Velocity (m/s)
50
40
30
20
10
0
A
Velocity vs Time
B
C
1 2 3 4 5 6 7 8 9 10
Time (s)
Use the information presented in the graph to answer
the questions.
Which segments show acceleration?
Which segment indicates that the object is slowing
down?
What is the velocity of segment B?
What is the acceleration of segment B?

Answer:

2kg brick

Explanation:

2kg brick at 25 degrees.

Answer: A 2kg brick at 20*C

Explanation:

Explanation:

This case is inertia.

Mass is directly proportional to inertia

The difference between polar covalent bonds and nonpolar covalent bonds using these two molecules - H2 and HCl are HCl is a polar covalent compound because the chloride ion is extra electronegative than the hydrogen ion.

HCl is a polar molecule. This is due to the fact the Chlorine (Cl) atom withinside the HCl molecule is extra electronegative and does now no longer proportion the bonding electrons similarly with Hydrogen (H). But H2 And Cl2 are nonpolar because of comparable electronegativity of each the atoms withinside the molecule H2 And Cl2 .

Hydrogen chloride is a diatomic molecule, such as a hydrogen atom H and a chlorine atom Cl related with the aid of using a polar covalent bond. The chlorine atom is an awful lot extra electronegative than the hydrogen atom, which makes this bond polar.

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a)The velocity of the object when it hits the ground will be 1.12 × 10⁵ m/sec

b) The time taken for the object to reach the ground will be 11.4 × 10³ sec.

The motion of an item hurled or projected into the air, subject only to gravity's acceleration, is known as projectile motion.

The velocity in the x-direction;

[tex]\rm v_x = (vcos \theta)^2 +2gh \\\\ v_x = (50 cos 30)^2+ 2 \times 9.81 \times 4000 \\\\ v_x = 80355 m/sec[/tex]

Velocity in y-direction;

[tex]\rm v_ y = (vsin \theta)^2 + 2gh \\\\ v_ y = (25)^2+2 \times 9.81 \times 4000 \\\\ v_y =79105 \ m/sec[/tex]

The resultant velocity is found as 1.12 × 10⁵ m/sec.

The time taken to reach the ground is found as;

[tex]\rm v = u+gt \\\\ 1.12 \times 10^5 \ m/sec = 50 \ m/sec + 9.81 \times t \\\\ t = 11.4 \times 10^3 \ sec[/tex]

Hence, the velocity of the object when it hits the ground will be 1.12 × 10⁵ m/sec, and the time taken for the object to reach the ground will be 11.4 × 10³ sec.

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Answer:

false

Explanation:

it cant defined the messy and clean states

Answer:

given -

initial velocity = 4.09 m/s

acceleration = 1.01 m/s²

distance = 23.4 m

time = ?

using second formula of motion,

s = ut + 1/2 at².

where, s = distance

u = initial velocity

t = time

a = acceleration

23.4 = 4.09(t) + 1/2(1.01)(t) ²

23.4 = 4.09t + 2.02t²

2.02t² + 4.09t - 23.4 = 0

solving the equation by using quadratic formula

Use the standard form, ax² + bx + c = 0 , to find the coefficients of our equation, :

a = 2.02

b = 4.09

c = -23.4

we get t=2.539 or t= -4.563

time cannot be negative so

t= 2.539 sec = 2.6 Sec is the answer

(a)  The charge stored in the 5.00-μF capacitor is 37.2  μC.

(b) The potential difference across the 10.0-μF capacitor is 3.72 V.

The capacitance of a capacitor is defined as the ratio of the charge stored and the potential difference between the capacitor.

The capacitance of a capacitor is denoted by C and expressed as

C = Q/V

Given, 5.00 μF, 10.0 μF, and 50.0 μF capacitors are connected in series across a 12.0-V battery.

(a) The equivalent capacitance is

1 / Ceq = 1 / C₁ +1 / C₂ + 1/ C₃

Substitute the values, we get

Ceq = 3.1  μF

The charge stored in 5.00-μF capacitor is

Q  = Ceq x V

Q = 3.1  μF x 12 V

Q = 37.2  μC

Thus, the charge stored in the 5.00-μF capacitor is 37.2  μC

(b) The potential difference across the 10.0-μF capacitor is given by

V = Q/C₂

Put the values, we get

V = 37.2 / 10

V = 3.72 V

Thus, the potential difference across the 10.0-μF capacitor is 3.72 V.

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The force is reasonable for making fusion possible in the Sun is heat energy.

When the slow moving neutrons are bombarded with the heavy radioactive nuclei, the product is the more number of neutrons are produced with the large amount of energy. This multiplying process is called nuclear fusion.

The amount of energy produced in such a reaction can be calculated using the equivalence of mass and energy relationship.

E = mc²

The same happens in nuclear fusion where large amount of energy is needed to make more heavy nuclei.

Thus, fusion requires heat energy to continue the reaction.

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The work function in eV for the given cutoff frequency is  5.05 eV.

The work function is related to the frequency as

Wo = h x fo

where, fo = cutoff frequency and h is the Planck's constant

Given is the cutoff frequency for a certain element is 1.22 x 10¹⁵ Hz

Wo = 6.626 x 10⁻³⁴ x  1.22 x 10¹⁵ Hz / 1.6 x 10⁻¹⁹

Wo = 5.05 eV

Thus, the work function is  5.05 eV

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The specific heat capacity of the alcohol will be 3.72  kJ/kg°C.

The amount of heat required to increase a substance's temperature by one degree Celsius is known as its "specific heat capacity."

Similarly, heat capacity is the relationship between the amount of energy delivered to a substance and the increase in temperature that results.

Given data;

Mass of liquid sample of Alcohol  m₁ = 200-gram

The temperature of alcohol, T₁ =  -6°C.

Mass of liquid sample of water  m₂ = 400-gram

The temperature of the water, T₂=  20°C.

The specific heat capacity of the alcohol, S₁=?

The specific heat capacity of water is, S₂=4.19 kJ/kg.°C

As we know that;

[tex]\rm Q_{gain}= Q{loss} \\\\ Q_{alcohol} =Q_{water} \\\\\ m_1s_1\triangle T_1 = m_2S_2 \triangle S_2 \\\\ 200 \times 10^{-3} \times S_1 [ (12-(-6) ] = 40 \times 10^{-3} \times 4.19 \times 10^{-3} \times (20-12)\\\\S_1 = 2 \times 4.19 \times 10^3 \times \frac {8}{18} \\\\ S_1 = 3.72 \ kJ /kg ^0 C[/tex]

Hence the specific heat capacity of the alcohol will be 3.72  kJ/kg°C.

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Answer:

CABD

.....................

The image is cut and jumbled and placed into different places, so the correct order of the image that it can be back together is CABD.

A specific piece of something is an image. It can provide information to the optical system in two dimensions, three components, or in another way. An item that mimics a subject, such as a picture or other two-dimensional image, might be considered an image. An image in the context of signal analysis is a dispersed color amplitude.

A graphical image need not make use of the complete visual system. A common illustration of this is a grayscale image, which does not employ color, but instead relies on the visual game's sensitivity to brightness across all wavelengths. Even if it doesn't make proper use of the visual system, a black-and-white and white visual depiction of something is nonetheless an image.

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You are watching a television show about Navy pilots. The narrator says that when a Navy jet takes off, the average force on the jet is due to the catapult is mathematically given as

What is the average force on the jet is due to the catapult?

Generally, the equation for acceleration is mathematically given as

[tex]a=\frac{vf^2-vi^2}{2s}\\\\\Therefore\\\\a=\frac{69.29^2-0^2}{2(90}\\\\a=26.673m/s^2[/tex]

In conclusion, The force

F=m*a

F=15100*26.673

F=40272.3N

F 402 KN

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(a) The required magnitude of the electric field when the point charge is an electron is 5.57 x 10⁻¹¹ N/C.

(b) The required magnitude of the electric field when the point charge is an proton is 1.02 x 10⁻⁷ N/C.

The magnitude of electric field is given by the following equation.

F = qE

But F = mg

mg = qE

E = mg/q

where;

E = (9.11 x 10⁻³¹ x 9.8)/(1.602 x 10⁻¹⁹)

E = 5.57 x 10⁻¹¹ N/C

E = (1.67 x 10⁻²⁷ x 9.8)/(1.602 x 10⁻¹⁹)

E = 1.02 x 10⁻⁷ N/C

Thus, the required vertical electric field is greater when the charge is proton.

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The I3 will be 158 A.

To calculate I3 firstly, V4 has to be calculated,

[tex]V_{4} =I_{4} R_{4}[/tex]

[tex]V_{4} = V_{2} / R_{4} + R_{5} * R_{4}[/tex]

[tex]V_{4} = 12 * 135 / 135+61[/tex]

[tex]V_{4} = 8.26V[/tex]

For I3,

[tex]I_{3} = R_{1} /(R1+R3 + (R1+R3)(R2+R6) * (V2 - V1 (R1+R2+R6/R1)[/tex]

[tex]I3=(61)/((61)(50)+(61+50)(141+141)) (12 -18 (1+(141+141)/61)) = -.158 A[/tex]

Hence, the current through I3 will be 158 A.

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A beta particle or an electron is released during beta decay. The charges can be positive or negative. Co changes to Ni, Fe to Mn, Pb to Tl, and Pu to Am.

Beta-decay is the radioactive decay that involves the release of the beta particle or the positron or the electron. The larger nucleus splits during nuclear fission and releases smaller nuclei.

The nuclear reactions are shown as:

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Answer:

Approximately [tex]325[/tex] (rounded down,) assuming that [tex]g = 9.81\; {\rm N \cdot kg^{-1}}[/tex].

The number of repetitions would increase if efficiency increases.

Explanation:

Ensure that all quantities involved are in standard units:

Energy from the cookie (should be in joules, [tex]{\rm J}[/tex]):

[tex]\begin{aligned} & 53\; {\rm kCal} \times \frac{1\; {\rm kJ}}{4.184\; {\rm kCal}} \times \frac{1000\; {\rm J}}{1\; {\rm kJ}} \approx 2.551 \times 10^{5}\; {\rm J} \end{aligned}[/tex].

Height of the weight (should be in meters, [tex]{\rm m}[/tex]):

[tex]\begin{aligned} h &= 2\; {\rm dm} \times \frac{1\; {\rm m}}{10\; {\rm dm}} = 0.2\; {\rm m}\end{aligned}[/tex].

Energy required to lift the weight by [tex]\Delta h = 0.2\; {\rm m}[/tex] without acceleration:

[tex]\begin{aligned} W &= m\, g\, \Delta h \\ &= 100\; {\rm kg} \times 9.81\; {\rm N \cdot kg^{-1}} \times 0.2\; {\rm m} \\ &= 196\; {\rm N \cdot m} \\ &= 196\; {\rm J} \end{aligned}[/tex].

At an efficiency of [tex]0.25[/tex], the actual amount of energy required to raise this weight to that height would be:

[tex]\begin{aligned} \text{Energy Input} &= \frac{\text{Useful Work Output}}{\text{Efficiency}} \\ &= \frac{196\; {\rm J}}{0.25} \\ &=784\; {\rm J}\end{aligned}[/tex].

Divide [tex]2.551 \times 10^{5}\; {\rm J}[/tex] by [tex]784\; {\rm J}[/tex] to find the number of times this weight could be lifted up within that energy budget:

[tex]\begin{aligned} \frac{2.551 \times 10^{5}\; {\rm J}}{784\; {\rm J}} &\approx 325 \end{aligned}[/tex].

Increasing the efficiency (the denominator) would reduce the amount of energy input required to achieve the same amount of useful work. Thus, the same energy budget would allow this weight to be lifted up for more times.

The answer to the first part of the question is 24 V and the answer to the second part of the question is 48 V.

The formula which relates Charge, Capacitance and Voltage is

Charge = Capacitance × Voltage

Q = CV

where Q denotes the charge, C denotes the capacitance and V denote the voltage.

Using the above formula the charge on the capacitance will be

Q = CV

C = 10 mF

V = 24 V

Q = 240 mC

Charge on capacitance is 240 mC.

Now after we disconnect the battery

C = 10 mF (will remain same)

Q = 240 mC

V = Q/C

V = 240/10 V

V = 24 V

So after we removed the battery, The voltage will remain same.

Now we know that parallel plate capacitor formula is

C = ε(A/d)

from here

C ∝ 1/d, and as there is no loss of charge we can say that V ∝ 1/d

Form C ∝ 1/d and V ∝ 1/C we can state that

V ∝ d   {where d is the measurement of separation between the plates}

so if we double the distance between the plates then, the voltage will also get double.

Previously our voltage was 24 V, now if we double the distance between the plates the voltage will also get double, and it will become 48 V.

So, the voltmeter will take a reading of 24 V if voltmeter is connected across the two plates without discharging them, and if we double the plate separation then it will take 48 V as reading.

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The mass and tension due to the system are as follows:

Tension is a type of pulling force due transmitted by means of a string or cable.

a) The mass of the child and scale = 104/9.81 = 10.6 kg

b) The tension T, in the cord attaching the child to the scale = scale reading = 104N

c) The tension T', in the cord attaching the scale to the ceiling = scale reading + weight of scale

T' = 104 + (0.5 * 9.81)

T' = 108.9 N

d) The sketch is attached in the picture

In conclusion, the tension is force exerted on the cord due to the weight of the scale and the baby.

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The force the rod exerts on the ball at the given angle is determined as 3.94 N.

The force exerted is calculated as follows;

F = ma

F = mv²/r

F = mω²r

where;

r = 2cosθ

ωf² = ωi² + 2αθ

where;

ωf² = (1)² + 2(2)(15 x π/180)

ωf² = 2.04

ωf = √2.04

ωf = 1.428 rad/s

F = mω²(2cosθ)

F = (1)(1.428)²(2 x cos15)

F = 3.94 N

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Hello!

We can use the following equation for magnetic force on a charged particle:
[tex]F_B = qv \times B[/tex]

[tex]F_B[/tex] = Magnetic force (N)

q = Charge of particle (1.6 × 10⁻¹⁹ C)
v = velocity of particle (5.2 × 10⁷ m/s)

B = Magnetic field strength (1.4 T)

This is a cross-product, so the equation can be rewritten to F = qvBsinφ where φ is the angle between the magnetic field and particle velocity vectors.  

Since the proton's velocity vector and the magnetic field vector are perpendicular, sin(90) = 1. We can reduce the equation to:

[tex]F_B = qvB[/tex]

Plug in the known values.

[tex]F_B = (1.6*10^{-19})(5.2*10^7)(1.4) = \boxed{1.1648 *10^{-11} N}[/tex]

The work done is positive and is equal to 20000 J

Work done is defined as the product of force and the distance moved by the force.

Mathematically:

The work done by the force = 20 * 1000 = 20000J

The work done is positive and is equal to 20000 J

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The velocity of the air relative to the runner is 5 m/s.

We must recall that velocity is a vector quantity and the relative velocity must be obtained vectorially. Thus we know that;

Velocity of the runner = 4m/s. due west

Velocity of the wind =  3m/s due south

The relative velocity is;

Vr = √(4)^2 + (3)^2

Vr = 5 m/s

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A 3rd harmonic of a tube open at both ends will have displacement antinodes at both ends.

In a tube of length L with two open ends, the longest standing wave has displacement antinodes (pressure nodes) at both ends. The fundamental or first harmonic is what it is known as. The second harmonic is the longest standing wave in a tube of length L with two open ends.

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Answer:

Given - Force = 21.4 N

Work done = 1974 J

To find - Distance

Solution -

Work done = Force * displacement

1974 j = 21.4 N * displacement

1974/21.4 = displacement

92.24

[tex]\\ \rm\Rrightarrow Work=Force\times Displacement [/tex]

[tex]\\ \rm\Rrightarrow 1974=21.4×Displacement[/tex]

[tex]\\ \rm\Rrightarrow Displacement=92.24m[/tex]

Answer:

1 W

Explanation:

Power is the ratio of the total work done by the body to the time it will takes to do the work.

Total Work done = Potential energy (P.E) = mgh

Total work done = mgh

Total work done = 1kg × 10m/s² × 1m

Total work done = 10 J

Therefore Total Work done is 10J.

And then time is 10 s.

Power = Total Work done divided by time

Power = 10 J÷ 10 s = 1 W

Here is your answer mate,

Question,

[tex]Determine\: the\: amount\\ \: of\: power\:used\: in\\\: holding\: a\: 25\: kg\: box\:\\ , \: 1.5\: meters \: above\: the\: floor\\\: for\: 60\: seconds[/tex]

Answer,

[tex][/tex]

Solution,

[tex][/tex]

Given,

[tex]MASS \: IS\: 25\: KG\: \\ and \: HEIGHTIS\: 1.5m\: [/tex]

[tex][/tex]

WORK DONE (done against gravity) =

mass×acceleration due to gravity ×height

WORK = 25× 10× 1.5

[tex]\: \: \: \: \: \: \: \:\: \: \: \: \: \: \: [/tex]= 375 Nm

[tex][/tex]

Now

POWER =

[tex] \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:\frac{work}{time} [/tex]

POWER

[tex]\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:= \frac{375}{60} Watt [/tex]

[tex]\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: =6.25[/tex]

[tex]Therfore\: your \: answer\: is\: 6.25[/tex]

[tex][/tex]

Check this,

[tex]Acceleration\: due\: to \: gravity\\\: can\: be\: 9.8\: m/s²\: \\As\: nothing\: mentioned\\\: in\: question\: \\I \: took \: it \: as \: 10[/tex]

[tex][/tex]

Have a good day

Answer:

See the answer Explain why aircraft are carefully designed so that parts do not resonate. Expert Answer This virtually takes place, however maximum usually in small piston-engined airplanes, in particular dual-engined airplanes. The resonant frequency of the fuselage of a small plane goes to have numerous nodes, withinside the low loads of hertz.

Answer:

D

Explanation:

15m/ s

i hope this helps

Answer:

Answer is C because light travels in a sight line but when light pass through a refractor the light from the source changes direction when passes through a refractor