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2 years ago

A force of 124 newtons pushed a 12kg object forward 13 meters. How much work is being done?

Physics

1 answer:

MariettaO [177]2 years ago

5 0

Answer:

Work = Force * Distance

Here,

force = 124 N

distance = 13m

Thus, Work = 124 * 13 = 1612 joules(J)

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During a compaction test in the lab a cylindrical mold with a diameter of 4in and a height of 4.58in was filled. The compacted s

Ray Of Light [21]

Answer:

part a : The dry unit weight is 0.0616  $lb/in^3$ 

part b : The void ratio is 0.77

part c : Degree of Saturation is 0.43

part d : Additional water (in lb) needed to achieve 100% saturation in the soil sample is 0.72 lb

Explanation:

Part a

Dry Unit Weight

The dry unit weight is given as

$\gamma_{d}=\frac{\gamma}{1+\frac{w}{100}}$

Here

$\gamma_d$ is the dry unit weight which is to be calculated
γ is the bulk unit weight given as

$\gamma =weight/Volume \\\gamma= 4 lb / \pi r^2 h\\\gamma= 4 lb / \pi (4/2)^2 \times 4.58\\\gamma= 4 lb / 57.55\\\gamma= 0.069 lb/in^3$

w is the moisture content in percentage, given as 12%

Substituting values

$\gamma_{d}=\frac{\gamma}{1+\frac{w}{100}}\\\gamma_{d}=\frac{0.069}{1+\frac{12}{100}} \\\gamma_{d}=\frac{0.069}{1.12}\\\gamma_{d}=0.0616 lb/in^3$

The dry unit weight is 0.0616  $lb/in^3$ 

Part b

Void Ratio

The void ratio is given as

$e=\frac{G_s \gamma_w}{\gamma_d} -1$

Here

e is the void ratio which is to be calculated

$\gamma_d$ is the dry unit weight which is calculated in part a
$\gamma_w$ is the water unit weight which is 62.4 $lb/ft^3$ or 0.04 $lb/in^3$

G is the specific gravity which is given as 2.72

Substituting values

$e=\frac{G_s \gamma_w}{\gamma_d} -1\\e=\frac{2.72 \times 0.04}{0.0616} -1\\e=1.766 -1\\e=0.766$

The void ratio is 0.77

Part c

Degree of Saturation

Degree of Saturation is given as

$S=\frac{G w}{e}$

Here

e is the void ratio which is calculated in part b

G is the specific gravity which is given as 2.72

w is the moisture content in percentage, given as 12% or 0.12 in fraction

Substituting values

$S=\frac{G w}{e}\\S=\frac{2.72 \times .12}{0.766}\\S=0.4261$

Degree of Saturation is 0.43

Part d

Additional Water needed

For this firstly the zero air unit weight with 100% Saturation is calculated and the value is further manipulated accordingly. Zero air unit weight is given as

$\gamma_{zav}=\frac{\gamma_w}{w+\frac{1}{G}}$

Here

$\gamma_{zav}$ is the zero air unit weight which is to be calculated

$\gamma_w$ is the water unit weight which is 62.4 $lb/ft^3$ or 0.04 $lb/in^3$

G is the specific gravity which is given as 2.72

w is the moisture content in percentage, given as 12% or 0.12 in fraction

$\gamma_{zav}=\frac{\gamma_w}{w+\frac{1}{G}}\\\gamma_{zav}=\frac{0.04}{0.12+\frac{1}{2.72}}\\\gamma_{zav}=\frac{0.04}{0.4876}\\\gamma_{zav}=0.08202 lb/in^3\\$

Now as the volume is known, the the overall weight is given as

$weight=\gamma_{zav} \times V\\weight=0.08202 \times 57.55\\weight=4.72 lb$

As weight of initial bulk is already given as 4 lb so additional water required is 0.72 lb.

4 0

3 years ago

Which person would experience the greatest force of impact in a sudden collision? A 200 lb. man traveling 20 mph b 150 lb. woman

Alisiya [41]

Answer:

The 150 lb woman at 30 mph would experience the greatest force of impact in a sudden collision.

Explanation:

Momentum

The force of impact exerted on an moving object that suddenly stops or changes its movement is measures by the physics magnitude called Impulse, which can be computed with the formula

$I=F.t$

Where F is the force and t is the time that force acts to produce the impact on the object. The impulse is also defined as the change in the momentum of the object:

$I=\Delta p=p_2-p_1$

Or equivalently

$I=m(v_2-v_1)$

The question describes four situations where different persons and object suffer impact that make them stop from their moving state. Thus $v_2=0$ and the impulse is

$I=-m.v_1$

We are only interested in the relative magnitudes of each case, so we won't consider the sign in the calculations

Case 1: A 200 lb. man traveling 20 mph

$I_1=200(20)=4000\ lb.mph$

Case 2: A 150 lb. woman at 30 mph

$I_2=150(30)=4500\ lb.mph$

Case 3: A 35 lb. infant at 75 mph

$I_3=35(75)=2625\ lb.mph$

Case 4: A 75 lb. child at 55 mph

$I_4=75(55)=4125\ lb.mph$

By comparing the results, we can see that the 150 lb woman at 30 mph would experience the greatest force of impact in a sudden collision.

7 0

3 years ago

Put the phase of mitosis or mitosis in order from first to last

vladimir1956 [14]

The phases of mitosis in order from first to last are: Interphase, prophase, metaphase, anaphase, and telophase & cytokinesis.

7 0

3 years ago

Read 2 more answers

Zero, a hypothetical planet, has a mass of 5.3 x 1023 kg, a radius of 3.3 x 106 m, and no atmosphere. A 10 kg space probe is to

Goryan [66]

(a) $3.1\cdot 10^7 J$

The total mechanical energy of the space probe must be constant, so we can write:

$E_i = E_f\\K_i + U_i = K_f + U_f$ (1)

where

$K_i$ is the kinetic energy at the surface, when the probe is launched

$U_i$ is the gravitational potential energy at the surface

$K_f$ is the final kinetic energy of the probe

$U_i$ is the final gravitational potential energy

Here we have

$K_i = 5.0 \cdot 10^7 J$

at the surface, $R=3.3\cdot 10^6 m$ (radius of the planet), $M=5.3\cdot 10^{23}kg$ (mass of the planet) and m=10 kg (mass of the probe), so the initial gravitational potential energy is

$U_i=-G\frac{mM}{R}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{3.3\cdot 10^6 m}=-1.07\cdot 10^8 J$

At the final point, the distance of the probe from the centre of Zero is

$r=4.0\cdot 10^6 m$

so the final potential energy is

$U_f=-G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{4.0\cdot 10^6 m}=-8.8\cdot 10^7 J$

So now we can use eq.(1) to find the final kinetic energy:

$K_f = K_i + U_i - U_f = 5.0\cdot 10^7 J+(-1.07\cdot 10^8 J)-(-8.8\cdot 10^7 J)=3.1\cdot 10^7 J$

(b) $6.3\cdot 10^7 J$

The probe reaches a maximum distance of

$r=8.0\cdot 10^6 m$

which means that at that point, the kinetic energy is zero: (the probe speed has become zero):

$K_f = 0$

At that point, the gravitational potential energy is

$U_f=-G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{8.0\cdot 10^6 m}=-4.4\cdot 10^7 J$

So now we can use eq.(1) to find the initial kinetic energy:

$K_i = K_f + U_f - U_i = 0+(-4.4\cdot 10^7 J)-(-1.07\cdot 10^8 J)=6.3\cdot 10^7 J$

4 0

3 years ago

A film of soapy water on top of a plastic cutting board has a thickness of 255 nm. What wavelength and color is most strongly re

Alona [7]

Answer:

the reflected wavelength is lano = 4.55 10⁻⁷ m which corresponds to the blue color

Explanation:

This is a case of reflection interference, we must be careful

* There is a 180º phase change when light passes from the air to the soap film (n = 1,339), but there is no phase change when passing from the pomp to the plastic (n = 1.3)

* the wavelength within the film is modulated by the refractive index

λₙ = λ₀ / n

if we consider these relationships the condition for constructive interference is

2 t = (m + ½) λₙ

2t = (m + ½) λ₀ / n

λ₀ = 2t n / (m + ½)

we substitute the values

λ₀= 2 255 10⁻⁹ 1,339 / (m + ½)

λ₀ = 6.829 10⁻⁷ (m + ½)

let's calculate the wavelength for various interference orders

m = 0

λ₀ = 6.829 10⁻⁷/ ( 0 + ½ )

λ₀ = 13.6 10⁻⁷

it is not visible

m = 1

λ₀ = 6,829 10⁻⁷/ (1 + ½)

λ₀ = 4.55 10⁻⁷

color blue

m = 2

λ₀ = 6.829 10⁻⁷ / (2 + ½)

λ₀ = 2,7 10⁻⁷

it is not visible

therefore the reflected wavelength is lano = 4.55 10⁻⁷ m which corresponds to the blue color

5 0

3 years ago