A. Tick (1) the best alternatives. 1. What is the acceleration due to gravity on the surface of moon ? (a) 9.8m/s (b)1.6m/s2 (c) 6.67x10-1m/s (d) 9.8m/s? ​

Answer:

Acceleration of this electron: [tex]-5.60 \times 10^{15}\; \rm m \cdot s^{-2}[/tex].

Time taken: approximately [tex]9.39 \times 10^{-9}\; \rm s[/tex].

Explanation:

Apply the SUVAT equation that does not involve time:

[tex]v^{2} - u^{2} = 2\, a \, x[/tex].

Equivalently:

[tex]\begin{aligned}a &= \frac{v^{2} - u^{2}}{2\, x}\end{aligned}[/tex].

By this equation, the acceleration of this electron would be:

[tex]\begin{aligned}a &= \frac{v^{2} - u^{2}}{2\, x} \\ &= \frac{(7.72 \times 10^{7}\; \rm m \cdot s^{-1})^{2} - (2.46 \times 10^{7} \; \rm m \cdot s^{-1})^{2}}{2 \times 0.478\; \rm m} \\ &\approx -5.60 \times 10^{15}\; \rm m \cdot s^{-2}\end{aligned}[/tex].

The speed of this electron has changed from [tex]u = 7.72 \times 10^{7}\; \rm m\cdot s^{-1}[/tex] to [tex]v = 2.46 \times 10^{7}\; \rm m \cdot s^{-1}[/tex]. Calculate the time required to achieve this change at a rate of [tex]a \approx -5.60 \times 10^{15}\; \rm m\cdot s^{-2}[/tex]:

[tex]\begin{aligned}t &= \frac{v - u}{a} \\ &\approx \frac{2.46\times 10^{7}\; \rm m \cdot s^{-1} - 7.72 \times 10^{7}\; \rm m\cdot s^{-1}}{-5.60 \times 10^{15}\; \rm m\cdot s^{-2}} \\ &\approx 9.39 \times 10^{-9}\; \rm s\end{aligned}[/tex].

Answer:

B illustrates refraction

KE = (0.5) m v²

given that : v = speed of the bottle in each case = 4 m/s when m = 0.125 kg

KE = (0.5) m v² = (0.5) (0.125) (4)² = 1 J

when m = 0.250 kg KE = (0.5) m v² = (0.5) (0.250) (4)² = 2 J

when m = 0.375 kg KE = (0.5) m v² = (0.5) (0.375) (4)² = 3 J

when m = 0.0.500 kg KE = (0.5) m v² = (0.5) (0.500) (4)² = 4 J

Answer:

True

Explanation:

The tensile stress, σ, on a solid cylindrical wire is given by the following relationship;

[tex]\sigma = \dfrac{F_t}{A_o}[/tex]

Where;

[tex]F_t[/tex] = The tensile force

[tex]A_o[/tex] = The original cross sectional area of the cylindrical wire = π·R²

R = The radius of the wire

Therefore;

[tex]F_t[/tex] = σ × [tex]A_o[/tex] = σ × π × R²

Therefore, the tensile force is directly proportional to the square of the radius of the cylindrical wire, and as the radius of the wire increases, which is by increasing the thickness of the wire, the tensile force is largely increased

The correct option is; True.

Answer:

2min

Explanation:

i think it will be clear from photo

Answer: The formula for current is charge/time

so here we have to change subject so we are asked to look for time  

Explanation: so it would be T=Q/I

T=4/2

T=2s

hope this helpss

Answer:

 ΔR = 5 s

Explanation:

The absolute uncertainty or error in an expression is

       ΔR = | [tex]\frac{dR}{dB}[/tex] | ΔB + | [tex]\frac{dR}{dA}[/tex] | ΔA

the absolute value guarantees to take the unfavorable case, that is, the maximum error.

We look for the derivatives

       [tex]\frac{dR}{dB}[/tex] = 1

       [tex]\frac{dR}{dA}[/tex] = -1

we substitute

       ΔR = 1 ΔB + 1 ΔA

of the data

       ΔB = 3 s

       ΔA = 2 s

       ΔR = 3 + 2

       ΔR = 5 s

Answer:

see the explanation below

Explanation:

Momentum is a product of the mass of a particle and its velocity.

and also, momentum is a vector quantity; i.e. it has both magnitude and direction.

Now a plane in the air has both magnitude and velocity

When the plane lands the velocity will amount to zero although the mass is still very much intact

Now the mass* zero velocity= zero

Hence when a plane lands the momentum is zero

Answer:

This question is misleading since

1/C = 1/C1 + 1/C2 + 1/C3 + 1/C4

so it would have to be less than the smallest individual capacitance

Answer:

30.5mm

Explanation:

There are 300 pages, so to find the number of pages we do 0.1mm*300 to get 30mm. There are two covers in the book (the front and the back). 0.25mm*2 = 0.5mm. To determine the total thickness of the book we do 30mm+0.5mm to get a total of 30.5mm

Your answer is 30.5mm.

Answer:

15.5mm

Explanation:

there are 150 pages(one page has 2 sides so 300/2 = 150) and 2 covers

page's thickness is 0.1mm, cover's thickness is 0.25

so books thickness = 150*0.1+0.25*2 = 15.5mm

Answer:

Frequency of the tuning fork[second resonance] = 512 Hz (Approx.)

Explanation:

Given:

Speed of sound = 340 m/s

Length of resonance position above a closed air column = 49.8 cm = 0.498 m

Find:

Frequency of the tuning fork

Computation:

Frequency of the tuning fork[second resonance] = 3v / 4l

Frequency of the tuning fork[second resonance] = 3(340) / 4(0.498)

Frequency of the tuning fork[second resonance] = 512.04

Frequency of the tuning fork[second resonance] = 512 Hz (Approx.)

Answer:

The frequency is 512 Hz.

Explanation:

speed, v = 340 m/s

length, L = 49.8 cm = 0.498 m

let the frequency is f.

[tex]f =\frac{3 v}{4 L}\\\\f = \frac{3 \times 340 }{4\times 0.498}\\\\f = 512 Hz[/tex]

Answer:

average speed is 60km/h

Explanation:

you sum up the speed attained in each distance covered and divide it by 2 to get your answer

Answer:

Increasing

Explanation:

I Hope it Helps

Answer:

c

Explanation:

Explanation:

125 is your answer........

Explanation:

→ Volume of cone = πr² × h/3

Here,

→ Volume of cone = π(13)² × 27/3 cm³

→ Volume of cone = 169π × 9 cm³

→ Volume of cone = 1521π cm³

→ Volume of cone = 1521 × 22/7 cm³

→ Volume of cone = 33462/7 cm³

→ Volume of cone = 4780.28 cm³

Answer:

4,778.4 is correct

Explanation:

Answer:

A. moving with uniform velocity.

Explanation:

Was this hepfull

Answer:

A

Explanation:

Equilibrium is a state in which opposing forces or influences are balanced hence A being correct

Answer:

a) Pascal is a derived unit because it is derived from the unit of force and area

b)Mass is a fundamental quantity because it doesn't depends upon others physical quantity and made up of only one unit

c) unit of power is a derived unit because they are dependent quantities

D) unit of length is a fundamental unit because it cannot be expressed in terms of another quantity.

Answer:

This may refer to a situation like:

"one person pushes a box, if there is equal and opposite reaction why the box moves and the person does not?"

Remember the second Newton's law:

F = m*a

suppose that the mass of the person is 3 times the mass of the box.

So, if the box has a mass M, the person will have a mass 3*M

Then the Newton's equation for the box when the person pushes with a force F is:

F = M*a

solving for the acceleration, we get:

F/M = a

While the person is also pushed by the box with a force with the same magnitude, then the equation for the person is:

F = (3*M)*a'

Solving for the acceleration, we get:

F/(3M) = a'

Now we can compare the acceleration of the box (F/M) with the acceleration of the person (F/3M).

Is easy to see that the acceleration of the box is 3 times the acceleration of the person.

So regardless of the fact that both the box and the person experience a force with the same magnitude, the box will move more due to this force.

This is why in situations like this, the forces do not seem balanced.

Answer:

cant see picture

Explanation:

Answer:

please add picture so i can help you

Explanation:

Answer:

Explanation:

We need to find the x-components of each of these vectors and then add them together, then we need to find the y-components of these vectors and then add them together. Let's get to that point first. That's hard enough for step 1, dontcha think?

The x-components are found by multiplying the magnitude of the vectors by the cosine of their respective angles, while the y components are found by multiplying the magnitude of the vectors by the sine of their respective angles.

Let's do the x-components for all the vectors first, so we get the x-component of the resultant vector:

[tex]F_{1x}=12 cos0[/tex] and

[tex]F_{1x}=12[/tex]

[tex]F_{2x}=9cos90[/tex] and

[tex]F_{2x}=0[/tex]

[tex]F_{3x}=15 cos126.87[/tex] and

[tex]F_{3x}=-9.0[/tex]  (the angle of 126.87 is found by subtracting the 53.13 from 180, since angles are to be measured from the positive axis in a counterclockwise fashion).

That means that the x-component of the resultant vector, R, is 3.0

Now for the y-components:

[tex]F_{1y}=12sin0[/tex] and

[tex]F_{1y}=0[/tex]

[tex]F_{2y}=9sin90[/tex] and

[tex]F_{2y}=9[/tex]

[tex]F_{3y}=15sin126.87[/tex] and

[tex]F_{3y}=12[/tex]

That means that the y-component of the resultant vector, R, is 21.

Put them together in this way to find the resultant magnitude:

[tex]R_{mag}=\sqrt{(3.0)^2+(21)^2}[/tex] which gives us

[tex]R_{mag}=21[/tex] and now for the angle. Since both the x and y components of the resultant vector are positive, our angle will be where the x and y values are both positive in the x/y coordinate plane, which is Q1.

The angle, then:

[tex]tan^{-1}(\frac{21}{3.0})=82[/tex] degrees, and since we are QI, we do not add anything to this angle to maintain its accuracy.

To sum up: The resultant vector has a magnitude of 21 N at 82°

Answer:

ΔP = 1875 Pa,   P₂ = P₁ - 1875

Explanation:

Let's use Bernoulli's equation, with the subscript 1 for the widest Mars and the subscript 2 for the narrowest part, suppose that the pipe is horizontal

          P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

          P₁ -P₂ = ½ ρ (v₂² - v₁²)

suppose the fluid is water

          P₁ - P₂ = ½ 1000 (2² - 0.5²)

         ΔP = 1875 Pa

this is the pressure difference between the two sections

the pressure in the narrowest section is

           P₂ = P₁ - 1875

Answer:

It is simile uses "like" or "as": metaphor does not

Explanation:

it just it

Answer:

third one

Explanation:

i cant think of anymore

Answer:

The time required is 0.45 s.

Explanation:

Height, h = 1 m

initial velocity, u = m/s

Let the time is t.

Use second equation of motion

[tex]h = u t + 0.5 at^2\\\\1 = 0 +0.5 \times 9.8 \times t^2\\\\t = 0.45 s[/tex]

The gravitational force between 2 bodys decreases with distance between the two bodies.

f=G m1m2/r2

Answer:

Answer:

The position is 8.18cm from the mirror.

Nature is b=virtual

Size is 1.82cm

Explanation:

Note that for a convex mirror, the image distance and the focal length are negative;

Given

Object height H0 = 4cm

object distance u = 18cm

Radius of curvature R = 30cm

Since f = R/2

f = 30/2

f = -15cm

Recall that:

[tex]\frac{1}{f} =\frac{1}{u}+ \frac{1}{v}\\\frac{1}{-15}=\frac{1}{18}+\frac{1}{v} \\\frac{1}{v} =\frac{1}{-15} -\frac{1}{18}\\ \frac{1}{v} = \frac{-18-15}{270}\\\frac{1}{v} = \frac{-33}{270}\\v=\frac{-270}{33}\\v=-8.18cm[/tex]

Since the image distance is negative, this shows that the image is a virtual image.

To get the size:

[tex]\frac{H_1}{H_0}=\frac{v}{u}\\\frac{H_1}{4}=\frac{8.18}{18}\\18H_i=32.72\\H_i=\frac{32.72}{18}\\H_i= 1.82cm[/tex]