of 9.8 m/s².
Substituting the values in the above equation, we get:
[tex]\begin{gathered} P_{net}=25\text{ m}\times1.03\times10^3\text{ kg/m}^3\times9.8\text{ m/s}^2 \\ \\ P_{net}=252350\text{ N/m}^2\text{..........\lparen3\rparen} \end{gathered}[/tex]The area of the hatch can e calculated as:
[tex]\begin{gathered} A=\pi r^2 \\ \\ A=\pi\times(\frac{0.25}{2})^2 \\ \\ A=0.0491\text{ m}^2..........(4) \end{gathered}[/tex]Substituting equation (3) and equation (4) in equation (1) and rearranging it, we get:
[tex]\begin{gathered} 252350\text{ N/m}^2=\frac{F_{net}}{0.0491\text{ m}^2} \\ \\ F=252350\text{ N/m}^2\times0.0491\text{ m}^2 \\ \\ F=12390.39\text{ N} \end{gathered}[/tex]A force of 12390.39 Newtons acts on the hatch by the seawater above it. To open the hatch from the inside, an equal magnitude of force is needed to apply from inside the submarine. Thus, a force of 12390 Newtons is required to open the hatch from the inside.
Final answer:
The magnitude of force pressing the hatch from the outside by the seawater is 12390.39 Newtons.
The magnitude of force required to open the hatch from inside the submarine is 12390.39 Newtons.
(A) The magnitude of the force in newtons pressing on the hatch from outside by the seawater given it is circular and 0.25 m in diameter is 1.24 × 10 ⁴ N.
(B) The force applied needed to open the hatch from the inside is 1.24 × 10 ⁴ N.
What is force?In the body, force is the result of either a push or a pull. The three main categories of forces are friction, nuclear, and gravitational. For instance, when a hand strikes a wall, the hand puts force on the wall and the wall also exerts a force on the hand. Newton was given a variety of laws to help him understand force.
Given :
The density of seawater, ρ = 1.03 x 10³ kg / m³ ,
The height of the hatch of the submarine, h= 25 m,
The gravitational acceleration, g = 9.8 m / s² ,
The radius of the hatch, r = 0.25 m
Calculate the area of the hatch by the formula given below,
A = π × r² = π × (0.125 m)² [r = d / 2]
A = 4.9 x 10 ⁻² m²
(A) Calculate the force of the water by the following formula,
[tex]P = F / A[/tex]
F = P × A (P = ρ × g × h)
F = ρ × g × h × A
Substitute the values,
F = 1.03 x 10³ kg / m³ × 9.8 m / s² × 25 m × 4.9 x 10 ⁻² m²
F = 12365.15 N or 1.24 × 10 ⁴ N
(B) The force needed to open the hatch = The Force applied by water on the hatch.
Therefore, the magnitude of the force in newtons pressing on the hatch from outside by the seawater given it is circular and 0.25 m in diameter is 1.24 × 10 ⁴ N, and the force applied needed to open the hatch from the inside is 1.24 × 10 ⁴ N.
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Consider the diagram of a combination circuit below on the left. In the middle, the resistors in the two parallel branches have been replaced by a single resistor (R4) with an equivalent resistance to the overall branch resistors. On the right, all three resistors have been replaced by a single resistor (R5) with an equivalent resistance as all three original resistors. Suppose that you know that:R1 = 24.8ΩR2 = 24.8ΩR3 = 12.7Ω What must R4 and R5 be in order for the two circuits to have the same equivalent resistance? R4 = ------- Ω R5 = ----- Ω
Given:
• R1 = 24.8Ω
,• R2 = 24.8Ω
,• R3 = 12.7Ω
From the diagram, let's find R4 and R5.
We can see that the 3 resistors R1, R2, and R3 are connected in parallel.
Where:
R1 + R2 = R4
To solve for R4, we have:
[tex]\frac{1}{R_4}=\frac{1}{R_1}+\frac{1}{R_2}[/tex]Thus, we have:
[tex]\begin{gathered} \frac{1}{R_4}=\frac{1}{24.8}+\frac{1}{24.8} \\ \\ \frac{1}{R_4}=\frac{1+1}{24.8} \\ \\ \frac{1}{R_4}=\frac{2}{24.8}=\frac{1}{12.4} \\ \\ R_4=12.4\text{ \Omega} \end{gathered}[/tex]Now, to solve for R5 since R3 and R4 are now in series, we have:
[tex]\begin{gathered} R_5=R_3+R_4 \\ \\ R_5=12.7+12.4 \\ \\ R_5=25.1Ω \end{gathered}[/tex]Therefore, we have:
R4 = 12.4 Ω
R5 = 25.1 Ω
ANSWER:
• R4 = 12.4 ,Ω
,• R5 = 25.1 ,Ω
If a bullet with mass 2.7 grams is fired from a gun with a speed of 202
m/s into a block of wood. If the kinetic energy is transformed to heat,
what is the increase in temperature of the bullet in degrees C,
assuming the specific heat of the bullet is 234 J/kg/C.
The increase in temperature of the bullet in degrees C is 87.17
Define Specific Heat Capacity?
When a material's temperature rises by 1 K (or 1 °C), or when its mass increases by 1 kg, the specific heat capacity is defined as the amount of heat (J) absorbed per unit mass (kg).
This measurement is expressed as J/(kg K) or J/(kg °C).
The heat that is transferred to the bullet as a result of the conversion of kinetic energy to heat is what causes the bullet's temperature to rise.
Q = cm T,
where Q is the energy transferred, c is the substance's specific heat, m is its mass, and T is the temperature change, describes this relationship.
Given,
mass = 2.7 x 10^-3 kg
speed = 202 m/s
C = 234 J/kg/C.
The kinetic energy of the travelling bullet is,
1/2 m(v^2) = 1/2 x 2.7 x 10^-3 x 202 x 202
= 55.08 J
Energy that raises the temperature = 55.08 J
This is equal to mCΔT
Equalising the 2 energies,
55.08 = mCΔT
ΔT = 55.08 / mC
= 55.08 / 2.7 x 10^-3 x 234
= 87.17
Hence, the increase in temperature of the bullet in degrees C is 87.17
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1. The pupil brought the needle to one pole of the magnetic pointer. The needle attracted the pole of the pointer. Can this be evidence that the needle is magnetized? Why?
Yes, it is an evidence that the needle is magnetized.
Explanation:Magnetism describes the ability of an object to be attracted or repelled by a magnet.
Since the needle was attracted when brought to one pole of the magnet, we can conclude that the pin has properties that make it to be magnetized.
Therefore, the above is an evidence that the needle is magnetized.
The maximum possible efficiency of a reversible heat engine is 79.46% when the cold temperature is 224.0°C. What is the hot temperature in degrees Celsius?
We have
A heat engine is 79.46%
TC= 224.0°C
TH=?
we have the next formula
[tex]\eta(\text{\%)}=1-\frac{T_C}{T_H}\times100[/tex]we substitute the values
[tex].7946=1-\frac{224}{T_H}\times100[/tex]then we isolate TH
[tex]\begin{gathered} .07946-1=-\frac{T_C}{T_H}\times1 \\ -0.2054=-\frac{T_C}{T_H}\times1 \\ \frac{T_C}{T_H}\times1=-0.2054 \\ \frac{T_C}{T_H}=\frac{0.2054}{1} \\ \frac{T_H}{T_C}=\frac{1}{0.2054} \\ T_H=4.8685\cdot224 \\ T_H=1090.55\text{ \degree{}C} \end{gathered}[/tex]the hot temperature is 1090.55 degrees Celsius.
Suppose a 1.90 N force can rupture an eardrum having an area of 1.14 cm².(a) Calculate the maximum tolerable gauge pressure inside the eardrum (in the middle ear) in N/m². (Pressures in themiddle ear may rise when an infection causes a fluid buildup. Use 13.6 x 10³ kg/m³ as the density of mercury.) submit answer in N/m²(a) part 2: Convert this value to mm Hg.mm Hg(b) At what depth in fresh water would this person's eardrum rupture, assuming the gauge pressure in the middle ear iszero?Submit Answer in m
Given:
The force is
[tex]F=1.90\text{ N}[/tex]The area of the eardrum is
[tex]\begin{gathered} A=1.14\text{ cm}^2 \\ =1.14\times10^{-4}\text{ m}^2 \end{gathered}[/tex]To find:
The maximum tolerable gauge pressure inside the eardrum
a) the pressure in mm of Hg
b) At what depth in freshwater would this person's eardrum rupture
Explanation:
The pressure at the eardrum is
[tex]\begin{gathered} P=\frac{F}{A} \\ =\frac{1.90}{1.14\times10^{-4}} \\ =16.67\times10^3\text{ N/m}^2 \end{gathered}[/tex]Hence, the pressure is
[tex]16.67\times10^3\text{ N/m}^2[/tex]a)
We know,
[tex]1\text{ N/m}^2=0.0075\text{ mm of Hg}[/tex]So,
[tex]\begin{gathered} 16.67\times10^3\text{ N/m}^2=0.0075\times16.67\times10^3\text{ mm of Hg} \\ =125.02\text{ mm of Hg} \end{gathered}[/tex]Hence, the pressure is 125.02 mm of Hg.
b)
The depth of fresh water is,
[tex]\begin{gathered} h=\frac{P}{dg} \\ Here,\text{ d=1000 kg/m}^3 \\ g=9.8\text{ m/s}^2 \end{gathered}[/tex]So,
[tex]\begin{gathered} h=\frac{16.67\times10^3}{1000\times9.8} \\ =1.70\text{ m} \end{gathered}[/tex]Hence, the depth of water is 1.70 m.
Which of the following is NOT a requirement for a planet? A. must orbit a star, but is not a star or satellite of another planetB. must be roundC. must clear its orbit of debrisD. must have a moon
Answer:
D. must have a moon
Explanation:
Scientists say that a planet must orbit a star, must be round, and must be big enough so its gravity clears its orbit of other objects. Therefore, the statement that is not a requirement for a planet is
D. must have a moon
For example, Mercury and Venus are planets with no moons.
Explain what inertia is. What causes inertia in an object that is at rest? In an object that is moving?
Inertia is the ability of the body to remain in the state or in uniform motion unless some external force acts on the body.
The inertia of the body is direly proportional to the mass of the body.
Thus, the greater the mass of the body greater is its inertia.
Three types of inertia:
In the state of rest, the inertia of the body resists the cause of motion. This inertia is known as resting inertia.
In the state of uniform motion, the inertia resists the cause of changing of motion. This inertia is known as motion inertia.
Some external force is applied to the body to change the direction of the motion of the body, the inertia that opposes this change in direction is known as directional inertia.
Which option gives an object's volume in SI units?O A. 2,6 m3OB. 4.3 kgOC. 3.4 LD. 5.5 K
ANSWER:
A. 2.6 m^3
STEP-BY-STEP EXPLANATION:
The volume of an object is given by the cube of its length.
In SI the unit for length is the meter, therefore, in the case of volume it would be cubic meters.
This means that the correct answer is: A. 2.6 m^3
A rectangular container measuring 34 cm ⨯ 38 cm ⨯ 55 cm is filled with water. What is the mass of this volume of water in kilograms and grams?
We will have the following:
*Kg:
[tex](\frac{1kg}{m^3})(0.34m\ast0.38m\ast0.55m)=0.07106kg/m^3[/tex]*G:
[tex](\frac{1000g}{m^3})(0.34m\ast0.38m\ast0.55m)=71.06g/m^3[/tex]when a car driver reached the crest of a hill, she saw the stop light turn red and slammed on the brakes. the car slowed down as it moved to the bottom of the hill. is the KE and PE increasing, decreasing, or staying the same?
As the brakes are opposing the motion of the car and trying to decrease the velocity of the car while moving down the hill.
Thus, the kinetic energy of the car is decreasing as the car is moving down the hill.
The height of the car is also decreasing with the motion of the car down the hill, thus, the potential energy of the car is also decreasing.
The brakes are applying the frictional force on the car which is converting their energy into heat energy. The thermal energy of the car increases as the car moves down the hill.
Thus, the kinetic energy and potential energy of the car decreases while moving down the hill.
What is the momentum of a 7.30 kg bowling ball going down the alley with a speed of 20.0 m/s?
Given data
*The given mass of the bowling ball is m = 7.30 kg
*The given speed is v = 20.0 m/s
The formula for the momentum of a 7.30 kg bowling ball is given as
[tex]p=mv[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} p=(7.30)\times(20.0) \\ =146\text{ kg.m/s} \end{gathered}[/tex]Hence, the momentum of a 7.30 kg bowling ball is p = 146 kg.m/s
How do i solve this problem? Hint: The cannonball is being launched vertically upwards, therefore, there is no initial horizontal speed. The given initial speed will also be the initial vertical speed.
The initial velocity of the ball is given as 36.0 m/s.
The horizontal component of velocity of ball is given as,
[tex]v_x=v\cos \theta[/tex]The ball is projected vertically, therefore, the angle made by ball is 90 degree.
Plug in the known values,
[tex]\begin{gathered} v_x=(36.0m/s)cos90^{\circ} \\ =(36.0\text{ m/s)(0)} \\ =0\text{ m/s} \end{gathered}[/tex]Therefore, the initial horizontal velocity of ball is 0 m/s.
The vertical component of velocity of ball is given as,
[tex]v_y=v\sin \theta[/tex]Plug in the known values,
[tex]\begin{gathered} v_y=(36.0m/s)\sin 90^{\circ} \\ =(36.0\text{ m/s)(1)} \\ =36.0\text{ m/s} \end{gathered}[/tex]Therefore, the initial vertical velocity of the ball is 36.0 m/s.
If you have ever been to the beach you will find that the sand heats up and cools down faster than the water. What can you predict about the heat capacity of water and sand based on this result.
Water has a higher heat capacity than sand
The change in temperature is faster in sand than in water
Good morning I really need some help on this question!
Sound travels faster in water than air
Here, C is the correct option.
‼️‼️IF ANSWERED W WORK SHOWN I WILL GIVE BRAINLIEST‼️‼️The electric field intensity between two large, parallel metal plates is 6000 N/C. The plates are 0.05 mapart. What is the electric potential difference between them?
We have the next information
E= 6000 N/C
d=0.05m
In order to find the electric potential difference we have the next formula
[tex]\Delta V=Ed[/tex]then we substitute the data given in the formula
[tex]\Delta V=(6000\frac{N}{C})(0.05m)=300\frac{J}{C}=300V[/tex]The electric potential difference between them is 300 V
The position of a particle moving along the x axis is given by x = (21 + 22t - 6.0t 2)m, where t is in s. What is the average velocity during the time interval t = 1.0 s to t = 3.0 s?
Using the concept of linear motion, we got the average velocity as -2m/s.
Since the particle is moving in the x direction and we are given an equation
x = (21 + 22t - 6.0t 2)m
using the concept of differentiation we got
velocity (v) = dx/dt
Now distance x for t=1 s
x=21+22(1)-6
x=37m .....equation (1)
similarly distance at t=3 s
x=21+22(3)-54
x=33m ....equation (2)
so average velocity is
average velocity = total distance/total time
v=(33-37)/(3-1)
v=-2m/s
So average velocity for a given time interval is -2m/s
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A grocery shopper tosses a 6 kg bag of rice into a stationary 27 kg grocery cart. The bag hits the cart with a horizontal speed of 8.0 m/s toward the front of the cart. What is the final speed of the cart and the bag? Round to the hundredths place
Given,
The mass of the bag, m=6 kg
The mass of the cart, M=27 kg
The initial velocity of the cart, u=0 m/s
The initial velocity of the bag, v=8.0 m/s
From the law of conservation of momentum, the sum momentum of the bag and the momentum of the cart before the bag hits the cart must be equal to the momentum of the bag and the cart after the bag hits the cart.
Thus,
[tex]mv+Mu=(M+m)V[/tex]Where V is the velocity of the bag and the cart after the bag hits the cart.
On rearranging,
[tex]V=\frac{mv+Mu}{(M+m)}[/tex]On substituting the known values,
[tex]\begin{gathered} V=\frac{6\times8.0+0}{(27+6)} \\ =\frac{48}{33} \\ =1.45\text{ m/s} \end{gathered}[/tex]Thus the final speed of the cart and the bag is 1.45 m/s
A laboratory cart with a mass of 7 kilograms moves a distance of 0.5 meters in 3 seconds.Calculate the momentum.
Given that the mass of the cart is m = 7 kg.
The distance is d = 0.5 m.
The time is t = 3 s.
We need to find the momentum, p.
The formula to calculate momentum is
[tex]\begin{gathered} p=mass\times velocity \\ =m\times\frac{d}{t} \end{gathered}[/tex]
Substituting the values, the momentum will be
[tex]\begin{gathered} p=7\times\frac{0.5}{3} \\ =1.167kg\text{ m/s} \end{gathered}[/tex]Thus, the momentum is 1.167 kg m/s.
A gardener pushes a lawn roller through a distance of 20m if he applies a force of 20kg weight in a direction inclined at 70 degree to ground find the work done by himG=9.8m/s^2
The total force applied to the roller can be obtained as,
[tex]\begin{gathered} F=(20\text{ kg-wt)(}\frac{9.8\text{ N}}{1\text{ kg-wt}}) \\ =196\text{ N} \end{gathered}[/tex]The work done by the gardner can be given as,
[tex]W=Fd\cos \theta[/tex]Substitute the known values,
[tex]\begin{gathered} W=(196N)(20m)(\frac{1\text{ J}}{1\text{ Nm}})cos70^{\circ} \\ =(3920\text{ J)(}0.342) \\ =1340.64\text{ J} \end{gathered}[/tex]Thus, the work done by the gardener is 1340.64 J.
What would happen to the pressure and temperature of a box as gas is added to it
When gas is added to the box, the number of molecules in the box will increase.
This will result in an increase in pressure as volume is fixed or constant.
As the box has a fixed volume, this will be an isochoric process.
In an isochoric process, the pressure is directly proportional to temperature, so the temperature will also increase.
Thus, pressure and temperature will increase when gas is added to a box.
hello I need help with question 7 a and b please
Given data:
* The mass of the child is 40kg.
* The reading of the scale is 30 N.
Solution:
(a). The free body diagram of the given system is,
(b). The weight of the child is,
[tex]\begin{gathered} W=40\times9.8 \\ W=392\text{ N} \end{gathered}[/tex]The Scale reading is,
[tex]N=30\text{ Newton}[/tex]The net force acting on the box in terms of the weight of the box is,
[tex]\begin{gathered} F_{\text{net}}=N-mg \\ F_{\text{net}}=30-392 \\ F_{\text{net}}=-362\text{ N} \end{gathered}[/tex]According to the Newton's second law,
[tex]undefined[/tex]Romeo and Juliet are sitting on a balcony 1.5 meters apart. If Romeo has a mass of 61.6 Kg and Juliet has a mass of 48.8 kg. What is the attractive force between them?
A spaceship and an asteroid are moving in the same direction away from Earth with speeds of 0.8 c and 0.45 c, respectively. What is the relative speed between the spaceship and the asteroid?
Answer:
0.35c
Explanation:
The relative speed between the spaceship and the asteroid can be calculated as the difference between their speeds, so it is equal to
Relative speed = 0.8c - 0.45c
relative speed = 0.35c
Therefore, the answer is
0.35c
A bicycle with a mass of 10.0 kg is pushed up an incline of 30 ° with respect to the horizon. Friction is assumed to be negligible. a) Draw the free body diagram. b) It determines the intensity of the force which, applied horizontally, makes the bicycle move forward at a constant speed. c) Determines the intensity of the force which, applied horizontally, moves the bicycle forward with an acceleration of 100 m / s²
a) The free body diagram is drawn in the explanation section
b) The intensity of the force applied horizontally which makes the bicycle move forward at a constant speed = 49.05 N
c) The intensity of the force applied horizontally which makes the bicycle move forward with an acceleration of 100 m/s² = 1049.05N
Explanation:The mass of the bicycle, m = 10.0 kg
Angle of inclination, θ = 30°
The free body diagram of the illustration is drawn below
The bicyce is pushed up an incline of 30°
The weight of the bicycle acts downward (because the weight of every object acts downward)
Since the bicycle is pushed up an inclined of 30°, the force has to be resolved to the vertical(mgcosθ) and horizontal(mgsinθ)
b) The net force on the bicycle is:
[tex]\sum F=mg\sin \theta+ma[/tex]If the bicycle moves at constant speed, the acceleration is 0 m/s²
a = 0m/s²
[tex]\begin{gathered} \sum F=10(9.81)\sin 30+10(0) \\ \sum F=10(9.81)(0.5) \\ \sum F=49.05N \end{gathered}[/tex]c) The intensity of the force applied horizontally which moves the bicycle forward with an acceleration of 100 m/s²
[tex]\begin{gathered} \sum F=10(9.81)\sin 30+10(100) \\ \sum F=49.05+1000 \\ \sum F=1049.05N \end{gathered}[/tex]In Denver, children bring their old jack o lanterns to the top of a tower and compete for accuracy in hitting a target on the ground. suppose that the tower is 9.0m high and that the bullseye is a horizontal distance of 3.5m from the launch point.if the pumpkin is thrown horizontally what is the launch speed needed to hit the ground? (b) If the jack o lantern is given an initial horizontal speed of 3.3m/s, what are the direction and magnitude of its velocity (c) 0.75s later and (d) just before it lands.
a)
In order to calculate the required horizontal speed, first let's calculate the falling time, using the free-fall formula:
[tex]d=\frac{gt^2}{2}[/tex]For d = 9 and g = 9.8, we have:
[tex]\begin{gathered} 9=4.9t^2 \\ t^2=\frac{9}{4.9} \\ t=1.355\text{ s} \end{gathered}[/tex]Then, let's use this time in the following formula for horizontal distance, so we can calculate the horizontal speed:
[tex]\begin{gathered} \text{distance}=\text{speed}\cdot\text{time} \\ 3.5=\text{speed}\cdot1.355 \\ \text{speed}=\frac{3.5}{1.355}=2.58\text{ m/s} \end{gathered}[/tex]b)
Let's calculate the vertical velocity after 0.75 seconds, using the formula:
[tex]\begin{gathered} V=V_0+a\cdot t \\ V=0+9.8\cdot0.75 \\ V=7.35\text{ m/s} \end{gathered}[/tex]The magnitude can be calculated using the Pythagorean Theorem with the horizontal and vertical velocities:
[tex]\begin{gathered} V^2=7.35^2+3.3^2 \\ V^2=54.0225+10.89 \\ V^2=64.9125 \\ V=8.06\text{ m/s} \end{gathered}[/tex]And the direction is given by the arc tangent of the vertical velocity divided by the horizontal velocity (for this, let's use a negative value of vertical velocity, since it points downwards)
[tex]\theta=\tan ^{-1}(-\frac{7.35}{3.3})=\tan ^{-1}(-2.2272)=-65.82\degree[/tex](d)
To find the vertical velocity, let's use Torricelli's equation:
[tex]\begin{gathered} V^2=V^2_0+2\cdot a\cdot d \\ V^2=0^2+2\cdot9.8\cdot9 \\ V^2=176.4 \\ V=13.28\text{ m/s} \end{gathered}[/tex]Calculating the final speed magnitude and orientation, we have:
[tex]\begin{gathered} V^2=13.28^2+3.3^2 \\ V^2=187.2484 \\ V=13.68\text{ m/s} \\ \\ \theta=\tan ^{-1}(-\frac{13.68}{3.3})=\tan ^{-1}(-4.145)=-76.44\degree \end{gathered}[/tex]Force can be given as ?
The force is, momentum/time.
option 1
A cyclist is freewheeling (not exerting additional force) down a 7 degree angle hill. The cyclists weight is 75N. What acceleration is the cyclist experiencing? I have to do the following:1. Draw a free body diagram2. Identify Givens and Unknowns3. Identify the Equations4. Set up the equation using the givens and unknowns5. Solve
The free body diagram in shown below:
From the diagram and the problem we have that:
• The weight and angle of the inclined plane are given.
,• The normal force and the components of the weight are unknown (this implies that the acceleration is unknown too); we also notice that the mass is not given the it is also an unknown.
We know that Newton's second law states that:
[tex]\vec{F}=m\vec{a}[/tex]where F is the resultant force and a is the acceleration. Since this is a vector equation we can decomposed it in two scalar equations (in this case we only need two scalar equations since the forces are coplanar), then we have:
[tex]\begin{gathered} Wx=ma_x \\ N-W_y=ma_y \end{gathered}[/tex]Since we don't expect the cyclist to move in the y direction (otherwise he will surely fall) the equations above would reduce to:
[tex]\begin{gathered} W_x=ma \\ N-W_y=0 \end{gathered}[/tex]From the first equation we can solve the acceleration, to do this we use the triangle to get the x-component of the weight:
[tex]\begin{gathered} W_x=ma \\ W\sin \theta=ma \\ a=\frac{W\sin \theta}{m} \end{gathered}[/tex]Since the weight is given but not the mass we use the fact that the weight is:
[tex]W=mg[/tex]to get the mass, then we have:
[tex]\begin{gathered} m=\frac{W}{g} \\ m=\frac{75}{9.8} \\ m=7.65 \end{gathered}[/tex]hence the mass of the cylcist is 7.65 kg.
Now that we have all the values we need we plug them in the expression for the acceleration:
[tex]\begin{gathered} a=\frac{75\sin 7}{7.65} \\ a=1.19 \end{gathered}[/tex]Therefore the acceleration of the cyclist is 1.19 meters per second per second.