The distance travelled by the object in the first 3 seconds will be 29.4 m.
What is acceleration?The rate of change of velocity with respect to time is known as acceleration. According to Newton's second law, the eventual effect of all forces applied to a body is its acceleration.
The equation of the motion along the inclined plane;
[tex]\rm mg sin \theta - F_k = ma \\\\ 2 \times 9.81 \times sin 60^0 -3.929 = 2a \\\\ a= 6.533 \ m/sec ^2[/tex]
The distance traveled by the block for the given time interval ,t=3 sec is ;
[tex]\rm S= ut+ \frac{1}{2}at^2 \\\\ S= 0 \times 3 + \frac{1}{2} \times 6.533 \times (3)^2\\\\ S=29.4 \ m[/tex]
Hence,the distance travelled by the object in the first 3 seconds will be 29.4 m.
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9
Robin chases a squirrel moving 6 meters from the porch to a tree in 1.2 seconds. Wha
was his rate of movement?
Answer:
5 m/s
Explanation:
Rate of movement is distance / time
6 m / 1.2 s = 5 m/s
3. Which vehicle has more momentum? *
1 point
A.
B
VA= 25 m/s
Ug = 20 m/s
ma= 15000 Kg me = 15000 Kg
Ο Α
О.
B
Answer:
the one with v = 25 m/s
Explanation:
Momentum = m * v
if they both have the same mass (15000 kg) , then the one with the higher v has more momentum...I think A= 25 m/s
20. Which of the following situations is reaction time exhibited?
A. Jenny and Tina plays jumping rope
B. Ben and Ted plays soccer and football C. Cina and Horia attend swimming sessions
D. Bovet and Cris perform acrobic exercises
Jenny and Tina play jumping rope,Ben and Ted plays soccer and football and Bovet and Cris perform aerobic exercises are the following situations is reaction time exhibited. Option A,B and D are correct.
What is reaction time?The reaction time of an organism is a measurement of how quickly it responds to a stimulus.
The time interval between the presentation of the stimulus and the manifestation of a suitable voluntary response in the subject is defined as RT.
In the following situations is reaction time exhibited is;
A. Jenny and Tina play jumping rope.
B. Ben and Ted play soccer and football .
D. Bovet and Cris perform aerobic exercises.
Hence,option A,B and D are correct.
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3. An assassin who climbed to the top of the catwalk balcony of a mansion, to get a
good view of his target who was 6.1 ft tall. The assassin pulled the trigger when
the target is 294 ft away, discharging the bullet at a 63° angle to the level of the
victim's head. Find the height that the assassin shot the target from.
The height of the mansion that the assassin shot the target from is determined as 1,164.9 ft.
Time of motion of the bullet
The time of motion of the bullet is calculated as follows;
-h = vsinθ(t) - ¹/₂gt²
-6.1 = (vt) sin63 - ¹/₂(32)t²
-6.1 = 0.89vt - 16t² ---(1)
From horizontal motionX = vcosθ t
294 = (vt)cos63
294 = 0.454vt
vt = 294/0.454
vt = 647.577 ---(2)
solve (1) and (2) together;
-6.1 = 0.89(647.577) - 16t²
16t² = 576.344 + 6.1
16t² = 582.44
t² = 582.44/16
t² = 36.4
t = √36.4
t = 6.03 s
Initial velocityv = 647.577/t
v = 647.577/6.03
v = 107.4 ft/s
Height of the mansionh = vt + ¹/₂gt²
h = (107.4 x sin63 x 6.03) + ¹/₂(32)(6.03)²
h = 1,158.8 ft
H = 1,158.8 ft + 6.1 ft = 1,164.9 ft
Thus, the height of the mansion that the assassin shot the target from is determined as 1,164.9 ft.
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Throughout the reflection, make sure you have a
down menus to complete the statements.
copy of the Student Guide and your data tables. Use the drop-
In Part I of this experiment, the
was intentionally manipulated. This was the independent
variable. The dependent variable measured was the
In Part II of this experiment, the
was intentionally manipulated. This was the
independent variable. The dependent variable measured was the
Answer:
The dependent variable is the variable that is studied while the independent variable is the variable that is being manipulated
Answer: In this experiment, the
✔ fan speed
was intentionally manipulated. This was the independent variable.
The dependent variable measured was the
✔ acceleration
.
Explanation:
5. A person on a trampoline jumps straight up. When the person is 3 m above the surface and
with upward velocity of 2.5 m/s, drops a ball. a) What is the speed of the ball at the time of
impact? b) How long before the ball impacts the surface of trampoline?
a)The speed of the ball at the time of impact will be 8.1 m/sec.
b)The time taken by the ball impacts the surface of trampoline will be 1.1 second.
What is speed?Speed is defined as the rate of change of the distance or the height attained. it is a time-based quantity. Its SI unit is m/sec.
Given data;
g( gravitational acceleration) = 9.81 m/s2
u (initial velocity of ball) = 2.5 m/s
h₁(initial height of the ball) = 3.0 m
v (speed of the ball at the time of impact ) ?
The time taken by ball for upward journey = u / g
t₁ = u / g
t₁ = (2.5 m/s) / (9.81 m/s2) =
t₁ =0.2548 s
From second equation of motion,upward vertical distance covered by ball is;
[tex]\rm h_2= \frac{1}{2} gt^2\\\\ h_2=\frac{1}{2} \times 9.81 \times (0.2548)^2\\\\ h_2=0.3184[/tex]
The total distance in the downward direction travelled;
[tex]\rm h_3=h_1+h_2 \\\\ h_3=3.0+0.3184 \\\\ h_3=3.3184 \ m[/tex]
The time taken for the downward motion is;
[tex]\rm t_2 = (\frac{2 \times h_3 }{g} )^\frac{1}{2}\\\\ t_2 = 0.825 \ m/sec[/tex]
The speed of the ball at the time of impact is found as;
[tex]\rm V=gt_2 \\\\ V= 9.81 \times 0.8225 \\\\ V= 8.07 \ m/sec[/tex]
The total time of the flight of ball is;
[tex]\rm t_3 = t_1+t_2 \\\\ t_3=0.2548+0.8225 \\\\ t_3=1.0774 \ sec[/tex]
Hence,the speed of the ball at the time of impact will be 8.1 m/sec and time taken by the ball impacts the surface of trampoline will be 1.1 second.
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QUESTION 4 A ball is dropped from a height. If it takes 0.2 s to cross the last 6 m before hitting the ground, find the approximate height from which it is dropped.
Answer:
S = V t + 1/2 g t^2 where the ball has a speed of V and falls for .2 sec
6 = V t + 4.9 * .04 = V * .2 + .2
V = (6 - .2) / .2 = 29 m/s speed entering area
T = V / g = 29 / 9.8
T = 2.96 sec time to reach speed V
H = 1/2 g t^2 time to fall a distance H
H = 4.9 * 2.96^2 = 42.9 m height from which ball was dropped
Check using first equation:
S = 29 * .2 + 4.9 + .2^2 = 6 m
If you drive 2 hours at 50mph and then drive 4 hours at 65mph, how far did you drive in all?
Answer:
i think its 400 miles driven
Explanation:
attempted to do it...
How many times is the motion of the object stopped?
Answer:
4
Explanation:
We can say the motion of the object is stopped when its velocity is 0. That means we just have to check how many times the line of the graph intersects the x-axis. Hence, we can say that the object of the motion is stopped 4 times.
Three resistances of 12 , 16 , and 20 are connected in parallel. What resistance must be connected in series with this combinations to give a total resistance of 25 ?
Answer:
Total resistance in parallel is Rp
So (1/Rp) = (1/12)+(1/16)+(1/20)
Rp = ?
impedance is 25
25 = Rp + Rs
put the value of Rp you'll get the answer.
Describe what will happen to
the molecules below if the
temperature is increased by
20°C.
To measure the vast distance in space ,scientists use a unit called a ______, which is how far travels in one year.
The closest star to Earth is ______.
(Look at the picture)
Answer:
"light year" - about 6 million million miles
The nearest star is alpha Centauri - about 24 million million miles away
How long will it take you to ride your bike 120 miles if you are riding at 5mph?
Which phenomenon provides evidence that the hydrogen atom has discrete energy levels?
The phenomenon known as the Emission Spectra.
Emission spectrum is the phenomenon provides evidence that the hydrogen atom has discrete energy levels.
What is emission spectrum?Emission spectrum is the spectrum of the different energy levels. The occurrence of emission spectrum gives proof that the hydrogen atom has distinct energy levels.
Emission spectrum is the phenomenon provides evidence that the hydrogen atom has discrete energy levels
Hence, the phenomenon is known as emission spectrum.
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Two spherical objects have masses of 7.5 x 10 kg and 9.2 x 10'kg. Their centers are
separated by a distance of 2.5 x 103 m. What is the magnitude of the gravitational force
between them?
Answer: 7.364 x 10^-4 N
Explanation:
[tex]$$Given:$$\begin{aligned}m_{1} &=7.5 \times 10^{5} \mathrm{~kg} \\m_{2} &=9.2 \times 10^{7} \mathrm{~kg} \\r &=2.5 \times 10^{3} \mathrm{~m}\end{aligned}$$Here, $m_{1}, m_{2}$, and $r$ are the first mass, the second mass, and the distance[/tex]
[tex]$$The gravitational force exerted between the masses is determined as follows.$$F=G \frac{m_{1} m_{2}}{r^{2}}$$Here, $G$ is the gravitational constant.Substitute the known values.$$\begin{aligned}F &=\left(6.67 \times 0^{-11} \mathrm{~N} \mathrm{~m}^{2} / \mathrm{kg}^{2}\right) \frac{\left(7.5 \times 10^{5} \mathrm{~kg}\right)\left(2.5 \times 10^{7} \mathrm{~kg}\right)}{\left(2.5 \times 10^{3} \mathrm{~m}\right)^{2}} \\&\bold{=7.364 \times 10^{-4} \bold{\mathrm{~N}}}\end{aligned}$$[/tex]
in the R-L-C series circuit shown, suppose R = 200 Ω, L = 70 mH, C = 2.0 μF, Vmax = 80V and the angular frequency of the voltage source ω = 2,000 rad/s. (a) Find the reactances and , and the impedance . (3 marks) (b) Find the maximum current amplitude, Imax and the maximum voltage amplitude across each circuit element. (4 marks). (c) Explain why the maximum voltage across each element add up to more than 80 V. (3 marks)
Hello!
A)
We can solve for the reactances of each element using the following equations.
Capacitive reactance:
[tex]X_C = \frac{1}{\omega C}[/tex]
[tex]X_C[/tex] = Capacitive reactance (Ω)
ω = Angular frequency (2000 rad/sec)
C = Capacitance (2 μF)
Plug in the given values and solve.
[tex]X_C = \frac{1}{(2000)(0.000002)} = \boxed{250 \Omega}[/tex]
Inductive reactance:
[tex]X_L = \omega L[/tex]
[tex]X_L[/tex] = Inductive reactance (Ω)
L = Inductance (70 mH)
Solve:
[tex]X_L = 2000 * 0.07 = \boxed{140 \Omega}[/tex]
We can solve for impedance using the following equation:
[tex]Z = \sqrt{R^2 + (X_L - X_C)^2}[/tex]
Z = Impedance (Ω)
R = Resistance (200Ω)
Find the impedance using the values above and the given resistance:
[tex]Z = \sqrt{200^2 + (140 - 250)^2} = \boxed{228.254 \Omega}[/tex]
B)
The maximum current amplitude is found using Ohm's law:
[tex]I_{Max} = \frac{V_{Max}}{Z}[/tex]
[tex]V_{Max}[/tex] = 80V
Z = 228.254 Ω
Solve.
[tex]I_{Max} = \frac{80}{228.254} = \boxed{0.3505 A}[/tex]
To find the max voltages across each element, we can also use Ohm's Law.
For the resistor:
[tex]V_{R, max} = i_{Max} R\\\\V_{R, max} = (0.3505)(200) = \boxed{70.097 V}[/tex]
For the capacitor:
[tex]V_{C, max} = i_{Max}X_C\\\\V_{C, max} = (0.3505)(250) = \boxed{87.622 V}[/tex]
For the inductor:
[tex]V_{L, max} = i_{Max}X_L\\\\V_{L, max} = (0.3505)(140) = \boxed{49.068 V}[/tex]
C)
The maximum voltage across each element is greater than 80 V because of the transfer of energy between the capacitor and inductor that occur at different times (energy is stored and released) and being in addition to the voltage of the power source.
At left A red ball in a box with arrows pointing away from the ball in all directions. In the middle, a blue ball in a box with arrows pointing toward the ball from all directions. At right, a grey ball in a box. Which object represents a negatively charged particle? Which object represents a positively charged molecule? Which object represents an uncharged molecule? Which object will not move when in an electric field?
Answer: The answers are, 1) B. 2) A. 3) C. 4) C.
here is the proof maybe u will understand it more than the numbers and letters lm.ao
Explanation:
The red ball with arrows pointing away is a negatively charged particle, the blue ball is a positively charged molecule, and the grey ball is an uncharged molecule. Electric fields don't move uncharged molecules.
Analyse each scenario to identify representations:
A negatively charged particle is the red ball in a box with all arrows pointing away from it. The red ball generates an electric field, and negatively charged particles radiate electric field lines. The arrows indicate how a positive test charge would move in the electric field.
A positively charged molecule is the blue ball in a box with all arrows pointing to it. An electric field converges on the blue ball, signifying its destination. The electric field lines point towards positively charged particles, indicating the direction a positive test charge would move.
A boxed grey ball depicts an uncharged molecule. The ball has no arrows. Uncharged molecules do not emit or experience electric fields. The grey ball is electrically neutral because it lacks electric pitch lines.
An electric field will not move the grey ball in a box (uncharged molecule). Electric fields do not interact with uncharged molecules. Thus, an electric field will keep the grey ball still.
Science and engineering require understanding these representations and their electric field behaviour. Scientists and engineers can create and manipulate electrical systems, electronics, and other technologies that use electric charge interactions by recognising charged particles, uncharged molecules, and electric field effects.
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someone help me answer this.
Bug Walking on Pivoted Ring A ring of radius R and mass my lies on its side on a frictionless table. It is pivoted to the table at its rim. A bug of mass my walks on the ring with constant speed v relative to the ring, starting at the pivot, when the ring is initially at rest. Take k to point out of the page. top view frictionless table ring bug Pluot Point (a) What is the angular velocity of the ring when the bug is halfway around? Express you answer in terms of some or all of the following: mi, m2, u, R and k. (b) What is the angular velocity of the ring when the bug is back at the pivot? Express you answer in terms of some or all of the following: mı, m2, v, R and k.
The angular velocity of the ring when the bug is halfway around and the angular velocity of the ring when the bug is back at the pivot is [m₂v / {(2m₁ +m₂)R}].
What is angular velocity?The velocity of a particle when moving in the circular path.
Let speed of the bug with respect to ground is u.
Speed of bag with respect to ring will be
v = u - (- Rω) =
Then, u = v- Rω...............(1)
Angular momentum of ring and bug will remain conserved.
Initial momentum: L ring + Lbug =0
Final momentum: -2m₁ R²ω + m₂uR =0...............(2)
Using equation (1) and (2), the angular velocity expression will be
ω =[m₂v / {(2m₁ +m₂)R}] in positive z direction
Thus, the angular velocity of the ring when the bug is halfway around and the angular velocity of the ring when the bug is back at the pivot is [m₂v / {(2m₁ +m₂)R}].
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On a frictionless horizontal air table, puck A (with mass 0.251 kg ) is moving toward puck B (with mass 0.371 kg ), which is initially at rest. After the collision, puck A has velocity 0.120 m/s to the left, and puck B has velocity 0.654 m/s to the right.
What was the speed vAi of puck A before the collision?
Calculate ΔK , the change in the total kinetic energy of the system that occurs during the collision.
(a) The speed of puck A before the collision is determined as 0.847 m/s.
(b) The change in the total kinetic energy of the system that occurs during the collision is 0.009 J.
Speed of puck A before the collisionThe speed of puck A before the collision is calculated from the principle of conservation of linear momentum as follows;
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
0.251u₁ + 0.371(0) = 0.251( -0.12) + 0.371(0.654)
0.251u₁ = 0.2125
u₁ = 0.2125/0.251
u₁ = 0.847 m/s to the right
Change in the total kinetic energy of the systemInitial kinetic energy = ¹/₂ x 0.251 x (0.847)² + ¹/₂ x 0.371 x (0) = 0.09 J
Final kinetic energy = ¹/₂ x 0.251 x (-0.12)² + ¹/₂ x 0.371 x (0.654)² = 0.081 J
ΔK.E = 0.081 - 0.09 = 0.009 J ≈ 0 J
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5. How do you determine the strength of the electric field of a charge?
A. Putting a small test charge close to the electric field.
B. Using a voltmeter.
C. Using an ammeter.
D. Putting a large test charge close to the electric field.
A sprinter starting to run from rest, reaches a speed of 8m/s in 2 second, what is the average power needed to accelerate like this ?
Answer:
okay
Explanation:
power is rate of doing work
power is workdone over time taken
but you don't have information concerning the force
Power = work/time
Work = force x distance
Force = mass x acceleration
So Power = (mass x acceleration x distance) / (time)
acceleration = 8m/s / 2sec = 4 m/s²
distance = (avg speed) x (time) = 8 m
time = 2 sec
we don't know the runner's mass
Power = (mass) x (4 m/s²) x (8 m) / (2 sec)
Power = (16 x mass in kg) watts
=============================================
Slightly easier way to do it:
Power = (runner's kinetic energy at the end) / (time)
Power = (1/2 m v²) / (time)
Power = (1/2 mass) (64 m²/s²) / (2 sec)
Power = (16 x runner's mass in kg) watts
Find the wavelength of light given off by a hydrogen atom when its electron drops from the n = 4 to n = 1 energy level
Answer: 1.55 x 10 -19J on top
Explanation:
Find energy
[tex]\\ \rm\Rrightarrow E=-13.6\left(\dfrac{1}{6^2}-\dfrac{1}{4^2}\right)[/tex]
[tex]\\ \rm\Rrightarrow E=-13.6\left(\dfrac{1}{36}-\dfrac{1}{16}\right)[/tex]
[tex]\\ \rm\Rrightarrow E=-13.6(0.0278-0.0625)[/tex]
[tex]\\ \rm\Rrightarrow E=-13.6(-0.347)[/tex]
[tex]\\ \rm\Rrightarrow E=0.47eV[/tex]
Now
[tex]\\ \rm\Rrightarrow E=h\nu[/tex]
[tex]\\ \rm\Rrightarrow \nu=\dfrac{E}{h}[/tex]
[tex]\\ \rm\Rrightarrow \nu=\dfrac{0.47\times 10^{-19}}{6.626\times 10^{-34}}[/tex]
[tex]\\ \rm\Rrightarrow \nu=0.071\times 10^{15}[/tex]
[tex]\\ \rm\Rrightarrow \nu=7.1\times 10^{13}Hz[/tex]
Do lighting filters use additive or subtractive properties to bring light of a certain color to our eyes?
Answer:
Different colour lights (RBG) uses additive properties to bring light of a certain color to our eyes.
Explanation:
:))
can someone help with the vocabulary of the relative age of rocks
Gamma rays can cause cancer, but they can also be used to treat cancer. How
would you answer a sixth grader who asks you, "Is radiation good or bad?"
Answer:
I think it depends on the purpose of using Gamma rays. It can be both good and bad
A ball of mass 0.4kg is dropped from a height of 30m. Determine the potential energy on impact to the ground
How does wind help you to know gases are matter
A ball rolls 60 centimeters along a sidewalk in 5 seconds what is the speed of the ball.
Answer:
12 cm/s
Explanation:
Quite simply, you are looking for cm/s
so 60 cm / 5s = 12 cm/s
Answer:
12 cm/second
Explanation:
To calculate the speed of an object, simply obtain the distance traveled by the ball and divide it by the total time it took to travel 60 centimeters.
⇒ Distance/Time = Speed
In this case, the ball traveled 60 centimeters in 5 seconds.
Therefore, the speed of the ball is;
60 centimeters (Distance)/5 seconds (Time) = 12 cm/second
What is the potential difference across the source?
60 V
220 V
440 V
120 V
Answer:
120 v
Explanation:
The two resistors have an equivalent of 20 * 30 /(20+30) = 12 ohms
10 amps of current in the circuit
v = ir
= 10 * 12 = 120 volts
Here is another way:
The two resistors are in prallel so the voltae across both is the same
use the one on the right v = ir = 4 x 30 = 120 v
A (1.28×10-5 C) charge is moving at (5.63×107 m/s) perpendicular to a magnetic field
of (8.91×10-4 T). What is the magnitude of the magnetic force on the charge?
A. 6.42×10 -1 N
B. 5.02×10 4 N
C. 7.21×10 2 N
D. 1.14×10 -8 N
Hello!
We can use the following equation for magnetic force on a moving particle:
[tex]F_B = qv \times B[/tex]
[tex]F_B[/tex] = Magnetic Force (? N)
q = Charge of particle (1.28 × 10⁻⁵C)
v =velocity of particle (5.63× 10⁷m/s)
B = Strength of magnetic field (8.91 × 10⁻⁴T)
This is a cross-product, so the magnetic force depends on the SINE of the angle between the particle's velocity vector and the magnetic field vector.
Since the charge is moving PERPENDICULAR to the field, the angle between the velocity and magnetic field is 90°. The sine of 90° = 1, so we can simplify the equation to:
[tex]F_B = qvB(1) = qvB\\[/tex]
Plug in the values and solve.
[tex]F_B = (1.28*10^{-5})(5.63*10^7)(8.91*10^{-4}) = \boxed{\text{A. } 0.6421 N}[/tex]
