Find a parametric representation for the surface. The plane through the origin that contains the vectors i - j and j - k

Answer:

hope it helps

much as you can

Answer:

Option "D" is the correct answer to the following question.

Explanation:

The gravitational potential energy of an item is determined by its mass, elevation, and gravitational acceleration. As a result, angular momentum and energy are preserved. The gravitational potential energy, on the other hand, varies with distance. When a consequence, kinetic energy varies during each orbit, resulting in a faster speed as a planet approaches the Sun.

Answer:

SPEED AND MASS

Explanation:

TOOK THE TEST

Answer:

See explanation

Explanation:

a) maximum height of a projectile = u sin^2θ/2g

H= 600 × (sin 30)^2/2 × 10

H= 7.5 m

b) Time of flight

t= 2u sinθ/g

t= 2 × 600 sin 30/10

t= 60 seconds

Range

R= u^2sin2θ/g

R= (600)^2 × sin2(30)/10

R= 31.2 m

Answer:

The velocity is 2661.5 m/s.

Explanation:

Radius, horizontal distance, d = 309 km

height, h = 25 km

acceleration due to gravity on moon, g =3.71 m/s^2

Let the time taken is t and the horizontal velocity is u.

horizontal distance = horizontal velocity x time

309 x 1000 = u t .... (1)

Use second equation of motion in vertical direction.

[tex]h = u_yt +0.5 gt^2\\\\25000 = 0 + 0.5\times 3.71\times t^2\\\\t =116.1 s[/tex]  

So, put in (1)

309 x 1000 = u x 116.1

u = 2661.5 m/s

Answer:

0.5

Explanation:

because the block is attached to the pulley of the string

Explanation:

For engine 1,

Energy removed = 239 J

Energy added = 567 J

[tex]\eta_1=\dfrac{239}{567}\cdot100=42.15\%[/tex]

For engine 2,

Energy removed = 457 J

Energy added = 789 J

[tex]\eta_2=\dfrac{457}{789}\cdot100=57.92\%[/tex]

For engine 3,

Energy removed = 422 J

Energy added = 1038 J

[tex]\eta_3=\dfrac{422}{1038}\cdot100=40.65\%[/tex]

So, the engine 2 has the highest thermal efficiency.

Answer:

how to solve this problem ???????

The magnitude of the resultant force is 919.6 N and the value of angle θ is 36.87⁰.

The resultant of the two forces is determined by resolving the force into x and y component as shown below;

[tex]F_1_x + F_2x_x = F_R_x \ --- (1) \\\\F_1_y + F_2_y = F_R_y\ ---(2)[/tex]

where;

The value of Angle θ is determined from equation (1)

-500sinθ + 600sin(30) = 0

500sinθ = 600sin(30)

500sinθ = 300

sinθ = 3/5

θ = 36.87⁰

The resultant of the forces is determined using the second equation;

500cosθ + 600cos(30) = R

500 x cos(36.87) + 600 x cos(30) = R

919.6 N = R

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Answer:

A will attract

B will repare

Answer:

Toaster = I = 14.67 A

Frying Pan = 11.83 A

Lamp = 0.71 A

Explanation:

The electric power is given as:

[tex]P = VI\\\\I = \frac{P}{V}[/tex]

where,

I = current

P = Power

V = Voltage = 120 V

FOR TOASTER:

P = 1760 W

Therefore,

[tex]I = \frac{1760\ W}{120\ V}[/tex]

I = 14.67 A

FOR FRYING PAN:

P = 1420 W

Therefore,

[tex]I = \frac{1420\ W}{120\ V}[/tex]

I = 11.83 A

FOR LAMP:

P = 85 W

Therefore,

[tex]I = \frac{85\ W}{120\ V}[/tex]

I = 0.71 A

Answer:

[tex]M_b=6kg[/tex]

Explanation:

From the question we are told that:

Coefficient of restitution [tex]\mu=1.00[/tex]

Mass A [tex]M_a=6kg[/tex]

Initial Velocity of A [tex]U_a=6m/s[/tex]

Initial Velocity of B [tex]U_b=0m/s[/tex]

Generally the equation for Coefficient of restitution is mathematically given by

 [tex]\mu=\frac{V_b-V_a}{U_a-U_b}[/tex]

 [tex]1=\frac{v_B}{6}[/tex]

 [tex]V_b=6*1[/tex]

 [tex]V_b=6m/s[/tex]

Generally the equation for conservation of linear momentum  is mathematically given by

 [tex]M_aU_a+M_bU_b=M_aV_a+M_bV_b[/tex]

 [tex]6*6+=M_b*6[/tex]

 [tex]M_b=6kg[/tex]

Answer: [tex]29.85\ W/m^2[/tex]

Explanation:

Given

Power [tex]P=60\ W[/tex]

Distance from the light source [tex]r=0.4\ m[/tex]

Intensity is given by

[tex]I=\dfrac{P}{4\pi r^2}[/tex]

Inserting values

[tex]\Rightarrow I=\dfrac{60}{4\pi (0.4)^2}\\\\\Rightarrow I=\dfrac{60}{2.010}\\\\\Rightarrow I=29.85\ W/m^2[/tex]

Answer:

29.85 W/ m^2

Explanation:

Answer:

[tex]T=66262.4s[/tex]

Explanation:

From the question we are told that:

Altitude [tex]A=2.90 *10^7[/tex]

Mass [tex]m=5.97 * 10^{24} kg[/tex]

Radius [tex]r=6.38 *10^6 m.[/tex]

Generally the equation for Satellite Speed is mathematically given by

[tex]V=(\frac{GM}{d} )^{0.5}[/tex]

[tex]V=(\frac{6.67*10^{-11}*5.97 * 10^{24}}{6.38 *10^6+2.90 *10^7} )^{0.5}[/tex]

[tex]V=3354.83m/s[/tex]

Therefore

Period T is Given as

[tex]T=\frac{2 \pi *a}{V}[/tex]

[tex]T=\frac{2 \pi *(6.38 *10^6+2.90 *10^7}{3354.83}[/tex]

[tex]T=66262.4s[/tex]

Answer:

f = 7.06 N

Explanation:

The maximum frictional force on the knee joint of the person can be given by the following formula:

[tex]f = \mu R = \mu W \\[/tex]

where,

f = maximum frictional force = ?

μ = static friction coefficient = 0.016

W = Weight load on knee = mg

m = mass supported by knee = 45 kg

g = acceleration due to gravity = 9.81 m/s²

Therefore,

[tex]f = \mu mg\\f = (0.016)(45\ kg)(9.81\ m/s^2)\\[/tex]

f = 7.06 N

Answer:

D

The answer cannot be found until it is known whether q is greater than, less than, or equal to 45°.

Explanation:

Since block moves with constant speed

So, frictional force

f = FCosq

Work done by friction

W = - fd

W = - fd Cos q

The answer may be greater or less than - fdSinq. It depends on the value of q which is less than, or equal to 45°.

Answer:

Answer:

because it look more impressive than empty dark sky .

Answer:

a) Initial Temperature = 609.4 K and Final Temperature = 325.7 K

b) the change in internal energy is -18279.78 J

c) heat lost by the gas is zero or 0

d) the work done on the gas is -18279.78 J

Explanation:

Given the data in the question;

P[tex]_i[/tex] = 1 atm = 101325 pascal

P[tex]_f[/tex] = ?

V[tex]_i[/tex] = 0.1550 m³

V[tex]_f[/tex] = 0.742 m³

we know that for an adiabatic process  γ = 1.4

P[tex]_i[/tex]V[tex]_i^Y[/tex] = P[tex]_f[/tex]V[tex]_f^Y[/tex]

P[tex]_f[/tex] = P[tex]_i[/tex][tex]([/tex] V[tex]_i[/tex] / V[tex]_f[/tex] [tex])^Y[/tex]

we substitute

P[tex]_f[/tex] = 1 × [tex]([/tex] 0.1550  / 0.742  [tex])^{1.4[/tex]

= [tex]([/tex] 0.2088948787 [tex])^{1.4[/tex]

= 0.11166 atm

a) the initial and final temperatures

Initial temperature

T[tex]_i[/tex] = P[tex]_i[/tex]V[tex]_i[/tex] / nR

given that n = 3.10 mol

= ( 101325 × 0.1550 ) / ( 3.10 × 8.314 )

= 15705.375 / 25.7734

T[tex]_i[/tex]  = 609.4 K

Final temperature

T[tex]_f[/tex] = P[tex]_f[/tex]V[tex]_f[/tex] / nR

= ( 0.11166 × 101325 × 0.742 ) / ( 3.10 × 8.314 )

= 8394.95 / 25.7734

= 325.7 K

Therefore, Initial Temperature = 609.4 K and Final Temperature = 325.7 K

b) the change in internal energy

ΔE[tex]_{int[/tex] = nC[tex]_v[/tex]ΔT

here, C[tex]_v[/tex] = ( 5/2 )R

ΔE[tex]_{int[/tex] = 3.10 × ( 5/2 )8.314 × ( 325.7 - 609.4 )

= -18279.78 J

Therefore, the change in internal energy is -18279.78 J

c) the heat lost by the gas

Since its an adiabatic process,

Q = 0

Therefore, heat lost by the gas is zero or 0

d)  the work done on the gas

W = ΔE[tex]_{int[/tex] - Q

= -18279.78 J - 0

W = -18279.78 J

Therefore, the work done on the gas is -18279.78 J

a) The Initial Temperature and Final Temperature of gas are 601.68 K and 321.61 K respectively.

b) The change in internal energy is -18279.78 J.

c) The heat lost by the gas is zero.

d) The work done on the gas is -18279.78 J.

Given data:

The moles of sample is, n = 3.10 mol.

The initial volume of sample is, [tex]V_{1}=0.1550 \;\rm m^{3}[/tex].

The final volume of sample is, [tex]V_{2}=0.742 \;\rm m^{3}[/tex].

The initial pressure of the sample is, [tex]P_{1}=1.00 \;\rm atm[/tex].

(a)

We know that the relation between the pressure and volume for an adiabatic process is as follows,

[tex]P_{1}V_{1}^{\gamma} = P_{2}V_{2}^{\gamma}[/tex]

Here, [tex]\gamma[/tex]  is a adiabatic index. And for air, its value is 1.41.

Solving as,

[tex]P_{2}=P_{1} \times\dfrac{V_{1}^{\gamma}}{V_{2}^{\gamma}}\\\\\\P_{2}=1.00 \times\dfrac{0.1550^{1.41}}{0.742^{1.41}}\\\\\\P_{2} = 0.11166 \;\rm atm[/tex]

Now, calculate the final temperature using the ideal gas equation as,

[tex]P_{2}V_{2}=nRT_{2}\\\\T_{2}= \dfrac{P_{2} \times V_{2}}{nR}\\\\T_{2}= \dfrac{0.11166 \times 10^{5}\times 0.742}{3.10 \times 8.31}\\\\T_{2}=321.61 \;\rm K[/tex]

Similarly, calculate the initial temperature as,

[tex]P_{1}V_{1}=nRT_{1}\\\\T_{1}= \dfrac{P_{1} \times V_{1}}{nR}\\\\T_{1}= \dfrac{1 \times 10^{5}\times 0.1550}{3.10 \times 8.31}\\\\T_{1}=601.68 \;\rm K[/tex]

Thus, we can conclude that the initial and final temperature of the gas is 601.68 K and 321.61 K respectively.

(b)

The change in internal energy is given as,

ΔE = nCΔT

here, C = ( 5/2 )R

ΔE = 3.10 × ( 5/2 )8.314 × ( 325.7 - 609.4 )

      = -18279.78 J

Therefore, the change in internal energy is -18279.78 J.

c)

The heat lost by the gas . Since its an adiabatic process, so there will be no heat interaction.  

Q = 0

Therefore, heat lost by the gas is zero or 0

d)  

The work done on the gas

W = ΔE - Q

W = -18279.78 J - 0

W = -18279.78 J

Therefore, the work done on the gas is -18279.78 J.

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Answer:

The mass and atomic numbers don't change

Explanation:

An excited atom relaxes to the ground state emitting a photon...called a gamma ray.

The answer is that the mass and atomic numbers don't change.

In gamma decay, the product refers to the nucleus resulting from the emission of a gamma ray. Gamma decay occurs when an excited atomic nucleus releases excess energy in the form of a high-energy photon called a gamma ray.

To find the product of gamma decay, you need to identify the nucleus before and after the decay process. The product nucleus is determined by the parent nucleus that undergoes gamma decay.

During gamma decay, the number of protons and neutrons in the nucleus remains unchanged. Therefore, the identity of the element remains the same, but the energy state of the nucleus is altered.

The product nucleus is typically represented by the same chemical symbol as the parent nucleus, followed by a superscript indicating the mass number (total number of protons and neutrons) and a subscript indicating the atomic number (number of protons).

For example, if a parent nucleus with an atomic number of Z and a mass number of A undergoes gamma decay, the product nucleus will have the same atomic number Z and mass number A.

It's important to note that gamma decay does not involve the emission or absorption of any particles, only the release of electromagnetic radiation (gamma ray).

Thus, the product nucleus remains unchanged in terms of atomic number and mass number.

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#SPJ4

Answer:

The magnitude of the maximum possible torque exerted on the coil is 5.73 x 10⁻³ Nm

Explanation:

Given;

number of turns of the circular coil, N = 49.5 turns

radius of the coil, r = 5.10 cm = 0.051 m

magnitude of the magnetic field, B = 0.535 T

current in the coil, I = 26.5 mA = 0.0265 A

The magnitude of the maximum possible torque exerted on the coil is calculated as;

τ = NIAB

where;

A is the area of the coil

A = πr² = π(0.051)² = 0.00817 m²

Substitute the given values and solve for the maximum torque

τ = (49.5) x (0.0265) x (0.00817) x (0.535)

τ = 0.00573 Nm

τ = 5.73 x 10⁻³ Nm

Answer:

Plastic

Explanation:

Shear Modulus can be defined as the ratio of shear stress to shear strain with respect to a physical object.

This ultimately implies that, Shear Modulus arises as a result of the application of a shear force on an object or body which eventually leads to its deformation. Thus, this phenomenon is simply used by scientists to measure or determine the rigidity of an object or body.

Fluids in which the shear stress must reach certain minimum value (yield stress) before flow commences are called plastic. Thus, a plastic would only begin to flow when its shear stress attain a certain minimum value (yield stress). The unit of measurement of yield stress is usually mega pascal (MPa).

Answer:

19N to the south

Explanation:

F =10N + 5N + 4N

Answer:

Explanation:

6cos28

=5.3 N

Answer:

Graph (c) would be created by a pendulum with the greatest amplitude.

Explanation:

The amplitude of a wave is the greatest displacement covered by an object. It refers to the maximum amount of displacement of a particle on the medium from its rest position. It is the distance from rest to crest.

Out of three graphs, the amplitude is greatest in graph 3 as the distance from rest is crest in this case is maximum. Hence, the correct option is (c).

Answer:

mercury and alcohol

ii) used to test temperatures

i) It is a good conductor of heat and therefore the whole liquid reaches the temperature of the surroundings quickly.

ii) It does not wet (cling to the sides of) the tube.

Alcohol:

i) Alcohol has greater value of temperature coefficient of expansion than mercury.

ii) it's freezing point is below –100°C.

Answer:

Quantum Mechanics

Explanation:

Well, that's what I think personally.

Answer:

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Explanation:

rshyyjfshfsgfshfsyhrsyhuydtufhr6ra6yris7toe7r9w7rr6w996ryrowosotusuogsuoufsutot

Answer:

7.84 m/s

Explanation:

Height, h = 2 m

Initial velocity, u = 10 m/s

Angle, A = 80°

(a) Let the time taken to go to the net is t.

Use second equation of motion

[tex]h = u t + 0.5 at^2\\\\- 2 = - 10 sin 80 t - 4.9 t^2\\\\4.9 t^2 + 9.8 t - 2 = 0 \\\\t= \frac{- 9.8\pm\sqrt{9.8^2 + 4\times 4.9\times 2}}{9.8}\\\\t = \frac{- 9.8 \pm 11.6}{9.8}\\\\t = - 2.2 s , 0.2 s[/tex]

Time cannot be negative.

So, t = 0.2 s

The vertical velocity at t = 0.2 s is

v = u + at

v = 10 sin 80 - 9.8 x0.2

v = 9.8 - 1.96 = 7.84 m/s

Answer:

(a) 1 : 2

(b) same

Explanation:

Let the mass of puck A is m and the mass of puck B is 2 m.

initial speed for both the pucks is same as u and the distance is same for both is s.

let the tension is T for same.

The kinetic energy is given by

[tex]K = 0.5 mv^2[/tex]

(a) As the speed is same, so the kinetic energy depends on the mass.

So, kinetic energy of A : Kinetic energy of B = m : 2m  = 1 : 2

(b) A the distance s same so the final velocities are also same.

(a)  The kinetic energy of puck B is 2 times the kinetic energy of puck A.

(b)  The final speed of both the puck A and B are same.

Let the mass of puck A is m and the mass of puck B is 2 m.

Initial speed for both the pucks is same as u and the distance is same for both is s.

Let the tension is T for same.

Then, the kinetic energy is given as,

[tex]KE = \dfrac{1}{2}mv^{2}[/tex]

(a)

As the speed is same, so the kinetic energy depends on the mass.

Then,

[tex]\dfrac{KE_{A}}{KE_{B}} = \dfrac{1/2 \times mv^{2}}{1/2 \times (2m)v^{2}}\\\\\\\dfrac{KE_{A}}{KE_{B}} =\dfrac{1}{2}[/tex]

So, kinetic energy of A : Kinetic energy of B = 1 : 2.

Thus, we can conclude that the kinetic energy of puck B is 2 times the kinetic energy of puck A.

(b)

The final speed for the puck is given as,

v = s/t

here, s is the distance covered.

Since, both pucks are pulled the same distance across frictionless ice. Then, the final speed of each puck is also same.

Thus, we can conclude that the final speed of both the puck A and B are same.

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Answer: The force on a negative charge that is placed midway between two equal positive charges is zero when all charges have the same magnitude.

Explanation:

Let us assume that

[tex]q_{1} = q_{2} = +q[/tex]

[tex]q_{3} = -q[/tex]

As [tex]q_{3}[/tex] is the negative charge and placed midway between two equal positive charges ([tex]q_{1}[/tex] and [tex]q_{2}[/tex]).

Total distance between [tex]q_{1}[/tex] and [tex]q_{2}[/tex] is 2r. This means that the distance between [tex]q_{1}[/tex] and [tex]q_{3}[/tex], [tex]q_{2}[/tex] and [tex]q_{3}[/tex] = d = r

Now, force action on charge [tex]q_{3}[/tex] due to [tex]q_{1}[/tex] is as follows.

[tex]F_{31} = k(\frac{q_{1} \times q_{3}}{d^{2}})[/tex]

where,

k = electrostatic constant = [tex]9 \times 10^{9} Nm^{2}/C^{2}[/tex]

Substitute the values into above formula as follows.

[tex]F_{31} = k(\frac{q_{1} \times q_{3}}{d^{2}})\\= 9 \times 10^{9} (\frac{q \times (-q)}{r^{2}})\\= - 9 \times 10^{9} (\frac{q^{2}}{r^{2}})[/tex] ... (1)

Similarly, force acting on [tex]q_{3}[/tex] due to [tex]q_{1}[/tex] is as follows.

[tex]F_{32} = k \frac{q_{2}q_{3}}{d^{2}}\\= -9 \times 10^{9} \frac{q^{2}}{r^{2}}\\[/tex]   ... (2)

As both the forces represented in equation (1) and (2) are same and equal in magnitude. This means that the net force acting on charge [tex]q_{3}[/tex] is zero.

Thus, we can conclude that the force on a negative charge that is placed midway between two equal positive charges is zero when all charges have the same magnitude.

Answer:

A change of 160.819 atmospheres is required to keep water from expanding when it is heated from 10.9 °C to 40.0 °C.

Explanation:

The bulk modulus of water ([tex]B[/tex]), in newtons per square meters, can be estimated by means of the following model:

[tex]B = \rho_{o}\cdot \frac{\Delta P}{\rho_{f} - \rho_{o}}[/tex] (1)

Where:

[tex]\rho_{o}[/tex] - Water density at 10.9 °C, in kilograms per cubic meter.

[tex]\rho_{f}[/tex] - Water density at 40 °C, in kilograms per cubic meter.

[tex]\Delta P[/tex] - Pressure change, in pascals.

If we know that [tex]\rho_{o} = 999.623\,\frac{kg}{m^{3}}[/tex], [tex]\rho_{f} = 992.219\,\frac{kg}{m^{3}}[/tex] and [tex]B = 2.2\times 10^{9}\,\frac{N}{m^{2}}[/tex], then the bulk modulus of water is:

[tex]\Delta P = B\cdot \left(\frac{\rho_{f}}{\rho_{o}}-1 \right)[/tex]

[tex]\Delta P = \left(2.2\times 10^{9}\,\frac{N}{m^{3}} \right)\cdot \left(\frac{992.219\,\frac{kg}{m^{3}} }{999.623\,\frac{kg}{m^{3}} }-1 \right)[/tex]

[tex]\Delta P = -16294943.19\,Pa \,(-160.819\,atm)[/tex]

A change of 160.819 atmospheres is required to keep water from expanding when it is heated from 10.9 °C to 40.0 °C.