A thin aluminum meter stick hangs from a string attached to the 50.0 cm mark of the stick. From the 0.00 cm mark on the meter stick hangs a 5.202 kg concrete block. From the 75.0 cm mark on the meter stick hangs a 7.99 kg steel ball. At what mark on the meter stick must a 2.46 kg wooden block be attached so that the meter stick balances horizontally when lowered into fresh water? Assume the densities of concrete, steel, and wood are 2500.0 kg/m3, 8000.0 kg/m3, and 500 kg/m3 respectively. A. 57.6 cm
B. 57.4 cm
C. 57.2 cm
D. 57.8 cm

Answers

Answer 1

Answer:

The wooden block must be attached at the 57.6 cm mark on the meter stick.

What is the value

We can solve this equation for the distance of the wooden block from the pivot point, which is the mark on the meter stick where it must be attached so that the meter stick balances horizontally when lowered into fresh water.

The density of aluminum is 2700 kg/m³ The volume of the meter stick is 0.01 m³ (length times width times thickness). The acceleration due to gravity is 9.81 m/s²

Substituting these values into the equation above, we get:

distance of wooden block from pivot point = -44.89 Nm / [(2700 kg/m³)(0.01 m³)(9.81 m/s²)]

= 0.576 m

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Answer 2

Answer:

Answer:

Stop cheating in exam

Explanation:

Shame!!!!

I am sorry but I will have to refer you to the student conduct at UTA.

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An unstable atomic nucleus of mass 1.82 10-26 kg initially at rest disintegrates into three particles. One of the particles, of mass 5.18 10-27 kg, moves in the y direction with a speed of 6.00 106 m/s. Another particle, of mass 8.50 10-27 kg, moves in the x direction with a speed of 4.00 106 m/s. (a) Find the velocity of the third particle.

Answers

Answer:

Explanation:

Using Conservation of momentum (total final momentum of system is)

m1•v1f + m2•v2 f + m3•v3 f=0

and it must be zero to equal the original momentum( since the original body is at rest).

Given that

original mass M=1.82×10^-26

First disintegrate mass m1=5.18×10^-27kg

In y direction V1f=6×10^6 I'm/s

Second disintegrate mass m2=8.5×10^-27kg

In x direction V2f=4×10^6 im/s

Then the third disintegrate will be

m3=M-m1-m2

m3=1.82×10^-26-5.18×10^-27-8.5×10^-27

m3=4.52×10^-27

And the velocity is unknown

Now using the formula above

m1•v1f + m2•v2 f + m3•v3 f=0

m3•V3f= - m1•v1f - m2•v2 f

4.52E-27V3f=-5.18E-27×6E6j - 8.5E-27×4E6 i

Divide thorough by 4.52E-27

V3f= - 6.88×10^6j - 7.52×10^6i

V3f= - 7.52×10^6i - 6.88×10^6j

The final velocity of the third mass disintegrate is 6.88×10^6j - 7.52×10^6i m/s

In the standing broad jump, one squats and then pushes off with the legs to see how far one can jump. Suppose the extension of the legs from the crouch position is 0.55 m and the acceleration achieved during the time the jumper is extending their legs is 1.2 times the acceleration due to gravity, g .How far can they jump? State your assumptions. (Increased range can be achieved by swinging the arms in the direction of the jump.)

Answers

Answer:

1.32 m.

Explanation:

Below is an attachment containing the solution.

The electric field at the distance of 3.5 meters from an infinite wall of charges is 125 N/C. What is the magnitude of the electric field 1.5 meters from the wall?

Answers

Explanation:

It is given that,

Distance, r = 3.5 m

Electric field due to an infinite wall of charges, E = 125 N/C

We need to find the electric field 1.5 meters from the wall, r' = 1.5 m. Let it is equal to E'. For an infinite wall of charge the electric field is given by :

$E=(\lambda)/(2\pi \epsilon_o r)$

It is clear that the electric field is inversely proportional to the distance. So,

$(E)/(E')=(r')/(r)$

$E'=(Er)/(r')$

$E'=(125* 3.5)/(1.5)$

E' = 291.67 N/C

So, the magnitude of the electric field 1.5 meters from the wall is 291.67 N/C. Hence, this is the required solution.

*PLEASE HELP*A baseball is pitched with a horizontal velocity of 25.21 m/s. Mike Trout hits the ball, sending it in the opposite direction (back toward the pitcher) at a speed of -50.67 m/s. The ball is in contact with the bat for 0.0014 seconds. What is the
acceleration of the ball?

Answers

Answer:

-54,200 m/s^2

Explanation:

a=(vf-vi)/t

4 A wheel starts from rest and has an angular acceleration of 4.0 rad/s2. When it has made 10 rev determine its angular velocity.]

Answers

The rate of change of angulardisplacement is defined as angular velocity. The angular velocity will be 22.41rad/s.

What is angular velocity?

The rate of change of angular displacement is defined as angular velocity. Its unit is rad/sec.

ω = θ t

Where,

θ is the angle of rotation,

tis the time

ω is the angular velocity

The given data in the problem is;

u is the initialvelocity=0

α is the angularacceleration = 4.0 rad/s²

t is the time period=

n is the number of revolution = 10 rev

From Newton's second equation of motion in terms of angular velocity;

$\rm \omega_f^2 - \omega_i^2 = 2as \n\n \rm \omega_f^2 - 0 = 2* 4 * 62.83 \n\n \rm \omega_f= 22.41 \ rad/sec$

Hence the angular velocity will be 22.41 rad/s.

To learn more about angularvelocity refer to the link

brainly.com/question/1980605

Answer:

$w_f$ = 22.41rad/s

Explanation:

First, we know that:

a = 4 rad/s^2

S = 10 rev = 62.83 rad

Now we know that:

$w_f^2-w_i^2=2aS$

where $w_f$ is the final angular velocity, $w_i$ the initial angular velocity, a is the angular aceleration and S the radians.

Replacing, we get:

$w_f^2-(0)^2=2(4)(62.83)$

Finally, solving for $w_f$ :

$w_f$ = 22.41rad/s

As in problem 80, an 76-kg man plans to tow a 128000-kg airplane along a runway by pulling horizontally on a cable attached to it. Suppose that he instead attempts the feat by pulling the cable at an angle of 6.7° above the horizontal. The coefficient of static friction between his shoes and the runway is 0.87. What is the greatest acceleration the man can give the airplane? Assume that the airplane is on wheels that turn without any frictional resistance.

Answers

In order to solve this problem it is necessary to apply the concepts related to Newton's second law and the respective representation of the Forces in their vector components.

The horizontal component of this force is given as

F_x = Fcos(6.7)

While the vertical component of this force would be

F_y = Fsin(6.7)

In the vertical component, the sum of Force indicates that:

$\sum F_y= 0$