Explain whether the unit of work is a fundamental or derived unit

Answer:

23s

Explanation:

s=ut+1/2at^2

the distance (s) is 400, initial velocity (u) is 0, acceleration (a) is 1.5 therefore

400=0t+1/2(1.5)t^2

400/0.75=0.75t^2/0.75

t^2=√533.33

t=23s

I hope this helps and sorry if it's wrong

Answer:

Magnitude = 140 x 10⁻¹² Cm

Direction = upwards

Explanation:

A pair of two equal and opposite point charges forms an electric dipole.

The magnitude of the moment of such dipole is the product of the magnitude of any of the charges (since the charges are the same in magnitude) and the distance of separation between them. i.e

p = q x d          ----------(i)

Where;

p = dipole moment

q = magnitude of any of the charges

d = distance between the charges.

The direction of the dipole moment is from the negative charge to the positive charge.

(a) From the question, the charges are +35 nC and -35 nC, and the distance between them is 4.00mm.

This implies that;

q = 35 nC = 35 x 10⁻⁹C

d = 4.00mm = 4.0 x 10⁻³ m

Substitute the values of q and d into equation (i) to give;

p = 35 x 10⁻⁹C x 4.00 x 10⁻³ m

p = (35 x 4.0) x (10⁻⁹ x 10⁻³) C m

p = 140 x 10⁻¹² Cm

The magnitude of the dipole moment is 140 x 10⁻¹² Cm

(b) From the question, the +35nC charge is above the -35nC charge on a vertical line as shown below;

                         o   +35nC

                          |

                          |

                          |

                          |

                          |

                          |

                         o    -35nC

Since the direction should point from the negative charge to the positive charge, this means that the direction of the dipole moment of the two charges is upwards (due North).

                         o   +35nC

                         ↑

                          |

                          |

                          |

                          |

                          |

                          |

                         o    -35nC

Explanation:

We can write Newton's 2nd law as applied to the sliding mass [tex]m_1[/tex] as

[tex]T - m_1g\sin38 = m_1a\:\:\:\:\:\:\:(1)[/tex]

For the hanging mass [tex]m_2,[/tex] we can write NSL as

[tex]T - m_2g = -m_2a\:\:\:\:\:\:\:(2)[/tex]

We need to solve for a first before we can solve the tension T. So combining Eqns(1) & (2), we get

[tex](m_1 + m_2)a = m_2g - m_1g\sin38[/tex]

or

[tex]a = \left(\dfrac{m_2 - m_1\sin38}{m_1 + m_2}\right)g[/tex]

[tex]\:\:\:\:= 0.30\:\text{m/s}^2[/tex]

Using this value for the acceleration on Eqn(2), we find that the tension T is

[tex]T = m_2(g - a) = (2.6\:\text{kg})(9.51\:\text{m/s}^2)[/tex]

[tex]\:\:\:\:=24.7\:\text{N}[/tex]

Answer:

uninstell this apps nobody will give u ans its happening to me alsoi was in exam hall i thought this app will give answer but no

Answer:

The correct solution is "37.5 km".

Explanation:

Given:

Distance between the trains,  

d = 75 km

Speed of each train,

= 15 km/h

The relative speed will be:

= [tex]15 + (-15)[/tex]

= [tex]30 \ km/h[/tex]

The speed of the bird,

V = 15 km/h

Now,

The time taken to meet will be:

[tex]t=\frac{Distance}{Relative \ speed}[/tex]

  [tex]=\frac{75}{30}[/tex]

  [tex]=2.5 \ h[/tex]

hence,

The distance travelled by the bird in 2.5 h will be:

⇒ [tex]D = V t[/tex]

        [tex]=15\times 2.5[/tex]

        [tex]=37.5 \ km[/tex]

Answer:

The simplest kinetic model is based on the assumptions that: (1) the gas is composed of a large number of identical molecules moving in random directions, separated by distances that are large compared with their size; (2) the molecules undergo perfectly elastic collisions (no energy loss) with each other and with the walls of the container, but otherwise do not interact; and (3) the transfer of kinetic energy between molecules is heat.

Answer:

asdasd dsa dasdasd sadas dasd asdasd asd asd dsa asdd 223 aasd ada dasd sa dasd dsaa sd adsd asasd

Explanation:

Answer:

Magnification = 1

Explanation:

given data

radius of curvature r = - 0.983 m

image distance u = - 0.155

solution

we get here first focal length that is

Focal length, f = R/2     ...................1

f = -0.4915 m

we use here formula that is

[tex]\frac{1}{v} + \frac{1}{u} + \frac{1}{f}[/tex]      .................2

put here value and we get

[tex]\frac{1}{v} = \frac{1}{0.155} - \frac{1}{4915}[/tex]  

so

Magnification will be here as

m = [tex]- \frac{v}{u}[/tex]

m =  [tex]\frac{0.155}{0.155}[/tex]

Answer:

The magnification is 1.5.

Explanation:

radius of curvature, R = - 0.983 m

distance of object, u = - 0.155 m

Let the distance of image is v.

focal length, f = R/2 = - 0.492 m

Use the mirror equation

[tex]\frac{1}{f}=\frac{1}{v}+\frac {1}{u}\\\\\frac{-1}{0.492}=\frac{1}{v}-\frac{1}{0.155}\\\\\frac{1}{v}=\frac{1}{0.155}-\frac{1}{0.492}\\\\\frac{1}{v}=\frac{0.492-0.155}{0.155\times 0.492}\\\\\frac{1}{v}=\frac{0.337}{0.07626}\\ \\v = 0.226 m[/tex]

The magnification is given by

m = - v/u

m = 0.226/0.155

m = 1.5

Answer:

The correct answer is - 8.99N/C

Explanation:

[tex]dE=k\dfrac{dq}{x^2}\\ dq=\lambda{dx}\\ \lambda=2nC/m\\ dq=2dxnC\\ dE=k\dfrac{2dx}{x^2}\\ E=2k\int_1^2\dfrac{dx}{x^2}\\ E=2k(\frac{-1}{x})_1^2=k\times10^{-9}N/C\\ E=8.99\times10^9\times10^{-9}N/C\\ E=8.99N/C\\dE=k[/tex]

Answer:

3.1 kg

Explanation:

Applying,

R = m(g-a)..................... Equation 1

Where R = weight of the scale when the elevator is coming down, a = acceleration of the elevator, g = acceleration due to gravith.

From the question,

Given: m = 5 kg, a = 3.8 m/s²

Constant: g = 9.8 m/s²

Substitute these values into equation 1

R = 5(9.8-3.8)

R = 5(6)

R = 30 N

Hence the spring scale is

m' = R/g

m' = 30/9.8

m' = 3.1 kg

Answer:

A hypothesis for the period of a pendulum is:

"The period of the pendulum varies with its length"

Explanation:

A hypothesis for the period of a pendulum is:

"The period of the pendulum varies with its length"

To test this hypothesis we can carry out a measurement of a simple pendulum keeping the angle fixed, in general the angle used is about 5º since when placing this value in radiand and the sine of this angle they differ little <5%. therefore measured the time of some oscillations, for example about 10 oscillations, changing the length of the pendulum to test the hypothesis.

If the hypothesis and the model used is correct, the relationship to be tested is

              T² =(4π² /g)   L  

by making a graph of the period squared against the length if obtaining, os a line, the hypothesis is tested.

Answer:

Explanation:

kinematic equation (g will have a negative value if we assume UP is positive)

v² = u² + 2as

a) v = √(0² + 2(g)(y - h))

b) v = √(vi² + 2(g)(y - h))

they offer fewer luxury services because they lack the money to provide them.

Answer:

B. is subject to a smaller net force but same acceleration.

Explanation:

F = m*a

So because our force applied is constant from the women pulling on the rope which means the acceleration is the same on both the 4kg create and the 6kg create. The only thing that changes here is the mass of the creates, so there is more tension force between the women and the 6kg create then there is between the 4kg create and the 6kg. It takes less force to move the 4kg create therefore the tension force is less between the two creates.

The net force on both crates is the same and the acceleration of both crates is the same.

The given parameters;

The force on the 4kg crate is calculated as follows;

[tex]F_{4kg } = T + F[/tex]

The force on the 6kg crate is calculated as follows;

[tex]F_{6 kg} = -T + F[/tex]

The net force on both crates is calculated as follows;

[tex]\Sigma F= -T + F - (T + F)\\\\\Sigma F= -2T[/tex]

Thus, we can conclude that the net force on both crates is the same and the acceleration of both crates is the same.

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Answer:

the speed of the block at the given position is 21.33 m/s.

Explanation:

Given;

spring constant, k = 3500 N/m

mass of the block, m = 4 kg

extension of the spring, x = 0.2 m

initial velocity of the block, u = 0

displacement of the block, d =1.3 m

The force applied to the block by the spring is calculated as;

F = ma = kx

where;

a is the acceleration of the block

[tex]a = \frac{kx}{m} \\\\a = \frac{(3500) \times (0.2)}{4} \\\\a = 175 \ m/s^2[/tex]

The final velocity of the block at 1.3 m is calculated as;

v² = u² + 2ad

v² = 0 + 2ad

v² = 2ad

v = √2ad

v = √(2 x 175 x 1.3)

v = 21.33 m/s

Therefore, the speed of the block at the given position is 21.33 m/s.

The speed of the block at a height of 1.3 m above the starting position is 21.33 m/s

To solve this question, we'll begin by calculating the acceleration of the block.

F = Ke

Also,

F = ma

Thus,

ma = Ke

Divide both side by m

a = Ke / m

a = (3500 × 0.2) / 4

a = 175 m/s²

v² = u² + 2as

v² = 0² + (2 × 175 × 1.3)

v² = 455

Take the square root of both side

v = √455

v = 21.33 m/s

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Answer:

The number of neutrons or electrons in an atom can change without changing the identity of the element.

Answer:

variable

Explanation:

ANSWER-:

1/2 mv²

K.E = 1/2 mv²

K.E = 0.01 J.

Hence, the kinetic energy of a body is 0.01 Joule.

!! HOPE ITS HELP U !!

Answer:

endothermic

Explanation:

An endothermic is any process with an increase in the enthalpy H (or internal energy U) of the system. In such a process, a closed system usually absorbs thermal energy from its surroundings, which is heat transfer into the system.

The question is incomplete, the complete question is;

how much charge is stored on the parallel-plate capacitors by the 12.0 V battery? One is filled with air, and the other is filled with a dielectric for which k = 3.00; both capacitors have a plate area of 5.00×10 −3  m 2  and a plate separation of 2.00 mm.

The capacitance of the capacitor is the quantity of charge stored by the capacitor.

Given that;

C1= εo k * A/d

εo = permittivity of free space

C1 = 8.85 x 10-12 farad per meter *1 * 5.00×10 −3  m 2/2 * 10^-3

= 2.21 * 10^-11 F

C2 = 8.85 x 10-12 * 3 * 5.00×10 −3  m 2/2 * 10^-3

= 6.63 * 10^-11 F

q1 = C1V1 = 2.21 * 10^-11 C * 12 V

= 2.65 * 10^-10 C

q2 = C2V2 = 6.63 * 10^-11 F * 12 V

= 7.96 * 10^-10 C

qtotal = 2.65 * 10^-10 C + 7.96 * 10^-10 C

qtotal = 1.061 * 10^-9C

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Answer:

Here, m=9×10

−31

kg,

q=1.6×10

−19

C,v=3×10

7

ms

−1

,

b=6×10

−4

T

r=

qB

mv

=

(1.6×10

−19

)(6×10

−4

)

(9×10

−31

)×(3×10

7

)

=0.28m

v=

2πr

v

=

2πm

Bq

=

2×(22/7)×9×10

−31

(6×10

−4

)×(1.6×10

−19

)

=1.7×10

7

Hz

Ek=

2

1

mv

2

=

2

1

×(9×10

−31

)×(3×10

7

)

2

J

=40.5×10

−17

J=

1.6×10

−16

40.5×10

−17

keV

=2.53keV

THERE ARE COMMONLY THREE SYSTEMS OF UNIT. THEY ARE:-

• CGS System- (Centimeter-Gram-Second system) A metric system of measurement that uses the centimeter, gram and second for length, mass and time.

• FPS System- (Foot–Pound–Second system).

The system of units in which length is measured in foot , mass in pound and time in second is called FPS system. It is also known as British system of units.

• MKS System- (Meter-Kilogram-Second system) A metric system of measurement that uses the meter, kilogram, gram and second for length, mass and time. The units of force and energy are the "newton" and "joule."

The work done by [tex]\vec F[/tex] along the given path C from A to B is given by the line integral,

[tex]\displaystyle \int_C \mathbf F\cdot\mathrm d\mathbf r[/tex]

I assume the path itself is a line segment, which can be parameterized by

[tex]\vec r(t) = (1-t)(-2\,\vec\imath + 5\,\vec\jmath) + t(4\,\vec\imath+7\,\vec\jmath) \\\\ \vec r(t) = (6t-2)\,\vec\imath+(2t+5)\,\vec\jmath \\\\ \vec r(t) = x(t)\,\vec\imath + y(t)\,\vec\jmath[/tex]

with 0 ≤ t ≤ 1. Then the work performed by F along C is

[tex]\displaystyle \int_0^1 \left(6x(t)^3\,\vec\imath-4y(t)\,\vec\jmath\right)\cdot\frac{\mathrm d}{\mathrm dt}\left[x(t)\,\vec\imath + y(t)\,\vec\jmath\right]\,\mathrm dt \\\\ = \int_0^1 (288(3t-1)^3-8(2t+5)) \,\mathrm dt = \boxed{312}[/tex]

Answer:

6000 lines/cm

Explanation:

From the question we are told that:

Grating 1=4000 lines/cm

Grating 2=6000 lines/cm

Generally The Spread of fringes is Larger when the Grating are closer to each other

Therefore

Grating 2 will spread the the light through a larger angle in the first-order pattern because its the closest with 6000 lines/cm

Answer:

[tex]T=1.54s[/tex]

Explanation:

From the question we are told that:

Speed of Motor 1 [tex]\omega_1=825rpm=>2 \pi 13.75[/tex]

Speed of Motor 2 [tex]\omega_1=786rpm=>2 \pi 13.1[/tex]

Therefore

Frequency of Motor 1 [tex]f_1=13.75[/tex]

Frequency of Motor 2  [tex]f_2= 13.1[/tex]

Generally the equation for Time Elapsed is mathematically given by

[tex]T=\frac{1}{df}[/tex]

Where

[tex]df=f_1-f_2[/tex]

[tex]df=13.75-13.1[/tex]

[tex]df=0.65Hz[/tex]

Therefore

[tex]T=\frac{1}{65}[/tex]

[tex]T=1.54s[/tex]

Answer:

366 N

Explanation:

  τ = Iα

FR = ½mR²α

  F = ½mR(Δω/t)

  F = ½(207)(1.50)(0.75)(2π) /2.00

  F = 365.79919...

Answer:

196 lbf

Explanation:

v² = u² + 2as

1420² = 0² + 2a(24/12)

a = 1420²/4 = 504,100 ft/s²

F = ma = 0.2oz(1lb/16 oz)(1slug/32.2 lb)(504,100) = 195.6909...

Answer:

D = .25g/cm³

Explanation:

D = m/V

D = 120g/480cm³

D = .25g/cm³

Answer:

The current is more in the parallel combination than in the series combination.

Explanation:

two resistances, R1 and R2 are connected to a battery of voltage V.

When they are in series,

R = R1 + R2

In series combination, the current is same in both the resistors, and it is given by Ohm's law.

V = I (R1 + R2)

[tex]I = \frac{V}{R_1 + R_2}[/tex]..... (1)

When they are connected in parallel.

the voltage is same in each resistor.

The effective resistance is R.

[tex]R = \frac{R_1R_2}{R_1 + R_2}[/tex]

So, the current is

[tex]I = \frac{V(R_1+R_2)}{R_1 R_2}[/tex]..... (2)

So, the current is more is the parallel combination.