HELP PLEASE ASAP!!!
I CAN'T FAIL PHYSICS
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Answer:

um.......I don't know soory

The Moment of Inertia of the Disc is represented by [tex]I = \frac{15}{32}\cdot M\cdot R^{2}[/tex]. (Correct answer: A)

Let suppose that the Disk is a Rigid Body whose mass is uniformly distributed. The Moment of Inertia of the element is equal to the Moment of Inertia of the entire Disk minus the Moment of Inertia of the Hole, that is to say:

[tex]I = I_{D} - I_{H}[/tex] (1)

Where:

Then, this formula is expanded as follows:

[tex]I = \frac{1}{2}\cdot M\cdot R^{2} - \frac{1}{2}\cdot m\cdot \left(\frac{1}{2}\cdot R^{2} \right)[/tex] (1b)

Dimensionally speaking, Mass is directly proportional to the square of the Radius, then we derive the following expression for the Mass removed by the Hole ([tex]m[/tex]):

[tex]\frac{m}{M} = \frac{R^{2}}{4\cdot R^{2}}[/tex]

[tex]m = \frac{1}{2}\cdot M[/tex]

And the resulting equation is:

[tex]I = \frac{1}{2}\cdot M\cdot R^{2} -\frac{1}{2}\cdot \left(\frac{1}{4}\cdot M \right) \cdot \left(\frac{1}{4}\cdot R^{2} \right)[/tex]

[tex]I = \frac{1}{2} \cdot M\cdot R^{2} - \frac{1}{32}\cdot M\cdot R^{2}[/tex]

[tex]I = \frac{15}{32}\cdot M\cdot R^{2}[/tex]

The moment of inertia of the Disc is represented by [tex]I = \frac{15}{32}\cdot M\cdot R^{2}[/tex]. (Correct answer: A)

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Answer:

Help meh in this question

Explanation:

Answer:

Submultiples of kilogram are decagram ,centigram etc. And multiples of kilogram is megagram(1 tonne) , gigagram etc.

hope this helps you:)

Answer:

bnkjlji

Explanation:

Answer :

Population composition is the description of a,population according to characteristics such as age and sex . These data are often compared over time using population pyramids.

Answer:

don't know what class are you which subject is this

Answer:

I hope it's helpful for you ☺️

Answer:

Nó

Answer:

14.6 m/s

Explanation:

The total run time was 5 + 60 = 65 minute or 65(60) = 3900 s

At his average velocity, emu ran 15 m/s(3900 s) = 58,500 m

Which is a heck of a running distance for ANY emu.

In the first 5 minutes the emu traveled 20 m/s(5 min)(60 s/min) = 6000 m

So in the last hour (3600 s) the emu traveled 58,500 - 6000 = 52,500 m

at a speed of 52,500 m /3600 s = 14.583333333... m/s

The emu was running at a speed of 14.58 m/s when he was tired.

To solve this question, we'll begin by calculating the distance travelled in the first part of the trip. This can be obtained as follow:

Time (t₁) = 5 min = 5 × 60 = 300 s

Speed 1 (S₁) = 20 m/s

Speed = distance / time

S₁ = d₁ / t₁

20 = d₁ / 300

Cross multiply

d₁ = 20 × 300

Next, we shall determine  the total distance travelled by the emu.

Average speed = 15 m/s

Time 1 (t₁) = 300 s

Time 2 (t₂) = 1 h = 60 mins = 60 × 60 = 3600 s

Total time (T) = t₁ + t₂ = 300 + 3600 = 3900 s

Average speed = Total distance / total time

15 = D / 3900

Cross multiply

D = 15 × 3900

Next, we shall determine the distance travelled in the second part (i.e when he was tired) of the trip.

Total distance (D) = 58500 m

Distance 1 (d₁) = 6000 m

D = d₁ + d₂

58500 = 6000 + d₂

Collect like terms

58500 – 6000 = d₂

Finally, we shall determine the speed of the emu in the second part of the trip.

Distance 2 (d₂) = 52500 m

Time 2 (t₂) = 3600 s

Speed = distance / time

S₂ = 52500 / 3600

Therefore, the emu was running at a speed of 14.58 m/s when he was tired.

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Answer:

Force is 2 × 10^-4 N

Explanation:

[tex]{ \bf{F = \frac{ \gamma_{o} I _{1} I _{2} l}{2\pi r} }}[/tex]

F is force

I is current

r is separation distance

gamma is a constant

l is wire length

[tex]F = \frac{(4\pi \times {10}^{ - 7}) \times (20) \times (10) \times 1 }{2\pi \times 0.2} \\ \\ F = 2 \times {10}^{ - 4} \: \: newtons[/tex]

The magnitude of the force on 1m length of the 20A wire is 2×10⁻⁴N.

To determine the answer, we need to know about magnetic field due to a current carrying wire and magnetic force.

The magnetic field due to a straight wire having current I₁ at a perpendicular distance (say) 'd'  is μ₀I₁ / 2[tex]\pi[/tex]d.

Magnetic force on a current carrying wire of length 'L' and having current I₂ is I₂(L×B). Where B is the magnetic field at that wire.

  ( Here L and B are perpenicular to each other so L×B= L·B)    

                              = 20×1×10⁻⁵

                              =2×10⁻⁴N

Thus, we can conclude that the magnitude of the force on 1m length of the 20A wire is 2×10⁻⁴N.

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2: For a sample of data [tex]x_1,x_2,\ldots,x_N[/tex], the mean of this sample denoted by [tex]\overline x[/tex] is the sum of the data divided by the number of data points,

[tex]\overline x = \dfrac{x_1+x_2+\cdots+x_N}N = \displaystyle\frac1N\sum_{i=1}^N x_i[/tex]

As an example, consider [tex]x_1=-1[/tex], [tex]x_2=1[/tex], and [tex]x_3=3[/tex]. Then

[tex]\overline x = \dfrac{-1+1+3}3 = 1[/tex]

3: Standard deviation is a measure of how dispersed a given data sample is relative to the mean. Consult the plot: for a normal distribution, approximately 68% of it lies within 1 standard deviation of the mean, approx. 95% within 2 standard deviations, and approx. 99.7% within 3 standard deviations.

For instance, if the data is pulled from a normally distributed population with mean 0 and standard deviation 1, if you were to randomly select any data from the population, then 68% of the time it will fall in the range (-1, 1); 95% of the time it will fall within (-2, 2); 99.7% of the time it fall within (-3, 3).

To compute the standard deviation for a sample, for each [tex]x_i[/tex] in [tex]x_1,x_2,\ldots,x_N[/tex], you

• take the difference between [tex]x_i[/tex] and the mean [tex]\overline x[/tex]

• square this difference

• sum all the squared differences

• divide the sum by N - 1 (for a sample) or N (for a population)

• take the square root

Here the standard deviation is denoted [tex]\sigma^x_{N-1}[/tex], which I would read as "the sample standard deviation of the data x" - sample because of the N - 1 subscript.

Continuing with the previous example, we'd have

[tex]\sigma^x_{N-1} = \displaystyle \sqrt{\frac{\left(-1-1\right)^2+\left(1-1\right)^2+\left(3-1\right)^2}{3-1}} = \sqrt4 = 2[/tex]

4: Not much more to say here, the standard error is basically a measure of how accurate a given estimate is about the population based on the sample data. It's analogous to uncertainty in measuring length with a ruler, for instance.

In our example,

[tex]\alpha^x = \dfrac2{\sqrt3}[/tex]

5: If x, y, and z are random variables, then I suppose ρ is meant to denote a function of these random variables (so that ρ itself is just another random variable). For instance, you could have ρ = x + 3y - 2z. Then [tex]\overline\rho[/tex] is the sample mean of ρ.

I'm not entirely sure about the notation [tex]x(\overline x,\sigma^x_{N-1},\alpha^x)[/tex], but I suspect it's just referring to sample x with mean [tex]\overline x[/tex] and standard deviation [tex]\sigma^x_{N-1}[/tex] with standard error [tex]\alpha^x[/tex].

∆ρ is just the differential of ρ, essentially capturing how ρ changes with respect to small changes in x, y, and z. The expression you see here follows from the chain rule for differentiation.

The formula you see for [tex]\sigma^\rho_{N-1}[/tex] is the sample standard deviation of ρ. Think of ∆ρ as a vector with 3 components. Then [tex]\sigma^\rho_{N-1}[/tex] is the magnitude of this vector.

Similarly, [tex]\alpha^\rho[/tex] is the standard error for ρ, and corresponds to the magnitude of the vector whose components are the standard errors of x, y, and z.

In order for these statistics to make sense, each of x, y, and z must be samples of the same number of data. Say we take x as before [tex](x_1=-1,x_2=1,x_3=3)[/tex], along with [tex]y_1=0,y_2=4,y_3=-2[/tex] and [tex]z_1=-3,z_2=\frac12,z_3=10[/tex]. Suppose ρ = x + 3y - 2z. Then

• the sample means of y and z :

[tex]\overline y = \dfrac{0+4-2}3 = \dfrac23 \\\\ \overline z = \dfrac{-3+\frac12+10}3 = \dfrac52[/tex]

• the standard deviations of y and z :

[tex]\sigma^y_{N-1} = \sqrt{\dfrac{\left(0-\frac23\right)^2+\left(4-\frac23\right)^2+\left(2-\frac23\right)^2}{3-1}} = 2\sqrt{\dfrac73} \approx 3.06\\\\ \sigma^z_{N-1} = \sqrt{\dfrac{\left(-3-\frac52\right)^2+\left(\frac12-\frac52\right)^2+\left(10-\frac52\right)^2}{3-1}} = \dfrac{\sqrt{181}}2 \approx 6.73[/tex]

• the values of ρ :

[tex]\rho_1 = x_1+3y_1-2z_1 = -1+2\times0-2\times(-3) = 5 \\\\ \rho_2 = x_2+3y_2-2z_2 = 1+3\times4-2\times\dfrac12=12 \\\\ \rho_3 = x_3+3y_3-2z_3 = 3+3\times(-2)-2\times10 = -23[/tex]

• the sample mean of ρ :

[tex]\overline\rho = \dfrac{5+12-23}3 = -2[/tex]

• by the chain rule,

[tex]\Delta\rho = \Delta x+3\Delta y-2\Delta z[/tex]

so the standard deviation of ρ :

[tex]\sigma^\rho_{N-1} = \sqrt{\left(\sigma^x_{N-1}\right)^2 + \left(3\sigma^y_{N-1}\right)^2 + \left(-2\sigma^z_{N-1}\right)^2} \\\\\sigma^\rho_{N-1}= \sqrt{2^2 + 9\left(2\sqrt{\dfrac73}\right)^2 + 4\left(\dfrac{\sqrt{181}}2\right)^2} = \dfrac12\sqrt{\dfrac{703}3} \approx 7.65[/tex]

• the standard errors of y and z :

[tex]\alpha^y = \dfrac{2\sqrt{\frac73}}{\sqrt3} = \dfrac23\sqrt7 \approx 1.76 \\\\ \alpha^z = \dfrac{\frac{\sqrt{181}}2}{\sqrt3} = \dfrac12\sqrt{\dfrac{181}3} \approx 3.88[/tex]

• the standard error of ρ :

[tex]\alpha^\rho=\sqrt{\left(\alpha^x\right)^2+\left(3\alpha^y\right)^2+\left(-2\alpha^z\right)^2}\\\\\alpha^\rho=\sqrt{\left(\dfrac2{\sqrt3}\right)^2+9\left(\dfrac23\sqrt7\right)^2+4\left(\dfrac12\sqrt{\dfrac{703}3}\right)^2}=\sqrt{269}\approx16.40[/tex]

Newton's third law of motion sates that force is directly proportional to the rate of change of momentum produced

(a) The final speeds of the ice sleds is approximately 0.49 m/s each

(b) The impulse on the cat is 11.0715 kg·m/s

(c) The average force on the right sled is 922.625 N

The reason for arriving at the above values is as follows:

The given parameters are;

The masses of the two ice sleds, m₁ = m₂ = 22.7 kg

The initial speed of the ice, v₁ = v₂ = 0

The mass of the cat, m₃ = 3.63 kg

The initial speed of the cat, v₃ = 0

The horizontal speed of the cat, v₃ = 3.05 m/s

(a) The required parameter:

The final speed of the two sleds

For the first jump to the right, we have;

By the law of conservation of momentum

Initial momentum = Final momentum

∴ m₁ × v₁ + m₃ × v₃ = m₁ × v₁' + m₃ × v₃'

Where;

v₁' = The final velocity of the ice sled on the left

v₃' = The final velocity of the cat

Plugging in the values gives;

22.7 kg × 0 + 3.63 × 0 = 22.7 × v₁' + 3.63 × 3.05

∴  22.7 × v₁'  = -3.63 × 3.05

v₁' =  -3.63 × 3.05/22.7 ≈ -0.49

The final velocity of the ice sled on the left, v₁' ≈ -0.49 m/s (opposite to the direction to the motion of the cat)

The final speed ≈ 0.49 m/s

For the second jump to the left, we have;

By conservation of momentum law,  m₂ × v₂ + m₃ × v₃ = m₂ × v₂' + m₃ × v₃'

Where;

v₂' = The final velocity of the ice sled on the right

v₃' = The final velocity of the cat

Plugging in the values gives;

22.7 kg × 0 + 3.63 × 0 = 22.7 × v₂' + 3.63 × 3.05

∴  22.7 × v₂'  = -3.63 × 3.05

v₂' =  -3.63 × 3.05/22.7 ≈ -0.49

The final velocity of the ice sled on the right = -0.49 m/s (opposite to the direction to the motion of the cat)

The final speed ≈ 0.49 m/s

(b) The required parameter;

The impulse of the force

The impulse on the cat = Mass of the cat × Change in velocity

The change in velocity, Δv = Initial velocity - Final velocity

Where;

The initial velocity = The velocity of the cat before it lands = 3.05 m/s

The final velocity = The velocity of the cat after coming to rest =

∴ Δv = 3.05 m/s - 0 = 3.05 m/s

The impulse on the cat = 3.63 kg × 3.05 m/s = 11.0715 kg·m/s

(c) The required information

The average velocity

Impulse = [tex]F_{average}[/tex] × Δt

Where;

Δt = The time of collision = The time it takes the cat to finish landing = 12 ms

12 ms = 12/1000 s = 0.012 s

We get;

[tex]F_{average} = \mathbf{\dfrac{Impulse}{\Delta \ t}}[/tex]

∴ [tex]F_{average} = \dfrac{11.0715 \ kg \cdot m/s}{0.012 \ s} = 922.625 \ kg\cdot m/s^2 = 922.625 \ N[/tex]  

The average force on the right sled applied by the cat while landing, [tex]\mathbf{F_{average}}[/tex] = 922.625 N

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Answer:

HOPE THIS ANSWER WILL HELP YOU

Un proyectil se define como cualquier objeto que traza una trayectoria parabólica. Los parámetros importantes en el movimiento de proyectiles son; Tiempo de vuelo, rango y altura máxima.

a) 1 segundo

b) 1,25 m

c) 10 m

Deje que el tiempo de vuelo sea T

T = 2usinθ / g

u = velocidad inicial

g = aceleración debida a la gravedad

θ = ángulo

T = 2 * 10 m / s * sin (30) / 10

T = 1 segundo

Sea la altura máxima H

H = u ^ 2 sin ^ 2 θ / 2g

H = (10) ^ 2 (sin30) ^ 2/2 * 10

Alto = 1,25 m

Sea el rango R

R = u ^ 2sin 2θ / g

R = (10) ^ 2 sin 2 (30) / 10

R = 8.66 m

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a) El tiempo que tarda el proyectil en llegar al piso es:

[tex]t_{vuelo}=1\: s[/tex]  

b) La máxima altura alcanzada por el proyectil es:

[tex]y_{max}=1.25\: m[/tex]  

c) La máxima distancia alcanzada por el proyectil es:

[tex]x_{max}=8.66\: m[/tex]

a)

Para calcular el tiempo que le toma al proyectil llegar al piso, se puede usar la siguiente ecuación de tiro parabólico.

[tex]y=y_{i}+v_{iy}t-0.5gt^{2}[/tex] (1)

Donde:

Recordemos que la componente de la velocidad incial (v(i)=10 m/s) en el eje y esta dada por:

[tex]v_{iy}=v_{i}sin(30)[/tex]

[tex]v_{iy}=10sin(30)[/tex]

Sabemos que y(i) = 0 y ademas haciendo que y = 0, sabremos el tiempo total de vuelo. De la ecuación (1):

[tex]0=0+10sin(30)t-0.5(10)t^{2}[/tex]

Despejando t.

[tex]10sin(30)=0.5(10)t[/tex]

[tex]t=\frac{10sin(30)}{5}[/tex]  

El tiempo de vuelo sera:

[tex]t_{vuelo}=1\: s[/tex]  

b)

Para calcular la máxima altura, usamos la siguiente ecuación.

[tex]v_{fy}^{2}=v_{iy}^{2}-2gy[/tex]

Si hacemos que veocidad final en y v(fy) sea 0, encontraremos la máxima altura.

[tex]0=v_{iy}^{2}-2gy_{max}[/tex]

[tex]0=(v_{i}sin(30))^{2}-2gy_{max}[/tex]

Despejamos y(max):

[tex]y_{max}=\frac{(v_{i}sin(30))^{2}}{2g}[/tex]

[tex]y_{max}=\frac{(10sin(30))^{2}}{2(10)}[/tex]  

La máxima altura alcanzada por el proyectil será:

[tex]y_{max}=1.25\: m[/tex]  

c)

Sabemo que la velocidad del proyectil en la dirección x es constante, por lo tanto, la ecuacion cinemática será:

[tex]x=vt[/tex]

La máxima distancia se determina con el tiempo de vuelo:

[tex]x_{max}=v_{ix}t_{vuelo}[/tex]

La componente de la velocidad en la direccion x es:

[tex]v_{ix}=v_{i}cos(30)[/tex]

Por lo tanto, el máximo desplazamiento será:

[tex]x_{max}=10cos(30)*1[/tex]

[tex]x_{max}=8.66\: m[/tex]

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Answer:

The correct choice is option D. glass

Glass's optical density is very great. Which is why speed of light will be minimum when passing glass.

Answer:

glass

Explanation:

i think, sana makatulong

Answer:

b

Explanation:

Given

[tex]\vec F = (4xz+y^2)\,\vec\imath + 2xy\,\vec\jmath + 2x^2\,\vec k[/tex]

its curl would be

[tex]\nabla\times\vec F = \left(\dfrac{\partial(2xy)}{\partial z} - \dfrac{\partial(2x^2)}{\partial y}\right)\,\vec\imath - \left(\dfrac{\partial(4xz+y^2)}{\partial z} - \dfrac{\partial(2x^2)}{\partial x}\right)\,\vec\jmath + \left(\dfrac{\partial(4xz+y^2)}{\partial y} - \dfrac{\partial(2xy)}{\partial x}\right)\,\vec k[/tex]

which reduces to the zero vector. Since the curl is zero, and [tex]\vec F[/tex] doesn't have any singularities, [tex]\vec F[/tex] is indeed a gradient field.

To determine what it is a gradient of, we look for a scalar function f(x, y, z) such that [tex]\nabla f = \vec F[/tex]. This entails solving for f such that

[tex]\dfrac{\partial f}{\partial x} = 4xz+y^2 \\\\ \dfrac{\partial f}{\partial y} = 2xy \\\\ \dfrac{\partial f}{\partial z} = 2x^2[/tex]

Integrate both sides of the first equation with respect to x, which gives

[tex]f(x,y,z) = 2x^2z+xy^2 + g(y,z)[/tex]

Differentiate both sides of this with respect to y :

[tex]\dfrac{\partial f}{\partial y} = 2xy = 2xy+\dfrac{\partial g}{\partial y} \\\\ \implies \dfrac{\partial g}{\partial y} = 0 \implies g(y,z) = h(z)[/tex]

Differentiate f with respect to z :

[tex]\dfrac{\partial f}{\partial z} = 2x^2 = 2x^2 + \dfrac{\mathrm dh}{\mathrm dz} \\\\ \implies \dfrac{\mathrm dh}{\mathrm dz} = 0\implies h(z) = C[/tex]

So it turns out that [tex]\vec F[/tex] is the gradient of

[tex]f(x,y,z) = 2x^2z+xy^2+C[/tex]

Answer:

Below

Explanation:

To find the speed of an object you can use this formula

     speed = displacement / elapsed time

For the first one

24 m / 0.6 seconds = 40 m/s

For the second one

0.3 m / 0.01 seconds = 30 m/s

Hope that helped

Answer:

Explanation:

Did this come with a Picture?

Answer:

Reactivity

Explanation:

Explanation:

Her cells will not work well when they have low levels of oxygen.

Answer:

molecule, a group of two or more atoms that form the smallest identifiable unit into which a pure substance can be divided and still retain the composition and chemical properties of that substance

Explanation:

Here are examples of common molecules:

H2O (water)

N2 (nitrogen)

O3 (ozone)

CaO (calcium oxide)

C6H12O6 (glucose, a type of sugar)

NaCl (table salt)

Explanation:

molecule, a group of two or more atoms that form the smallest identifiable unit into which a pure substance can be divided and still retain the composition and chemical properties of that substance

(A) The total distance covered by the plane is 500,000 m

(B) The plane's total displacement is 300,000 m

(C) The average velocity of the plane is 100 m/s

The given parameters:

initial displacement of the plane = 400 km Dallas

final displacement of the train = 100 km Huntsville

the time taken for the initial displacement = 120 min

final time for the second trip = 70 min

A sketch of the plane's trip is as follows:

                                         Dallas

                              400 km ↑ ↓

                                            ↑  ↓ Huntsville  100 km

                                             ↑

                                             ↑

                                        Houston

(A) The total distance covered by the plane:

The total distance is the total path travelled by the plane.

Total distance = 400 km + 100 km = 500 km, = 500,000 m

(B) The plane's total displacement:

The total displacement is the change in the position of the plane.

The displacement = 400 km - 100 km = 300 km = 300,000 m

(C) The average velocity of the plane

[tex]Average \ velocity = \frac{change \ in \ displacement }{change \ in \ time} \\\\Average \ velocity = \frac{400 \ km - 100 \ km}{120 \ \min - \ 70 \ \min} \\\\Average \ velocity =\frac{300 \ km}{2\ hr - 1.167 \ hr} \\\\Average \ velocity = \frac{300 \ km}{0.833 \ hr} \\\\Average \ velocity = 360.14 \ km/hr\\\\Average \ velocity = \frac{360.14 \ km/h}{3.6 \times \frac{km/hr}{m/s} } = 100 \ m/s[/tex]

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Air pressure or atmospheric pressure is the pressure as the force exerted by the collisions of particles in the air.

The key difference between air pressure and liquid pressure is that air pressure allows the gaseous state of matter to be compressible, whereas liquid pressure makes a liquid incompressible

Answer:

Atmospheric pressure, also known as barometric pressure, is the pressure within the atmosphere of Earth. The standard atmosphere is a unit of pressure defined as 101,325 Pa, which is equivalent to 760 mm Hg, 29.9212 inches Hg, or 14.696 psi.

Liquid pressure is the increase in pressure at increasing depths in a liquid. This pressure increases because the liquid at lower depths has to support all of the water above it. We calculate liquid pressure using the equation liquid pressure = mass x acceleration due to g density x depth in fluid.

Answer:

Please mark brainliest

Answer:

[tex]\huge\boxed{\sf F = 147 N}[/tex]

Explanation:

Given:

Mass = m = 12 g = 0.012 kg

Final Velocity = Vf = 35 m/s

Initial Velocity = Vi = 0 m/s

Distance = S = 5 cm = 0.05 m

Required:

Force = F = ?

Formula:

3rd equation of motion

2aS = Vf²-Vi²

Solution:

2aS = Vf²-Vi²

a = (35)²- (0)² / 2S

a = 1225 / 2(0.05)

a = 1225 / 0.1

a = 12250 m/s²

Force = Mass x Acceleration

F = 0.012 x 12250

F = 147 N

[tex]\rule[225]{225}{2}[/tex]

Hope this helped!

Answer:

12.5

Explanation:

Its 5m per second times 2.5 seconds so it's 12.5m

Answer:

[tex]\boxed {\boxed {\sf 12.5 \ centimeters}}[/tex]

Explanation:

We are asked to find how far a pulse travels or the distance. Distance is the product of speed and time.

[tex]d= s \times t[/tex]

The speed of the plus is 5 centimeters per second and the time is 2.5 seconds.

Substitute the values into the formula.

[tex]d= 5 \ cm/s \times 2.5 \ s[/tex]

Multiply. Note that the units of seconds will cancel each other out.

[tex]d= 12.5 \ cm[/tex]

The pulse travels 12.5 centimeters.

Answer:

Anyway, space exploration absolutely does give us a direct benefit. When space technology has advanced far enough, we will be able to leave this planet in large numbers and live among the stars. This will solve our population/environment/resource/energy problems for a long, long time. Even a fraction of the money spent annually on space exploration could save millions of people in poverty-stricken countries, and improve living conditions for future generations. The foundations of the world we live in are largely based on science and it is indeed vital to extend our knowledge of the universe.

                                            Thank You

                                        Please mark me brainliest

Answer:

If u dig a hole 3 feet how big is that hole

Explanation:

It’s about 20 feet

Answer:

N = 167 Newtons

R = 727 Newtons

Explanation:

i) For static equilibrium, moments about any convenient point must sum to zero.

A moment is the product of a force and a moment arm length. Only the force acting perpendicular to a moment arm passing through the pivot point makes a moment.

ii) I will ASSUME the two moment arms are 0.05m and 0.15 m

CCW moments about the fulcrum are

190 N(0.2 m) + 280 N(0.05 m) = 52 N•m

CW moments are (N)N(0.15 m + 90 N(0.3 m) = 27 + 0.15N N•m

For static equilibrium, these must be equal

27 + 0.15N = 52

       0.15N = 25

              N = 166.6666666...

Sum moments about N to zero

(Same as saying CW and CCW moments must balance)

190(0.2 + 0.15) + 280(0.05 + 0.15) - R(0.15) - 90(0.3 - 0.15) = 0

R = 726.6666666...

We could verify this by summing vertical forces to zero.

R - 190 - 280 - 166.666666 - 90 = 0

R = 726.6666666...