The Linear speed of the sphere is mathematically given as
v = 4.1 m/s
What is the linear speed of the sphere when it reaches the bottom of the ramp?Question Parameter(s):
A solid sphere of mass 4.0 kg and radius of 0.12 m starts from rest at the top of a ramp inclined 15°
Generally, the equation for the conservation of energy is mathematically given as
[tex]mgh = \frac{1}{2}mv^2 + \frac{1}{2}Iw^2[/tex]
Therefore
[tex]mgh = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mR^2)(\frac{v^2}{R^2})\\\\mgh = \frac{7}{10}mv^2[/tex]
In conclusion, the Speed is
[tex]v^2 = \sqrt{\frac{10}{7}(9.8)(1.2)}[/tex]
v = 4.1 m/s
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The linear speed of the sphere when it reaches the bottom of the ramp is approximately[tex]\( 4.85 \, \text{m/s} \).[/tex]
At the top of the ramp, the potential energy U of the sphere is given by:
[tex]\[ U = mgh \][/tex]
where m is the mass of the sphere, g is the acceleration due to gravity (approximately[tex]\( 9.8 \, \text{m/s}^2[/tex], and h is the height of the ramp.
At the bottom of the ramp, the kinetic energy K of the sphere is given by:
[tex]\[ K = \frac{1}{2}mv^2 \][/tex]
where v is the linear speed of the sphere.
Since energy is conserved, we have:
U = K
[tex]\[ mgh = \frac{1}{2}mv^2 \][/tex]
We can solve for v by canceling the mass m from both sides and multiplying through by 2:
[tex]\[ 2gh = v^2 \][/tex]
Taking the square root of both sides gives us the linear speed v :
[tex]\[ v = \sqrt{2gh} \][/tex]
Given that the height h is 1.2 m and the acceleration due to gravity g is [tex]\( 9.8 \, \text{m/s}^2 \)[/tex], we can plug in these values:
[tex]\[ v = \sqrt{2 \times 9.8 \, \text{m/s}^2 \times 1.2 \, \text{m}} \][/tex]
[tex]\[ v = \sqrt{23.52 \, \text{m}^2/\text{s}^2} \][/tex]
[tex]\[ v \ =4.85 \, \text{m/s} \][/tex]
Calculate the force (in pounds) ona 4.0 ft by 7.0 ft door due to a 0.4 psi pressure difference.
Answer:
7174.09 N
Explanation:
Area = 7 ft x 4 x ft = 28 ft^2 = 2.60129 m^2
Pressure = 0.4 psi = 2757.9 Pa
Force = Pressure x Area
Force = 2757.9 x 2.60129 = 7174.09 N
A thin conducing plate 2.3 m on a side is given a total charge of −20.0 µC. (Assume the upward direction is positive.) (a) What is the electric field (in N/C) 1.0 cm above the plate? (Indicate the direction with the sign of your answer.)
Answer:
The electric field is [tex]-2.14\times10^{5}\ N/C[/tex]
Explanation:
Given that,
Distance = 2.3 m
Charge [tex]q= -20.0\ \muC[/tex]
We need to calculate the electric field
Using formula of electric field
[tex]E=\dfrac{\sigma}{2\epsilon_{0}}[/tex]
Where, [tex]\sigma=\dfrac{Q}{A}[/tex]
Q = charge
A = area
Put the value into the formula
[tex]E=\dfrac{Q}{2A\epsilon_{0}}[/tex]
[tex]E=\dfrac{-20.0\times10^{-6}}{2\times2.3\times2.3\times8.85\times10^{-12}}[/tex]
[tex]E=-213599.91\ N/C[/tex]
[tex]E=-2.14\times10^{5}\ N/C[/tex]
Negative sign shows the direction of the electric field.
The direction of electric field is toward the plates.
Hence, The electric field is [tex]-2.14\times10^{5}\ N/C[/tex]
Final answer:
The electric field 1.0 cm above a thin conducting plate with a side length of 2.3 m and a total charge of -20.0 µC is -213,559.32 N/C, indicating a downward direction.
Explanation:
A thin conducting plate 2.3 m on a side is given a total charge of -20.0 µC. To find the electric field 1.0 cm above the plate, we use the concept of surface charge density (σ) and the formula for the electric field near an infinite plane sheet of charge, E = σ / (2ε0), where ε0 is the vacuum permittivity (ε0 = 8.85 x [tex]10^-^1^0[/tex] C2/N·m²). The surface charge density is σ = Q / A, with Q being the charge and A the area of the plate.
The area of the plate, A, is 2.3 m × 2.3 m = 5.29 m2. Therefore, σ = (-20.0 x 10-6 C) / (5.29 m2) = -3.78 x [tex]10^-^6[/tex] C/m². Substituting σ into the electric field equation yields E = -3.78 x [tex]10^-^6[/tex] C/m² / (2 × 8.85 x [tex]10^-^1^2[/tex] C2/N·m2) = -213,559.32 N/C, indicating the direction is downward due to the negative sign.
A soccer player swings his leg forward to kick the ball. If his leg has a moment of inertia of 0.9 kg m2, and is initially at rest, how much torque must he generate with his hip flexor muscle in order to reach an angular velocity of -8 rad/s in 0.5 seconds?
Answer:
14.4 Nm
Explanation:
Moment of Inertia, I = 0.9 kg m^2, w0 = 0, w = 8 rad/s, t = 0.5 second
Use first equation of motion for rotational motion
w = w0 + α t
where, α be the angular acceleration
8 = 0 + α x 0.5
α = 16 rad/s^2
Now Torque = Moment of inertia x angular acceleration
τ = I x α
τ = 0.9 x 16 = 14.4 Nm
An angle has a value of 2.2 radians. Find its value in degrees.
Answer:
The value of 2.2 radians in degree is 126.11°.
Explanation:
Given that,
Angle = 2.2 radians
We need to calculate the value in degrees
We know that,
[tex]1\ radians = \dfrac{180}{\pi}[/tex]
[tex]2.2\ radians = \dfrac{180}{3.14}\times2.2[/tex]
[tex]2.2\ radians = 126.11^{\circ}[/tex]
Hence, The value of 2.2 radians in degree is 126.11°.
A sled is pulled along a level path through snow by a rope. A 20-lb force acting at an angle of 40° above the horizontal moves the sled 90 ft. Find the work done by the force. (Round your answer to the nearest whole number.)
Answer:
Work one by the force, W = 1378.87 lb-ft
Explanation:
It is given that,
Force acting on the sled, F = 20 lb
Angle between the force and the horizontal, θ = 40°
Distance moved, d = 90 ft
We need to find the work done by the force. We know that the work done can be calculated as :
[tex]W=Fd\ cos\theta[/tex]
[tex]W=20\ lb\times 90\ ft\ cos(40)[/tex]
W = 1378.87 lb-ft
So, the work done by the force is 1378.87 lb-ft. Hence, this is the required solution.
An elevator is moving downward when someone presses the emergency stop button. The elevator comes to rest a short time later. Give the signs for the velocity and the acceleration of the elevator after the button has been pressed but before the elevator has stopped. Enter the correct sign for the elevator's velocity and the correct sign for the elevator's acceleration, separated by a comma. For example, if you think that the velocity is positive and the acceleration is negative, then you would enter ,- . If you think that both are zero, then you would enter 0,0 .
Answer:
Velocity: +ve, Acceleration: -ve
Explanation:
Here I've considered downward direction as positive direction.
Answer:
The answer is -,+ that is minus, plus
Explanation:
In the question, the elevator was described as moving downward, therefore its direction is negative. (-)
From the question we could tell the elevator is decelerating, so the acceleration vector should be pointing upward, in contrast with the motion of the elevator.(+)
VELOCITYVelocity is a vector quantity that indicates how fast an object is moving and in what direction, it has to do with and object’s displacement, time, and direction. The SI unit of velocity is meter per second (m/s). It should not be confused with speed which is a scalar quantity and measures on distance moved without stating what direction it moves.
For instance, it would not be enough to say that the car has a velocity of 50 miles/hr. the direction in which the car moves must be included to fully describe the velocity of the car. The correct way would be the car has a velocity of 50 miles/hour East.
ACCELERATIONIn physics, acceleration is defined as the rate of change of velocity. By altering an object’s speed or direction which changes its velocity hence its acceleration. Just like velocity, acceleration is a vector quantity. The SI unit of acceleration is meter per second squared (m/s^2)
What is the magnitude of a the vertical electric field that will balance the weight of a plastic sphere of mass that has been charged to -3.0 nC? g
Answer:
Explanation:
Let E be the strength of electric field and m be the mass of plastic sphere.
charge on plastic sphere, q = - 3 n C = - 3 x 10^-9 C
The force due to electric field on the charged plastic sphere = q E
The weight of the plastic sphere = m g
where, g is the acceleration due to gravity.
Now the force due to electrostatic field is balanced by teh weight of the plastic sphere.
q E = m g
E = m g / q
E = m x 9.8 / (3 x 10^-9) = 3.26 x 10^9 x m
Substitute the value of m and get the value E.
The magnitude of the vertical electric field required to balance the weight of the charged plastic sphere with a mass of 2.1 g and a charge of -3.0 nC is [tex]6.86 * 10^6 N/C[/tex].
To find the magnitude of the vertical electric field that will balance the weight of a charged plastic sphere, we need to equate the electrostatic force on the sphere due to the electric field with the weight of the sphere.
Given information:
- Mass of the plastic sphere, m = 2.1 g = [tex]2.1 * 10^-3 kg[/tex]
- Charge on the plastic sphere, q = -3.0 nC = [tex]-3.0 * 10^-9 C[/tex]
- Acceleration due to gravity, g = [tex]9.8 m/s^2[/tex]
- Coulomb's constant, k = 1/(4πε₀) =[tex]9.0 * 10^9 N.m^2/C^2[/tex]
Step 1: Calculate the weight of the plastic sphere.
Weight of the sphere = mg
Weight = [tex](2.1 * 10^-3 kg)[/tex] × [tex](9.8 m/s^2)[/tex]
Weight = 2.058 × [tex]10^{-2}[/tex] N
Step 2: Equate the electrostatic force on the sphere to its weight.
Electrostatic force = qE
Weight = qE
[tex]2.058 * 10^{-2} N = (-3.0 * 10^{-9} C) * E[/tex]
Step 3: Rearrange the equation to find the electric field magnitude E.
Equation to find the electric field magnitude E.
E = (2.058 × [tex]10^{-2}[/tex] N) / (-3.0 × [tex]10^{-9}[/tex]C)
[tex]E = 6.86 * 10^6 N/C[/tex]
Complete question:
what's the magnitude of a vertical electric field that will balance the weight of a plastic sphere of mass 2.1g that has been charged to -3.0nC? (k=[tex]1/4\pi[/tex], [tex]E_0[/tex]=[tex]9.0*10^9 N*m^2/C^2[/tex])
Two parallel wires are separated by 6.65 cm, each carrying 3.25 A of current in the same direction. (a) What is the magnitude of the force per unit length between the wires?
Answer:
The magnitude of the force per unit of lenght between the wires are of F/L= 3.17 * 10⁻⁵ N/m.
Explanation:
d=0.0665m
I1=I2= 3.25A
μo= 4π * 10⁻⁷ N/A²
F/L= (μo * I1 * I2) / (2π * d)
F/L= 3.17 * 10⁻⁵ N/m
A person pushes a 16.0 kg lawn mower at constant speed with a force of 87.5 N directed along the handle, which is at an angle of 45.0 to the horizontal. Part D friction force? What force must the person exert on the lawn mower to accelerate it from rest to 1.3 m/s in 2.5 seconds, assuming the same Express your answer to three significant figures and Include the appropriate units
The person must exert a force of 8.32 N to accelerate the lawn mower from rest to 1.3 m/s in 2.5 seconds.
When an object is moving at a constant speed, the net force acting on it is zero.
In this case, the force applied by the person along the handle is balanced by the frictional force opposing the motion.
So, the frictional force [tex](\(F_{\text{friction}}\))[/tex] is equal in magnitude but opposite in direction to the force applied by the person.
Given:
- Force applied by the person = 87.5 N
- The angle of the handle = 45.0 degrees to the horizontal
To find the horizontal component of the force applied by the person, use trigonometry:
[tex]\[F_{\text{person-horizontal}} = F_{\text{person}} \cdot \cos(\theta)\][/tex]
[tex]\[F_{\text{person-horizontal}} = 87.5 \, \text{N} \cdot \cos(45.0^\circ)\][/tex]
[tex]= 87.5 \, \text{N} \cdot 0.707 = 61.29 \, \text{N}[/tex]
So, the person exerts a horizontal force of 61.29 N.
Since the lawn mower moves at a constant speed, the frictional force must be equal in magnitude and opposite in direction to this horizontal force:
[tex]\[F_{\text{friction}} = -61.29 \, \text{N}\][/tex]
To calculate this force [tex](\(F_{\text{acceleration}}\))[/tex], we can use Newton's second law:
[tex]\[F = m \cdot a\][/tex]
Given:
Mass of the lawn mower = 16.0 kg
Acceleration = [tex]\(\frac{\Delta v}{\Delta t}\)[/tex],
Now, calculate [tex]\(\Delta v\):[/tex]
[tex]\[\Delta v = 1.3 \, \text{m/s} - 0 \, \text{m/s} = 1.3 \, \text{m/s}\][/tex]
Now, using Newton's second law:
[tex]\[F_{\text{acceleration}} = m \cdot a = (16.0 \, \text{kg}) \cdot \left(\frac{1.3 \, \text{m/s}}{2.5 \, \text{s}}\right)\][/tex]
[tex]\[F_{\text{acceleration}} = 8.32 \, \text{N}\][/tex]
Thus, the person must exert a force of 8.32 N.
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To accelerate a 16.0 kg lawn mower from rest to 1.3 m/s in 2.5 seconds, a person needs to exert a force of 136 N along the handle, assuming that the same friction force is present as when the mower is pushed at constant speed.
Explanation:In order to find the force that the person must exert on the lawn mower to accelerate it from rest to 1.3 m/s in 2.5 seconds, we can use the second law of motion that states that the net force acting on an object is equal to its mass times its acceleration: F = m*a.
In this case, the acceleration would be the change in speed over time, which is (1.3 m/s - 0 m/s) / 2.5 s = 0.52 m/s². Consequently, F = (16.0 kg) * (0.52 m/s²) = 8.3 N.
It's important to mention this force is in addition to the friction force and the force to move the mower horizontally. Asume that the 87.5 N force previously exerted was just overcoming friction, then the total horizontal force required would be 87.5 N (for friction) + 8.3 N (for acceleration) = 95.8 N. However, since this force makes an angle of 45.0 with the horizontal, the person must exert a larger force along the handle, about 135.5 N. Hence, the correct answer, expressed to three significant figures, is 136 N.
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Two resistors connected in series have an equivalent resistance of 724.5 Ω, when they are connected in parallel, their equivalent resistance is 135.2 Ω. Find the resistance of each resistor.
Answer:
544.65 ohm and 179.84 ohm
Explanation:
Hello
Let
Resistor 1 (R1) and Resistor 2 (R2)
in series [tex]R_{eq} =R1+R2[/tex]
in parallel
[tex]\frac{1}{R_{eqp} } =\frac{1}{R1} +\frac{1}{R2}\\\\R1=R_{eqs} -R2\\\\R1 =724.5-R2 (equation 1)\\ R_{eqp}}=\frac{R1*R2}{R1+R2}\\\\\ 135.2 = \frac{R1*R2}{R1+R2}(equation 2)\\\\\\\\replacing 1 in \ 2\\\\probema data[/tex]
[tex]135.2=\frac{(724.5-R2)(R2)}{724.5-R2+R2} }\\135.2=\frac{(-R_{2} ^{2}+724.5R2) }{724.5}\\ 135.2*724.5=-R_{2} ^{2}+724.5R_{2}\\\\R_{2} ^{2} -724.5R_{2} +97952.4 =0\\R_{2} =\frac{724.5 \±\sqrt{(-724.5^{2})-4(1)(97952.4) } }{2(1)}\\R_{2} =\frac{724.5\±\sqrt{(133092.25) } }{2(1)}\\\\R_{2} =\frac{724.5+364.81810}{2} \\R_{2} =544.65 ohm\\\\R_{2} =\frac{724.5-364.81810}{2} =179.84 ohm[/tex]\\\\
let R2=544.65 and replace in equation 1
R1=724.5-544.65
R1=179.85
so, the resistors are 544.65 ohm and 179.84 ohm
Have a great day
Is finding the distance between two points in the rectangular coordinate system related to the Pythagorean theorem? Why or why not? Provide an example in your answer.
Answer:
Yes it is related
Explanation:
Consider 2 points in a rectangular co-ordinate system
[tex](x_{1},y_{1}),(x_{2},y_{2})[/tex]
The distance between them can be found by
d = [tex]\sqrt{(x_{1}-x_{2})^{2}-(y_{2}-y_{1})^{^{2}}}[/tex]
This is clear from the figure attached below
A ray of light traveling in air is incident on the flat surface of a piece of glass at an angle of 65.9° with respect to the normal to the surface of the glass. If the ray refracted into the glass makes an angle of 34.8° with respect to the normal, what is the refractive index of the glass?
Answer:
[tex]n_{glass}[/tex] = 1.6
Explanation:
[tex]\theta _{i}[/tex] = Angle of incidence = 65.9°
[tex]\theta _{r}[/tex] = Angle of refraction = 34.8°
[tex]n_{air}[/tex] = Index of refraction of air = 1
[tex]n_{glass}[/tex] = Index of refraction of glass = ?
Using Snell's law
[tex]n_{air}[/tex] Sin[tex]\theta _{i}[/tex] = [tex]n_{glass}[/tex] [tex]\theta _{r}[/tex]
(1) Sin65.9 = [tex]n_{glass}[/tex] Sin34.8
[tex]n_{glass}[/tex] = 1.6
Since the index of refraction of air is 1, the refractive index of the glass is 1.6 approximately
Refractive IndexRefractive Index is the measure of refraction or bending when light passes from one medium to another.
Given that a ray of light traveling in air is incident on the flat surface of a piece of glass at an angle of 65.9° with respect to the normal to the surface of the glass. If the ray refracted into the glass makes an angle of 34.8° with respect to the normal, the refractive index of the glass can be calculated with the formula below
n = sin i / sin r
Where
n = refractive indexi = angle of incidencer = angle of refractionSubstitute all the parameters
n = sin 65.9 / sin 34.8
n = 0.913 / 0.5707
n = 1.599
n = 1.6 approximately
Therefore, the refractive index of the glass is 1.6 approximately.
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An asteroid is moving along a straight line. A force acts along the displacement of the asteroid and slows it down. The asteroid has a mass of 3.5× 104 kg, and the force causes its speed to change from 6600 to 5700m/s. (a) What is the work done by the force? (b) If the asteroid slows down over a distance of 1.5× 106 m determine the magnitude of the force.
Try this option; answers are marked with red colour: a) 19.215 GJ; b) 12.81 kN.
All the details are in the attached picture.
Consider two objects on the Moon's surface that can just be resolved by one of the 10.0 m telescopes at the Keck Observatory. What is the separation of the two objects? Assume the resolving power is only limited by diffraction effects. The distance from the Earth to the Moon is 3.8 x 105km, assume the wavelength of the light is 550 nm.
Answer: 25 m
Explanation:
The resolving power of a telescope is given by:
[tex] \theta = 1.22\frac{\lambda}{diameter}[/tex]
Diameter of the telescope = 10.0 m
Wavelength of the telescope = 550 nm = 550 ×10⁻⁹m
Thus, the resolving power of one of telescopes at the Keck Observatory is:
[tex] \theta = 1.22 \frac{550\times 10^{-9}}{10.0} =671 \times 10^{-10} rad = 3.85 \times 10^{-6} deg[/tex]
Also, [tex] tan\theta = \frac {\text {separation between two objects}}{\text{distance to the objects}} [/tex]
separation between objects = tan (3.85×10⁻⁶)° × 3.8×10⁵km = 2.5 × 10⁻²km =25 m
Cart A, with a mass of 0.4 kg, travels on a horizontal air track at 6 m/s and hits cart B, which has a mass of 0.8 kg and is initially at rest. After the collision, cart A has a velocity of 2 m/s in the opposite direction. What is the final kinetic energy of cart B
Answer:
6.4 J
Explanation:
For cart A
mA = 0.4 kg, uA = 6 m/s, vA = - 2 m/s
For cart B
mB = 0.8 kg, uB = 0, vB = ?
use the law of conservation of momentum
momentum of system before coliision = momentum of system after collision
mA uA + mB uB = mA vA + mB vB
0.4 x 6 + 0 = 0.4 x (- 2) + 0.8 x vB
vB = 4 m/s
Kinetic energy of Cart B after collision = 1/2 mB vB^2
= 1/2 x 0.8 x 4 x 4 = 6.4 J
As you move away from a positive charge distribution, the electric field:
Always decreases
Always increases
Must change
None of the above
please explain this thoroughly!
Answer:
The electric field always decreases.
Explanation:
The electric field due to a point charge is given by :
[tex]E=\dfrac{kq}{r^2}[/tex]
Where
k = electric constant
q = charge
r = distance from the charge
It is clear from the above equation that as the distance from the charge particle increases the electric field decreases. As you move away from a positive charge distribution, the electric field always decreases. Hence, the correct option is (c) "Always decreases".
Determine the energy required to accelerate an electron between each of the following speeds. (a) 0.500c to 0.900c MeV (b) 0.900c to 0.942c MeV
Answer:
The energy required to accelerate an electron is 0.582 Mev and 0.350 Mev.
Explanation:
We know that,
Mass of electron [tex]m_{e}=9.11\times10^{-31}\ kg[/tex]
Rest mass energy for electron = 0.511 Mev
(a). The energy required to accelerate an electron from 0.500c to 0.900c Mev
Using formula of rest,
[tex]E=\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{f}^2}{c^2}}}-\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{i}^2}{c^2}}}[/tex]
[tex]E=\dfrac{0.511}{\sqrt{1-\dfrac{(0.900c)^2}{c^2}}}-\dfrac{0.511}{\sqrt{1-\dfrac{(0.500c)^2}{c^2}}}[/tex]
[tex]E=0.582\ Mev[/tex]
(b). The energy required to accelerate an electron from 0.900c to 0.942c Mev
Using formula of rest,
[tex]E=\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{f}^2}{c^2}}}-\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{i}^2}{c^2}}}[/tex]
[tex]E=\dfrac{0.511}{\sqrt{1-\dfrac{(0.942c)^2}{c^2}}}-\dfrac{0.511}{\sqrt{1-\dfrac{(0.900c)^2}{c^2}}}[/tex]
[tex]E=0.350\ Mev[/tex]
Hence, The energy required to accelerate an electron is 0.582 Mev and 0.350 Mev.
A projectile is launched with initial speed v and angle ? over level ground. The projectile's maximum height H and horizontal range D are related by the equation D = 1.8H, what was the launch angle of the projectile?
Answer:
65.75 deg
Explanation:
v = initial speed of launch of projectile
θ = initial angle of launch
H = maximum height of the projectile
maximum height of the projectile is given as
[tex]H=\frac{v^{2}Sin^{2}\theta }{2g}[/tex] eq-1
D = horizontal range of the projectile
horizontal range of the projectile is given as
[tex]D=\frac{v^{2}Sin{2}\theta }{g}[/tex] eq-2
It is also given that
D = 1.8 H
using eq-1 and eq-2
[tex]\frac{v^{2}Sin{2}\theta }{g} = (1.8) \frac{v^{2}Sin^{2}\theta }{2g}[/tex]
[tex]Sin{2}\theta = (1.8) \frac{Sin^{2}\theta }{2}[/tex]
[tex]2 Sin\theta Cos\theta= (0.9) Sin^{2}\theta[/tex]
[tex]2 Cos\theta = (0.9) Sin\theta[/tex]
tanθ = 2.22
θ = 65.75 deg
The cheetah is the fastest land animal, reaching speeds as high as 33 m/s, or roughly 75 mph. What is the kinetic energy of a 65 kg cheetah running at top speed? Express your answer in SI units.
Answer:
Ek= 35.392 kJ or 35.392 10³ J
Explanation:
Formula for Kinetic energy:
Ek= [tex]\frac{1}{2}[/tex] × m × [tex]v^{2}[/tex]
since m= 65kg and v= 33m/s
Ek= [tex]\frac{1}{2}[/tex]× 65×[tex]33^{2}[/tex]
Ek= 35.392 kJ or 35.392 10³ J
Final answer:
The kinetic energy of a 65 kg cheetah running at its top speed of 33 m/s is calculated using the formula KE = 1/2 mv², resulting in a kinetic energy of 35442.5 Joules.
Explanation:
The question asks for the kinetic energy of a 65 kg cheetah running at its top speed, which is given to be 33 m/s. To calculate kinetic energy, we use the formula KE = 1/2 mv², where m is the mass of the object (in this case, the cheetah) and v is the velocity of the object. Substituting the given values, we get:
KE = 1/2 × 65 kg × (33 m/s)²
KE = 1/2 × 65 × 1089
KE = 1/2 × 70885
KE = 35442.5 Joules
Therefore, the kinetic energy of a 65 kg cheetah running at its top speed of 33 m/s is 35442.5 Joules.
A sound wave from a siren has an intensity of 100.0 point, what the sound intensity level of the sound?
Answer:
Sound intensity level of the sound = 140 dB
Explanation:
We have expression for sound intensity level
[tex]L=10log_{10}\left ( \frac{I}{I_0}\right )[/tex]
We have sound intensity, I = 100 W/m²
I₀ = 10⁻¹² W/m²
We need to find sound intensity level
Substituting
[tex]L=10log_{10}\left ( \frac{I}{I_0}\right )=10log_{10}\left ( \frac{100}{10^{-12}}\right )=10x14=140dB[/tex]
Sound intensity level of the sound = 140 dB
A ball is thrown from the top of one building toward a tall building 50 m away. The initial velocity of the ball is 20 m/s at 40° above the horizontal. How far above or below its original level will the ball strike the opposite wall?
Answer:
Ball hit the tall building 50 m away below 10.20 m its original level
Explanation:
Horizontal speed = 20 cos40 = 15.32 m/s
Horizontal displacement = 50 m
Horizontal acceleration = 0 m/s²
Substituting in s = ut + 0.5at²
50 = 15.32 t + 0.5 x 0 x t²
t = 3.26 s
Now we need to find how much vertical distance ball travels in 3.26 s.
Initial vertical speed = 20 sin40 = 12.86 m/s
Time = 3.26 s
Vertical acceleration = -9.81 m/s²
Substituting in s = ut + 0.5at²
s = 12.86 x 3.26 + 0.5 x -9.81 x 3.26²
s = -10.20 m
So ball hit the tall building 50 m away below 10.20 m its original level
The distance the ball will strike the opposite wall is 32.79 m.
Time of motion of the ball
The time of motion of the ball from the given height is calculated as follows;
h = vsinθ(t) + ¹/₂gt²
50 = 20 x sin(40)t + 0.5(9.8)t²
50 = 12.86t + 4.9t²
4.9t² + 12.86t - 50 = 0
solve the quadratic equation using formula method,
t = 2.14 s
Horizontal distance of the ballThe horizontal distance of the ball from the initial position is calculated as follows;
X = vcosθ(t)
X = 20 x cos(40) x 2.14
X = 32.79 m
Thus, the distance the ball will strike the opposite wall is 32.79 m.
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A parachutist who weighs 200lbs is falling at 120 miles/hour when his parachute opens. His speed is reduced to 15 miles/hour in a vertical distance of 120ft. What force did the parachute exert on the jumper?
Answer:
F = 3482.9 N
Explanation:
Change in velocity of the Parachutist is given as
[tex]v_f = 15 mph = 6.675 m/s[/tex]
[tex]v_i = 120 mph = 53.4 m/s[/tex]
now it is given as
[tex]\Delta v = v_f - v_i [/tex]
[tex]\Delta v = 120 - 15 = 105 mph[/tex]
[tex]\Delta v = 46.7 m/s[/tex]
now the acceleration of the parachutist is given as
[tex]a = \frac{v_f^2 - v_i^2}{2d}[/tex]
distance moved by the parachutist is given as
[tex]d = 120 ft = 36.576 m[/tex]
now we have
[tex]a = \frac{6.675^2 - 53.4^2}{2(36.576)}[/tex]
[tex]a = - 38.4m/s^2[/tex]
Now the mass of parachutist is given as
[tex]m = 200 lb = 90.7 kg[/tex]
now we have
[tex]F = ma[/tex]
[tex]F = (90.7 kg)(38.4) = 3482.9 N[/tex]
A toroid of mean radius 26 cm and circular cross section of radius 1.9 cm is wound with a superconducting wire of length 1000 m that carries a current of 410 A. (a) What is the number of turns on the coil?
Answer:
8377 turns
Explanation:
Given:
Mean radius of toroid = 26 cm
Circular cross-section radius, r = 1.9 cm = 0.019
Length of the wire, l = 1000 m
The number of turns (n) on the coil is given by
[tex]n = \frac{l}{s}[/tex]
where,
l is the length of the wire
s is the circumference of the circular wire
⇒[tex]n = \frac{1000}{2\pi r}[/tex]
or
[tex]n = \frac{1000}{2\pi \times 0.019}[/tex]
or
n = 8376.57 or 8377 turns
hence, the number of turns on the coils is 8377
The number of turns of the coil is determined as 8377 turns.
Number of turnsThe number of turns of the superconducting wire is calculated as follows;
n = L/2πr
where;
L is the length of superconducting wire = 1000 mr is the radius of the circular cross section = 1.9 cm = 0.019 mSubstitute the given parameters and solve for the number of turns of the coil,
n = 1000/(2π x 0.019)
n = 8377 turns
Thus, the number of turns of the coil is determined as 8377 turns.
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A boy flies a kite with the string at a 30∘ angle to the horizontal. The tension in the string is 4.5 N. Part A Part complete How much work does the string do on the boy if the boy stands still?
Answer:
Work done is zero
Explanation:
given data
Angle of kite with horizontal = 30 degree
tension in the string = 4.5 N
WE KNOW THAT
Work = force * distance
horizontal force = [tex]Tcos\theta = 4.5*cos30 = 3.89 N[/tex]
DISTANCE = 0 as boy stands still. therefore
work done = 3.89 *0 = 0
Using a 681 nm wavelength laser, you form the diffraction pattern of a 0.109 mm wide slit on a screen. You measure on the screen that the 14th dark fringe is 9.51 cm away from the center of the central maximum. How far is the screen located from the slit?
Answer:
The distance between the screen and slit is 1.08 m
Explanation:
Given that,
Wavelength = 681 nm
Width a= 0.109 mm
Number of fringe n = 14
Distance from the center of the central maximum d= 9.51 cm
We need to calculate the distance between the screen and slit
Using formula of distance
[tex]D=\dfrac{d\times a}{n\lambda}[/tex]
Where, a = width
d = distance from the center of the central maximum
[tex]\lambda[/tex] = wavelength
Put the value into the formula
[tex]D=\dfrac{9.51\times10^{-2}\times0.109\times10^{-3}}{14\times681\times10^{-9}}[/tex]
[tex]D = 1.08\ m[/tex]
Hence, The distance between the screen and slit is 1.08 m
If the gravitational force between two masses located a distance 100 m apart is F, what would be the force between these objects if the distance between them increases to 800 m?
Answer:
[tex] \frac{F}{64}[/tex]
Explanation:
m = product of masses of two objects
r = distance between the two objects = 100 m
F = initial force between the two object
r' = new distance between the two objects = 800 m
F' = new force between the two objects = ?
k = constant of proportionality
initial force between the two object is given as
[tex]F=\frac{km}{r^{2}}[/tex] eq-1
new force between the two objects is given as
[tex]F'=\frac{km}{r'^{2}}[/tex] eq-2
Dividing eq-2 by eq-1
[tex]\frac{F'}{F}=\frac{r^{2}}{r'^{2}}[/tex]
Inserting the values
[tex]\frac{F'}{F}=\frac{100^{2}}{800^{2}}[/tex]
[tex]F' = \frac{F}{64}[/tex]
Determine the position in which a solid cylindrical block of wood of diameter 0.3 m and length 0.4 m will float in water. Take specific gravity of wood as 0.5
Answer:
0.2 m
Explanation:
Diameter = 0.3 m
radius, r = 0.15 m
Length, H = 0.4 m
density of wood, d = 0.5 g/cm^3 = 500 kg/m^3
density of water, d = 1000 kg/m^3
Let h be the depth of cylinder immersed in water.
By the principle of floatation.
Buoyant force = Weight of cylinder
Volume immeresed x density of water x g = Volume of cylinder x density of wood x g
A x h x 1000 x g = A x H x 500 x g
1000 h = 500 x 0.4
h = 0.2 m
A car starts from rest and speeds up at a constant rate of 2.03 m/s^2 in a straight line until it reaches a speed of 21.1 m/s. The car then slows down at half the rate that it sped up until it stops. How far (in m) does the car move from start to stop?
Answer:
The car moves from start to stop 328.88 m in total.
Explanation:
Vo= 0 m/s
V1= 21.1 m/s
V2= 0 m/s
a1= 2,03 m/s²
a2= -1.015 m/s²
Speed Up:
Speed up time:
V1= Vo + a1 * t1
t1= V1/a1
t1= 10.39 sec
total distance of speed up:
d1= Vo * t1 + (a1 * t1²)/2
d1= 109.57m
Slow Down:
V2= V1 - a2 * t2
t2= V1/a2
t2= 20.78 sec
total distance of slow down:
d2= V1 * t2 - (a2 * t2²)/2
d1= 219.31m
Total Distance:
TD= d1+d2= 109.57m + 219.31m
TD= 328.88 m
Electric field intensity, E can be defined as the electric (Coulomb) force on a unit test (1C) charge. Find the electric field intensity at a distance, d from the center of a +Q charged proton. use k as Coulombe constant.
Explanation:
Electric field intensity E is defined as the electric (Coulomb) force on a unit test (1 C) charge. Mathematically, it is given by :
[tex]E=\dfrac{F}{q}[/tex]
The electric force is given by :
[tex]F=k\dfrac{qQ}{d^2}[/tex]
Where
Q and q are electric charges
d is the distance between charges
The electric field intensity at a distance d from the center is given by :
[tex]E=\dfrac{k\dfrac{qQ}{d^2}}{q}[/tex]
So,
[tex]E=\dfrac{kQ}{d^2}[/tex]
Hence, this is the required solution.
The amplitude of a wave decreases gradually as the wave travels down a long, stretched string. What happens to the energy of the wave when this happens? (The frequency is the same).
Answer:
energy decreases
Explanation:
The energy of a wave is directly proportional to the square of amplitude of a wave.
Amplitude is defined as the maximum displacement of the wave particle from its mean position.
So, as the amplitude goes on decreasing, the energy of the wave is also decreasing.
If the amplitude of a wave decreases gradually as the wave travels down, the energy of the wave increases.
What is a wave?A wave can be defined as a disturbance in a medium that progressively transport energy from a source to another location, especially without the transportation of matter.
The characteristics of a wave.In Science, some of the characteristics of a wave include the following:
FrequencyWavelengthAmplitudeSpeedGenerally, the energy of a wave is directly proportional to the square of wave amplitude and the square of its frequency. Thus, the energy of the wave increases when the amplitude of a wave decreases gradually as the wave travels down.
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