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castortr0y [4]
2 years ago
8

A plastic box has an initial volume of 2.00 m 3 . It is then submerged below the surface of a liquid and its volume decreases to

1.96 m 3. what is the volume strain on the box
Physics
1 answer:
nikitadnepr [17]2 years ago
6 0

Answer:

Volume strain is 0.02

Explanation:

Volume strain is defined as the change in volume to the original volume.

It is given that,

Initial volume of the plastic box is 2 m³

It is then submerged below the surface of a liquid and its volume decreases to 1.96 m³

We need to find the volume strain on the box. It is defined as the change in volume divided by the original volume. So,

\delta V=\dfrac{V_f-V_i}{V_i}\\\\\delta V=\dfrac{1.96-2}{2}\\\\\delta V=0.02

So, the volume strain on the box is 0.02.

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Calculate the average speed and average velocity of a complete round-trip in which the outgoing 250 km is covered at 95 km/h fol
ser-zykov [4K]

To solve this problem we will apply the concepts related to the kinematic equations of linear motion. From them we will consider speed as the distance traveled per unit of time. Said unit of time will be cleared to find the total time taken to travel the given distance. Later with the calculated average times and distances, we will obtain the average speed.

PART A)

The time taken to travel a distance of 250km with a speed of 95km/h is

t = \frac{d}{v}

t = \frac{250km}{95km/h}

t = 2.63h

Time taken for the lunch is

t = 1h

The time taken travel a distance of 250km with a speed of 55km/h

t = \frac{d}{v}

t = \frac{250}{55}

t = 4.54h

The total time taken is

t = t_{outgoing}+t_{lunch}}t_{return}

t = 2.63+1+4.54

t = 8.17h

The average speed is the ratio of total distance and total time

v = \frac{250+250}{8.17}

v = 61.15km/h

PART B)

As the displacement is zero the average velocity is zero.

5 0
2 years ago
The initial velocity of a 4.0-kg box is 11 m/s, due west. After the box slides 4.0 m horizontally, its speed is 1.5 m/s. Determi
ankoles [38]

Answer:

F = - 59.375 N

Explanation:

GIVEN DATA:

Initial velocity = 11 m/s

final velocity = 1.5 m/s

let force be F

work done =  mass* F = 4*F

we know that

Change in kinetic energy = work done

kinetic energy = = \frac{1}{2}*m*(v_{2}^{2}-v_{1}^{2})

kinetic energy = = \frac{1}{2}*4*(1.5^{2}-11^{2}) = -237.5 kg m/s2

-237.5 = 4*F

F = - 59.375 N

7 0
2 years ago
carbon-14 has a half-life of approximately 5,700 years. a fossil shell contain 25% of the original amount of its carbon-14. appr
denis23 [38]

The half-life equation m=m_{0} (\frac{1}{2})^n in which <em>n </em>is equal to the number of half-lives that have passed can be altered to solve for <em>n.</em>

<em>n = \frac{log(\frac{m}{m_{0}} )}{log(\frac{1}{2})}</em>

<em>\frac{log(\frac{.25}{1} )}{log(\frac{1}{2})} = 2</em>

Then, the number of half-lives that passed can be multiplied by the length of a half-life to find the total time.

<em>2 * 5700 =  </em>11400 yr

3 0
2 years ago
A 900-kg car traveling east at 15.0 m/s collides with a 750-kg car traveling north at 20.0 m/s. The cars stick together. Assume
dalvyx [7]

Explanation:

It is given that,

Mass of the car 1, m_1=900\ kg

Initial speed of car 1, u_1=15i\ m/s (east)

Mass of the car 2, m_2=750\ kg

Initial speed of car 2, u_1=20j\ m/s (north)

(b) As the cars stick together. It is a case of inelastic collision. Let V is the common speed after the collision. Using the conservation of momentum as :

m_1u_1+m_2u_2=(m_1+m_2)V

900\times 15i +750\times 20j=(900+750)V

13500i+15000j=1650V

V=(8.18i+9.09j)\ m/s

The magnitude of speed,

|V|=\sqrt{8.18^2+9.09^2}

V = 12.22 m/s

(b) Let \theta is the direction the wreckage move just after the collision. It is given by :

tan\theta=\dfrac{v_y}{v_x}

tan\theta=\dfrac{9.09}{8.18}

\theta=48.01^{\circ}

Hence, this is the required solution.

4 0
2 years ago
Assume that you stay on the earth's surface. what is the ratio of the sun's gravitational force on you to the earth's gravitatio
Pachacha [2.7K]
First, let's determine the gravitational force of the Earth exerted on you. Suppose your weight is about 60 kg. 

F = Gm₁m₂/d²
where
m₁ = 5.972×10²⁴ kg (mass of earth)
m₂ = 60 kg
d = 6,371,000 m (radius of Earth)
G = 6.67408 × 10⁻¹¹ m³ kg⁻¹ s⁻²

F = ( 6.67408 × 10⁻¹¹ m³ kg⁻¹ s⁻²)(60 kg)(5.972×10²⁴ kg)/(6,371,000 m )²
F = 589.18 N

Next, we find the gravitational force exerted by the Sun by replacing,
m₁ = 1.989 × 10³⁰<span> kg
Distance between centers of sun and earth = 149.6</span>×10⁹ m
Thus,
d = 149.6×10⁹ m - 6,371,000 m = 1.496×10¹¹ m

Thus,
F = ( 6.67408 × 10⁻¹¹ m³ kg⁻¹ s⁻²)(60 kg)(1.989 × 10³⁰ kg)/(1.496×10¹¹ m)²
F = 0.356  N

Ratio = 0.356  N/589.18 N
<em>Ratio = 6.04</em>
5 0
2 years ago
Read 2 more answers
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