Answer:
a) a = - 1.76 m / s², b) F = 20.5 N, c) w = 3.55 rad / s
Explanation:
a) a simple harmonic motion is described by the expression
x = A cos (wt + Ф)
in this case they give us the frequency
w = 2π f
w = 2π 0.40
w = 2.51 rad / s
as the maximum elongation is 0.65 m this corresponds to the amplitude of the movement
A = 0.65 m
to find the phase constant (Ф) we use the initial condition that for t = 0 v = 0 and x = A, we substitute
A = A cos (0+ Ф)
cos Ф = 1
Ф = 0
the resulting equation is
x = 0.65 cos (2.51 t)
Let's find the time it takes to get to x = 0.28 m
0.28 = 0.65 cos 2.51 t
2.51 t = cos-1 (0.28 / 0.65)
remember angles are in radians
t = 1.1254 / 2.51
t = 0.448 s
the acceleration is
[tex]a = \frac{dv}{dt} = \frac{d^2x}{dt^2}[/tex]
a = -A w² cos wt
we subtitle
a = - 0.65 2.51² cos (2.51 0.448)
a = - 1.76 m / s²
b) the maximum acceleration occurs when the cosine is ±1
a = A w²
a = 0.65 2.51²
a = 4.10 m / s²
Let's use Newton's second law
F = m a
F = 5 4.1
F = 20.5 N
c) The angular velocity is given by
w² = k / m
let's find the spring constant
k = m w²
k = 5 2.51²
k = 31.5 N / m
therefore if the block is exchanged for another with mass m'= 2.5 kg
w = √(31.5 / 2.5)
w = 3.55 rad / s
A 20 kg box has an initial velocity of 2 m/s starting at the bottom of a 30-degree inclined plane. A person pushes on the box directly up the frictionless inclined plane so that it travels up the inclined plane at a constant velocity of 2 m/s. Calculate the how much is done by the person after 5 seconds have past.
Answer:
Explanation:
The box is moving with constant velocity so acceleration of box is zero . That means net force on the box is zero .
The weight component acting on box parallel to incline plane
= mg sin 30⁰ = 20 x 9.8 x sin 30 = 98 N
This force is acting down the plane , hence to make the net force zero acting on box , force exerted by person will also be 98 N up the incline .
Force exerted by person = 98 N
distance travelled in 5 s
= velocity x time
= 2 x 5 = 10 m
Work done by person
= 98 x 10
= 980 J .
An accelerator produces a beam of protons with a circular cross section that is 2.0 mm in diameter and has a current of 1.0 mA. The current density is uniformly distributed through the beam. The kinetic energy of each proton is 20 MeV. The beam strikes a metal target and is absorbed by the target. (a) What is the number density of the protons in the beam
Answer:
the number density of the protons in the beam is 3.2 × 10¹³ m⁻³
Explanation:
Given that;
diameter D = 2.0 mm
current I = 1.0 mA
K.E of each proton is 20 MeV
the number density of the protons in the beam = ?
Now, we make use of the relation between current and drift velocity
I = MeAv ⇒ 1 / eAv
The kinetic energy of protons is given by;
K = [tex]\frac{1}{2}[/tex][tex]m_{p}[/tex]v²
v = √( 2K / [tex]m_{p}[/tex] )
lets relate the cross-sectional area A of the beam to its diameter D;
A = [tex]\frac{1}{4}[/tex]πD²
now, we substitute for v and A
n = I / [tex]\frac{1}{4}[/tex]πeD² ×√( 2K / [tex]m_{p}[/tex] )
n = 4I/π eD² × √([tex]m_{p}[/tex] / 2K )
so we plug in our values;
n = ((4×1.0 mA)/(π(1.602×10⁻¹⁹C)(2mm)²) × √(1.673×10⁻²⁷kg / 2×( 20 MeV)(1.602×10⁻¹⁹ J/ev )
n = 1.98695 × 10¹⁸ × 1.6157967 × 10⁻⁵
n = 3.2 × 10¹³ m⁻³
Therefore, the number density of the protons in the beam is 3.2 × 10¹³ m⁻³
Help me I don't know what I'm doing
Answer:
C the metal handle because it is a good conductor
Answer:
D.
Explanation:
Although the metal handle will last longer, if heated up enough it could burn her hand.
You have three objects of varying shapes and sizes: Object 1 is a rectangular block of tin. Object 2 is a cube of aluminum. Object 3 is a sphere of copper.
a. the density of tin is 5.75g/cm2. What is the mass of object 1 in kg if the rectangular block has a volume of 1.34L?
b. what is the volume in cubic inches of object 2 if the cube of aluminum 7.58 inches on a side?
c. what is the mass in kg of object 2? the density of aluminum is 2.70g/cm3
d. what is the volume in cm3 of object 3 if the sphere of copper has a diameter 8.62cm? the volume of the sphere is 4 {pi}^3/3
e. what is the mass in kg of object 3? Copper has a density of 8.96g/cm3
Answer: a. m = 7.7 kg
b. V = 435.52 in³
c. m = 1927 kg
d. V = 335.37 cm³
e. m = 3 kg
Explanation: Density is the ratio of mass per volume, i.e., it's the measure of an object's compactness. Its representation is the greek letter ρ.
The formula for density is
[tex]\rho=\frac{m}{V}[/tex]
Density's unit in SI is kg/m³, but it can assume lots of other units.
Some unit transformations necessary for the resolution of the question:
1 L = 1 dm³ = 1000 cm³
1 in³ = 16.3871 cm³
1 g = 0.001 kg
a. V = 1.34 L = 1340 cm³
[tex]\rho=\frac{m}{V}[/tex]
[tex]m=\rho.V[/tex]
m = 5.75 * 1340
m = 7705 g => 7.705 kg
Mass of object 1 with volume 1.34L is 7.7 kg.
b. A cube's volume is calculated as V = side³
V = 7.58³
V = 435.52 in³
Volume of object 2 is 435.52 in³.
c. Using 1 in³ = 16.3871 cm³ to change units:
V = 435.52 * 16.3871
V = 713689.4 cm³
Then, mass will be
[tex]m=\rho.V[/tex]
m = 2.7 * 713689.4
m = 1926961.4 g => 1927 kg
Mass of object 2 is 1927 kg.
d. Volume of a sphere is calculated as [tex]V=\frac{4}{3}.\pi.r^{3}[/tex]
Diameter is twice the radius, then r = 4.31 cm.
Volume is
[tex]V=\frac{4}{3}.\pi.(4.31)^{3}[/tex]
V = 335.37 cm³
Volume of object 3 is 335.37 cm³.
e. [tex]m=\rho.V[/tex]
m = 8.96 * 335.37
m = 3004.91 g => 3 kg
Mass of object 3 is 3 kg.
Astronaut 1 has a mass of 70 kgkg. Astronaut 2 has a mass of 80 kgkg. Astro 1 and 2 want to travel to separate planets, but they want to experience the same weight (in NN). Astro 1 visits a planet with gravitational acceleration 7.0 m/s2m/s2. What must be Astro 2 planet's agag to equal Astro 1's weight
Answer:
g₂ = 39 m / s²
Explanation:
Mass is the amount of matter that a body has, so it does not change, but the weight is the force with which the planet attracts this matter, this value does change
W = m g
let's use subscript 1 for the first planet and subscript 2 for the second planet
W₁ = m₁ g₁
W₂ = m₂ g₂
They give us the values of the masses and the value of g₁ = 7.0 m / s, let's find the value of the acceleration of gravity on Planet 2
They tell us that the weight of the two astronauts is the same therefore
W₁ = W₂ = W
let's match the expressions
m₁ g₁ =m₂ g₂
g₂ = [tex]\frac{m_1}{m_2} \ g_1[/tex]
let's calculate
g₂ = [tex]\frac{70}{80} \ 7.0[/tex]
g₂ = 39.2 m / s²
g₂ = 39 m / s²
The open-circuit voltage of a car battery is measured to be 12 V. During engine startup, the battery delivers 10 A to the starter motor, and the battery voltage drops to 11.7 V. Draw the Thévenin equivalent circuit for the battery. How much power does the battery deliver to the starter motor?
Answer:
- the Thevenin equivalent circuit for the battery is uploaded below
- the battery delivered 117 watts of power to the starter motor
Explanation:
Given the data in the equation
diagram of the Thevenin equivalent circuit for the battery is uploaded below.
Current I = 10 A
Voltage 1 = 12 V
voltage 2 = 11.7 v
R = (V1 - V2) / I
R = (12-11.7)/10
R = 0.3 / 10
R = 0.03Ω
Thevenin equivalent circuit
[tex]R_{L}[/tex] = V2 / I = 11.7 / 10
[tex]R_{L}[/tex] = 1.17Ω
so, power delivered to the starter motor will be;
p = (V2)² / [tex]R_{L}[/tex]
P = ( 11.7 V )² / 1.17Ω
p = 136.89 / 1.17
p = 117 watts
Therefore, the battery delivered 117 watts of power to the starter motor
I NEED HELP WITH THIS QUESTION ASAP PLEASE!
Most new jobs in the United States will be in the _____.
in the service producing sector
A sprinter practicing for the 200-m dash accelerates uniformly from rest at A and reaches a top speed of 35 km/h at the 67-m mark. He then maintains this speed for the next 88 meters before uniformly slowing to a final speed of 32 km/h at the finish line. Determine the maximum horizontal acceleration which the sprinter experiences during the run. Where does this maximum acceleration value occur
Answer:
0.705 m/s²
Explanation:
a) The sprinter accelerates uniformly from rest and reaches a top speed of 35 km/h at the 67-m mark.
Using newton's law of motion:
v² = u² + 2as
v = final velocity = 35 km/h = 9.72 m/s, u = initial velocity = 0 km/h, s = distance = 67 m
9.72² = 0² + 2a(67)
134a = 94.484
a = 0.705 m/s²
b) The sprinter maintains this speed of 35 km/h for the next 88 meters. Therefore:
v = 35 km/h = 9.72 m/s, u = 35 km/h = 9.72 m/s, s = 88 m
v² = u² + 2as
9.72² = 9.72² + 2a(88)
176a = 9.72² - 9.72²
a = 0
c) During the last distance, the speed slows down from 35 km/h to 32 km/h.
u = 35 km/h = 9.72 m/s, v = 32 km/h = 8.89 m/s, s = 200 - (67 + 88) = 45 m
v² = u² + 2as
8.89² = 9.72² + 2a(45)
90a = 8.89² - 9.72²
90a = -15.4463
a = -0.1716 m/s²
The maximum acceleration is 0.705 m/s² which is from 0 to 67 m mark.
A centrifuge is a device that rotates an object to produce an acceleration many times that of gravity Pilots are trained in such devices because they can simulate the same Accelerations that are experienced in certain types of flight. At which angular velocit would a space station of 30 m radius have to rotate to simulate Earth graivty?
Answer:
ω = 0.571 rad/s
Explanation:
given data
radius = 30 m
solution
we take here g = 9.8 m/s²
and g is express as
g = r × ω² ....................1
put here value and we get
9.8 = 30 × ω²
solve it we get
ω = 0.571 rad/s
Beams of high-speed protons can be produced in "guns" using electric fields to accelerate the protons. (a) What acceleration would a proton experience if the gun's electric field were 2.95 × 104 N/C? (b) What speed would the proton attain if the field accelerated the proton through a distance of 1.26 cm?
Answer:
(A) the acceleration experienced by the proton 2.821 x 10¹² m/s²
(B) the speed of the proton is 2.67 x 10⁵ m/s
Explanation:
Given;
electric field experienced by the proton, E = 2.95 x 10⁴ N/C
charge of proton, Q = 1.6 x 10⁻¹⁹ C
mass of proton, m = 1.673 x 10⁻²⁷ kg
distance moved by the proton, d = 1.26 cm = 0.0126 m
(a)
The force experienced by the proton is calculated as;
F = ma = EQ
where;
a is the acceleration experienced by the proton
[tex]a = \frac{EQ}{m} \\\\a = \frac{2.95\times 10^4 \ \times \ 1.6\times 10^{-19}}{1.673 \times 10^{-27}} \\\\a = 2.821 \times 10^{12} \ m/s^2[/tex]
(b) the speed of the proton is calculated;
v² = u² + 2ad
v² = 0 + (2 x 2.821 x 10¹² x 0.0126)
v² = 7.109 x 10¹⁰
v = √7.109 x 10¹⁰
v = 2.67 x 10⁵ m/s
77. A drag racing vehicle travels from 0 to 100 mph in 5 seconds north. What is the acceleration?
a).004
s2
b).0056 m/s2 c).0079"
d).01 m/s2
M
m
Answer:
a
Explanation:
i just took the test
a bus initially at rest accelerated of 4m/S2 at the end of 10 second find average velocity
Answer:
V = 20m/s
Explanation:
Given the following data;
Acceleration = 4m/s²
Time = 10 secs
Initial velocity = 0m/s (since it's at rest).
To find the average velocity, we would use the first equation of motion;
[tex] V = U + at[/tex]
Where;
V is the final velocity.
U is the initial velocity.
a is the acceleration.
t is the time measured in seconds.
Substituting into the equation, we have;
V = 0 + 4*10
V = 40m/s
To find the average velocity;
Average velocity = (U + V)/2
Average velocity = (0 + 40)/2
Average velocity = 40/2
Average velocity = 20m/s
please help!!!
When a switch is turned from the off to the on position, it is changing the circuit in which of the following ways? O An open circuit is being changed into a closed circuit. A closed circuit is being changed into an open circuit. O A parallel circuit is being changed into a series circuit. A series circuit is being changed into a parallel circuit.
Answer:
i Believe the correct answer is "An open circuit being changed into a closed circuit"
Explanation:
Solve this chemical equation: CH3CH2OH+__O2=CO2+__H2O
Answer:
some kind of chemical of which i do not know
Explanation:
WILL MARK BRAINLIEST IF CORRECT. Worth Lots Of Points. NEEDS TO BE WITH EXPLANATION PLEASE. PLEASE HELP!
Calculate the displacement and velocity at times of (a) 0.500, (b) 1.00, (c) 1.50, (d) 2.00, and (e) 2.50 s for a rock thrown straight down with an initial velocity of 14.0 m/s from the Verrazano Narrows Bridge in New York City. The roadway of this bridge is 70.0 m above the water
Answer:
See explanation below
Explanation:
This is a typical exercise of free falling. In this case a rock thrown straight down from the bridge, and we are asked to determine the final velocity of the rock and it's displacement at those given times.
First, just for this problem, as the rock is going straight down, we'll say that the downward direction is positive, therefore, the following expressions to calculate velocity and speed will be:
V = V₀ + gt (1)
X = V₀t + gt²/2 (2)
In this case, g = 9.8 m/s²
Now, let's see the displacement and velocity for each given time:
a) For t = 0.5 s
V = 14 + (9.8)*0.5
V = 18.9 m/s
X = (14*0.5) + (9.8)(0.5)²/2
X = 7 + 1.225
X = 8.225 m
b) For t = 1.00 s
V = 14 + (9.8)*1
V = 23.8 m/s
X = (14*1) + (9.8)(1)²/2
X = 14 + 4.9
X = 18.9 m
c) For t = 1.5 s
V = 14 + (9.8)*1.5
V = 28.7 m/s
X = (14*1.5) + (9.8)(1.5)²/2
X = 21 + 11.025
X = 32.025 m
d) For t = 2 s
V = 14 + (9.8)*2
V = 33.6 m/s
X = (14*2) + (9.8)(2)²/2
X = 28 + 19.6
X = 47.6 m
e) For t = 2.50 s
V = 14 + (9.8)*2.5
V = 38.5 m/s
X = (14*2.5) + (9.8)(2.5)²/2
X = 35 + 30.625
X = 65.625 m
Hope this helps
when air mass is caught between two cold fronts the result is a _______ front.
Answer choices
A.occluded
B.warm
C.cold
D.stationary
Two people stand across from one another at the top edges of identical buildings, 50 meters above the ground. One person throws a rock straight upwards with an initial speed of 15 m/s, while the other person throws their rock straight downwards with an initial speed of 15 m/s. If both of the rocks miss the building and continue down to the street,
a. How long will it take for the rock thrown upwards to reach the ground?
b. How long will it take for the rock thrown downwards to reach the ground?
c. How fast will the upward thrown rock be travelling just before it hits the ground? The downward thrown rock?
Answer:
C
Explanation:
Hope this helps!!!!
anteaters are animals with long tongues that they used to wear a ant they sometimes free by poking their tongues into a bed and licking the ants up if an anteater is born with shortened tongue this would be an example of a
Answer: an ant eater who can’t eat
Explanation:
Which of the following is not a true statement?
A
B
C
D
Answer:
I think A
Explanation:
because I don't think a unknowned number can be a division problem
A hockey puck with mass 0.30 kg is sliding along the ice with initial speed of 12.68 m/s. A hockey player is heading toward the puck with his stick in hand. After the player strikes the puck, the puck reverses its direction and is traveling at double its speed before the strike. If the collision occurs in 0.05 s, what is the magnitude of the force the hockey player's stick applied to the puck
Answer:
F = 228.24 N
Explanation:
According Newton's 2nd Law, the impulse on one object is equal to the change in momentum of that object.I = F*Δt = Δp = pf - po (1)where pf = final momentum = m*vf
p₀ = initial momentum = m*v₀
Since after the strike, the puck reverses its direction and travels at double its speed before the strike, that means that vf = -2*v₀.Replacing in the right side of (1), we have:[tex]m*v_{f} - m*v_{o} = -2*v_{o} -m*v_{o} = -3*m*v_{o} = -3*0.3kg*12.68m/s = -11.41m/s (2)[/tex]
Replacing Δt = 0.05s, and solving for F in (1):[tex]F_{net} = \frac{-11.41m/s}{0.05s} = -228.24 N (3)[/tex]
which means that the force is applied in a direction opposite to the initial velocity of the puck.The magnitude of the force is just 228.24 N.
What is the main cause of tides on Earth? *
1. Sun's gravity
2. Moon's gravity
Two motorcycles are traveling due east with different velocities. However, 5.68 seconds later, they have the same velocity. During this 5.68-second interval, motorcycle A has an average acceleration of 3.87 m/s2 due east, while motorcycle B has an average acceleration of 18.2 m/s2 due east. (a) By how much did the speeds differ at the beginning of the 5.68-second interval, and (b) which motorcycle was moving faster?
Answer:
The answer is below
Explanation:
Let a be the initial velocity of motorcycle A and b be the initial velocity of motorcycle B.
After 5.68 seconds, both motorcycle had the same velocity (v), therefore for motorcycle A:
(a - v) / 5.68 = 3.87
a - v = 21.9816
v = a - 21.9816
For motorcycle B:
(b - v) / 5.68 = 18.2
b - v = 103.376
v = b - 103.376
Therefore:
a - 21.9816 = b - 103.376
b - a = -21.9816 + 103.376
b - a = 81.3944
a) The difference between their speeds at the beginning was 81.3944 m/s
b) Since b - a = 81.3944. This means that the initial velocity of motorcycle B is greater than that of motorcycle A by 81.3944 m/s.
Therefore motorcycle B was moving faster
calculate the force needed to push the ball up a 4 m ramp if the work is equal to 16 joules.
Why do people eat bo oty
Answer: I don't know my dude
Explanation:
NEED HELP ASAP
How does radiation help treat cancer?
(a) It cools down the cancer cells and weakens them.
(b) It cools down the non-cancerous cells and strengthens them.
(c) It heats up the cancer cells and weakens them.
(d) It heats up the non-cancerous cells and strengthens them.
THANK YOU
Answer:
c heats up cancer cells and weakens them.
Answer:
I think it is C
Explanation:
because
a body accelerates uniformly from rest at 2m/s^2 for 5 seconds. Calculate its averege velocity in this time
HERE IS YOUR ANSWER!
A wire has a length of 0.50 m and measures about 0.50 mm in its cross-sectional radius. At normal temperature, what is its resistance in Ohms, if Aluminum has a resistivity of 2.82x10^-8 Ohms*meter
Answer:
Explanation:
For resistance , the expression is as follows .
R = ρ L / S where ρ is specific resistance , L is length of wire and S is cross sectional area .
cross sectional area = π x ( .5 x 10⁻³ )²
S = .785 x 10⁻⁶ m²
Putting the values
R = 2.82 x 10⁻⁸ x .50 / .785 x 10⁻⁶
= 1.796 x 10⁻² ohm .
kenneth ran a marathon (26.2 miles) in 5.5 hours. What was Kenneth’s average speed? (Round your answer to the nearest tenth.)
0.2 mph
4.8 mph
5.5 mph
144.1 mph
it is actually science i couldn't find the word science
Answer:
4.8mph
Explanation:
Speed= Distance/time
Speed= 26.2/5.5
= 4.76mph
( To the nearest tenth ) = 4.8mph
Answer:
38.7 mph
Explanation:
I just add all the numbers together then divided them by 4, which is the amount of numbers you gave.