The slope of the graph current vs voltage can be explained with the next
[tex]\begin{gathered} V=IR \\ I=\frac{V}{R} \end{gathered}[/tex]In this case, current vs voltage means that current is y axis, and voltage is x axis. Based n that, the equation is the second one
[tex]I=\frac{V}{R}[/tex]And the slope is the inverse of the resistance. Dont forget that current - voltage is different to voltage - current
I just started a new lesson and have this study guide, but the material is unfamiliar to me and
2.79 eV
Explanation:Given
The wavelength, λ = 445 nm
The planck's constant, h = 6.626 x 10^-34 m² kg/s
The speed of light, c = 3 x 10^8 m/s
The energy of the photon is calculated as:
[tex]\begin{gathered} E=\frac{hc}{\lambda} \\ \\ E=\frac{6.626\times10^{-34}\times3\times10^8}{445\times10^{-9}} \\ \\ E=\frac{6.626\times3}{445}\times10^{-34+8+9} \\ \\ E=0.0447\times10^{-17} \\ \\ E=4.47\times10^{-2}\times10^{-17} \\ \\ E=4.47\times10^{-19}J \end{gathered}[/tex]But, 1 eV = 1.6 x 10^-19J
[tex]\begin{gathered} E=\frac{4.47\times10^{-19}}{1.6\times10^{-19}} \\ \\ E=2.79eV \end{gathered}[/tex]What is the relationship of the force on the spring and stretch of the spring.(This is the spring we are looking at weights can be attached to it)
We will have the following:
We know that for a simple spring the following is true:
[tex]\begin{cases}F=-kx \\ \\ F=ma\end{cases}[/tex]So:
[tex]-kx=ma[/tex]So, the stretch of the spring is directly proportional to the force.
It takes 225 kJ of work to accelerate a car from 20.1 m/s to 28.1 m/s. What is the car's mass?
Answer:
THE REMAINIG WILL BE 75
Explanation:
HOPE IT HELPS YOU
The tortoise and the hare A tortoise and a hare run to the East. The hare knows that it is faster, so it gives the tortoise a 30-meter head start. The tortoise is moving east at 1 m/s and the hare is moving east at 4 m/s.
The time taken for the hare to catch up with the tortoise is 10 seconds.
What is the time taken for the hare to catch the tortoise?
The time taken for the hare to catch up with the tortoise is calculated by applying the principle of relative velocity as shown below.
Distance travelled by hare + Distance travelled by tortoise = Total distance
V₁t + V₂t = d
where;
V₁ is the velocity of hareV₂ is the velocity of tortoiset is the time taken for them to meetd is the distance between themSince they are moving in the same direction, the relative velocity becomes
V₁t - V₂t = d
(V₁ - V₂)t = d
(4 - 1)t = 30
3t = 30
t = 30/3
t = 10 seconds
Thus, the time taken for the hare to catch up with the tortoise is 10 seconds.
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The complete question is below:
A tortoise and a hare run to the East. The hare knows that it is faster, so it gives the tortoise a 30-meter head start. The tortoise is moving east at 1 m/s and the hare is moving east at 4 m/s. At what time does the hare catch up with the tortoise?
mick took his friend out to dinner the bill was $40 but he applied a coupon then the total price was $33 what was the % off?
ANSWER
17.5 %
EXPLANATION
We have to find the percent change, given that the initial price was $40 and the final price was $33.
[tex]\text{ \% }change=\frac{final.price-initial.price}{initial.price}\times100[/tex][tex]\text{\% }change=\frac{33-40}{40}\times100=\frac{-7}{40}\times100=-0.175\times100=-17.5\text{\%}[/tex]The coupon was for 17.5% off
A net force of 43.1 N causes a mass to accelerate at a rate of 0.2 m/s2. Determine the mass in kilograms.
The mass of the an object caused to accelerate at 0.2m/s² by a 43.1 Newton force is 215.5 kilograms.
What is the mass of the object?A force is simply referred to as either a push or pull of an object resulting from the object's interaction with another object.
Force, according to Newton's Second Law is expressed as;
F = m × a
Where a is acceleration and m is the mass.
Given the data in the question;
Force applied F = 43.1N = 43.1kgm/s²Acceleration a = 0.2m/s²Mass m = ?Plug the given values into the formula above and solve for m.
F = m × a
43.1kgm/s² = m × 0.2m/s²
m = 43.1kgm/s² / 0.2m/s²
Mass m = 215.5kg
Therefore, the mass of the object is 215.5 kilograms.
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Students were experimenting with objects in a collision. Ball A moves with a constant acceleration by sliding down a frictionless incline plane before colliding with Ball B. Students used motion detectors to measure the velocity of Ball A at various points. Ball A (mass of 1.00 kg) began from rest, at position 1, on a ramp at a height of 1.25 m. At position 2, ball A was moving at 5.00 m/s. Ball A continues to roll at a constant 5.00 m/s when it collides with Ball B (mass of 1.00 kg) at position 3. Ball A comes to a complete stop. Ball B moves at a constant velocity after the collision.FORMULAS: PE = m•g•h (g=9.8m/s²) KE = 1/2•m•v² momentum = m•v
Given data:
* The mass of ball A is m_1 = 1 kg.
* The mass of ball B is m_2 = 1 kg.
* The initial velocity of ball A is u_1 = 5 m/s.
* The final velocity of ball A is v_1 = 0 m/s.
* The initial velocity of the ball B is u_2 = 0 m/s.
Solution:
According to the law of conservation of momentum, the net momentum of the system before the collision is equal to the net momentum of the system after the collision.
Thus,
[tex]m_1u_1+m_2u_2=m_1v_1+m_2v_2[/tex]Substituting the known values,
[tex]\begin{gathered} 1\times5+1\times0=1\times0+p_2 \\ 5+0=0+p_2 \\ p_2=5\text{ kgm/s} \end{gathered}[/tex]where p_2 is the momentum of the ball B,
Thus, the momentum of ball B after the collision is 5 kgm/s.
Hence, the third option is the correct answer.
n Fig. P9.82, the cylinder Figure P9.82 and pulley turn without friction about stationary horizontal axles that pass through their centers. A light rope is Pulley wrapped around the cylinder, passes over the pulley, and has a 3.00 kg box Cylinder Box suspended from its free end. There is no slipping between the rope and the pulley surface. The uniform cylinder has mass 5.00 kg and radius 40.0 cm. The pulley is a uniform disk with mass 2.00 kg and radius 20.0 cm. The box is released from rest and descends as the rope unwraps from the cylinder. Find the speed of the box when it has fallen 2.50 m.
The speed of the box when it has fallen 2.50 m is 4.22 m/s.
What is the speed of the box?
The speed of the box when it has fallen through the given height is calculated as follows;
Apply the principle of conservation of energy to determine the speed of the box.
ΔK.E = ΔP.E
K.Ef - K.Ei = mg(hf - hi)
K.Ef - 0 = mg(hf - 0)
K.Ef = mghf
where;
K.E is the final kinetic energy = rotational + translational kinetic energyhf is the final height of the box¹/₂mv² + ¹/₂I_pω² + ¹/₂I_cω²= mghf
¹/₂mv² + ¹/₂(I_p + I_c)ω² = mghf
where;
I_p is moment of inertia of the pulleyI_c is the moment of inertia of the cylinderω is the angular speed of the boxm is mass of the boxI_p = ¹/₂MR²
where;
M is mass of the pulleyR is the radius of the pulleyI_p = ¹/₂(2)(0.2)² = 0.04 kgm²
I_c = MR²
I_c = (5)(0.4)²
I_c = 0.8 kgm²
¹/₂mv² + ¹/₂(I_p + I_c)(v/r)² = mghf
¹/₂mv² + ¹/₂r²(I_p + I_c)v² = mghf
¹/₂v²[m + 1/r²(I_p + I_c)] = mghf
v²[m + 1/r²(I_p + I_c)] = 2mghf
v² [3 + 1/0.4²(0.04 + 0.8)] = 2(3)(9.8)(2.5)
v² [3 + 1/0.4²(0.04 + 0.8)] = 2(3)(9.8)(2.5)
8.25v² = 147
v² = 147/8.25
v² = 17.8
v = √17.8
v = 4.22 m/s
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The figure shows a standing wave oscillating at 100 Hz on a string. What is the wave speed ?
Answer:
The wavelength is 60 cm.
Explanation:
Speed = frequency x wavelength = 100 x 60 = 6000 cm/s
or if you wanted the answer in m/s
Speed = frequency x wavelength= 100 x 0.60 = 60 m/s
Set the cannon to have an initial speed of 20 m/s. For which situation do you think the cannon ball will go father: if it is set at a 60-degree angle, or if it is set at a 70-degree angle?
Question 2 options:
60 degrees
70 degrees
The cannon ball will go farther if the the angle of projection is set at 60 degrees
How to determine which angle will result in farther distance
Case 1:
Initial velocity (u) = 20 m/sAngle of projection (θ) = 60 ° Acceleration due to gravity (g) = 9.8 m/s²Horizontal distance (R) =?R = u²Sine(2θ) / g
R = 20² × Sine (2×60) / 9.8
R = 346.41 / 9.8
R = 35.35 m
Case 2:
Initial velocity (u) = 20 m/sAngle of projection (θ) = 70 ° Acceleration due to gravity (g) = 9.8 m/s²Horizontal distance (R) =?R = u²Sine(2θ) / g
R = 20² × Sine (2×70) / 9.8
R = 257.12 / 9.8
R = 26.24 m
From the above calculations, we can conclude that the ball will go farther, if the angle is 60 °
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Information givenknown: mass of Christine=60 kgmass of cart= 22 kg mass of hailey=69The two girls on the cart to the left pushing off of Conner take .3833 s to reach a distance of 0.3m. Conner reaches the same distance in 0.2333s. What is the mass of Conner?
We will have the following:
We use conservation of momentum to solve, that is:
[tex]\begin{gathered} (60kg+22kg+69kg)(0.3m/0.3833s)=(m+22kg)(0.3m/0.233s) \\ \\ \Rightarrow118.1841899kg\ast m/s=(m+22kg)(\frac{300}{233}m/s) \\ \\ \Rightarrow m+22kg=91.78968976kg\Rightarrow m=69.78968976... \\ \\ \Rightarrow m\approx69.8 \end{gathered}[/tex]So, Conner's mass is approximately 69.8 kg.
In a physics lab a student discovers that the magnitude of the magnetic field in a specific location near a long wire is 21.919 microTesla. If the wire carries a current of 35.483 A, what is the distance from the wire to that location ?
We will have the following:
First, we have that the permeability of free space is:
[tex]\mu_0=4\pi\ast10^{-7}Tm/A[/tex]Then:
[tex]\begin{gathered} B=\frac{\mu_0I}{2r\pi}\Rightarrow r=\frac{\mu_0I}{2B\pi} \\ \\ \Rightarrow r=\frac{(4\pi\ast10^{-7}Tm/A)(35.483A)}{2\pi(2.1919\ast10^{-5}T)}\Rightarrow r=0.3237647703...m \\ \\ \Rightarrow r\approx0.32m \end{gathered}[/tex]So, the distance is approximately 0.32 m.
make G the subject of the formula
F = GMM²/1²
HENCE WRITE THE DIMENSION FOR G
The value of G is FR²/M₁M₂. and the dimension of G is [M⁻¹L³T⁻²]..
The provided formula is of gravitational force F between two objects,
F = GM₁M₂/R²
Where M₁ is the mass of first object and M₂ is the mas of the other object while R is the distance between there centers and G is the universal gravitation constant.
To find the dimension of G, making G the subject of formula,
G = FR²/M₁M₂.
As we know, unit of mass is Kilogram (Kg), unit of force is Newton (N) and unit of distance is Meter (M).
Putting all the values, Units in the place of quantities,
G = N.R²/Kg.Kg
Now, using Dimensional analysis, and writing the dimensions of all the other units,
G = [MLT⁻²][L²]/[M][M]
G = [ML³T⁻²]/[M²]
G = [M⁻¹L³T⁻²]
The dimensions of G are [M⁻¹L³T⁻²].
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A tensile load of 190 kN is applied to a round metal bar with a diameter of 16mm and a gage length Of 50mm. Under this load the bar elastically deforms so that the gage length increases to 50.1349 mm and the diameter decreases to 15.99 mm. Determine the modulus of elasticity and Poisson s ratio for this metal.
The elasticity and Poisson's ratio for this metal is 0.232.
What is ratio?
The realation between two numbers which shows how much bigger one quantity is than another.
Sol-
As per the given question
P=190KN
d=16 mm
Lo=50mm
X=50.1349-50=0.1349mm
Y=15.99-16=-0.01mm
The formula-
E=ó/€
Ó=P/A
A=r/4 d^2 =π/4(16)^2=201.062 mm
ó={190(1000)}201.062=944.982 Mpa
E=944.982/0.002698=350.253 GPa
€y=-0.000625
v=0.232(answer)
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A 12 N force acts at a 25 degrees and an 8 N force acts at 65 degrees. Determine the magnitude and direction (include angle) of the resultant . Scale is 1cm = 1N
NEEDDD HELLPP ASAPPPPP
A 12 N force acts at a 25-degree angle, while an 8 N force acts at a 65-degree angle. The magnitude and direction (including angle) of the resultant force are 21.43.
Consider the formula for a force, F=M×Cos∝
As M is a Mass of an object and, ∝ Is the angle at which force is acting on an object.
[tex]F_{1}[/tex]=M × Cos ∝
M=12N, and Cos∝= Cos25 = 0.906
[tex]F_{1}[/tex] = 12× 0.906
∴ [tex]F_{1}[/tex]= 10.87N
For [tex]F_{2}[/tex] = 25×Cos65
= 10.56N
According to Superposition resulting force is the Sum of total forces.
[tex]F_{RESULTANT} = F_{1} + F_{2}[/tex]
= 10.87+ 10.56
=21.43N
∴The resultant force is 21.43.N
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Number 1. Part b: what are the final kinetic energy of the system
Given that there is a cart of mass, m = 0.12 kg moving with initial speed of, u1 = 0.45 m/s and it collides with another cart of mass, m = 0.12 kg with initial speed, u2 = 0 m/s
We have to find the initial and final kinetic energy.
(a) Initial kinetic energy,
[tex]\begin{gathered} K\mathrm{}E.1=\frac{1}{2}mv^2 \\ =\frac{1}{2}\times0.12\times(0.45)^2 \\ =0.012\text{ J} \end{gathered}[/tex]According to the conservation of linear momentum,
[tex]mu1+mu2=2mv[/tex]Here, v is the final speed.
[tex]\begin{gathered} 0.12\times0.45=2\times0.12\times v \\ v=\frac{0.45}{2} \\ =0.225\text{ m/s} \end{gathered}[/tex]Here, the final speed is 0.225 m/s.
(b) The formula to find kinetic energy is
[tex]K\mathrm{}E\mathrm{}=\frac{1}{2}(2m)v^2[/tex]Substituting the values, we get
[tex]\begin{gathered} K\mathrm{}E\mathrm{}=0.12\times(0.225)^2 \\ =6.075\times10^{-3}\text{ J} \end{gathered}[/tex]Hence the kinetic energy is 6.075 x 10^(-3) J.
Diffraction is:A.the difference in density of the compression and rarefaction parts of a sound wave.B.the change of frequency heard by an observer when sound waves come from a moving source.C.when waves suddenly appear in a medium without a source.D.the apparent bending of sound waves around obstacles.
Diffraction is a phenomenon that occurs when a wave hits an object or it passes through a small gap.
The propagation of the wave will have a circular pattern after the diffraction effect:
Therefore the correct option is D.
A football player kicks a football off a tee with a speed of 16 m/s at an angle of 63°. How far is the ball from the football player when it lands? How much farther would the ball go if he kicked it with the same speed, but at a 45° angle? Which ball will land first: the ball kicked at 16 m/s and at a 63° angle, or one kicked at 9 m/s and at a 45° angle?
A football player kicks a football off a tee with a speed of 16 m/s at an angle of 63°
The horizontal and vertical speed of the ball is given by
[tex]\begin{gathered} v_x=v\cos (\theta) \\ v_y=v\sin (\theta) \end{gathered}[/tex]Where v = 16 m/s and θ = 63°
[tex]\begin{gathered} v_x=16\cos (63\degree)=7.26\; \frac{m}{s} \\ v_y=16\sin (63\degree)=14.26\; \frac{m}{s} \end{gathered}[/tex]How far is the ball from the football player when it lands?
The range of the ball is given by
[tex]x=v_x\times t[/tex]Where t is the time the ball remains in the air.
The time (t) can be found as
[tex]y=v_yt+\frac{1}{2}at^2[/tex]y = 0 when the ball is in the air.
The acceleration is due to gravity (-9.8 m/s²)
[tex]\begin{gathered} 0=14.26t+\frac{1}{2}(-9.8)t^2 \\ 0=14.26t-4.9t^2 \\ 0=t(14.26-4.9t) \\ 0=14.26-4.9t \\ 4.9t=14.26 \\ t=\frac{14.26}{4.9} \\ t=2.91\; s \end{gathered}[/tex]Finally, the range is
[tex]x=v_x\times t=7.26\times2.91=21.13\; m[/tex]Therefore, the ball will land 21.13 meters far from the football player.
How much farther would the ball go if he kicked it with the same speed, but at a 45° angle?
We need to repeat the above calculations
The horizontal and vertical speed of the ball is given by
[tex]\begin{gathered} v_x=v\cos (\theta)=16\cos (45\degree)=11.31\; \frac{m}{s} \\ v_y=v\sin (\theta)=16\sin (45\degree)=11.31\; \frac{m}{s} \end{gathered}[/tex]The time (t) is given by
[tex]\begin{gathered} y=v_yt+\frac{1}{2}at^2 \\ 0=11.31_{}t+\frac{1}{2}(-9.8)t^2 \\ 0=11.31_{}t-4.9t^2 \\ 0=11.31_{}-4.9t \\ 4.9t=11.31_{} \\ t=\frac{11.31_{}}{4.9} \\ t=2.31\; s \end{gathered}[/tex]Finally, the range is
[tex]x=v_x\times t=11.31\times2.31=26.13\; m[/tex]Therefore, the ball will land 26.13 meters far from the football player.
Which ball will land first: the ball kicked at 16 m/s and at a 63° angle, or one kicked at 9 m/s and at a 45° angle?
The ball kicked at 16 m/s and at a 63° angle takes 2.91 s to land.
The ball kicked at 9 m/s and at a 45° angle will take
[tex]v_y=9\sin (45\degree)=6.36\; \frac{m}{s}[/tex][tex]\begin{gathered} y=v_yt+\frac{1}{2}at^2 \\ 0=6.36t+\frac{1}{2}(-9.8)t^2 \\ 0=6.36t-4.9t^2 \\ 0=6.36-4.9t \\ 4.9t=6.36 \\ t=\frac{6.36}{4.9} \\ t=1.30\; s \end{gathered}[/tex]So, the ball kicked at 9 m/s and at a 45° angle takes 1.30 s to land.
Therefore, the ball kicked at 9 m/s and at a 45° angle will land first.
Given two 2.00μC charges on the horizontal axis are positioned at x=0.8m, and the
other at x=-0.8m, and a test charge q = 1.28x10-18C at the origin.
(a) What is the net force exerted on q by the two 2.00μC charges? [5]
(b) What is the electric field at the origin due to 2.00μC charges? [5]
(c) What is the electric potential at the origin due to the two 2.00μC charges
Answer:Question 1
Given q1=2µC
q2=2µC
q= 1.2×10^-18C at origin
Net force exerted by two charges on q
F_1= force on q due to q1
F_2= force on q due to q2
F_net= F_(1-) F_2
= (Kqq_1)/r^2 - (kqq_2)/r^2 Then q_1=q_2=〖2×10〗^(-6)
F_net=0N
b) The electric field at charge q
E_net= E_1- E_2
= (kq_1)/r^2 - (kq_2)/r^2
Then q_1=q_2
E)_net= 0 N/C
c) The electric potential at origin due to two charge
V_net= V_(1 )- V_2
= (kq_1)/r - (kq_2)/r
Then q1= q2
V_net= 0 V
Explanation:
A resistor uses energy at a rate of 2.50W when there is a current of 4.00A passing through it. What must be the potential difference across the resistor?1.25V1.50V0.625V10.0V
As we know
[tex]P=\text{ V}\times I;[/tex]Where,
P= electric power= 2.50 W
V= potential difference= ?
I= current = 4.00A
Using above formula we get,
[tex]\begin{gathered} P=V\times I; \\ \therefore2.50=\text{ V}\times4; \\ V=\text{ }\frac{2.50}{4}=\text{ 0.625V} \end{gathered}[/tex]Final answer is :- 0.625 V
A rope pulls a 72.5 kg skier upa 21.7°slope with /k = 0.120.The rope is parallel to the slope,and exerts a force of 383 N. What isthe acceleration of the skier?(Unit = m/s?)Enter
A rope pulls a 72.5 kg skier upa 21.7°slope with /k = 0.120.The rope is parallel to the slope,and exerts a force of 383 N. The acceleration of the skier is approximately 2.97 m/s².
To find the acceleration of the skier, we need to use Newton's second law of motion and consider the forces acting on the skier.
Identify the forces acting on the skier:
The forces acting on the skier are the force of gravity (mg) and the force exerted by the rope (T), where m is the mass of the skier (72.5 kg) and g is the acceleration due to gravity (9.8 m/s²). The force exerted by the rope is parallel to the slope and can be calculated using the given value of 383 N.
Resolve the forces:
Since the rope is parallel to the slope, we need to resolve the force of gravity into components parallel and perpendicular to the slope. The component parallel to the slope is m * g * sin(21.7°).
Apply Newton's second law of motion:
Newton's second law states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration. Mathematically, Fnet = ma.
Determine the net force:
The net force acting on the skier is the difference between the force exerted by the rope and the component of the force of gravity parallel to the slope. Fnet = T - m * g * sin(21.7°).
Calculate the acceleration:
Using Newton's second law, we can rearrange the equation Fnet = ma to solve for the acceleration (a). a = Fnet / m.
Substitute the values and solve:
Substitute the known values into the equation to find the acceleration.
Therefore, the acceleration of the skier is approximately 2.97 m/s².
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Answer:
about 0.567 m/s²
Explanation:
You want the acceleration of a 72.5 kg skier up a 21.7° slope with µk = 0.120, towed by a rope exerting a force of 383 N.
ForcesThe force up the slope is 383 N.
The forces down the slope will be the sum of the force due to gravity and the friction force.
GravityThe force down the slope due to gravity is ...
F = m·g·sin(θ) = (72.5 kg)(9.8 m/s²)(sin(21.7°) ≈ 262.7 N
FrictionThe force due to friction will be the product of µk and the force normal to the slope:
F = m·g·cos(θ)·µk = (72.5 kg)(9.8 m/s²)cos(21.7°)·0.120 ≈ 79.22 N
Net ForceThe net force up the slope is ...
383 N -262.7 N -79.22 N ≈ 41.08 N
This will accelerate a mass of 72.5 kg in the amount of ...
A = F/m = 41.08 N/(72.5 kg) ≈ 0.567 m/s²
The acceleration of the skier is about 0.567 m/s² up the slope.
__
Additional comment
We don't have to figure the forces. Rather we can figure the gross acceleration due to the tow rope, then subtract the accelerations due to gravity and friction. This saves a few math operations as we don't have to multiply, then divide, by 72.5 kg.
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3. A boulder rolls with speed of 3.5 m/s off a cliff. It hits the ground 2.25 m from the base ofthe ledge. A) How high is the ledge? B) How long did it take the boulder to fall to the bottomof the cliff?DrawingVerticalHorizontal
Given data
*The given distance from the base of the ledge is R = 2.25 m
*The given speed is v = 3.5 m/s
The diagram is given below
(a)
Let (h) be the height of the edge
The formula for the distance from the base of the ledge is given as
[tex]\begin{gathered} R=v\times t \\ R=v\times\sqrt[]{\frac{2h}{g}} \\ h=\frac{R^2\times g}{2v^2} \end{gathered}[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} h=\frac{(2.25)^2\times9.8}{2\times(3.5)^2} \\ =2.025\text{ m} \end{gathered}[/tex]Hence, the height of the ledge is h = 2.025 m
(b)
The formula for the time taken by the boulder to fall to the bottom of the cliff is given as
[tex]t=\sqrt[]{\frac{2h}{g}}[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} t=\sqrt[]{\frac{2\times2.025}{9.8}} \\ =0.642\text{ s} \end{gathered}[/tex]Hence, the time taken by the boulder to fall to the bottom of the cliff is t = 0.642 s
009 (part 1 of 2) 10.0 points When a water gun is fired while being held horizontally at a height of 1.19 m above ground level, the water travels a horizontal distance of 1.98 m. Find the initial velocity of the water. The acceleration of gravity is 9.81 m/s^2
Answer in units of m/s.
010 (part 2 of 2) 10.0 points A child, holding the same gun in a horizontal
position, slides down a 33.0◦incline at a constant speed of 1.40 m/s. The child fires the gun when it is 4.36 m above the ground and the water takes 0.868 s to reach the ground. How far will the water travel horizontally?
Answer in units of m.
Answer:
Speed = 4 m/s
Explanation:
Given:
009 (part 1 of 2)
H =1 .19 m
L = 1.98 m
g = 9.81 m/c²
____________
V₀ - ?
Equation of motion horizontally:
L = V₀*t (1)
Equation of vertical motion:
H = g*t² / 2 (2)
From equation (2):
Time:
t = √ (2*H / g) = √ ( 2*1.19 / 9.81) ≈ 0,49 s
From equation (1):
Horizontal speed:
V₀ = L / t = 1.98 / 0.49 ≈ 4 m/s
When is a secondary source more helpful than a primary source?
A. When you want to confirm the conclusion with your own experiment
B. When you are not an expert in the field being studied in the experiment
C. When you do not need to know the conclusion of the experiment
D. When you want to know the exact data values
Option A is the correct answer: A secondary source is more helpful than a primary source when you want to confirm the conclusion with your own experiment.
Primary and secondary sources work well together to support the argument you are trying to make. Although secondary sources demonstrate how your work connects to earlier research, primary sources are more reliable as evidence. When you want to confirm the conclusion of your experiment, secondary sources prove to be more useful.
Original research is built on primary sources. You are able to discover novel information, offer solid justification for your claims, and provide reliable facts on your subject.
Secondary sources are useful for getting a comprehensive perspective of your subject and learning other researchers' methods. They frequently combine numerous primary materials that would take a lot of time and effort to obtain independently.
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Calculate the depth in the ocean at which the pressure is three times the atmospheric pressure
ANSWER:
20.17 meters
STEP-BY-STEP EXPLANATION:
Given:
Pressure = Po = 1.013x10^5 Pa
Pressure at depth = P = 3Po
Density of sea water = 1025 kg/m^3
We can calculate the depth as follows:
[tex]\begin{gathered} P=P_o+d\cdot g\cdot h \\ \text{ we solve for h} \\ dgh=P-P_o \\ h=\frac{P-P_o}{d\cdot g} \end{gathered}[/tex]We replacing and calculate the depth:
[tex]\begin{gathered} h=\frac{3\cdot P_o-P_o}{1025\cdot9.8}=\frac{2P_o}{10045}=\frac{2\cdot1.013\cdot10^5}{10045} \\ h=20.17\text{ m} \end{gathered}[/tex]Therefore, the depth is equal to 20.17 meters
You throw an object up with an initial velocity of Voy = 11 m/s from a height of y = 25 m. How long, in seconds, does it take for the object to reach the ground? What is the objects final velocity, in meters per second, as it impacts the ground? Find the time, in seconds, if you instead threw the object DOWN with the same velocity Voy
Calculate the ball's greatest height using the vertical motion model, h = -16t2 + vt + s, where v is the beginning velocity in feet/second and s is the height in feet.
What does a ball being thrown upwards accelerate to?A ball is thrown into the air, where it gradually loses speed until it abruptly comes to a rest at the peak of the motion. The body is travelling upward against gravity at the top, hence the acceleration there is 9.8 ms2. For example, g=9.8 ms2 is the formula for the acceleration caused by gravity.
Only at the greatest point of a body being hurled vertically upwards would velocity be zero because of the constant downward acceleration brought on by the gravitational force. As a result, velocity is zero throughout the rest of the motion.
A ball is originally moving upward when it is tossed into the air, for instance.
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Suppose you walk 16 m straight east and then 22 m straight south. At what angle, in degrees South of East, is a line connecting your starting point to your final position?
53.9 angle, in degrees South of East and 27.20 m is a line connecting your starting point to your final position.
What is initial position and final position?The distance in decimeters between the starting point and the ending position is in a straight line. The distance between an object's original position and its ultimate position is known as displacement.
Briefing:You walk 16 m straight east and then 22 m straight south. This forms a right angled triangle with a horizontal distance of 16 m, a vertical distance of 22 m and the hypotenuse is the distance between the ending and starting point. Let x represent the distance between the ending and starting point. Using Pythagoras theorem:
x² = 16² + 22²
x² = 256 + 484
x² = 740
Taking square root of both sides:
√x² = √740
x = √740
x = 27.20 m = distance between the ending and starting point.
Now use trigonometry:
sinθ=B/R
sinθ=22/27.20
sinθ= 0.808
θ = 53.9 degree. This is your angle.
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5. Was you hypothesis supported by the data collected for question 2? Why or why not ?
Answer:
Your hypothesis was not supported by the data collected
Explanation:
On question 2 you said that the point with the greatest kinetic energy would be point 5. However, when you analyze the data, the point with the greatest velocity was point 2 which means that this is the point with the greatest kinetic energy.
Therefore, your hypothesis was not supported by the data collected because based on the picture, it is hard to say where is the lowest point of the roller coaster.
A dog running of constant speed of 3m/s increases it's speed to 7m/s upon seeing a lion. if the mass of the dog is 20kg, the work it does in achieving the new speed is.......
A dog running of constant speed of 3m/s increases it's speed to 7m/s upon seeing a lion. if the mass of the dog is 20kg, the work it does in achieving the new speed is....... 400 J.
What is Speed?
Speed is the time rate at which an object is travelling along a path, whereas velocity is the pace and direction of an object's movement. In other words, velocity is a vector, whereas speed is a scalar valu
Work = Change in Kinetic energy
= ½m(v² - u²)
= ½ × 20 kg × [(7 m/s)² - (3 m/s)²]
= 10 kg × 40 m²/s²
= 400 J
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A superelastic collision is one in which1) kinetic energy before the collision equals kinetic energy after the collision.2) kinetic energy after the collision is zero.3) kinetic energy before the collision is less than kinetic energy after the collision.4) kinetic energy before the collision is greater than kinetic energy after the collision
A superelastic collision is the the collision in which the potential energy of the system is converted into the kinetic energy such the kinetic energy of the system after the collision is more than the kinetic energy of the system before the collision.
Hence, 3rd option is the correct answer.