Answer: Mature red blood cells don't have nuclei.
Explanation: " Platelets are irregularly shaped, have no nucleus, and typically measure only 2–3 micrometers in diameter. "
Hope this Helps!!
9
os End
bilo
b) A battery of unknown emf is connected across resistances, as shown in fig below. The voltage
drop across the 8ohms resistor is 20V. What will be the current reading in the ammeter? What is the ammeter?what is the emf of the battery?
Answer is in the file below
tinyurl.com/wpazsebu
A 64.0-kg person holding two 0.900-kg bricks stands on a 2.00-kg skateboard. Initially, the skateboard and the person are at rest. The person now throws the two bricks at the same time so that their speed relative to the person is 19.0 m/s. What is the recoil speed of the person and the skateboard relative to the ground, assuming the skateboard moves without friction?
Answer:
0.518 m/s
Explanation:
To solve this question, we would use the Law of conservation of momentum.
It is stated that the skateboard and the person were at rest initially, this means that their initial momentum is 0. Since they aren't moving.
And from the law if conservation of momentum, we know that initial momentum must be equal to final momentum.
The final momentum of the bricks will be 2* 0.9 * 19 = 34.2 kg m/s.
This means that the momentum of the person and that of the skateboard has to be 34.2 kg m/s also, although, it will be in the opposite direction.
Mass of the person plus that of the skateboard is
64 + 2 = 66 kg,
If we divide the momentum of the person and that of the skateboard by the mass of the both of them, we get the speed at which they are moving after the bricks are thrown
Recoil speed = (34.2 kg m/s)/(66 kg) = 0.518 m/s.
The recoil speed of the person is 0.518 m/s
Calculation of the recoil speed of the person:
As per the conservation of momentum, the initial momentum should be equivalent to the final momentum
So, here the final momentum should be
= 2* 0.9 * 19
= 34.2 kg m/s.
Now the mass of the person should be
= 64 + 2
= 66 kg,
Now the recoil speed should be
Recoil speed = (34.2 kg m/s)/(66 kg)
= 0.518 m/s.
Learn more about speed here: https://brainly.com/question/23855783
Q1 what is the direction of current?
Answer:
the external circuit is directed away from the positive terminal and toward the negative terminal of the battery
PLEASE HELP
Section 1 - Question 6
Wave Movement Through Media
What could be happening to the wave as it travels from left to right?
A
It's moving through a medium whose density stays the same
B
It's moving from a low density medium to a high density medium.
С
It's moving from a high density medium to a low density medium.
D
It's moving from a low density medium, to a high density medium, and then back to a low density medium
Answer: B
Explanation:
g 4.86 Separators are used to separate liquids of diff erent densities, such as cream from skim milk, by rotating the mixture at high speeds. In a cream separator, the skim milk goes to the outside while the cream migrates toward the middle. A factor of merit for the centrifuge is the centrifugal acceleration force (RCF), which is the radial acceleration divided by the acceleration due to gravity. A cream separator can operate at 9000 rpm (rev/min). If the bowl of the separator is 20 cm in diameter, what is the centripetal acceleration if the liquid rotates as a solid body, and what is the RCF
Answer:
Centripetal Acceleration = 88826.44 m/s²
RCF = 9054.7
Explanation:
First, we will find the value of the centripetal acceleration by using the following formula:
[tex]Centripetal\ Acceleration = \frac{v^2}{r}\\[/tex]
where,
v = linear speed of liquid or separator = rω
ω = angular speed of liquid or separator = (9000 rpm)(2π rad/rev)(1 min/60 s) = 942.48 rad/s
r = radius of seperator = diameter/2 = 20 cm/2 = 10 cm = 0.1 m
Therefore,
[tex]Centripetal\ Acceleration = \frac{(r\omega)^2}{r}\\Centripetal\ Acceleration = r\omega^2\\Centripetal\ Acceleration = (0.1\ m)(942.48\ rad/s)^2\\[/tex]
Centripetal Acceleration = 88826.44 m/s²
Now, for the RCF:
[tex]RCF = \frac{Centripetal\ Acceleration}{g}\\RCF = \frac{88826.44\ m/s^2}{9.81\ m/s^2}\\[/tex]
RCF = 9054.7
. Assume that the batter does hit the ball. If the bat's instantaneous angular velocity is 30 rad/s at the instant of contact, and the distance from the sweet spot on the bat to the axis of rotation is 1.25 m, what is the instantaneous linear velocity of the sweet spot at the instant of ball contact
Answer:
37.5 m/s
Explanation:
Using,
Formula
v = ωr....................... Equation 1
Where ω = instantaneous angular velocity, v = instantaneous linear velocity, r = radius or distance from the sweet spot of the bat to the axis of rotation.
From the question,
Given: ω = 30 rad/s, r = 1.25 m
Substitute these values into equation 1
v = 30(1.25)
v = 37.5 m/s.
Hence the instantaneous linear velocity of the sweet spot at the instant of ball contact is 37.5 m/s
If the Moon did not rotate at the same rate that it revolved, which of the following would be true?
Answer:
There will be no tides
Explanation:
A railroad handcar is moving along straight, frictionless tracks with negligible air resistance. In the following cases, the car initially has a total mass (car and contents) of 170 kgkg and is traveling east with a velocity of magnitude 4.60 m/sm/s. Find the final velocity of the car in each case, assuming that the handcar does not leave the tracks.
This question is not complete, the complete question is;
A railroad handcar is moving along straight, frictionless tracks with negligible air resistance.
In the following cases, the car initially has a total mass (car and contents) of 170 kg and is traveling east with a velocity of magnitude 4.60 m/s.
Find the final velocity of the car in each case, assuming that the handcar does not leave the tracks.
Part A
An object with a mass of 22.0 kg is thrown sideways out of the car with a speed of 2.50 m/s relative to the car's initial velocity.
Part B
An object with a mass of 22.0 kg is thrown backward out of the car with a velocity of 4.60 m/s relative to the initial motion of the car.
Answer:
Part A) the final velocity of the car is 4.6 m/s
Part B) the final velocity of the car is 5.28 m/s
Explanation:
Given the data in the question;
Total mass (m₁+m₂) = 170 kg
velocity of magnitude Vx = 4.60 m/s
PART A)
An object with a mass of 22.0 kg is thrown sideways out of the car with a speed of 2.50 m/s relative to the car's initial velocity,
i.e
m₂ = 22.0 kg
so m₁ = 170 - 22 = 148 kg
so, we apply conservation of momentum
since the object thrown out of the car, it has nothing to do with the car's velocity.
(m₁+m₂)Vx = m₁Vx₁ + m₂Vx₂
we substitute
(170)4.60 = 148Vx₁ + 22(4.60)
782 = 148Vx₁ + 101.2
148Vx₁ = 782 - 101.2
148Vx₁ = 680.8
Vx₁ = 680.8 / 148
Vx₁ = 4.6 m/s
Therefore, the final velocity of the car is 4.6 m/s
Part B)
An object with a mass of 22.0 kg is thrown backward out of the car with a velocity of 4.60 m/s relative to the initial motion of the car.
Vx = V(m₁+m₂) / ((m₁+m₂) - m₁)
we substitute
Vx = 4.60(170) / ((170) - 22)
Vx = 782 / 148
Vx = 5.28 m/s
Therefore, the final velocity of the car is 5.28 m/s
What do you think will happen to the Lunar phases if the moon was hit by an asteroid?
A fireworks rocket is fired vertically upward. At its maximum height of 90.0 m , it explodes and breaks into two pieces, one with mass mA = 1.35 kg and the other with mass mB = 0.270 kg . In the explosion, 810 J of chemical energy is converted to kinetic energy of the two fragments.
Required:
a. What is the speed of each fragment just after the explosion?
b. It is observed that the two fragments hit the ground at the same time. What is the distance between the points on the ground where they land? Assume that the ground is level and air resistance can be ignored.
Answer:
Ai. Speed of the fragment with mass mA= 1.35 kg is 34.64 m/s
Aii. Speed of the fragment with mass mB = 0.270 kg is 77.46 m/s
B. 475.3 m
Explanation:
A. Determination of the speed of each fragment.
I. Determination of the speed of the fragment with mass mA = 1.35 kg
Mass of fragment (m₁) = 1.35 kg
Kinetic energy (KE) = 810 J
Velocity of fragment (u₁) =?
KE = ½m₁u₁²
810 = ½ × 1.35 × u₁²
810 = 0.675 × u₁²
Divide both side by 0.675
u₁² = 810 / 0.675
u₁² = 1200
Take the square root of both side.
u₁ = √1200
u₁ = 34.64 m/s
Therefore, the speed of the fragment with mass mA = 1.35 kg is 34.64 m/s
II. I. Determination of the speed of the fragment with mass mB = 0.270 kg
Mass of fragment (m₂) = 0.270 kg
Kinetic energy (KE) = 810 J
Velocity of fragment (u₂) =?
KE = ½m₂u₂²
810 = ½ × 0.270 × u₂²
810 = 0.135 × u₂²
Divide both side by 0.135
u₂² = 810 / 0.135
u₂² = 6000
Take the square root of both side.
u₂ = √6000
u₂ = 77.46 m/s
Therefore, the speed of the fragment with mass mB = 0.270 kg is 77.46 m/s
B. Determination of the distance between the points on the ground where they land.
We'll begin by calculating the time taken for the fragments to get to the ground. This can be obtained as follow:
Maximum height (h) = 90.0 m
Acceleration due to gravity (g) = 10 m/s²
Time (t) =?
h = ½gt²
90 = ½ × 10 × t²
90 = 5 × t²
Divide both side by 5
t² = 90/5
t² = 18
Take the square root of both side
t = √18
t = 4.24 s
Thus, it will take 4.24 s for each fragments to get to the ground.
Next, we shall determine the horizontal distance travelled by the fragment with mass mA = 1.35 kg. This is illustrated below:
Velocity of fragment (u₁) = 34.64 m/s
Time (t) = 4.24 s
Horizontal distance travelled by the fragment (s₁) =?
s₁ = u₁t
s₁ = 34.64 × 4.24
s₁ = 146.87 m
Next, we shall determine the horizontal distance travelled by the fragment with mass mB = 0.270 kg. This is illustrated below:
Velocity of fragment (u₂) = 77.46 m/s
Time (t) = 4.24 s
Horizontal distance travelled by the fragment (s₂) =?
s₂ = u₂t
s₂ = 77.46 × 4.24
s₂ = 328.43 m
Finally, we shall determine the distance between the points on the ground where they land.
Horizontal distance travelled by the 1st fragment (s₁) = 146.87 m
Horizontal distance travelled by the 2nd fragment (s₂) = 328.43 m
Distance apart (S) =?
S = s₁ + s₂
S = 146.87 + 328.43
S = 475.3 m
Therefore, the distance between the points on the ground where they land is 475.3 m
An 1120 kg car traveling at 17 m/s is brought to a stop while skidding 40 m. Calculate the work done on the car by friction forces.
Answer:
Approximately [tex](-1.6 \times 10^{5}\; \rm J)[/tex] (assuming that the car was on level ground.)
Explanation:
When an object of mass [tex]m[/tex] is moving at a speed of [tex]v[/tex], the kinetic energy of that object would be [tex](1/2) \, m \cdot v^{2}[/tex].
Initial kinetic energy of the car:
[tex]\begin{aligned}&\frac{1}{2}\, m \cdot v^{2} \\ &= \frac{1}{2}\times 1120\; \rm kg \times (17\; \rm m \cdot s^{-1})^{2} \\ &\approx 1.6\times 10^{5}\; \rm J \end{aligned}[/tex].
After the car comes to a stop, the kinetic energy of this car would be [tex]0\; \rm J[/tex] because the car would not be moving.
Change to the kinetic energy of the car: [tex]\text{Final KE} - \text{Initial KE} = -1.6 \times 10^{5}\; \rm J[/tex].
If the car is traveling on level ground, friction would be the only force that contributed to this energy change. Hence:
[tex]\text{Work of friction} = \text{Energy change} = -1.6\times 10^{5}\; \rm J[/tex].
The value of the work that friction did is negative. The reason is that at any instant before the car comes to a stop, friction would be exactly opposite to the direction of the movement of the car.
The work of a force on an object is the dot product of that force and the displacement of that object. The dot product of two vectors of opposite directions is negative. Hence, in this question, the work that friction did on the car would be negative because the friction vector would be opposite to the movement of the car.
Is there a way to see moon and the sun at once?
In a double-slit experiment, light from two monochromatic light sources passes through the same double slit. The light from the first light source has a wavelength of 640 nm. Two different interference patterns are observed. If the 10th order bright fringe from the first light source coincides with the 12th order bright fringe from the second light source, what is the wavelength of the light coming from the second monochromatic light source
Answer:
533.33 nm
Explanation:
Since dsinθ = mλ for each slit, where m = order of slit and λ = wavelength of light. Let m' = 10 th order fringe of the first slit of wavelength of light, λ = 640 nm and m"= 12 th order fringe of the second slight of wavelength of light, λ'.
Since the fringes coincide,
m'λ = m"λ'
λ' = m'λ/m"
= 10 × 640 nm/12
= 6400 nm/12
= 533.33 nm
A particle with an initial linear momentum of 2.00 kg-m/s directed along the positive x-axis collides with a second particle, which has an initial linear momentum of4.00 kg-m/s, directed along the positive y-axis. The final momentum of the first particle is 3.00 kg-m/s, directed 45.0 above the positive x-axis.
a. the magnitude and direction (angle expressed counter-clockwise with respect to the positive x-axis) of the final momentum for the second particle
b. assuming that these particles have the same mass, % loss of their total kinetic energy after they collided
Answer:
a) p₂ = 1.88 kg*m/s
θ = 273.4 º
b) Kf = 37% of Ko
Explanation:
a)
Assuming no external forces acting during the collision, total momentum must be conserved.Since momentum is a vector, their components (projected along two axes perpendicular each other, x- and y- in this case) must be conserved too.The initial momenta of both particles are directed one along the x-axis, and the other one along the y-axis.So for the particle moving along the positive x-axis, we can write the following equations for its initial momentum:[tex]p_{o1x} = 2.00 kg*m/s (1)[/tex]
[tex]p_{o1y} = 0 (2)[/tex]
We can do the same for the particle moving along the positive y-axis:[tex]p_{o2x} = 0 (3)[/tex]
[tex]p_{o2y} = 4.00 kg*m/s (4)[/tex]
Now, we know the value of magnitude of the final momentum p1, and the angle that makes with the positive x-axis.Applying the definition of cosine and sine of an angle, we can find the x- and y- components of the final momentum of the first particle, as follows:[tex]p_{f1x} = 3.00 kg*m/s * cos 45 = 2.12 kg*m/s (5)[/tex]
[tex]p_{f1y} = 3.00 kg*m/s sin 45 = 2.12 kg*m/s (6)[/tex]
Now, the total initial momentum, along these directions, must be equal to the total final momentum.We can write the equation for the x- axis as follows:[tex]p_{o1x} + p_{o2x} = p_{f1x} + p_{f2x} (7)[/tex]
We know from (3) that p₀₂ₓ = 0, and we have the values of p₀1ₓ from (1) and pf₁ₓ from (5) so we can solve (7) for pf₂ₓ, as follows:[tex]p_{f2x} = p_{o1x} - p_{f1x} = 2.00kg*m*/s - 2.12 kg*m/s = -0.12 kg*m/s (8)[/tex]
Now, we can repeat exactly the same process for the y- axis, as follows:[tex]p_{o1y} + p_{o2y} = p_{f1y} + p_{f2y} (9)[/tex]
We know from (2) that p₀1y = 0, and we have the values of p₀₂y from (4) and pf₁y from (6) so we can solve (9) for pf₂y, as follows:[tex]p_{f2y} = p_{o1y} - p_{f1y} = 4.00kg*m*/s - 2.12 kg*m/s = 1.88 kg*m/s (10)[/tex]
Since we have the x- and y- components of the final momentum of the second particle, we can find its magnitude applying the Pythagorean Theorem, as follows:[tex]p_{f2} = \sqrt{p_{f2x} ^{2} + p_{f2y} ^{2} } = \sqrt{(-0.12m/s)^{2} +(1.88m/s)^{2}} = 1.88 kg*m/s (11)[/tex]
We can find the angle that this vector makes with the positive x- axis, applying the definition of tangent of an angle, as follows:[tex]tg \theta = \frac{p_{2fy} }{p_{2fx} } = \frac{1.88m/s}{(-0.12m/s} = -15.7 (12)[/tex]
The angle that we are looking for is just the arc tg of (12) which measured in a counter-clockwise direction from the positive x- axis, is just 273.4º.b)
Assuming that both masses are equal each other, we find that the momenta are proportional to the speeds, so we find that the relationship from the final kinetic energy and the initial one can be expressed as follows:[tex]\frac{K_{f}}{K_{o} } = \frac{v_{f1}^{2} + v_{f2} ^{2}}{v_{o1}^{2} + v_{o2} ^{2} } = \frac{12.5}{20} = 0.63 (13)[/tex]
So, the final kinetic energy has lost a 37% of the initial one.Interactive Solution 8.29 offers a model for this problem. The drive propeller of a ship starts from rest and accelerates at 2.38 x 10-3 rad/s2 for 2.04 x 103 s. For the next 1.48 x 103 s the propeller rotates at a constant angular speed. Then it decelerates at 2.63 x 10-3 rad/s2 until it slows (without reversing direction) to an angular speed of 2.42 rad/s. Find the total angular displacement of the propeller.
Answer:
Δθ = 15747.37 rad.
Explanation:
The total angular displacement is the sum of three partial displacements: one while accelerating from rest to a certain angular speed, a second one rotating at this same angular speed, and a third one while decelerating to a final angular speed.Applying the definition of angular acceleration, we can find the final angular speed for this first part as follows:[tex]\omega_{f1} = \alpha * \Delta t = 2.38*e-3rad/s2*2.04e3s = 4.9 rad/sec (1)[/tex]
Since the angular acceleration is constant, and the propeller starts from rest, we can use the following kinematic equation in order to find the first angular displacement θ₁:[tex]\omega_{f1}^{2} = 2* \alpha *\Delta\theta (2)[/tex]
Solving for Δθ in (2):[tex]\theta_{1} = \frac{\omega_{f1}^{2}}{2*\alpha } = \frac{(4.9rad/sec)^{2}}{2*2.38*e-3rad/sec2} = 5044.12 rad (3)[/tex]
The second displacement θ₂, (since along it the propeller rotates at a constant angular speed equal to (1), can be found just applying the definition of average angular velocity, as follows:[tex]\theta_{2} =\omega_{f1} * \Delta_{t2} = 4.9 rad/s * 1.48*e3 s = 7252 rad (4)[/tex]
Finally we can find the third displacement θ₃, applying the same kinematic equation as in (2), taking into account that the angular initial speed is not zero anymore:[tex]\omega_{f2}^{2} - \omega_{o2}^{2} = 2* \alpha *\Delta\theta (5)[/tex]
Replacing by the givens (α, ωf₂) and ω₀₂ from (1) we can solve for Δθ as follows:[tex]\theta_{3} = \frac{(\omega_{f2})^{2}- (\omega_{f1}) ^{2} }{2*\alpha } = \frac{(2.42rad/s^{2}) -(4.9rad/sec)^{2}}{2*(-2.63*e-3rad/sec2)} = 3451.25 rad (6)[/tex]
The total angular displacement is just the sum of (3), (4) and (6):Δθ = θ₁ + θ₂ + θ₃ = 5044.12 rad + 7252 rad + 3451.25 rad ⇒ Δθ = 15747.37 rad.HURRY
Explain the theory of plate tectonics and how they have changed Earth’s surface over time. Include the role of plate tectonics in the creation of landforms.
Explanation:
Plate tectonics is the scientific theory explaining the movement of the earth's crust. ... The movement of these tectonic plates is likely caused by convection currents in the molten rock in Earth's mantle below the crust. Earthquakes and volcanoes are the short-term results of this tectonic movement.
Answer:
The theory of plate tectonics describes the movement of the plates across the Earth's lithosphere (the crust and upper mantle) through immense periods of time. Earth's litoshphere is composed of 7 (or 8, depending on how they are defined) major plates and many more minor plates.
The movement is attributed to different phenomena stemming from Earth's rotation, gravity, and mantle dynamics. All of these forces play a role in influencing the size, shape, and positioning of the different landmasses that currently shape our continents and islands.
calculate the resistance if a potential difference of 3v exists when a current of 2A flows through the conductor
Answer:
Resistance = 1.5 ohms
Explanation:
Given:
Potential difference = 3v
Current flow = 2 A
Find:
Resistance
Computation:
Resistance = Potential difference / Current flow
Resistance = 3 / 2
Resistance = 1.5 ohms
Why is copper used extensively in electrical and thermal
applications?
Answer:
It is the best conducter of thermal and electrical energy
Find the kinetic energy of a ball of mass 200 grams moving at a speed of 20 m/s
Answer:
40 J
Explanation:
[tex]KE = \frac{1}{2} mv^{2} \\KE = \frac{1}{2} (0.200 kg)(20 m/s)^{2} \\KE = 40 J[/tex]
Discuss how the pressure cooker is designed to achieve temperatures above 100°C.
With rising heat, the steam pressure inside the pot builds up beyond atmospheric pressure, allowing the temperatures to rise well above boiling point. This design enables to save time, energy, and resources. The temperature inside a pressure cooker can well go beyond 110° C, which reduces the time needed to cook food.
Why do you need to apply a force in order to get the box to move?
Please why does a simple pendulum graph fail to pass through the origin ?
Answer:
The friction of the string and its pivotal anchor point cannot be eliminated. The precise measurement of the length of the pendulum is difficult to take by using meter sticks or rulers. The value of the acceleration due to gravity g in the locality is not constant and must be obtained from reliable sources.
You cannot completely remove the friction between the string and its key anchor point. Using meter sticks or rulers makes it challenging to take an accurate measurement of the pendulum's length. Since it is not constant, it is necessary to get accurate information on the local gravity acceleration related to g.
What is a pendulum?The body is suspended from a central object to act as a pendulum that swings back and forth as a result of gravity. Because the period—the amount of time between each full oscillation—is constant, pendulums are employed to control the movement of clocks. The formula for a pendulum's period T is T = 2 Square root of L/g, where L is the pendulum's length and g is the acceleration brought on by gravity.
Galileo, an Italian scientist, made the first observation about the consistency of a pendulum's period by contrasting the motion of a swinging lantern in a cathedral in Pisa with his heart rate.
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A light bulb is shown below, shining into a concave mirror, with its original light lines visible. Which statement best explains why the image of the bulb appears behind the mirror, as shown? .
A. The original light comes from there.
B. The reflected light comes from there.
C. The original light appears to come from there. This is often indicated with dotted apparent light lines.
D. The reflected light appears to come from there. This is often indicated with dotted apparent light lines.
Answer:
the answer is c
Explanation:
see the light appears from there and with the dotted lines you can clearly see the green line touches the dot okay, then the light appears smaller because of water and light source
An atom of tin has an atomic number of 50 and a mass number of 119. How many protons, electrons, and neutrons are found in one neutral atom of tin?
O 50 protons, 69 electrons, 50 neutrons
O 50 protons, 50 electrons, 69 neutrons
69 protons, 50 electrons, 69 neutrons
69 protons, 69 electrons, 50 neutrons
Answer:
50 protons 50 electrons and 69 neutrons...
Explanation:
the number of protons is equal to number of electrons. then mass number is the sum of protons and neutrons in a nucleus so for we to get the number of neutrons we take the mass number subtract the protons number.
Which of the following types of mirrors has an optical axis?
a convex mirror
a concave mirror
a double convex mirror
All of the choices are correct.
Answer:
its B
Explanation:
In searching the bottom of a pool at night, a watchman shines a narrow beam of light from his flashlight, 1.3 m above the water level, onto the surface of the water at a point 2.5 m from his foot at the edge of the pool.
Where does the spot of light hit the bottom of the pool, measured from the bottom of the wall beneath his foot, if the pool is 2.1 m deep pool?
Answer:
4.4 m
Explanation:
We are told the light from his flashlight, 1.3 m above the water level. Thus; h1 = 1.3m
Also,we are told that the light shone 2.5 m from his foot at the edge of the pool. Thus, L1 = 2.5 m
Angle of incidence θ1 is given by;
tan θ1 = L1/h1
tan θ1 = 2.5/1.3
tan θ1 = 1.9231
θ1 = tan^(-1) 1.9231
θ1 = 62.53°
Using Snell's law, we can find the angle of refraction from;
Sin θ2 = (η_air/η_water) Sin θ1
Where;
η_air is Refractive index of air = 1
η_water is Refractive index of water = 1.33
Thus;
Sin θ2 = (1/1.33) × sin 62.53°
Sin θ2 = 0.6671
θ2 = sin^(-1) 0.6671
θ2 = 41.84°
We want to find where the spot of light hit the bottom of the pool if the pool is 2.1 m deep. Thus, h2 = 2.1 m
Now, the spot can be found from;
L = L1 + L2
Where L2 = (h2) tan θ2
L = 2.5 + 2.1 tan 48.84
L = 2.5 + (2.1 × 0.8954)
L ≈ 4.4 m
Please help me with this
10 points
all i have rn :/
thanks <3
Answer: c the last one
Explanation:
A scientist adds different amounts of salt to 5 bottles of water. She then measures how long it takes for the water to boil. What is the responding variable in this experiment
Answer:
the responding variable is the water boiling
Explanation:
a responding variable is the same thing as a dependent variable and an independent variable you change the independent variable is the amount of salt, the control group is how long water takes to boil without adding salt, and a constant is the same amount of water
12. Identify the Leader
A projectile is fired with an initial velocity of 120.0 m/s at an angle, θ, above the horizontal. If the projectile’s initial horizontal speed is 55 meters per second, then angle θ measures approximately
Answer:
algm sabe tô precisando muito