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2 years ago

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A solid block of mass m is suspended in a liquid by a thread. The density of the block is greater than that of the liquid. Initi

ally, the fluid level is such that the block is at a depth d and the tension in the thread is T. Then, the fluid level is decreased such that the depth is 0.5d. What is the tension in the thread when the block is at the new depth?

1 answer:

Tanzania [10]2 years ago

3 0

Answer:

(C) T

The tension T at equilibrium will be equal to the Buoyant force.

The Buoyant force is given by:

Fb = density x acceleration due to gravity x volume displaced

The change in height doesn't affect the Buoyant force and hence the tension.

Note: The figure of question is added in the attachment

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Consider the two moving boxcars in Example 5. Car 1 has a mass of m1 = 65000 kg and a velocity of v01 = +0.80 m/s. Car 2 has a m

Amiraneli [1.4K]

Answer:

1.034m/s

Explanation:

We define the two moments to develop the problem. The first before the collision will be determined by the center of velocity mass, while the second by the momentum preservation. Our values are given by,

$m_1 = 65000kg\\v_1 = 0.8m/s\\m_2 = 92000kg\\v_2 = 1.2m/s$

<em>Part A)</em> We apply the center of mass for velocity in this case, the equation is given by,

$V_{cm} = \frac{m_1v_1+m_2v_2}{m_1+m_2}$

Substituting,

$V_{cm} = \frac{(65000*0.8)+(92000*1.2)}{92000+65000}$

$V_{cm} = 1.034m/s$

Part B)

For the Part B we need to apply conserving momentum equation, this formula is given by,

$m_1v_1+m_2v_2 = (m_1+m_2)v_f$

Where here $v_f$ is the velocity after the collision.

$v_f = \frac{m_1v_1+m_2v_2}{m_1+m_2}$

$v_f = \frac{(65000*0.8)+(92000*1.2)}{92000+65000}$

$v_f = 1.034m/s$

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2 years ago

Which of the following expressions will have units of kg⋅m/s2? Select all that apply, where x is position, v is velocity, m is m

netineya [11]

Answer: $m \frac{d}{dt}v_{(t)}$

Explanation:

In the image attached with this answer are shown the given options from which only one is correct.

The correct expression is:

$m \frac{d}{dt}v_{(t)}$

Because, if we derive velocity $v_{t}$ with respect to time $t$ we will have acceleration $a$ , hence:

$m \frac{d}{dt}v_{(t)}=m.a$

Where $m$ is the mass with units of kilograms ( $kg$ ) and $a$ with units of meter per square seconds $\frac{m}{s}^{2}$ , having as a result $kg\frac{m}{s}^{2}$

The other expressions are incorrect, let’s prove it:

$\frac{m}{2} \frac{d}{dx}{(v_{(x)})}^{2}=\frac{m}{2} 2v_{(x)}^{2-1}=mv_{(x)}$ This result has units of $kg\frac{m}{s}$

$m\frac{d}{dt}a_{(t)}=ma_{(t)}^{1-1}=m$ This result has units of $kg$

$m\int x_{(t)} dt= m \frac{{(x_{(t)})}^{1+1}}{1+1}+C=m\frac{{(x_{(t)})}^{2}}{2}+C$ This result has units of $kgm^{2}$ and $C$ is a constant

$m\frac{d}{dt}x_{(t)}=mx_{(t)}^{1-1}=m$ This result has units of $kg$

$m\frac{d}{dt}v_{(t)}=mv_{(t)}^{1-1}=m$ This result has units of $kg$

$\frac{m}{2}\int {(v_{(t)})}^{2} dt= \frac{m}{2} \frac{{(v_{(t)})}^{2+1}}{2+1}+C=\frac{m}{6} {(v_{(t)})}^{3}+C$ This result has units of $kg \frac{m^{3}}{s^{3}}$ and $C$ is a constant

$m\int a_{(t)} dt= \frac{m {a_{(t)}}^{2}}{2}+C$ This result has units of $kg \frac{m^{2}}{s^{4}}$ and $C$ is a constant

$\frac{m}{2} \frac{d}{dt}{(v_{(x)})}^{2}=0$ because $v_{(x)}$ is a constant in this derivation respect to $t$

$m\int v_{(t)} dt= \frac{m {v_{(t)}}^{2}}{2}+C$ This result has units of $kg \frac{m^{2}}{s^{2}}$ and $C$ is a constant

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2 years ago

Takumi works in his yard for 45 minutes each Saturday. He works in the morning, and he wears sunscreen and a hat each time he wo

MrRa [10]

Explanation :

Takumi wears sunscreen and a hat each time he works in the yard. This is to protect himself with the strong radiation coming from the sun. UV rays that are coming from the sun are the main cause of skin cancer.

Stochastic effects are the effects that are caused by chance. Cancer is one of the main stochastic effects.

So, the correct option is (b) "the severity of stochastic effects, such as cancer".

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3 years ago

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When a surface is submerged in a fluid, the resultant pressure force on the body acts in what manner? Assume no shear forces are

Alex17521 [72]

Answer:

Perpendicular to the surface of the body.

Explanation:

When a surface is submerged in a fluid the resultant pressure force on the body acts perpendicular to the surface of the body. This is because fluids cannot withstand nor exert sideways forces. One could obtain this analytically considering that the origin of this force comes from the movement of the fluid molecules.

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2 years ago

As shown in the figure below, Justin walks from the house to his truck on a windy day. He walks 20 m toward

juin [17]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The velocity is $v =0.333 \ m/s$ in positive x -direction

The speed is $s = 0.733 \ m/s$

Explanation:

From the question we are told that

The distance from the house to truck is D = 20 m

The distance traveled back to retrieve wind-blown hat is d = 15

The distance from the wind-blown hat position too the truck is k = 20 m

The total time taken is t = 75 s

Generally when calculating the displacement the Justin's backward movement to collect his wind - blown hat is taken as negative

Generally Justin's displacement is mathematically represented as

$L = 20 - 15 + 20$

=> $L = 25 \ m$

Generally the average velocity is mathematically represented as

$v = \frac{L}{t}$

=> $v = \frac{25}{75}$

=> $v =0.333 \ m/s$

Generally the distance covered by Justin is mathematically represented as

$R = D+ d + k$

=> $R = 20 + 15 +20$

=> $R = 55 \ m$

Generally Justin's average speed over a 75 s period is mathematically represented as

$s = \frac{R}{ t}$

=> $s = \frac{55}{ 75}$

=> $s = 0.733 \ m/s$

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2 years ago