Answer:
1.32kgm/s
Explanation:
use the formula: p=mv
The momentum of a lab cart = -1.32 kg m/s
What is momentum?It is measure of the inertia of a body/ object .It can be calculated by multiplying mass with velocity .
General formula for momentum = M = m * v
given
mass = 0.60 kg
speed = 2.2 m/s
velocity = - 2.2 m/s ( answer is in negative , since mass is a scaler quantity but velocity is a vector quantity hence , velocity can be negative )
momentum = mass * velocity
= 0.60 * (-2.2 ) = -1.32 kg m/s
The momentum of a lab cart =c) -1.32 kg m/s
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____ is the study of things getting faster as they move.
A. Anatomy
B. Force
C. Physics
D. Dynamics
Answer: b force
Explanation:
yes because the world comin g up with more technique
Sally travels by car from one city to another. She drives for 30 min at 80 km/hr, 12.0 min at 105 km/hr, and 45 min at 40.0 km/hr. What was the total distance traveled?
Given :
Sally travels by car from one city to another. She drives for 30 min at 80 km/hr, 12.0 min at 105 km/hr, and 45 min at 40.0 km/hr.
To Find :
The total distance traveled.
Solution :
Total distance traveled is given as sum :
Distance = sum of product of speed and time
D = (v₁ × t₁) + (v₂ × t₂) + (v₃ × t₃)
Putting all given value in this we get :
[tex]D = 80\times \dfrac{30}{60} + 105 \times \dfrac{12}{60} + 40\times \dfrac{45}{60}\\\\D = 91\ km[/tex]
Hence, this is the required solution.
a train is traveling at 50km/h average .what is the displacement of the train per second?
The accepted speed of sound at atmospheric pressure and 0 *C is 331.5 m/s. The speed of sound increases 0.607 m/s for every *C. Calculate the speed of sound at the temperature of your room and compare your measured value to the accepted value.
Complete Question
The accepted speed of sound at atmospheric pressure and 0 *C is 331.5 m/s. The speed of sound increases 0.607 m/s for every *C. Calculate the speed of sound at the temperature of your room(70F) and compare your measured value to the accepted value.
Answer:
[tex]V_{Tc}=344.314m/s[/tex]
Explanation:
From the question we are told that:
Speed of sound at Temperature [tex]0 \textdegree[/tex] [tex]V_0=331.5m/s[/tex]
Rate of Speed increase [tex]\triangle V_{infty}=0.607[/tex]
Generally the equation for Temperature in Celsius is mathematically given by
[tex]Tc=\frac{100}{180}(T_f-32)[/tex]
[tex]Tc=0.56*38[/tex]
[tex]Tc=21.11 textdegree C[/tex]
Therefore speed at Tc
[tex]V_{Tc}=V_0+(Tc)( V_{infty})[/tex]
[tex]V_{Tc}=331.5+(21.11)(0.607)[/tex]
[tex]V_{Tc}=344.314m/s[/tex]
Two workers are sliding 450 kg kg crate across the floor. One worker pushes forward on the crate with a force of 380 NN while the other pulls in the same direction with a force of 230 NN using a rope connected to the crate. Both forces are horizontal, and the crate slides with a constant speed. What is the crate's coefficient of kinetic friction on the floor
Answer:
The coefficient of kinetic friction on the floor is 0.138
Explanation:
Given;
mass of the crate, m = 450 kg
force applied by the first worker, F₁ = 380 N
force applied by the second worker in the same direction as the first worker, F₁ = 230 N
frictional force opposing the motion of the box = -[tex]F_k[/tex]
Apply Newton's second law of motion;
∑F = ma
[tex]F_1 + F_2 - F_k = ma[/tex]
If the crate slides with constant speed, acceleration is zero (0).
[tex]F_1 + F_2 - F_k = ma = 0\\\\F_1 + F_2 - F_k = 0\\\\F_k = F_1 + F_2\\\\\mu _kmg= F_1 + F_2\\\\\mu _k = \frac{F_1 + F_2}{mg} \\\\\mu _k = \frac{380 + 230}{450 \times 9.8} \\\\\mu _k = 0.138[/tex]
Therefore, the coefficient of kinetic friction on the floor is 0.138
What is the wavelength of a wave with a frequency of 262 Hz and a speed of
343 m/s
Answer:
time perion=1/frequency
so here time period =1/262
let wavelength be x
speed of sound=wave length/time period
342=x/1/262
342=262x
342/262=x
1.30=x
therefore wavelength=1.30 metre
Explanation:
The function s(t)s(t) describes the position of a particle moving along a coordinate line, where ss is in feet and tt is in seconds. (a) Find the velocity and acceleration functions. (b) Find the position, velocity, specd, and acceleration at time t
Answer:
Explanation:
From the given information:
Let's assume that the missing function is:
s(t) = t³ - 6t², t ≥ 0
From part (b), we are to find the given required terms when time t = 2
So; from the function s(t) = t³ - 6t², t ≥ 0
[tex]velocity \ v(t) \ = \dfrac{d}{dt}s(t)[/tex]
[tex]velocity \ v(t) \ = \dfrac{d}{dt}(t^3 - 6t^2)[/tex]
[tex]velocity \ v(t) \ = 3t^2 - 12t[/tex]
[tex]acceleration a(t) = \dfrac{d}{dt}*v(t)[/tex]
[tex]acceleration a(t) = \dfrac{d}{dt}(3t^2 - 12 t)[/tex]
[tex]acceleration\ a(t) = 6t - 12[/tex]
At time t = 2
The position; S(2) = (2)² - 6(2)²
S(2) = 8 - 6(4)
S(2) = 8 - 24
S(2) = - 16 ft
v(2) = 3(2)² - 12 (2)
v(2) = 3(4) - 24
v(2) = 12 - 24
v(2) = - 12 ft/s
speed = |v(2)|
|v(2)| = |(-12)|
|v(2)| = 12 ft/s
acceleration = 6t - 12
acceleration = 6(2) - 12
acceleration = 12 - 12
acceleration = 0 ft/s²
In a double-slit arrangement, the slits are separated by a distance equal to 100 times the wavelength of the light passing through the slits. (a) Calculate the angular separation, !, in radians between the central maximum and the 1st order maximum
Solution :
The conditions for the maximum in the Young's experiment is :
d sin θ = m λ, where m = 0, 1, 2, 3, .....
The angle between the central maximum and the 1st order maximum can be determined by setting the m = 1. So,
d sin θ = λ
[tex]$\theta = \sin^{-1}\left(\frac{\lambda}{d}\right)$[/tex]
Given : d = 100 λ
[tex]$\theta = \sin^{-1}\left(\frac{\lambda}{100 \lambda}\right)$[/tex]
[tex]$\theta = \sin^{-1}\left(\frac{1}{100}\right)$[/tex]
[tex]$=0.573^\circ$[/tex]
= 0.01 rad
If two charged objects each have 2.5 C of charge on them and are located 100 m apart, how strong is the electrostatic force between them?
Answer:
5.619×10⁶ N
Explanation:
Applying,
F = kqq'/r²................... Equation 1
Where F = electrostatic force between the charges, k = coulomb's constant, q = first charge, q' = second charge, r = distance btween the charges
From the questiion,
Given: q = 2.5 C, q' = 2.5 C, r = 100 m
Constant: 8.99×10⁹ Nm²/C²
Substitute these values into equation 1
F = (2.5×2.5×8.99×10⁹)/100²
F = 56.19×10⁵
F = 5.619×10⁶ N
A 1.2 kg mass is suspended from the ceiling by a string. A second horizontal string holds the mass at rest next to the wall. The angle between the string and the ceiling is 65o. What is the tension force in the horizontal rope
Answer:
The tension in the horizontal string is 25.2 N.
Explanation:
mass, m = 1.2 kg
Angle, A = 65 degree
The tension in the string is T.
So, T cos A = m g
T x cos 65 = 1.2 x 9.8
T x 0.4226 = 11.76
T = 27.83 N
Tension in the horizontal string, T' = T sin A = 27.83 x sin 65 = 25.22 N
Answer:
Explanation:
See the figure attached . T₁ and T₂ are tension in the inclined and horizontal string .
The vertical component of T₁ will balance weight and horizontal component will balance the tension T₂.
T₁ sin 65 = mg and T₁ cos 65 = T₂
Dividing ,
Tan 65 = mg / T₂
T₂ = mg / tan 65
= 1.2 x 9.8 / 2.1445
= 5.5 N
A 4.9 A current is set up in a circuit for 4.7 min by a rechargeable battery with a 12 V emf. By how much is the chemical energy of the battery reduced
Answer:
E = 16581.6 J
Explanation:
Given that,
Current, I = 4.9 A
Time for which the current is set up, I = 4.7 min = 282 s
The voltage of the battery, V = 12 V
We need to find how much chemical energy of the battery reduced. Let It is E. We know that,
E = P t
Where
P is power of battery, P = VI
So,
[tex]E=VIt[/tex]
Put all the values,
[tex]E=12\times 4.9\times 282\\E=16581.6\ J[/tex]
So, 16581.6 J of chemical energy of the battery is reduced.
Dos cargas puntuales iguales y negativas, q1=q2=-24micro C se localizan en x=0 y y=38m y x=0 y y=-7m, respectivamente. Calcula la magnitud de la fuerza electrica total en N que ejercen estas dos cargas sobre una tercera, tambien puntual, Q=26micro C en y=0 y x=16m
Answer:
F_net = 9.87 10⁻⁴ N
Explanation:
Let's use that force is a vector magnitude
∑ F = F₁₃ + F₂₃
De bold arfe vectros. The force is the electric force, we use that charges of the same sign repel and when the charges are of a different sign they attract
the charges q1 and q2 are negative and the charge q3 is positive with the positions y1 = 38 m, y2 = -7m, y3 = 16 m
∑ F = F₁₃ - F₂₃
F_net = [tex]k \frac{q_1q_3}{r_{13}^2 } - k \frac{q_2q_3}{r_{23}^2 }[/tex]
in this case q₁ = q₂ = q
F_net = k q q₃ ( )
let's look for the distance
r₂₃ = y₂ - y₃
r₂₃ = -7 -16
r₂₃ = - 23 m
r₁₃ = 38 - 16
r₁₃ = 22 m
let's calculate
F_net = 9 10⁹ 24 26 10⁻¹² ( )
F_net = 5.616 ( 1.758 10⁻⁴ )
F_net = 9.87 10⁻⁴ N
different between pressure and force
Force is mass into acceleration
and pressure is force applied per unit area.
Coherent monochromatic light falls perpendicularly on two slits (each of width 0.10 mm) separated by 0.50 mm. In the resulting interference pattern on a screen 2.80 m away, adjacent bright fringes are separated by 2.80 mm. (a)What is the wavelength of the light that falls on the slits
Answer:
The correct answer is "[tex]0.5\times 10^{-6} \ m[/tex]".
Explanation:
Given:
[tex]\frac{\lambda D}{d} =2.8\times 10^{-3}[/tex]
[tex]d = 0.5\times 10^{-3}[/tex]
[tex]D = 2.80[/tex]
Now,
The wavelength will be:
⇒ [tex]\lambda = 2.8\times 10^{-3}\times \frac{d}{D}[/tex]
By putting the values, we get
⇒ [tex]=\frac{2.8\times 10^{-3}\times 0.5\times 10^{-3}}{2.8}[/tex]
⇒ [tex]=\frac{1.4\times 10^{-6}}{2.8}[/tex]
⇒ [tex]=0.5\times 10^{-6} \ m[/tex]
A 40-kg crate is being lowered with a downward acceleration is 2.0 m/s2 by means of a rope. (a) What is the magnitude of the force exerted by the rope on the crate
Answer:
F = 312 N
Explanation:
Given that,
The mass of a crate, m = 40 kg
Acceleration of the crate, a = 2 m/s²
As the carte is falling downward, the net force exerted by the rope on the carte is given by :
F = m(g-a)
Put all the values,
F = 40(9.8-2)
F = 312 N
Hence, the required force exerted by the rope on the crate is equal to 312 N.
Danny is riding his bike at 12m/s he tries to pop a wheelie but he fails hits a curb flies through the air and comes to a complete stop in 30 seconds what is Danny's deceleration
Answer:
a = -0.4m/s²
Explanation:
v_f = v_I + (a)(t)
a(t) = v_f-v_I
a = (v_f-v_I)/t
a = (0m/s-12m/s)/30s
a = -0.4m/s²
The cliff divers at Acapulco, Mexico, jump off a cliff m above the ocean. Ignoring air resistance, how fast are the divers going when they hit the water
Answer:
22.87 m/s
Explanation:
Since the value of the height at which the divers jump off the cliff is not given,
Assuming that the height of the cliff above the ocean at which the divers jump = 26.7 m
Then;
The speed of the divers when they hit the ocean water can be determined by using the formula:
[tex]v^2 = 2gh \\ \\ v= \sqrt{2gh} \\ \\ v = \sqrt{2 \times 9.8 \times (26.7)} \\ \\ v \simeq 22.87 \ m /s[/tex]
After You Read
Mini Glossary
atomic number: the number of protons in an atom of
an element
mass number: the sum of the number of protons and
neutrons in an atom
average atomic mass: the average mass of the element's
isotopes, weighted according to the abundance of each isotope
nuclear decay: a process that occurs when an unstable atomic
nucleus changes into another more stable nucleus by
emitting radiation
ion: an atom that is no longer neutral because it has gained or
lost electrons
radioactive: spontaneously emits radiation
isotope: an atom of the same element that has a different
number of neutrons
1. Review the terms and their definitions in the Mini Glossary. Write a sentence that
explains how to determine the number of neutrons in an isotope that has 6 protons and
a mass number of 13.
Answer:
protons+neutrons= mass number, so if the mass number is 13 and protons are 6 its 13-6=7 neutrons
Explanation:
mass number: the sum of the number of protons and
neutrons in an atom
this is key as it explains that protons+neutrons= mass number, so if the mass number is 13 and protons are 6 its 13-6=7 neutrons
Find the value of T1 if 1 = 30°, 2 = 60°, and the weight of the object is 139.3 newtons.
A.
69.58 newtons
B.
45.05 newtons
C.
25 newtons
D.
98.26 newtons
Answer:
Option A (69.56 newtons) is the appropriate solution.
Explanation:
According to the question,
On the X-axis,
⇒ [tex]T_1Cos30^{\circ}-T_2Cos60^{\circ}=0[/tex]
or,
[tex]T_1Cos 30^{\circ}=T_2Cos60^{\circ}[/tex]
On substituting the values, we get
[tex]T_1\times \frac{\sqrt{3} }{2}=T_2\times \frac{1}{2}[/tex]
[tex]T_1\times \sqrt{3} =T_2[/tex]....(equation 1)
On the Y-axis,
⇒ [tex]T_1Sin30^{\circ}+T_2Sin60^{\circ}=139.3 \ N[/tex]
[tex]\frac{T_1}{2} +\frac{\sqrt{3} }{2} =139.2 \ N[/tex]
[tex]T_1+\sqrt{3}T_2=139.2\times 2[/tex]
From equation 1, we get
[tex]T_1+\sqrt{3}\times \sqrt{3}T_1 =278.4 \ N[/tex]
[tex]T_1+3T_1=278.4 \ N[/tex]
[tex]4T_1=278.4 \ N[/tex]
[tex]T_1=\frac{278.4}{4}[/tex]
[tex]=69.6 \ N[/tex]
Answer:
69.58
Explanation:
Please helppppppp!!!!!!
Answer:
The resulting force on the first object is 800 N.
Explanation:
Given;
force exerted on one of the objects, F₁ = 400 N
let the first charge = q₁
let the second charge = q₂
The force of repulsion between the objects is calculated using Coulomb's law;
[tex]F =\frac{kq_1q_2}{r^2} \\\\\frac{k}{r^2} = \frac{F}{q_1q_2} = \frac{F_1}{q_1\times2q_2} \\\\F_1 = \frac{F(q_1\times2q_2)}{q_1q_2} \\\\F_1 = 2F\\\\F_1 = 2(400 \ N)\\\\F_1 = 800 \ N[/tex]
Therefore, the resulting force on the first object is 800 N.
Once a disk forms around a star, the process of planetary formation can begin. Rank the evolutionary stages for the formation of planets from earliest to latest.
a. Small clumps of matter stick together via the process of accrection to form plantesimals a few hundred kilometers in diameter
b. Dust keeps matter inside the disk cool long enough for planet formation to start
c. Planetisimals begin to accrete, forming protoplanets
d. Dust grains form condensation nuclei on which surrounding atoms condense to form small clumps of matter
e. A collection of a few planet-sized protoplanets remain in a fairly cleared out disk around the star
Answer: See explanation
Explanation:
The evolutionary stages for the formation of planets from earliest to latest will be:
1. Dust keeps matter inside the disk cool enough for planet formation to start
2. Dust grains form condensation nuclei on which surrounding atoms condense to form small clumps of matter.
3. Small clumps of matter stick together via the process of accretion to form planetesimals a few hundred kilometers in diameter.
4. Planetesimals begin to accrete, forming protoplanets.
5. A collection of a few planet-sized protoplanets remain in a fairly cleared out disk around the star
a. 2.00kg object is subject to three force that gives acceleration a =8m/s^2 i +6m/s j if two of three forces are f1 =(30.0N)+(16N )and f2 =-(12.0N)+(8.0N) find the third force
By Newton's second law, the net force on the object is
∑ F = m a
∑ F = (2.00 kg) (8 i + 6 j ) m/s^2 = (16.0 i + 12.0 j ) N
Let f be the unknown force. Then
∑ F = (30.0 i + 16 j ) N + (-12.0 i + 8.0 j ) N + f
=> f = (-2.0 i - 12.0 j ) N
Cuando Daniel hace oscilar un péndulo, este realiza 30,6 ciclos (completos) en 9 [s].
¿Cuál es la frecuencia del péndulo?
A )3,4 [Hz].
B )4,3 [Hz].
C )30 [Hz].
D )5 [Hz]
Two gerbils run in place with a linear speed of 0.55 m/s on an exercise wheel that is shaped like a hoop. Find the rotational kinetic energy of the wheel if the exercise wheel has a radius of 9.5 cm and a mass of 5.0 g. Each gerbil has a mass of 0.02 kg if you think that is important.
Answer:
K = 7.56 10⁻⁴ J
Explanation:
The rotational kinetic energy is
K = ½ I w²
They ask us for the kinetic energy of the wheel, which can be approximated as a thin ring, its moment of inertia is
I = M r²
the linear speed of the gerbils is equal to the linear speed of the wheel. The linear and rotational variables are related
v = w r
w = v / r
we substitute
K = ½ (M r²) v² / r²
K = ½ M v²
let's calculate
K = ½ 5 10⁻³ 0.55²
K = 7.56 10⁻⁴ J
an animal which is known as an ascendant of man
Answer:
spirit animal?
Explanation:
What is the relationship between the density of the equipotential lines, the density of the electric field lines and the strength of the electric field?
Answer:
I dont. understand the question, maybe insert the picture?
After turning on the power source connected to your two electrodes, we expect to see the microbeads moving through the solution. What forces are acting on the microbeads as they move (ignore vertical forces)
Answer:
the pearls have an electrical charge induced by contact with the ions of the solution and these charges are attracted by the electrode by a force electric
Explanation:
The pearls are suspended in a solution, when connecting the power source, it is subjected to an electric shock, the pearls have an electrical charge induced by contact with the ions of the solution and these charges are attracted by the electrode by a force electric
F = q E
Answer this
a) which ink is likely to be pure? Why?
b) What does the chromatography tell us about ink Y
c) Why are the three different spots separated out from ink Y found at different heights?
Answer:
a) Ink X is likely to be pure because it only contain 1 spot.
b) The chromatography tell us about ink Y that it is a mixture as it contain more than 1 spot.
c) The three different spots are separated out from ink Y at different heights beacaus different substance have different solubility.
The different spots from Y are found at various heights because they represent different compounds.
What is chromatography?The term chromatography has to do with a method of separating the component of a substance. The term chromatography originally means color writing.
We can see that the pure ink is the ink marked X. We can see from the chromatogram that Y is a mixture of colors. The different spots from Y are found at various heights because they represent different compounds.
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Explain how muscles are effected by space travel
Which of these is NOT an effect of humor?
strengthened immune system
reduced stress levels
reduced feelings of anxiety
feelings of jealousy and envy