Answer:
Carbon dioxide controls the amount of water vapor in the atmosphere and thus the size of the greenhouse effect. Rising carbon dioxide concentrations are already causing the planet to heat up
Explanation:
Hope it helps! Correct me if I am wrong
Im sure about my answer
An ideal horizontal spring-mass system has a mass of 1.0 kg and a spring with constant 78 N/m. It oscillates with a period of 0.71 seconds. When this same spring-mass system oscillates vertically instead, the period is _______ seconds. Enter 2 significant figures (a total of three digits) and use g = 10.0 m/s2 if necessary.
Answer:
T = 0.71 seconds
Explanation:
Given data:
mass m = 1Kg, spring constant K = 78 N/m, time period of oscillation T = 0.71 seconds.
We have to calculate time period when this same spring-mass system oscillates vertically.
As we know
[tex]T = 2\pi \sqrt{\frac{m}{K} }[/tex]
This relation of time period is true under every orientation of the spring-mass system, whether horizontal, vertical, angled or inclined. Therefore, time period of the same spring-mass system oscillating vertically too remains the same.
Therefore, T = 0.71 seconds
A battery is connected to a light bulb in a circuit. There is a current of 2 A in the light bulb.
The voltage of the battery is 1.5 V. What is the resistance of the light bulb?
V
Use R = to solve this problem.
Answer:
Resistance of the light bulb = 1.33 ohm (Approx.)
Explanation:
Given:
Current in electric bulb = 2 Amp
Voltage in battery = 1.5 Volt
Find:
Resistance of the light bulb
Computation:
Resistance = Voltage / Current
Resistance of the light bulb = Voltage in battery / Current in electric bulb
Resistance of the light bulb = 2 / 1.5
Resistance of the light bulb = 1.33 ohm (Approx.)
A car is travelling at 27m/s and decelerates at a=5m/s2 for a distance of 10m. Calculate its final velocity. (Hint does deceleration imply that the acceleration is positive or negative?)[
Answer:
use the formula to calculate acceleration and you'll get the answers
Which of the following lists the elements in order, from those having the least protons to those having the most protons in the atoms?
A. O, N, B, Li
B. Na, S, Al, Cl
C. O, S, Se, Te
D. Rb, K, Na, Li
Answer:
C
Explanation:
O has 8 protones,S tiene 16, Se 34 y Te 52.
Question 5 of 10
The graph below shows the downloads of two songs over time.
70
Song 1
60
50
40
Number of downloads
(spopunu)
Song 2
30
20
10
O
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Time (minutes)
Which term describes the slope of the graph of song 2 between minute 6 and
minute 7?
A. Positive
B. Zero
C. It is impossible to calculate
D. Negative
SUBMIT
Answer:
b
Explanation:
A wheel rotates about a fixed axis with an initial angular velocity of 20 rad/s. During a 5.0-s interval the angular velocity increases to 40 rad/s. Assume that the angular acceleration was constant during the 5.0-s interval. How many revolutions does the wheel turn through during the 5.0-s interval ( in revelations)? Hint: You need to use one of the equations of constant angular acceleration that independent of angular acceleration.
Answer:
The appropriate solution is "23.87 rev".
Explanation:
The given values are:
Initial angular velocity,
[tex]\omega_i=20 \ rad/s[/tex]
Final angular velocity,
[tex]\omega_f=40 \ rad/s[/tex]
Time taken,
[tex]t = 5.0 \ s[/tex]
If, α be the angular acceleration, then
⇒ [tex]\omega_f=\omega_i+\alpha t[/tex]
or,
⇒ [tex]\alpha t=\omega_f-\omega_i[/tex]
⇒ [tex]\alpha=\frac{\omega_f-\omega_i}{t}[/tex]
On substituting the values, we get
⇒ [tex]=\frac{40-20}{5.0}[/tex]
⇒ [tex]=\frac{20}{5.0}[/tex]
⇒ [tex]=4 \ rad/s^2[/tex]
If, ΔФ be the angular displacement, then
⇒ [tex]\Delta \theta=\omega_i t+\frac{1}{2} \alpha t^2[/tex]
On substituting the values, we get
⇒ [tex]=[(20\times 5.0)+(\frac{1}{2})\times 4\times (5.0)^2][/tex]
⇒ [tex]=100+50[/tex]
⇒ [tex]=150[/tex]
On converting it into "rev", we get
⇒ [tex]\Delta \theta=(\frac{150}{2 \pi} )[/tex]
⇒ [tex]=23.87 \ rev[/tex]
A 06-C charge and a .07-C charge are apart at 3 m apart. What force attracts them?
Answer:
the force of attraction between the two charges is 4.2 x 10⁹ N.
Explanation:
Given;
the magnitude of first charge, q₁ = 0.06 C
the magnitude of the second charge, q₂ = 0.07 C
distance between the two charges, r = 3 m
The force of attraction between the two charges is calculated as ;
[tex]F = \frac{Kq_1q_2}{r^2}[/tex]
where;
k is Coulomb's constant = 9 x 10⁹ Nm²/C²
[tex]F = \frac{Kq_1q_2}{r^2} \\\\F = \frac{(9\times 10^9)(0.06)(0.07)}{3^2} \\\\F = 4.2 \times 10^{6} \ N[/tex]
Therefore, the force of attraction between the two charges is 4.2 x 10⁹ N.
Pause
He
When an unbalanced force of 10 N is applied to an object whose mass is 4.0 kg, the acceleration of the object will be:
OA. 40 m/s
OB. 9.8 m/s2
OC 2.5 m/s2
OD. 0.40 m/s2
Answer:
C
Explanation:
a=f/m
10Kgm/s2/4kg
2.5m/s2
When an unbalanced force of 10 N is applied to an object whose mass is 4.0 kg, the acceleration of the object will be 2.5 m/s².
What is acceleration?The rate at which an item changes its velocity is known as acceleration, a vector quantity. If an object's velocity is changing, it is acceleration. The net acceleration that objects get as a result of the combined action of gravity and centrifugal force is known as the Earth's gravity, or g. It is a vector quantity whose strength or magnitude is determined by the norm and whose direction correlates with a plumb bob.
Given in the question, force 10 N and mass 4.0 Kg the acceleration is,
a = 10/4 = 2.5 m/sec²
When an unbalanced force of 10 N is applied to an object whose mass is 4.0 kg, the acceleration of the object will be 2.5 m/s².
To learn more about acceleration refer to the link:
brainly.com/question/12550364
#SPJ2
How much heat is needed to warm 365 mL of water in a baby bottle from 240C to 38C
Answer:98.6 degrees Fahrenheit
Explanation:
What is the magnitude of the gravitational force acting on a
1.0 kg object which is 1.0 m from another 1.0 kg object?
Ans[tex]^{}[/tex]wer and expl[tex]^{}[/tex]anation is in a fi[tex]^{}[/tex]le. Li[tex]^{}[/tex]nk below! Go[tex]^{}[/tex]od luck!
bit.[tex]^{}[/tex]ly/3a8Nt8n
12. By convention (agreement of the scientific community for consistency)
magnetic field lines...
A. always start on the north pole and terminate (end) on the South Pole
B. start at infinity and point toward each pole
C. start at each pole and go outward
D. always start on the south pole and terminate (end) on the north pole.
Answer:
. always start on the north pole and terminate (end) on the South Pole
Explanation:
A TMS (transcranial magnetic stimulation) device creates very rapidly changing magnetic fields. The field near a typical pulsed-field machine rises from 0 T to 2.5 T in 200 μs . Suppose a technician holds his hand near the device so that the axis of his 2.0-cm-diameter wedding band is parallel to the field.
Part A
What emf is induced in the ring as the field changes?
Express your answer to two significant figures and include the appropriate units.
ε = ___________
Part B
If the band is gold with a cross-section area of 4.0 mm2, what is the induced current? Assume the band is of jeweler's gold and its resistivity is 13.2 x 1010 Ω*m.
Express your answer to two significant figures and include the appropriate units.
I = ____________
Answer:
A. 3.9 V B. 1.9 fA
Explanation:
Part A
What emf is induced in the ring as the field changes?
Express your answer to two significant figures and include the appropriate units.
The induced emf ε = ΔΦ/Δt where ΔΦ = change in magnetic flux = ΔABcosθ where A = area of coil and B = magnetic field strength, θ = angle between A and B = 0 (since the axis of the ring is parallel )Δt = change in time
ε = ΔΦ/Δt
ε = ΔABcos0°/Δt
ε = AΔB/Δt
A = πd²/4 where d = diameter of ring = 2.0 cm = 2.0 × 10⁻² m, A = π(2.0 × 10⁻² m)²/4 = π4.0 × 10⁻⁴ m²/4 = 3.142 × 10⁻⁴ m², ΔB = change in magnetic field strength = B₁ - B₀ where B₁ = final magnetic field strength = 2.5 T and B₀ = initial magnetic field strength = 0 T. ΔB = B₁ - B₀ = 2.5 T -0 T = 2.5 T and Δt = 200 μs = 200 × 10⁻⁶ s.
So, ε = AΔB/Δt
ε = 3.142 × 10⁻⁴ m² × 2.5 T/200 × 10⁻⁶ s
ε = 7.854 × 10⁻⁴ m²-T/2 × 10⁻⁴ s
ε = 3.926 V
ε ≅ 3.9 V
Part B
If the band is gold with a cross-section area of 4.0 mm2, what is the induced current? Assume the band is of jeweler's gold and its resistivity is 13.2 x 1010 Ω*m.
Express your answer to two significant figures and include the appropriate units.
Since current, i = ε/R where ε = induced emf = 3.926 V and R = resistance of band = ρl/A where ρ = resistivity of band = 13.2 × 10¹⁰ Ωm, l = length of band = πd where d = diameter of band = 2.0 cm = 2.0 × 10⁻² m. So, l = π2.0 × 10⁻² m = 6.283 × 10⁻² m and A = cross-sectional area of band = 4.0 mm² = 4.0 × 10⁻⁶ m².
So, i = ε/R
= ε/ρl/A
= εA/ρl
= 3.926 V × 4.0 × 10⁻⁶ m²/(13.2 × 10¹⁰ Ωm × 6.283 × 10⁻² m)
= 15.704 × 10⁻⁶ V-m²/(82.9356 × 10⁸ Ωm²
= 0.1894 × 10⁻¹⁴ A
= 1.894 × 10⁻¹⁵ A
≅ 1.9 fA
Which is a property of bases?
A.
highly metal reactive
B.
sour to the taste
C.
slippery feel
D.
low pH
Answer:
C. Slippery feel
Explanation:
To enhance heat rejection from a spacecraft, an engineer proposes to attach an array of rectangular fins to the outer surface of the spacecraft and to coat all surfaces with a material that approximates blackbody behavior. Consider the U-shaped region between adjoining fins and subdivide the surface into components associated with the base (1) and the side (2). Obtain an expression for the rate per unit length at which radiation is transferred from the surfaces to deep space, which may be approximated as a blackbody at absolute zero temperature. The fins and the base maybe assumed to be isothermal at a temperature T. Comment on your result. Does the engineer's proposal have merit
Answer:
Attached below is the required diagram related to the question
answer :
q'3 = WбT^4
engineer's proposal has merit
Explanation:
Let : A'3 represent the deep space
A'1 represent the surface area , F13 and F23 represent the view factors
T1 , T2, T3 ; represent temperatures
q'3 represent net rate of heat radiation
Derive the expression for the rate per unit length at which radiation is transferred from the surfaces to deep space
derived expression ; q'3 = WбT^4
attached below is a detailed solution
Given that The emission is proportional to the area of the opening and the surfaces ( 1 and 2 ) have the same temperature hence this problem can be treated as a two surface enclosure. hence the engineer's proposal have merit .
attached below is a prove ( b )
What are the similarities between a resultant force equilibrant force?
Answer:
Explanation:
Resultant is a single force that can replace the effect of a number of forces. "Equilibrant" is a force that is exactly opposite to a resultant. Equilibrant and resultant have equal magnitudes but opposite directions.
In a little league baseball game, the 145 grams ball enters the strike zone with a speed of 11.0 meters per second. The batter hits the ball and it leaves his bat with a speed of 25.0 meters per second in exactly the opposite direction. If the bat is in contact with the ball for 1.0 m/s, what is the magnitude of the average force exerted by the bat on the ball?
Answer:
Force = 5.22 N
Explanation:
According to Newton's Second Law of motion:
[tex]Force = Rate\ of\ Change\ of\ Momentum\\\\Force = \frac{mv_f-mv_i}{t}\\[/tex]
where,
m = mass of ball = 145 g = 0.145 kg
vf = final speed of ball after hit = 25 m/s
vi = initial speed of ball before hit = - 11 m/s (negative sign due to opposite direction)
t = time of contact = 1 s
Therefore,
[tex]Force = \frac{(0.145\ kg)(25\ m/s)-(0.145\ kg)(-11\ m/s)}{1\ s} \\\\[/tex]
Force = 5.22 N
A student must analyze data collected from an experiment in which a block of mass 2M traveling with a speed v0 collides with a block of mass M that is initially at rest. After the collision, the two blocks stick together. What applications of the equation for the conservation of momentum represent the initial and final momentum of the system for a completely inelastic collision between the blocks?
Solution :
In the question, it is given that the collision is inelastic and the blocks stick together.
In an inelastic collision, the linear momentum is conserved but the kinetic energy is not conserved.
The linear momentum is given by :
[tex]$\vec p = m \vec v$[/tex] (mass x velocity)
So according to the conservation of linear momentum,
[tex]$\vec p_{(\text{before collision})}=\vec p_{(\text{after collision})}$[/tex]
Let the velocity after the collision is [tex]$v_F$[/tex]
[tex]$m_1v_0+m_2 \times 0 = m_1v_F+m_2v_F$[/tex]
Putting the values of [tex]$m_1 \text{ and}\ m_2$[/tex]
[tex]$m_1=2M \text{ and}\ m_2=M$[/tex]
∴ [tex]$2Mv_0=2Mv_F+Mv_F$[/tex], as the blocks stick together after the collision.
and [tex]$2MV_0=3Mv_F$[/tex], as the blocks stick together after the collision.
A man pushes a 10-kg block 10 m, along a rough, horizontal
surface with a 40-N force directed 37° below the horizontal. If
the coefficient of kinetic friction is 0.2, calculate the total
work done on the block
Answer:
hope you can understand
If you are in a car that is traveling at 60 mph and a balanced force is applied to the car what will happen to the motion of the car? /gen
Answer:
The car will stop moving
Explanation:
If a balanced force is applied to the car in motion, the car will stop accelerating and remain stationary because all the forces acting on the car are equal.
Also, you can say, since a balanced force is applied, the net force or resultant force on the car is zero. According to Newton's second law of motion, an object will move in the direction of the applied force. When the resultant force on the object is zero, it means the object will not move.
what is force,momentum,and velocity.
Answer:
A force is a push or pull upon an object resulting from the object's interaction with another object.
Momentum is force or speed of movement.
Velocity defines the path of the motion of the frame or the object
Determine the direction of the force on a charge.
a. along the line between the charge and the center of the square outward of the center
b. along the side of the square outward of the other charge that lies on the side
c. along the line between the charge and the center of the square toward the center
d. along the side of the square toward the other charge that lies on the side
Answer:
hi your question is incomplete below is the complete question
A charge of 2.15 mC is placed at each corner of a square 0.500 m on a side. Determine the direction of the force on a charge.
answer : Along the line between the charge and the center of the square outward of the center ( A )
Explanation:
The direction of the force on a charge is along the line between the charge and the center of the square outward of the center
Given that; Fnet = 3.17 * 10^-1 N ( calculated value )
The nature of the Force is repulsive in nature
attached below is a Pictorial representation of the direction of the force on a charge
A lead fishing weight of a mass of 0.20 kg is tied to a fishing line that is 0.50 m long. The weight is then whirled in a vertical circle. The finishing line will break if its tension exceeds 100.0 N.
Required:
a. If the weight is whirled at higher and higher speeds, at what point in the vertical circle will the string break (top, bottom, or random position)?
b. At what speed will the string break?
Answer:
The solution of the given question is summarized in the below section.
Explanation:
The given values are:
Tension,
T = 100 N
mass,
m = 0.2 kg
length,
l = 0.5 m
Now,
(a)
Somewhere at bottom, string or thread breaks since string voltage seems to be the strongest around this stage.
then,
⇒ [tex]T-mg=\frac{mv^2}{l}[/tex]
or,
⇒ [tex]T=mg+\frac{mv^2}{l}[/tex]
(b)
As we know,
⇒ [tex]\frac{mv^2}{l}=T-mg[/tex]
or,
⇒ [tex]v^2=\frac{(T-mg)l}{m}[/tex]
On substituting the values, we get
⇒ [tex]=\frac{(100-0.2\times 10)0.5}{0.2}[/tex]
⇒ [tex]=\frac{49}{0.2}[/tex]
⇒ [tex]v =\sqrt{245}[/tex]
⇒ [tex]=15.65 \ m/s[/tex]
An ultrasonic tape measure uses frequencies above 20 MHz todetermine dimensions of structures such as buildings. It does so byemitting a pulse of ultrasound into air and then measuring the timeinterval for an echo to return from a reflecting surface whosedistance away is to be measured. The distance is displayed as adigital read-out. A tape measure emits a pulse of ultrasound with afrequency of 25.0 MHz.
(a) What is the distance to an object fromwhich the echo pulse returns after 24ms when the air temperature is 26°C?
(b) What should be the duration of the emitted pulse if it is toinclude 10 cycles of the ultrasonic wave?
(c) What is the spatial length of such a pulse?
Answer:
a) 1m
b) 2μs
c) 3mm
Explanation:
Which of these is NOT an inherited trait of the plant?
Explanation:
I think you forgot to add the other part!
Answer:
uhm what plant...
Explanation:
How does the wave period relate to the frequency of a wave?
Answer:
its in the picture hope it helps make brainlliest ty
A 6kg object undergoes an acceleration of 2m/s, what is the magnitude of the resultant acting on it . If this same force is applied to a 4kg object, what acceleration is produced
Answer:
[tex]12\; \rm N[/tex].
[tex]3\; \rm m\cdot s^{-2}[/tex].
Explanation:
By Newton's Second Law, the acceleration of an object is proportional to the size of the resultant force on it, and inversely proportional to the mass of this object.
[tex]\displaystyle \text{acceleration} = \frac{\text{resultant force}}{\text{mass}}[/tex].
Rearrange this equation for the resultant force on the object:
[tex]\text{resultant force} = \text{acceleration} \cdot \text{mass}[/tex].
For the [tex]6\; \rm kg[/tex] object in this question:
[tex]\begin{aligned} F &= m \cdot a \\ &= 6\; \rm kg \times 2\; \rm m \cdot s^{-2} \\ &=12\; \rm N\end{aligned}[/tex].
When the resultant force on the [tex]4\; \rm kg[/tex] object is also [tex]12\; \rm N[/tex], the acceleration of that object would be:
[tex]\begin{aligned} a &= \frac{F}{m} \\ &= \frac{12\; \rm N}{4\; \rm kg} = 3\; \rm m \cdot s^{-2}\end{aligned}[/tex].
When light with a wavelength of 221 nm is incident on a certain metal surface, electrons are ejected with a maximum kinetic energy of J. Determine the wavelength of light that should be used to double the maximum kinetic energy of the electrons ejected from this surface
This question is incomplete, the complete question is;
When light with a wavelength of 221 nm is incident on a certain metal surface, electrons are ejected with a maximum kinetic energy of 3.28 × 10⁻¹⁹ J. Determine the wavelength (in nm) of light that should be used to double the maximum kinetic energy of the electrons ejected from this surface.
Answer:
the required wavelength of light is 161.9 nm
Explanation:
Given the data in the question;
Let us represent work function of the metal by W.
Now, using Einstein photoelectric effect equation;
[tex]E_{proton[/tex] = W + [tex]K_{max[/tex]
hc/λ = W + [tex]K_{max[/tex] ------- let this be equation 1
we solve for W
W = hc/λ - [tex]K_{max[/tex]
given that; λ = 221 nm = 2.21 × 10⁻⁷ m, [tex]K_{max[/tex]= 3.28 × 10⁻¹⁹ J
we know that speed of light c = 3 × 10⁸ m/s and Planck's constant h = 6.626 × 10⁻³⁴ Js.
so we substitute
W = [( (6.626 × 10⁻³⁴)(3 × 10⁸) )/2.21 × 10⁻⁷ ] - 3.28 × 10⁻¹⁹
W = 8.99457 × 10⁻¹⁹ - 3.28 × 10⁻¹⁹
W = 5.71457 × 10⁻¹⁹ J
Now, to determine λ for which maximum kinetic energy is double
so;
[tex]K'_{max[/tex] = double = 2( 3.28 × 10⁻¹⁹ J ) = 6.56 × 10⁻¹⁹ J
from from equation 1
we solve for λ'
λ' = hc / W + [tex]K'_{max[/tex]
we substitute
λ' = ( (6.626 × 10⁻³⁴)(3 × 10⁸) ) / ( (5.71457 × 10⁻¹⁹ J) + ( 6.56 × 10⁻¹⁹ J ))
λ' = 1.9878 × 10⁻²⁴ / 1.227457 × 10⁻¹⁸
λ' = 1.619 × 10⁻⁷ m
λ' = 161.9 nm
Therefore, the required wavelength of light is 161.9 nm
Active listening includes all of the following EXCEPT: A. paraphrasing B. clarifying C. ignoring D. empathizing Please select the best answer from the choices provided. A B C D
Answer:
C: Ignoring
Explanation:
On edge 2021
Answer:
It's C.
Explanation:
Which of these will be the correct relationship between work input and work output?
A) Work input = Work output + Work against friction
B) Work input = Work output – Work against friction
C) Work input = Work output * Work against friction
D) Work input = Work output / Work against friction
Answer:
Work input = Work output * Work against friction is your answer so C
Explanation:
I hope this helps you :)
Answer:
A) Work input = Work output + Work against friction
Explanation:
Two loudspeakers S1 and S2, 2.20 m apart, emit the same single-frequency tone in phase at the speakers. A listener L is located directly in front of speaker S1, in other words, the lines LS1 and S1S2 are perpendicular. L notices that the intensity is at a minimum when L is 5.50 m from speaker S1. What is the lowest possible frequency of the emitted tone
Answer:
the lowest possible frequency of the emitted tone is 404.79 Hz
Explanation:
Given the data in the question;
S₁ ← 5.50 m → L
↑
2.20 m
↓
S₂
We know that, the condition for destructive interference is;
Δr = ( 2m + [tex]\frac{1}{2}[/tex] ) × λ
where m = 0, 1, 2, 3 .......
Path difference between the two sound waves from the two speakers is;
Δr = √( 5.50² + 2.20² ) - 5.50
Δr = 5.92368 - 5.50
Δr = 0.42368 m
v = f × λ
f = ( 2m + [tex]\frac{1}{2}[/tex])v / Δr
m = 0, 1, 2, 3, ....
Now, for the lowest possible frequency, let m be 0
so
f = ( 0 + [tex]\frac{1}{2}[/tex])v / Δr
f = [tex]\frac{1}{2}[/tex](v) / Δr
we know that speed of sound in air v = 343 m/s
so we substitute
f = [tex]\frac{1}{2}[/tex](343) / 0.42368
f = 171.5 / 0.42368
f = 404.79 Hz
Therefore, the lowest possible frequency of the emitted tone is 404.79 Hz