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2 years ago

The CBR method of flexible pavement design gives an idea about the:

Physics

1 answer:

marta [7]2 years ago

3 0

Answer:

A. quality of road-making material

Explanation:

The CBR method of flexible pavement design gives an idea about the:

A. quality of road-making material

B. traffic intensities

C. characteristics of soil

D. All of the above

The California Bearing Ratio (CBR) test is a penetration test used to evaluate the subgrade strength of roads and pavements. The results of these tests are used with the curves to determine the thickness of pavement and its component layers. it is the measure of the resistance of a material against the penetration of a standard plunger.

This can be used to determine the quality of road making material. CBR is expressed as the percentage of the actual load to the standard load.

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Trained dolphins are capable of a vertical leap of 7.0 m straight up from the surface of the water - an impressive feat. Suppose

dmitriy555 [2]

Answer:14 m

Explanation:

Given

Vertical jump make by the dolphin is given by $h=7\ m$

Suppose the dolphin jump with an initial velocity of $u$

so u is given by $u^2=2\cdot g\cdot h$

If dolphin launches at an angle $\theta$ then maximum horizontal range is given by

assuming the of Dolphin to be Projectile so range is given by

$R=\frac{u^2\sin 2\theta }{g}$

substitute the value of $u^2$

$R=\frac{2\times 9.8\times 7\sin 2\theta }{9.8}$

$R=2h\sin 2\theta$

Range will be maximum for $\theta =45^{\circ}$

thus $R_{max}=2\times 7\times 1=14\ m$

3 0

2 years ago

What statements accurately describe sunspots? Check all that apply.

polet [3.4K]

Answer:

sunspots are storms on the Suns surface

Sunspots are marked by intense magnetic activity

Sunspots produce solar flares and hot gassy ejections.

Sunspots can affect Earth’s climate.

Explanation:

I just did this lesson

6 0

2 years ago

Because the soles of your shoes have cleats, you can exert a forward force of 100 N even on slippery ice. A 10-kg picnic cooler

Brilliant_brown [7]

Answer:

you must throw 3 snowballs

Explanation:

We can solve this exercise using the concepts of conservation of the moment, let's define the system as formed by the refrigerator and all the snowballs. Let's write the moment

Initial. Before bumping that refrigerator

p₀ = n m v₀

Where n is the snowball number

Final. When the refrigerator moves

pf = (n m + M) v

The moment is preserved because the forces during the crash are internal

n m v₀ = (n m + M) v

n m (v₀ - v) = M v

n = M/m v/(vo-v)

Let's look for the initial velocity of the balls, suppose the person throws them with the maximum force if it slides in the snow (F = 100N), let's use the second law and Newton

F = m a

a = F / m

The distance the ball travels from zero speed to maximum speed is the extension of the arm (x = 1 m), let's look kinematically for the speed of the balls when leaving the arm

v₁² = v₀² + 2 a x

v₁² = 0+ 2 (100/1) 1

v₁ = 14.14 m / s

This is the initial speed for the crash

v₀ = v = 14.14 m / s

Let's calculate

n = M/m v/ (v₀-v)

n = 10/1 3 / (14.14 -3)

n = 2.7 balls

you must throw 3 snowballs

7 0

2 years ago

Fields of Point Charges Two point charges are fixed in the x-y plane. At the origin is q1 = -6.00 nC . and at a point on the x-a

My name is Ann [436]

Answer:

Part A) Electric fields at the point due to q₁ and q₂:

E₁ = 33.75*10³ N/C (-j) , E₂= ( 6.48 (-i) + 8.64 (+j) )*10³ N/C

Part B) Net electric field at P (Ep)

Ep= (6.48*10³ (-i)+25.11 10³ (-j) )N/C

Explanation:

Conceptual analysis

The electric field at a point P due to a point charge is calculated as follows:

E = k*q/d²

E: Electric field in N/C

q: charge in Newtons (N)

k: electric constant in N*m²/C²

d: distance from charge q to point P in meters (m)

Equivalence

1nC= 10⁻⁹C

1cm= 10⁻²m

Data

k= 9*10⁹ N*m²/C²

q₁ = -6.00 nC = -6 *10⁻⁹C

q₂ = +3.00 nC = +3*10⁻⁹C

d₁ = 4cm = 4 *10⁻²m

$d_{2} =\sqrt{(4*10^{-2})^{2}+((3*10^{-2})^{2} }$

d₂ = 5 *10⁻²m

Part A) Calculation of the electric fields at the point due to q₁ and q₂

Look at the attached graphic:

E₁: Electric Field at point P(0,4) cm due to charge q₁. As the charge q₁ is negative (q₁-), the field enters the charge

E₂: Electric Field at point P(0,4) cm due to charge q₂. As the charge q₂ is positive (q₂+) ,the field leaves the charge

E₁ = k*q₁/d₁² = 9*10⁹ *6 *10⁻⁹/ (4 *10⁻²)² = 33.75*10³ N/C

E₂ = k*q₂/d₂²= 9*10⁹ *3*10⁻⁹/(5 *10⁻²)² = 10.8*10³ N/C

E₁ = 33.75*10³ N/C (-j)

E₂x=E₂cosβ = 10.8*(3/5) = 6.48*10³ N/C

E₂y=E₂sinβ = 10.8*(4/5) = 8.64*10³ N/C

E₂= ( 6.48 (-i) + 8.64 (+j) )*10³ N/C

Part B) Calculation of the net electric field at P (Ep)

The electric field at a point P due to several point charges is the vector sum of the electric field due to individual charges.

Ep=Epx (i) + Epy (j)

Epx= E₂x= 6.48*10³ N/C (-i)

Epy= E₁y+E₂y= (33.75*10³ (-j) + 8.64*10³ (+j) ) N/C=25.11 10³ (-j) N/C

Ep= (6.48*10³ (-i)+25.11 10³ (-j) )N/C

Ep= (6.48*10³ (-i)+25.11 10³ (-j) )N/C

3 0

2 years ago

Батискаф витримує тиск 60 МПа. Чи можна провести дослідження

Elanso [62]

1) Yes

2) $6.34\cdot 10^9 N$

Explanation:

To solve this part, we have to calculate the pressure at the depth of the batyscaphe, and compare it with the maximum pressure that it can withstand.

The pressure exerted by a column of fluid of height h is:

$p=p_0+\rho g h$

where

$p_0 = 101,300 Pa$ is the atmospheric pressure

$\rho$ is the fluid density

$g=10 m/s^2$ is the acceleration due to gravity

h is the height of the column of fluid

Here we have:

$\rho=1030 kg/m^3$ is the sea water density

h = 5440 m is the depth at which the bathyscaphe is located

Therefore, the pressure on it is

$p=101,300+(1030)(10)(5440)=56.1\cdot 10^6 Pa = 56.1 MPa$

Since the maximum pressure it can withstand is 60 MPa, then yes, the bathyscaphe can withstand it.

Here we want to find the force exerted on the bathyscaphe.

The relationship between force and pressure on a surface is:

$p=\frac{F}{A}$

where

p is hte pressure

F is the force

A is the area of the surface

Here we have:

$p=56.1\cdot 10^6 Pa$ is the pressure exerted

The bathyscaphe has a spherical surface of radius

r = 3 m

So its surface is:

$A=4\pi r^2$

Therefore, we can find the force exerted on it by re-arranging the previous equation:

$F=pA=4\pi pr^2 = 4\pi (56.1\cdot 10^6)(3)^2=6.34\cdot 10^9 N$

6 0

2 years ago