can radiowaves penetrate clouds​

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:

Mass, M = 4.859 Kg

Explanation:

Given the following data;

Radius, r = 0.225 m

Moment of inertia, I = 0.123 kgm²

To find the mass;

Mathematically, the moment of inertia is given by the formula;

I = ⅖Mr²

Making M the subject of formula, we have;

Cross-multiplying, we have;

2I = Mr²

[tex] M = \frac {2I}{r^{2}} [/tex]

Substituting into the formula, we have;

[tex] M = \frac {2*0.123}{0.225^{2}} [/tex]

[tex] M = \frac {0.246}{0.050625} [/tex]

Mass, M = 4.859 Kg

Explanation:

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

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:

Distance = 450 kilometres

Explanation:

Given the following data;

Speed = 90 km/h

Time = 5 hours

To find the distance covered by the train;

Mathematically, the speed of an object or body is given by the formula;

[tex] Speed = \frac {distance}{time} [/tex]

Making distance the subject of formula, we have;

Distance = speed * time

Substituting into the formula, we have;

Distance = 90 * 5

Distance = 450 kilometres

Answer:

c

Explanation:

Answer: temperature and number of particles are constant

Explanation:

In 1662, Robert Boyle put forth a theory that the volume of a gas will be inversely proportional to the pressure it exerts so long as the temperature is constant and the number of particles or mass is constant as well.

This means that at a constant temperature, the higher the volume of a gas, the less pressure it exerts which makes sense because more volume means that the gas particles are more dispersed which means that they would only be able to apply minimal pressure wherever they are because the particles are not numerous enough to have much effect.

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:

Increasing

Explanation:

I Hope it Helps

Answer:

cant see picture

Explanation:

Answer:

please add picture so i can help you

Explanation:

Answer:

a)    I = 13.38 kg m / s, b)    F = 1,373 10³ N

Explanation:

The impulse is given by the relation

          I = ∫ F dt = Δp

          I = p_f -p₀

          I = m (v_f - v₀)

take the ball's exit direction as positive, whereby the ball velocities

v₀ = -90mph, the final velocity v_f = + 54 m / s

Let's reduce the units to

         I = 0.142 [54- (-40.23) ]

      the SI system

        v₀ = - 90 mph (1609.34 m / 1 mile) (1h / 3600 s = -40.23 m / s

        m = 142 g (1kg / 1000) = 0.142 kg

we calculate  

          I = 0.142 [54- (-40) ]

          I = 13.38 kg m / s

b) let's use the definition of momentum

         I = ∫ F .dt

         I = F ∫ dt

         F = I / t

         F = 13.38 / 0.008

         F = 1,373 10³ N

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]

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:

reduce the velocity of collision

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:

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]

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:

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:

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:

Explanation:

a)The average acceleration of the sled will be 6.198 m/sec

b)The amount of time required to travel at the highway's 100 km/h speed limit will be 4.48 sec.

c)The distances covered when the vehicle hits its top speed of 270 km/h will be 453.77 m.

The rate of velocity change concerning time is known as acceleration. According to Newton's second law, the eventual effect of all forces applied to a body is its acceleration.

The pace at which a body's velocity varies is represented by acceleration, which is a vector quantity.

[tex]\rm a = \frac{v-u}{t} \\\\[/tex]

The given data in the problem is given by ;

u is the initial speed of the rocket=  0 km/h

v is the final speed of the rocket=  270 km/h

t is the time interval= 12.1 seconds

a is the average acceleration of the ball=? m/sec²

to is the time it takes to reach the speed of 100 km/h

s is the distance traveled when it reaches the final speed

a)

The average acceleration of the sled is found as;

a = (v-u)/t

a=(75-0)/12.1

a = 6.198 m/sec

b)

The amount of time required to travel at the highway's 100 km/h speed limit;

From Newton's second equation of motion;

v =u+at

27.7 = 0 + 6.198t

t = 27.7 / 6.198

t = 4.48 sec

c)

The distances covered when the vehicle hits its top speed of 270 km/h;

v²=u²+2as

75²=0+2(6.198)s

s=(75)²/2(6.198)

s=453.77 m

Hence the average acceleration of the sled, the amount of time required to travel at the highway's 100 km/h speed limit, and the amounts of time required to travel at the highway's 100 km/h speed limit will be  6.198 m/sec,4.48 sec and 453.77 m respectively.

To learn more about acceleration, refer to the link;

https://brainly.com/question/2437624

#SPJ2

"Rotating wheel" is meant to indcate that the wheel is already rotating at the start. Denote the initial angular velocity by ω₀. Then the angular displacement θ at time t is

θ = ω₀t + 1/2 αt ²

while the angular velocity ω is

ω = ω₀ + αt

It takes 5 s for the wheel to rotate 38 times, or turn a total of (2π rad/rev) (38 rev) = 76π rad, as well as to reach an angular velocity of 79 rad/s, so that

76π rad = ω₀ (5 s) + 1/2 α (5 s)²

79 rad/s = ω₀ + α (5 s)

Solve the second equation for ω₀ and substitute into the first equation, then solve for α :

ω₀ = 79 rad/s - α (5 s)

==>   76π rad = (79 rad/s - α (5 s)) (5 s) + 1/2 α (5 s)²

==>   76π rad = (79 rad/s) (5 s) - 1/2 α (5 s)²

==>   1/2 α (5 s)² = (79 rad/s) (5 s) - 76π rad

==>   α = ((79 rad/s) (5 s) - 76π rad) / (1/2 (5 s)²)

==>   α ≈ 12.5 rad/s²

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:

 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:

work done = mgh

350×10×2

7000J

The work done by the forklift truck as it lifts the box to the given height is 6860J.

Work is simply referred to as the displacement of an object when a push or pull force is applied to the object. It is the energy transferred from or to an object when force is applied to it along a displacement.

It is expressed as;

W = F × s

Where F is force and s is displacement

Given the data in the question;

We substitute our given values into the expression above.

W = F × s

But F = Weight = mass × acceleration due to gravity = mg

acceleration due to gravity ( g = 9.8m/s²)

Hence,

W = mg × s

W = 350kg × 9.8m/s² × 2m

W = 6860kgm²/s²

W =  6860J

Therefore, the work done by the forklift truck as it lifts the box to the given height is 6860J.

Learn more about Work: https://brainly.com/question/9942439

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