A plastic ball in a liquid is acted upon by its weight and by a buoyant force. The weight of the ball is 4 N. The buoyant force has a magnitude of 5 N and acts vertically upward. When the ball is released from rest, what is it's acceleration and direction? [2 pts] for a Free Body Diagram correctly labeled.

Answers

Answer 1

Answer:

Answer:

The acceleration is 2.448 meters per square second and is vertically upward.

Explanation:

The Free Body Diagram of the plastic ball in the liquid is presented in the image attached below. By Second Newton's Law, we know that forces acting on the plastic ball is:

$\Sigma F = F - m\cdot g = m\cdot a$ (1)

Where:

$F$ - Buoyant force, measured in newtons.

$m$ - Mass of the plastic ball, measured in kilograms.

$g$ - Gravitational acceleration, measured in meters per square second.

$a$ - Net acceleration, measured in meters per square second.

If we know that $F = 5\,N$ , $m = 0.408\,kg$ and $g = 9.807\,(m)/(s^(2))$ , then the net acceleration of the plastic ball is:

$a = (F)/(m) - g$

$a= 2.448\,(m)/(s^(2))$

The acceleration is 2.448 meters per square second and is vertically upward.

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Impulse is the______of the force and time of contact

Answers

Answer:

Product

Explanation:

Impulse is defined as the average force acting on an object times the time the force acts:

Impulse = F · Δt

If the absolute pressure of gas is 550.280 kPa, its gauge pressure is

Answers

pressure absolute = pressure gage + pressure atmosphere

Answer:

650.280

Explanation: 100kpa + 550.280kpa

Guitar string has an overall length of 1.22 m and a total mass of 3.5 g before being strung on a guitar. Once it is used on the guitar, there is a distance of 70 cm between fixed end points. The guitar string is tightened to a tension of 255 N.What is the frequency of the fundamental wave on the guitar string?

Answers

Answer:

Fundamental frequency= 174.5 hz

Explanation:

We know

fundamental frequency= $(velocity)/(2 *length)$

velocity = $\sqrt{(tension)/(mass per unit length) }$

mass per unit length= $(3.5)/(1000*1.22)$ =0.00427 $(kg)/(m)$

Now calculating velocity v= $\sqrt{(255)/(0.00427) }$

=244.3 $(m)/(sec)$

Distance between two nodes is 0.7 m.

Plugging these values into to calculate frequency

f = $(244.3)/(2 *0.7)$ =174.5 hz

Blocks A (mass 2.00 kg) and B (mass 6.00 kg) move on a frictionless, horizontal surface. Initially, block B is at rest and block A is moving toward it at 2.00 m/s. The blocks are equipped with ideal spring bumpers. The collision is head-on, so all motion before and after the collision is along a straight line. (a) Find the maximum energy stored in the spring bumpers and the velocity of each block at that time. (b) Find the velocity of each block after they have moved apart.

Answers

Answer:

av=0.333m/s, U=3.3466J

$v_(A2)=-1.333m/s,\n v_(B2)=0.667m/s$

Explanation:

a. let $m_A$ be the mass of block A, and $m_B=10.0kg$ be the mass of block B. The initial velocity of A, $\rightarrow v_A_1=2.0m/s$

-The initial momentum =Final momentum since there's no external net forces.

$pA_1+pB_1=pA_2+pB_2\n\nP=mv\n\n\therefore m_Av_A_1+m_Bv_B_1=m_Av_(A2)+m_Bv_(B2)$

Relative velocity before and after collision have the same magnitude but opposite direction (for elastic collisions):

$v_A_1-v_B_1=v_(B2)-v_(A2)$

-Applying the conservation of momentum. The blocks have the same velocity after collision:

$v_(B2)=v_(A2)=v_2\n\n2* 2+10* 0=2v_2+10v_2\n\nv_2=0.3333m/s$

#Total Mechanical energy before and after the elastic collision is equal:

$K_1+U_(el,1)=K_2+U_(el,2)\n\n#Springs \ in \ equilibrium \ before \ collision\n\nU_(el,2)=K_1-K_2=0.5m_Av_A_1^2-0.5(m_A+m_B)v_2^2\n\nU_(el,2)=0.5* 2* 2^2-0.5(2+10)(0.333)^2\n\nU_(el,2)=3.3466J$

Hence, the maxumim energy stored is U=3.3466J, and the velocity=0.333m/s

b. Taking the end collision:

From a above, $m_A=2.0kg, m_B=10kg, v_A=2.0,v_B_1=0$

We plug these values in the equation:

$m_Av_A_1+m_Bv_B_1=m_Av_(A2)+m_Bv_(B2)$

$2*2+10*0=2v_A_2+10v_B_2\n\n2=v_A_2+5v_B_2\n\n#Eqtn 2:\nv_A_1-v_B_1=v_(B2)-v_(A2)\n\n2-0=v_(B2)-v_(A2)\n\n2=v_(B2)-v_(A2)\n\n#Solve \ to \ eliminate \ v_(A2)\n\n6v_(B2)=2.0\n\nv_(B2)==0.667m/s\n\n#Substitute \ to \ get \ v_(A2)\n\nv_(A2)=(4)/(6)-2=1.333m/s$

A white blood cell has a diameter of approximately 12 micrometers or 0.012 um a model represents its diameter as 24 um what ratio of model size

Answers

Answer:

The ratio of the model size is 1 : 2000

Explanation:

Given

Real Diameter = 0.012 um

Scale Diameter = 24 um

Required

Determine the scale ratio

The scale ratio is calculated as follows;

$Scale = (Real\ Measurement)/(Scale\ Measurement)$

Substitute values for real and scale measurements

$Scale = (0.012\ um)/(24\ um)$

Divide the numerator and the denominator by 0012um

$Scale = (1)/(2000)$

Represent as ratio

$Scale = 1 : 2000$

Hence, the ratio of the model size is 1 : 2000

The ratio of the model size to the actual size is 1 : 2000. This means the model represents the white blood cell's diameter 2000 times larger than its actual size.

The ratio of the model size to the actual size can be calculated using the given measurements:

Actual Diameter = 0.012 um

Model Diameter = 24 um

Ratio = Model Diameter / Actual Diameter

Ratio = 24 um / 0.012 um

Ratio = 2000

So, the ratio of the model size to the actual size is 1 : 2000. This means the model represents the white blood cell's diameter 2000 times larger than its actual size.

Learn more about white blood from the link given below.

brainly.com/question/30712922

#SPJ3

a mass of .4 kg is raised by a vertical distance of .450 m in the earth's gravitational field. what is the change in its gravitational potential energy

Answers

Answer:

E = 1.76 J

Explanation:

Given that,

Mass of an object, m = 0.4 kg

It moves by a vertical distance of 0.45 m in the Earth's gravitational field.

We need to find the change in its gravitational potential energy. It can be given by the formula as follow :

$E=mgh\n\nE=0.4* 9.8* 0.45\n\nE=1.76\ J$

So, the change in its gravitational potential energy is 1.76 J.

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