Answer:
Explanation:
The fish will have horizontal velocity of 2.6 m/s which is also the velocity of eagle. Additionally , he will have vertical velocity due to fall under gravity .
v² = u² + 2 g H .
v² = 0 + 2 x 9.8 x 4.7 m
= 92.12
v = 9.6 m /s
The fish's final velocity will have two components
vertical component = 9.6 m/s downwards
Horizontal component = 2.6 m /s .
Resultant velocity = √ ( 9.6² + 2.6² )
= √ ( 92.16 + 6.76 )
= 9.9 m /s
Answer:
The speed of fish at the time of hitting the surface is 9.95 m/s.
Explanation:
Horizontal speed, u = 2.6 m/s
height, h = 4.7 m
Let the vertical velocity at the time of hitting to water is v.
Use third equation of motion
[tex]v^2 = u^2 - 2 gh \\\\v^2 = 0 + 2 \times 9.8\times 4.7\\\\v = 9.6 m/s[/tex]
The net velocity with which the fish strikes to the water is
[tex]v' = \sqrt{9.6^2 + 2.6^2 }\\\\v' = 9.95 m/s[/tex]
A playground merry-go-round has a mass of 120 kg and a radius of 1.80 m and it is rotating with an angular velocity of 0.500 rev/s. What is its angular velocity after a 22.0-kg child gets onto it by grabbing its outer edge
Answer:
I think it is of science is it true na i knew it bro dont take tension
A uniform circular disk has a radius of 34 cm and a mass of 350 g. Its center is at the origin. Then a circular hole of radius 6.8 cm is cut out of it. The center of the hole is a distance 10.2 cm from the center of the disk. Find the moment of inertia of the modified disk about the origin.
Answer:
u can ask it to the person who give ot to u i dont no
The average 8-18 year old spends how many hours per day average in front of a screen doing little physical activity
Nearly four hours every day, doing little to no physical activity.
Two resistors with resistance values of 4.5 Ω and 2.3 Ω are connected in series or parallel
across a 30 V potential difference to a light bulb.
a. Calculate the current delivered through the light bulb in the two cases.
b. Draw the circuit connection that will achieve the brightest light bulb.
Explanation:
Given that,
Two resistors 4.5 Ω and 2.3 Ω .
Potential difference = 30 V
When they are in series, the current through each resistor remains the same. First find the equivalent resistance.
R' = 4.5 + 2.3
= 6.8 Ω
Current,
[tex]I=\dfrac{V}{R'}\\\\I=\dfrac{30}{6.8}\\\\=4.41\ A[/tex]
So, the current through both lightbulb is the same i.e. 4.41 A.
When they are in parallel, the current divides.
Current flowing in 4.5 resistor,
[tex]I_1=\dfrac{V}{R_1}\\\\=\dfrac{30}{4.5}\\\\I_1=6.7\ A[/tex]
Current flowing in 2.3 ohm resistor,
[tex]I_2=\dfrac{V}{R_2}\\\\=\dfrac{30}{2.3}\\\\I_2=13.04[/tex]
In parallel combination, are brighter than bulbs in series.
How does an airpump work?
describe the movement of the man when the resultant horizontal force is 0 N
can anyone help in both questions please
Answer:
Force A newton Law first law
F = M.A which Force in 0 N as you Questions Above
Force B
Newton Law 3
Action = -Reaction
Hope you can explain this formula as you want to scribe to explaining
How much work does a supermarket checkout attendant do on a can of soup he pushes 0.420 m horizontally with a force of 4.60 N? Express your answer in joules and kilocalories.
We know
[tex]\boxed{\sf Work\:Done=Force\times Displacement} [/tex]
[tex]\\ \rm\longmapsto Work\:done=0.420\times 4.60[/tex]
[tex]\\ \rm\longmapsto Work\:done=1.932J[/tex]
When a player's finger presses a guitar string down onto a fret, the length of the vibrating portion of the string is shortened, thereby increasing the string's fundamental frequency. The string's tension and mass per unit length remain unchanged.
If the unfingered length of the string is l=65cm, determine the positions x of the first six frets, if each fret raises the pitch of the fundamental by one musical note in comparison to the neighboring fret. On the equally tempered chromatic scale, the ratio of frequencies of neighboring notes is 21/12
x1=
x2=
x3=
x4=
x5=
x6=
Answer:
Explanation:
For frequencies n generated in a string , the expression is as follows
n = 1 /2L√ ( T/m )
n is fundamental frequency , T is tension in string , m is mass per unit length and L is length of string.
If T and m are constant , then
n x L = constant , hence n is inversely proportional to L or length of string.
Frequencies increase by 21/12 = 1.75 , length must decrease by 1 / 1.75 times
Initial length of string is 65 cm .
x1 = 65 x 1 / 1.75 = 37.14 cm
x2 = 37.14 x 1/ 1.75 = 21.22 cm
x3 = 21.22 x 1 / 1.75 = 12.12 cm
x4= 12.12 x 1 / 1.75 = 6.92 cm
x5 = 6.92 x 1 / 1.75 = 3.95 cm
x6 = 3.95 x 1 / 1.75 = 2.25 cm
The barometer of a mountain hiker reads 980 mbars at the beginning of a hiking trip and 790 mbars at the end. Neglecting the effect of altitude on local gravitational acceleration, determine the vertical distance climbed. A
Complete Question
The barometer of a mountain hiker reads 980 mbars at the beginning of a hiking trip and 790 mbars at the end. Neglecting the effect of altitude on local gravitational acceleration, determine the vertical distance climbed. Assume an average air density of 1.20kg/m^2
Answer:
[tex]h=1614m[/tex]
Explanation:
From the question we are told that:
Initial Pressure [tex]P_1=980mbar=>98000Pa[/tex]
Final Pressure [tex]P_2=790mbar=>79000Pa[/tex]
Density [tex]\rho=1.20kg/m^2[/tex]
Generally the equation for Height climbed is mathematically given by
[tex]h=\frac{P_1-P_2}{\rho*g}[/tex]
[tex]h=\frac{P_1-P_2}{1.20*9.81}[/tex]
[tex]h=1614m[/tex]
A glass block in air has critical angle of 49. What will happen to a ray of light coming through the glass when it is incident at and angle of 50 at the glass air boundary? Illustrate with a diagram
Answer:
b
Explanation:
An inductive circuit contains resistance of 20 ohm and an inductance of 20 H. If an ac voltage of 120 V and frequency 60 Hz is applied to this circuit, the current would be
A 0.0159
A 0.017
A 0.02
A 0.16
Answer:
answer : option (b) 0.016 amp
explanation : resistance of resistor , R = 10 Ω
inductance of inductor , X_LX
L
= 20H
voltage of AC circuit , V = 120volts
frequency, ff =60Hz
so, angular frequency, \omega=2\pi fω=2πf = 2 × π × 60 = 120π rad/s
now, current , i=\frac{V}{\sqrt{R^2+\omega^2L^2}}i=
R
2
+ω
2
L
2
V
= 120/√{10² + (120π)² × 20²}
= 120/√{100 + 14400π² × 400}
after solving this we get, i = 0.016 amp
Hannah wants to create a record keeping system to track the inventory needed to efficiently run her lawn and landscape business, such as spare parts, gas cans, string trimmers, etc. Her crew manager will also be using the system. Hannah is considering whether to use Excel or Access. Which one of the following is NOT a benefit of using Access?
a. More data storage
b. Multiuser capability
c. Easier setup
d. Additional reporting features
Answer:
c). Easier setup
Explanation:
As per the question, 'easier setup' cannot be characterized as the advantage of using Access because it comprises of plenty of steps that must be followed in the sequential order to establishing a database or carrying transactions based on time. However, there are plenty of advantages of using Microsoft access like 'enhanced and increased storage of data,' 'hassle free database systems,' 'easy importing of data,' 'highly economical,' 'capability to allow multiple users,' 'extra features for reporting,' and much more. Hence, option c is the correct answer.
* 1a Average speed
Carl Lewis runs the 100 m sprint in about 10 s.
His average speed in units of m/s would be:
of
Answer:
Explanation:
[tex] \implies v_{av} = \dfrac{total \: displacement}{total \: time} [/tex]
[tex] \implies v_{av} = \dfrac{100}{10} [/tex]
[tex]\implies v_{av} =10 \: {ms}^{ - 1} [/tex]
A car is traveling at 50 mi/h when the brakes are fully applied, producing a constant deceleration of 22 ft/s2. What is the distance covered before the car comes to a stop
Answer:
The correct solution is "122.2211".
Explanation:
Given:
deceleration,
a = 22 ft/sec²
Initial velocity,
[tex]V_i=50 \ m/h[/tex]
Now,
[tex]V_i=50 \ m/h\times 5280 \ ft/m\times hr/3600 \ s[/tex]
[tex]=73.333 \ ft/sec[/tex]
Now,
Final velocity,
[tex]V_f=0[/tex]
Initial velocity,
[tex]V_{initial} = 73.333 \ ft/sec[/tex]
hence,
⇒ [tex]V_f^2=V_i^2+2aD[/tex]
By putting the values, we get
[tex]0=(73.333)^2+2\times( -22) D[/tex]
[tex]44D=(73.333)^2[/tex]
[tex]D=\frac{(73.333)^2}{44}[/tex]
[tex]=122.2211[/tex]
State what is meant by a gravitational potential at point A is -1·70 × 109 J kg-1.
Answer:
The energy stored in a body due to either it's position or change in shape is called gravitational potential energy.
An initially motionless test car is accelerated uniformly to 105 km/h in 8.43 s before striking a simulated deer. The car is in contact with the faux fawn for 0.635 s, after which the car is measured to be traveling at 60.0 km/h. What is the magnitude of the acceleration of the car before the collision?
acceleration before collision:
3.45
m/s2
What is the magnitude of the average acceleration of the car during the collision?
average acceleration during collision:
19.68
m/s2
What is the magnitude of the average acceleration of the car during the entire test, from when the car first begins moving until the collision is over?
105 km/h ≈ 29.2 m/s
60.0 km/h ≈ 16.7 m/s
Before the collision the test car has an acceleration a of
a = (29.2 m/s - 0) / (8.43 s) ≈ 3.46 m/s²
During the collision, the car is slowed to about 16.7 m/s, so that its (average) acceleration is
a = (16.7 m/s - 29.2 m/s) / (0.635 s) ≈ -19.7 m/s²
i.e. with magnitude about 19.7 m/s².
Overall, the car has an average acceleration of
a = (16.7 m/s - 0) / (8.43 s + 0.635 s) ≈ 1.84 m/s²
If four students separately measure the density of a rock, and they all have very low percent
differences between their measurements, what can you say for certain about the accuracy of their
results?
Answer:
Their measured results are closer to the exact or true value. Hence, their measured value is considered to be more accurate.
Explanation:
Considering the situation described above, the accuracy of a measured value depicts how closely a measured value is to the accurate value.
Hence, since the students' measured values have very low percent differences, it shows the similarity of computations or estimates to the actual values, which in turn offers a smaller measurement error.
Therefore, their measured results are closer to the exact or true value, which implies that their measured value is considered to be more accurate.
Consider an electromagnetic wave propagating through a region of empty space. How is the energy density of the wave partitioned between the electric and magnetic fields?
1. The energy density of an electromagnetic wave is 25% in the magnetic field and 75% in the electric field.
2. The energy density of an electromagnetic wave is equally divided between the magnetic and electric fields.
3. The energy density of an electromagnetic wave is entirely in the magnetic field.
4. The energy density of an electromagnetic wave is 25% in the electric field and 75% in the magnetic field.
5. The energy density of an electromagnetic wave is entirely in the electric field
Answer:
Option (2) is correct.
The energy density of an electromagnetic wave is equally divided between the magnetic and electric fields.
Explanation:
An electromagnetic waves are the waves which are produced when the oscillating electric and magnetic field are interact each other perpendicular to each other. The direction of propagation of electro magnetic waves is perpendicular to each electric and magnetic fields.
The energy associated with the electromagnetic waves is equally distributed in form of electric and magnetic fields.
So, the correct option is (2).
The energy density is equally distributed among the magnetic field and electric field. Hence, option (2) is correct.
The given problem is based on the concept and fundamentals of electromagnetic waves. The waves created as a result of vibrations between an electric field and a magnetic field is known as Electromagnetic waves.
In other words, an electromagnetic waves are the waves which are produced when the oscillating electric and magnetic field are interact each other perpendicular to each other. The direction of propagation of electro magnetic waves is perpendicular to each electric and magnetic fields.
Also, the energy associated with the electromagnetic waves is equally distributed in form of electric and magnetic fields. So, the energy density of an electromagnetic wave is equally divided between the magnetic and electric fields.
Thus, we can conclude that the energy density is equally distributed among the magnetic field and electric field.
Learn more about the electromagnetic waves here:
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A uniform magnetic field is directed at an angle of 30o with the plane of a circular coil of radius 4cm and 1000 turns. The magnetic field changes at a rate of 5 T per second. Calculate the following:
a. Area vector
b. Induced emf
Answer:
(a) The area vector is 0.00503 m² at 30⁰ from the magnetic field
(b) The induced emf is 12.58 V
Explanation:
Given;
angle between the magnetic field and the plane of the circular coil, = 30⁰
number of turns of the coil, N = 1000
radius of the coil, r = 4 cm = 0.04 m
change in the magnetic field with time, dB/dt = 5 T/s
(a) The area vector is calculated as;
A = πr²
A = π x (0.04)²
A = 0.00503 m²
The area vector is 0.00503 m² at 30⁰ from the magnetic field.
(b) The induced emf is calculated as;
[tex]emf = N\frac{\Delta \phi}{\Delta t} \\\\where;\\\\\phi = BAcos \theta\\\\emf = N\times \frac{dB}{dt} \times Acos (\theta)\\\\where;\\\\\theta \ is \ the \ angle \ between \ a \ perpendicular \ vector \ to \ the \ area\\ and\ the \ magnetic\ field\\\\\theta = 90 - 30 = 60^0\\\\emf = N\times \frac{dB}{dt} \times Acos (\theta)\\\\emf = (1000) \times (5 )\times (0.00503) \times cos (60)\\\\emf = 12.58 \ V[/tex]
Some copper wire has a resistance of 200 ohms at 20 degrees C . A current is then passed through the same wire and the temperature rises to 90 degrees C. Determine the resistance of the wire at 90 degrees correct to the nearest ohm assuming the coefficient of resistance is 0.004/degree C at 0 degrees
Answer:
256 ohms
Explanation:
Applying,
R = R'[1+α(T-T')]............. Equation 1
Where R = Final resistance of the wire, R' = Initial resistance of the wire, T = Final temperature, T' = Initial temperature, α = Temperature coefficient of resistance
From the question,
Given: R' = 200 ohms, T = 90 degrees, T' = 20 degrees, α = 0.004/degree
Substitute these values into equation 1
R = 200[1+0.004(90-20)]
R = 200[1+0.28]
R = 200(1.28)
R = 256 ohms
The resistance of the wire at 90 °C correct to the nearest ohm assuming the coefficient of resistance is 0.004 °C¯¹ is 256 ohm
Data obtained from the question Original resistance (R₁) = 200 ohmOriginal temperature (T₁) = 20 °C Coefficient of resistivity (α) = 0.004 °C¯¹New temperature (T₂) = 90 °C New resistance (R₂) =? How to determine the new resistanceα = R₂ – R₁ / R₁(T₂ – T₁)
0.004 = R₂ – 200 / 200(90 – 20)
0.004 = R₂ – 200 / 200(70)
0.004 = R₂ – 200 / 14000
Cross multiply
R₂ – 200 = 0.004 × 14000
R₂ – 200 = 56
Collect like terms
R₂ = 56 + 200
R₂ = 256 ohm
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The figure below shows a combination of capacitors. Find (a) the equivalent capacitance of combination, and (b) the energy stored in C3 and C4.
Answer:
A) C_{eq} = 15 10⁻⁶ F, B) U₃ = 3 J, U₄ = 0.5 J
Explanation:
In a complicated circuit, the method of solving them is to work the circuit in pairs, finding the equivalent capacitance to reduce the circuit to simpler forms.
In this case let's start by finding the equivalent capacitance.
A) Let's solve the part where C1 and C3 are. These two capacitors are in serious
[tex]\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_3}[/tex] (you has an mistake in the formula)
[tex]\frac{1}{C_{eq1}} = (\frac{1}{30} + \frac{1}{15}) \ 10^{6}[/tex]
[tex]\frac{1}{C_{eq1}}[/tex] = 0.1 10⁶
[tex]C_{eq1}[/tex] = 10 10⁻⁶ F
capacitors C₂, C₄ and C₅ are in series
[tex]\frac{1}{C_{eq2}} = \frac{1}{C_2} + \frac{1}{C_4} + \frac{1}{C_5}[/tex]
[tex]\frac{1}{C_{eq2} } = (\frac{1}{15} + \frac{1}{30} + \frac{1}{10} ) \ 10^6[/tex]
[tex]\frac{1}{C_{eq2} }[/tex] = 0.2 10⁶
[tex]C_{eq2}[/tex] = 5 10⁻⁶ F
the two equivalent capacitors are in parallel therefore
C_{eq} = C_{eq1} + C_{eq2}
C_{eq} = (10 + 5) 10⁻⁶
C_{eq} = 15 10⁻⁶ F
B) the energy stored in C₃
The charge on the parallel voltage is constant
is the sum of the charge on each branch
Q = C_{eq} V
Q = 15 10⁻⁶ 6
Q = 90 10⁻⁶ C
the charge on each branch is
Q₁ = Ceq1 V
Q₁ = 10 10⁻⁶ 6
Q₁ = 60 10⁻⁶ C
Q₂ = C_{eq2} V
Q₂ = 5 10⁻⁶ 6
Q₂ = 30 10⁻⁶ C
now let's analyze the load on each branch
Branch C₁ and C₃
In series combination the charge is constant Q = Q₁ = Q₃
U₃ = [tex]\frac{Q^2}{2 C_3}[/tex]
U₃ =[tex]\frac{ 60 \ 10^{-6}}{2 \ 10 \ 10^{-6}}[/tex]
U₃ = 3 J
In Branch C₂, C₄, C₅
since the capacitors are in series the charge is constant Q = Q₂ = Q₄ = Q₅
U₄ = [tex]\frac{30 \ 10^{-6}}{ 2 \ 30 \ 10^{-6}}[/tex]
U₄ = 0.5 J
gAn optical engineer needs to ensure that the bright fringes from a double-slit are 15.7 mm apart on a detector that is from the slits. If the slits are illuminated with coherent light of wavelength 633 nm, how far apart should the slits be
Answer:
d = 68.5 x 10⁻⁶ m = 68.5 μm
Explanation:
The complete question is as follows:
An optical engineer needs to ensure that the bright fringes from a double-slit are 15.7 mm apart on a detector that is 1.70m from the slits. If the slits are illuminated with coherent light of wavelength 633 nm, how far apart should the slits be?
The answer can be given by using the formula derived from Young's Double Slit Experiment:
[tex]y = \frac{\lambda L}{d}\\\\d =\frac{\lambda L}{y}\\\\[/tex]
where,
d = slit separation = ?
λ = wavelength = 633 nm = 6.33 x 10⁻⁷ m
L = distance from screen (detector) = 1.7 m
y = distance between bright fringes = 15.7 mm = 0.0157 m
Therefore,
[tex]d = \frac{(6.33\ x\ 10^{-7}\ m)(1.7\ m)}{0.0157\ m}\\\\[/tex]
d = 68.5 x 10⁻⁶ m = 68.5 μm
Wind instruments like trumpets and saxophones work on the same principle as the "tube closed on one end" that we examined in our last experiment. What effect would it have on the pitch of a saxophone if you take it from inside your house (76 degrees F) to the outside on a cold day when the outside temperature is 45 degrees F?
Answer:
The correct answer is - low pitch
Explanation:
Now for the case it is mentioned that the tube closed on one end frequency is:
f = v/2l
Where,
l = length of the tube
v = velocity of longitudinal wave of gas filled in the tube
if frequency increases then pitch will be increase as well as pitch depends on frequency.
Now increase with the temperature the density of the gas decreases and velocity v is inversely proportional to density of gas so velocity increases. So if there is an increase in frequency so pitch also increases.
As the temperature inside the house is at 750 F more than outsideat 450 Fso pitch is more inside and the pitch is low outside.
A CD is spinning on a CD player. In 12 radians, the cd has reached an angular speed of 17 r a d s by accelerating with a constant acceleration of 3 r a d s 2 . What was the initial angular speed of the CD
Answer:
The initial angular speed of the CD is equal to 14.73 rad/s.
Explanation:
Given that,
Angular displacement, [tex]\theta=12\ rad[/tex]
Final angular speed, [tex]\omega_f=17\ rad/s[/tex]
The acceleration of the CD,[tex]\alpha =3\ rad/s^2[/tex]
We need to find the initial angular speed of the CD. Using third equation of kinematics to find it such that,
[tex]\omega_f^2=\omega_i^2+2\alpha \theta\\\\\omega_i^2=\omega_f^2-2\alpha \theta[/tex]
Put all the values,
[tex]\omega_i^2=(17)^2-2\times 3\times 12\\\\\omega_i=\sqrt{217}\\\\\omega_i=14.73\ rad/s[/tex]
So, the initial angular speed of the CD is equal to 14.73 rad/s.
A boy with a mass of 140 kg and a girl with a mass of 120 kg are on a merry go round. Th merry go round has a radius of 5 meters and its moment of inertia is 986 kg m 2. Beginning from rest the merry go round accelerates with an angular acceleration of 0.040 rad/s2 for 30 seconds then has a constant angular speed.
1. How many revolutions do the kids make before the constant operational speed is reached ?
2. What's the angular speed and magnitude of the tangential of the kids if they are standing at a distance of 1.5m and 2.4 m from the center of the ride.
3. During the ride the kids switch places what is the angular speed and magnitude of the tangential velocities ?
Answer:
we all are the human being we all dont no the all of 5he answer dont take tension beacause other one will give your answer
A certain heating element is made out of Nichrome wire and used with the standard voltage source of V=120 V. Immediately after the voltage is turned on, the current running through the element is measured at I1=1.28 A and its temperature at T1=25°C. As the heating element warms up and reaches its steady-state (operating) temperature, the current becomes I2=1.229 A.
Required:
Find this steady-state temperature T2.
Answer:
T₁ = 232.5 ºC
Explanation:
For this exercise let's start by finding the value of the resistance for the two currents, using Ohm's law
V = i R
R = V / i
i₀ = 1.28 A
R₀ = 120 / 1.28
R₀ = 93.75 ohm
i₁ = 1.229 A
R₁ = 120 / 1.229
R₁ = 97.64 or
Resistance in a metal is linear with temperature
ΔR = α R₀ ΔT
where the coefficient of thermal expansion for Nichrome is α=0.0002 C⁻¹
ΔT = [tex]\frac{\Delta R}{\alpha R_o}[/tex]
ΔT = [tex]\frac{97.64 \ -93.75}{ 0.00020 \ 93.75}[/tex]
ΔT = 2,075 10² C
ΔT = T₁-T₀ = 2,075 10²
T₁ = T₀ + 207.5
T₁ = 25+ 207.5
T₁ = 232.5 ºC
The reason why a teacher is more important then a farmer
Answer:
A teacher is more important than a famer.
Explanation:
A teacher is more important than a famer because the knowledge of farming is gotten through the teacher. Thus, without a teacher; whether formal or informal, there cannot be farming, let alone farmers.
A bullet is fired vertically upward a velocity of 80m/s to what height will the bullet rise above the point of projection
Answer:
The bullet will rise 320 meters above the point of projection.
Explanation:
Assuming that air friction is negligent we can use the kinematic equation:
[tex]v_{2} ^2=v_{1} ^2+2(-a)d\\0\frac{m^2}{s^2} =6400\frac{m^2}{s^2} +2(-10\frac{m}{s^2} )d\\-6400\frac{m^2}{s^2} =(-20\frac{m}{s^2}) d\\320m=d[/tex]
*acceleration is negative (-a) as it is acting in the opposite direction of the motion of the bullet.*
The bullet rises to a height of 3600 m if a bullet is fired upward with a velocity of 80 m/s.
Assume the air friction is negligible, the kinematic equation:
[tex]v_f^2 = v_i^2 +2(-a) d[/tex]
Where,
[tex]v_i^2[/tex] - iinitial velocity = 80 m/s
[tex]v_f^2[/tex]- final velocity = 0
[tex]d[/tex]- distance= ?
[tex]a[/tex]- gravitational acceleration = 9.8 m/s² = 10 m/s²
Put the values in the formula,
[tex]\begin {aligne} 0 = (80)^2 + 2 (10)^2 d\\\\d = \dfrac {6400}{ 200}\\\\d &= 3600 \rm \ m\end {aligne}\\[/tex]
Therefore, the bullet rises to a height of 3600 m if a bullet is fired upward with a velocity of 80 m/s.
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when we jump on a concrete surface,the feet get injured.Why
Answer:
Explanation:
Bhjb
Explanation:
its because a concrete surface is a hard surface which doesn't absorb the energy of gravitation when we fall down so we get hurt more badly..
hope this helps
A 215 N sign is supported by two ropes. One rope pulls up and to the right 1=29.5∘ above the horizontal with a tension 1 , and the other rope pulls up and to the left 2=44.5∘ above the horizontal with a tension 2 , as shown in the figure. Find the tensions 1 and 2 .
The sign is held in equilibrium. Using Newton's second law, we set up the equations of the net forces acting on the sign in the horizontal and vertical directions:
∑ F (horizontal) = T₁ cos(29.5°) - T₂ cos(44.5°) = 0
(right is positive, left is negative)
∑ F (vertical) = T₁ sin(29.5°) + T₂ sin(44.5°) - 215 N = 0
(up is positive, down is negative)
Solve the system of equations. I use elimination here:
• Multiply the first equation by sin(29.5°) and the second by cos(29.5°):
sin(29.5°) (T₁ cos(29.5°) - T₂ cos(44.5°)) = 0
cos(29.5°) (T₁ sin(29.5°) + T₂ sin(44.5°) - 215 N) = 0
T₁ cos(29.5°) sin(29.5°) - T₂ cos(44.5°) sin(29.5°) = 0
T₁ cos(29.5°) sin(29.5°) + T₂ cos(29.5°) sin(44.5°) = (215 N) cos(29.5°)
• Subtract the first equation from the second to eliminate T₁ :
T₂ cos(29.5°) sin(44.5°) - (- T₂ cos(44.5°) sin(29.5°)) = (215 N) cos(29.5°)
• Solve for T₂ :
T₂ (cos(29.5°) sin(44.5°) + cos(44.5°) sin(29.5°)) = (215 N) cos(29.5°)
T₂ sin(74.0°) = (215 N) cos(29.5°)
… … … (using the fact that sin(x + y) = sin(x) cos(y) + cos(y) sin(x))
T₂ = (215 N) cos(29.5°) / sin(74.0°)
T₂ ≈ 195 N
• Solve for T₁ :
T₁ cos(29.5°) - T₂ cos(44.5°) = 0
T₁ cos(29.5°) = T₂ cos(44.5°)
T₁ = T₂ cos(44.5°) / cos(29.5°)
T₁ ≈ 160. N
