The centripetal acceleration of the hummingbird is [tex]\frac{(3.65)^2}{0.25} = 53.29 \ m/s^2[/tex]
The given parameters;
time of motion of the hummingbird, t = 0.43 sradius of the circle, r = 0.25 mnumber of revolution of the hummingbird = 1 rev per 0.43 sThe angular speed of the hummingbird is calculated as follows;
[tex]\omega = \frac{1 \ rev}{0.43 \ s} \times \frac{2\pi \ rad}{1 \ rev} =14.61 \ rad/s[/tex]
The linear speed of the hummingbird is calculated as follows;
v = ωr
v = 14.61 x 0.25
v = 3.65 m/s
The centripetal acceleration of the hummingbird is calculated as follows;
[tex]a_c = \frac{v^2}{r} \\\\a_c = \frac{(3.65)^2}{0.25} \\\\a_c = 53.29\ m/s^2[/tex]
Thus, the centripetal acceleration of the hummingbird is [tex]\frac{(3.65)^2}{0.25} = 53.29 \ m/s^2[/tex]
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PLEASE I NEED HELP WITH 6 and 8 !!!! I will love u sm
Answer:
Explanation:
006
They are acting in opposite directions. Therefore the net force is found by subtraction. The sign is the same as the larger number.
Net Force = 99.6 - 52.8 = 46.8 N acting in the same direction as the 99.6 which is upward.
008
If the two forces act in the same direction, the net force is found by addition.
Net Force = 99.6 + 52.8 = 152.4 N downward.
what is 4 differences between saturated unsaturated and supersaturated solutions
Answer:
Unsaturated Solution: Less amount of salt in water, clear solution, no precipitation. Saturated Solution: The maximum amount of salt is dissolved in water, Colour of the solution slightly changes, but no precipitation. Supersaturated Solution: More salt is dissolved in water, Cloudy solution, precipitation is visible.
A 0.750 kg hammer is moving horizontally at 7.00 m/s when it strikes a nail and comes to rest after driving it 1.00 cm into a board. (a) Calculate the duration of the impact in seconds. (Enter a number.) s (b) What was the average force in newtons exerted downward on the nail
Answer:nh
0.750 kg hammer is moving horizontally at 7.00 m/s when it strikes a nail and comes to rest after driving it 1.00 cm into a board.
A football is kicked with a velocity of 5 m/s at an angle of 53° above the horizontal. What is its speed at the
maximum height?
A) 3 m/s
B) 6 m/s
C) 9 m/s
D) 12 m/s
E) 15 m/s
define potential difference as used in electricity
Answer:
This ability of charged particles to do work is called an electric potential.
Explanation:
When two negative charges are brought close to each other, they also repel. But when a positive and a negative charge are brought close together, they attract each other. When these two opposite charges are combined, they can be used to work. This is why we need a positive (+) and a negative (–) to light a bulb or run any electrical tool, equipment, mobile phone or home appliance.
Potential difference is the difference in the amount of energy that charge carriers have between two points in a circuit. ... The energy is transferred to the electrical components in a circuit when the charge carriers pass through them. We use a voltmeter to measure potential difference (or voltage).
what force is represented by the vector
relationship between the length (l) and ohmic conductor, its thickness (d) and the opposition (R) it offers a current passage
Answer:
Ohm’s Law states that current is proportional to voltage; circuits are ohmic if they obey the relation V=IR.
Explanation:
LEARNING OBJECTIVES
Contrast shape of current-voltage plots for ohmic and non-ohmic circuits
in order to keep heat in or out, you need a(n)
Answer:
What is instillation
Explanation:
Light of 650 nm wavelength illuminates two slits that are 0.20 mm
apart. (Figure 1) shows the intensity pattern seen on a screen behind the slits.
What is the distance to the screen?
PLEAsE URGENT. thank you
We have that for the Question "Light of 650 nm wavelength illuminates two slits that are 0.20 mm apart. (Figure 1) shows the intensity pattern seen on a screen behind the slits.
What is the distance to the screen?" it can be said that the distance to the screen
d=1.168m
From the question we are told
Light of 650 nm wavelength illuminates two slits that are 0.20 mm
apart. (Figure 1) shows the intensity pattern seen on a screen behind the slits.
What is the distance to the screen?
Generally the equation for the distance is mathematically given as
[tex]d=\frac{0.33*10^{-2}*0.23*10^{-3}}{650*10^{-9}}\\\\d=1.160m[/tex]
d=1.168m
Therefore
The distance to the screen
d=1.168m
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An astronaut standing on a platform on a foreign planet drops a hammer. If the hammer falls 9.0 meters vertically in 2.5 seconds, what is the acceleration due to gravity on that planet?
Answer: 1.646 m/s²
Explanation:
The distance that is traveled by the astronaut given that the motion is free-fall can be calculated through the equation,
d = Vot + 0.5at²
where d is the distance, Vo is the initial velocity, t is the time, and a is the acceleration. Substituting the known,
6 = (0 m/s)(2.7 s) + 0.5(a)(2.7 s)²
Determining the value of a,
a = 1.646 m/s²
A diffraction grating, ruled with 300 lines per mm, is illuminated with a white light source at normal incidence.
(i) What is the angular separation, in the third-order spectrum, between the 400 nm and 600 nm lines? [5]
(ii) Water (of refractive index 1.33) now fills the whole space between the grating and the screen. What is the angular separation, in the first-order spectrum, between the 400 nm and 600 nm lines? [5]
the expression for diffraction grating allows to find the results for the questions for the angular separation are:
i) The third order is Δθ = 0.203 rad.
ii) The first order with water is Δθ = 0.046 rad.
The diffraction grating is a system formed by a large number of equally spaced lines whose diffraction is given by the expression.
d sin θ = m λ
Where d is the distance between two lines, θ is the angle of diffraction, the order of diffraction and λ is the wavelength.
i) Let's start by looking for the separation between two lines
Let's use a rule of direct proportions. If there are 300 lines in 1 mm, what distance is there between two lines.
d = 1 lines (1 mm / 300 lines) = 3,333 10⁻³ mm
d = 3.333 10⁻⁶ m
Let's find the angle of diffraction for the third order (m = 3) for each wavelength.
λ₁ = 400 nm = 400 10⁻⁹ m
sin θ₁ = [tex]\frac{m \ \lambda }{d}[/tex]m λ/ d
sin θ₁ = [tex]\frac{3 \ 400 \ 10^{-9} }{3.333 \ 10^{-6} }[/tex]
θ₁ = sin⁻¹ 0.3600
θ₁ = 0.368 rad
λ₂ = 600 nm = 600 10⁻⁹ m
sin θ₂ = [tex]\frac{3 \ 600 \ 10^{-9} }{3.333 \ 10^{-6} }[/tex]
θ₂ = sin⁻¹ 0.5401
θ₂ = 0.571 rad
The angular separation is
Δθ = θ₂ - θ₁
Δθ = 0.571 - 0.368
Δθ = 0.203 rad
ii) In this case, the separation between the network and the observation screen is filled with water.
When the rays leave the network they undergo a refraction process, for which they must comply with the relationship.
[tex]n_i \ sin \theta_1 = n_r \ sin \theta_r[/tex]
The incident side is in the air, therefore its refractive index is n_i = 1 and when it passes into the water with refractive index n_r = 1.33.
Let's start looking for the incident angles for the first order of diffraction.
m = 1
λ₁ = 400 nm
θ₁ = sin⁻¹ [tex]\frac{1 \ 400 \ 10^{-9}}{3.33 \ 10^{-6}}[/tex]
θ₁ = 0.120 rad
λ₂ = 600 nm
θ₂ = sin⁻¹¹ [tex]\frac{1 \ 600 \ 10^{-9} }{3.33 \ 10^{-6}}[/tex]
θ₂ = 0.181 rad
we use the equation of refraction.
[tex]\theta_r[/tex] = sin⁻¹ ([tex]\frac{n_i}{n_r} \ sin \ \theta_i[/tex] )
λ₁ = 400 nm
θ₁ = sin¹ ([tex]\frac{1 sin 0.120}{1.33}[/tex]
θ₁ = 0.090 rad
λ₂ = 600 nm
θ₂ =sin⁻¹ [tex]\frac{1 sin 0.181}{1.33}[/tex]
θ₂ = 0.1358 rad
The angular separation is
Δθ = 0.1358 - 0.090
Δθ = 0.046 rad.
In conclusion using the relation for the diffraction grating we can find the results for the questions about angular separation are:
i) The third order is Δθ = 0.203 rad.
ii) The first order with water is Δθ = 0.046 rad.
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1. As the angle of the ramp is increased, the normal force increases /decreases / remains the same and the friction-force increases /decreases / remains the same. [1 Point] 2. As the angle of the ramp is increased, the force parallel increases /decreases / remains the same. [1 Point] 3. The angle at which the force down the plane was equal to the force of friction (for the cabinet) was _____________. [1 Point] 4. Consider a very low (~ zero) friction, 5.0 kg skateboard on a ramp at an angle of 15o to the horizontal. What would be the net force that would cause acceleration when the skateboard is allowed to move
(1) As the angle of the ramp is increased, the normal force decreases.
(2) As the angle of the ramp is increased, the parallel force increases.
(3) The angle at which the force down the plane was equal to the force of friction is zero degree.
(4) The net force that would cause acceleration is 47.33 N.
Let the angle of inclination of the ramp = θ
(1)
The normal force on an object on the ramp inclined to the ramp is calculated as follows;
[tex]F_n = mgcos (\theta)[/tex]
when θ is 0;
[tex]F_n = mgcos (0)\\\\F_n = mg[/tex]
when θ is 90;
[tex]F_n = mgcos(90)\\\\F_n = 0[/tex]
Thus, as the angle of the ramp is increased, the normal force decreases.
(2)
The parallel force on an object on the ramp inclined to the ramp is calculated as follows;
[tex]F_x = mgsin(\theta)\\\\[/tex]
when θ is 0;
[tex]F_x = mgsin(\theta)\\\\F_x = mgsin(0) \\\\F_x = 0[/tex]
when θ is 90;
[tex]F_x = mgsin(90)\\\\F_x = mg[/tex]
Thus, as the angle of the ramp is increased, the parallel force increases.
(3)
The force of friction is calculated as follows;
[tex]F_n = \mu F_n[/tex]
[tex]F_k = \mu mgcos(\theta)[/tex]
[tex]F_k = \mu mg cos(0)\\\\F_k = \mu mg[/tex]
Thus, the angle is zero degree
(4)
The net force that would cause acceleration is calculated as follows;
[tex]F_k = Fn\\\\F_k = mg cos(\theta)\\\\F_k = 5 \times 9.8 \times cos(15)\\\\F_k = 47.33 \ N[/tex]
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Need help on this thank you
Answer:
TRUE - In any collision between two objects, the colliding objects exert equal and opposite force upon each other. This is simply Newton's law of action-reaction.
The first motor abilities a new born exhibits are
Answer:
here your answer
i am sorry if wrong
Content from magnets
Answer:
magnets are magnetic :)
Explanation:
(c) Changing water into vapour is condensation true or false
Answer:
true
Explanation:
The displacement of an object in SHM is described by the equation
[tex] x = cos\binom{2\pi}{3}t[/tex]
where x is in meters and t in seconds. Determine the velocity of the object at t = 0.6 s.
Answer:
[tex]-1.99\:\mathrm{m/s}[/tex]
Explanation:
Assuming that the equation is intended to be [tex]\displaystyle x=\cos\left(\frac{2\pi}{3}t\right)[/tex], we can find the velocity vs. time equation by taking the first derivative with respect to [tex]t[/tex]:
[tex]\displaystyle \frac{dx}{dt}=\frac{d}{dt}\left(\cos\left(\frac{2\pi}{3}t\right)\right)[/tex]
Recall the chain rule:
[tex]\displaystyle f(g(x))'=f'(g(x))\cdot g'(x)[/tex]
Therefore,
[tex]\displaystyle \frac{d}{dt}\left(\cos\left(\frac{2\pi}{3}t\right)\right)=-\sin\left(\frac{2\pi}{3}t\right)\cdot \frac{2\pi}{3}[/tex]
Therefore, the velocity vs. time equation of the object is [tex]\displaystyle v=-\sin\left(\frac{2\pi}{3}t\right)\cdot \frac{2\pi}{3}[/tex].
Substitute [tex]t=0.6\text{ s}[/tex] into this equation to find the velocity at that given time:
[tex]\displaystyle v=-\sin\left(\frac{2\pi}{3}(0.6)\right)\cdot \frac{2\pi}{3}\approx \boxed{-1.99\text{ m/s}}[/tex]
Một vật được ném xiên lên từ đỉnh của một tòa nhà cao 20m với góc 600 so với phương ngang, tốc độ ban đầu là 10m/s. Xác định độ tốc độ của vật lúc chạm đất?
Answer:
Một vật được ném xiên lên từ đỉnh của một tòa nhà cao 20m với góc 600 so với phương ngang, tốc độ ban đầu là 10m/s. Xác định độ cao cực đại của vật so với mặt đất?
Explanation:
1.Which term describes the sum of the number of protons and neutrons in an atom?
Your answer is → Mass number
The lantern rises up because the __________ of the _________inside the lantern is _________ compared to the air around it
liquid
Gas
Density
fire
less dense
the same density
more dense
paper
Answer:
gass fire and the same density
Explanation:
Technician a says that the piston engine starts and stops several times per 2nd technician B says that with the rotary engine each rotor has 3 working chambers so it acts like a 3 cylinder engine while it is at operating
Answer:
Explanation:
trfbehherebgqerfherbfhyqerwbfvhrbhferhgvr
A mover pushes a 45 kg crate across a level floor with a force of 300 N, but the crate accelerates at a rate of only 4.44 m/s2 because a friction force opposes the crate's motion. What is the magnitude of this force of friction? 300 N Force of Friction O A. 50 N N OB. 25 N C. 15 N 0 D. 100 N
By Newton's second law, the net force on the crate is
∑ F = 300 N - f = (45 kg) (4.44 m/s²)
where f is the magnitude of friction. Solve for f :
300 N - f = (45 kg) (4.44 m/s²)
f = 300 N - (45 kg) (4.44 m/s²)
f = 100.2 N ≈ 100 N (D)
what is an electrostatic phenomenon?
Answer:
Electrostatic phenomenon is a from the forces that electric charges exert on each other.
When people do a bungee jump theyare asked how much they weigh. Why?
Answer:
The elasticity of the bungee cord reduces the gravitational forces applied on the body during bungee jumping. For example, if a 100-pound individual jumps from a building and encounters 900 pounds of deceleration force, they will feel 9 "G's" of force.
hopefully this'll help
have a nice day!!! :D
A brick is thrown upward from the top of a building at an angle of 25 degrees to the horizontal with an initial speed of 15 m/s. If the brick is in flight for 4 s before it hits the ground, how tall is the building?
Answer:
First, find the maximum height, which according to the values given, can be stated as:
H=(u²sin²theta)/2g
u=15m/s, theta=25 degrees, g=9.8m/s²
H= (15² * (sin 25)²))/2*9.8
H= (225*0.179)/19.6
H= 40.275/19.6
=2.06m
To find the velocity at maximum height:
Use the formula
v²=u²-2gH
It's minus because the brick was thrown upwards
So plugging everything into the above formula:
v²=15²-2*9.8*2.06
v²=225–40.376
v²=184.624
v=√184.624
v=13.59m/s
Please HELP!!
Diagram's BELOW↓
1. What is the mass of the object experimented on in this situation?
a. 10 kg
b. 15 kg
c. 20 kg
d. 25 kg
2. What is the net force on this object?
a. 20 N to the left
b. 10 N up
c. 10 N down
d. 35 N to the right
Thank you in advance! Very appreciated! (I will mark brainliest) :)
The mass of the object experimented in the situation will be equal to 20 kg in each case. Hence, option C is correct.
What is Acceleration?Acceleration is the rate at which the speed and orientation of a moving object change over time. When a spot or object moves faster or slower along a straight line, it is said to be accelerated. Even though the speed is constant, motion on a circle accelerates so because the direction is always shifting. Both by and to acceleration for all other types of motion.
It is a vector quantity because it has both magnitude and direction.
The given data as per the question is,
Force, F = 30 N
Acceleration, a = 1.5 m/s²
[tex]Force = mass*acceleration[/tex]
30 N = m × 1.5
m = 30/1.5
m = 20 kg.
Therefore, the mass of the object is 20 kg.
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The figure shows a jet engine suspended beneath the wing of an airplane. The weight of the engine is 14900 N and acts as shown in the figure. In flight the engine produces a thrust of 61500 N that is parallel to the ground. The rotational axis in the figure is perpendicular to the plane of the paper. With respect to this axis, find the magnitude of the torque due to (a) the weight and (b) the thrust.
We have that for the Question, it can be said that With respect to this axis, the magnitude of the torque due to the weight and ,the thrust is
TW=19740N-mTT=130387.39N-mFrom the question we are told
The figure shows a jet engine suspended beneath the wing of an airplane. The weight of the engine is 14900 N and acts as shown in the figure. In flight the engine produces a thrust of 61500 N that is parallel to the ground. The rotational axis in the figure is perpendicular to the plane of the paper. With respect to this axis, find the magnitude of the torque due to (a) the weight and (b) the thrust.
a)
Generally the equation for the Torque due to weight is mathematically given as
[tex]TW=Engine weight*2.50*sin32\\\\TW=14900*2.50*sin32[/tex]
TW=19740N-m
b)
Generally the equation for the Torque due to thrust is mathematically given as
[tex]TT=Engine thrust*2.50*cos32\\\\TT=61500*2.50*cos32\\\\[/tex]
TT=130387.39N-m
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Which of the following statements about migration is true?
a. Migration is always from one region to another.
b. Animals always migrate within a region.
c. Migration always negatively impacts an ecosystem.
d. Animals migrate for various reasons.
Answer:
A and D
b and C are false
Answer:
D. Animals migrate for various reasons.
Explanation:
Got it right on Edge 2023. : )
i just need help plzzzz
Answer:
I think D it's the correct answer!! Let me know if i'm wrong
Explanation:
A string is wrapped around a solid cylinder with mass M and radius R. The free end of the string is held in place, allowing the cylinder to fall. Recall that the moment of inertia of a solid cylinder rotated about its center is given by MR2/2. All answers to this problem should be symbolic, purely in terms of the variables M, R, and g. (a) Find the linear acceleration (in m/s2) of the cylinder and the tension in the string (in Newtons) as the cylinder falls. (b) Now suppose the cylinder is hollow instead of solid. The moment of inertia of a hollow cylinder rotated about its center is given by MR2. What is the acceleration and tension in this case?
Answer:
I will use (a / R) for alpha the angular acceleration
T R = I a / R torque equals angular acceleration for cylinder
M g - T = M a linear acceleration of center of mass
T = M (g - a) = I a / R^2 from first equation
If I = 1/2 M R^2 then M ( g - a) = M a / 2 from above
or g = 3 a / 2 and a = 2 g / 3
Also we have T = M (g - a) = M (g - 2 g / 3) = g / 3
Substitute I = M R^2 for the hollow cylinder
Looks like a = g/2 for hollow cylinder