You have two identical pure silver ingots. You place one of them in a glass of water and observe it to sink to the bottom. You place the other in a container full of mercury and observe that it floats. Comparing the buoyant forces in the two cases you conclude that a.) the buoyant force in water is smaller than in mercury

b.) the buoyant force in the water is larger than that in mercury

c.) the buoyant force in the water is zero and that in mercury is non - zero

d.) the buoyant force in the water is equal to that in mercury

e.) no conclusion can be made about the respective values of the buoyant forces

Answers

Answer 1

Answer:

Answer: a)

Explanation:

The buoyant force, as stated by Archimedes’ principle, is equal to the weight of the liquid that occupies the same volumen as the submerged object, as follows:

Fb = δ.V.g

If this force is larger than the weight of the object (that means that the fluid is denser than the solid), the object floats, which is the case for silver and mercury.

Instead, silver density is larger than water density, which explains why the pure silver ingot sinks.

Finally, as mercury is denser than water, we conclude that for a same object, the buoyant force in mercury is larger than in water (exactly 13.6 times greater).

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1. The workpart in a turning operation is 88 min in diameter and 400 mm long. A feed of 0.25 mm/rev is used in this operation. If cutting speed is 3.5 m/s, the too should be changed in every 3 workparts, but if the cutting speed is 2.5 m/sec, the tool can be used to produce 20 pieces between the tool changes. Determine the cutting speed that will allow the tool to be used for 50 parts between tool changes.

Answers

Find the given attachments

A curve of radius 35 m isbanked; therefore, no friction
is required at a speed of 7
m/s of a car. What is the?
banking angle

Answers

Answer:

The banking angle is 23.98 degrees.

Explanation:

We have,

Radius of a curve is 35 m

Speed of a car is 7 m/s

It is required to find the banking angle. At equilibrium, net force is equal to the centripetal force between vehicle and the road such that the banking angle is given by :

$\tan\theta=(v^2)/(rg)$

g is acceleration due to gravity

$\tan\theta=(7^2)/(35* (22)/(7))\n\n\theta=\tan^(-1)\left(0.445\right)\n\n\theta=23.98^(\circ)$

So, the banking angle is 23.98 degrees.

A rubber ball is shot straight up from the ground with speed v0. Simultaneously, a second rubber ball at height h directly above the first ball is dropped from rest. a) At what height above the ground do the balls collide? Your answer will be a symbolic expression in terms of v0, h, and g .b) What is the maximum value of h for which a collision occurs before the first ball falls back to the ground?
Express your answer in terms of the variable v0 and appropriate constants.
c) For what value of h does the collision occur at the instant when the first ball is at its highest point?
Express your answer in terms of the variable v0 and appropriate constants.

Answers

In this exercise we have to have knowledge about the horizontal launch, so we have to use the known formulas to find that:

a) $h*(1 - 1/2 g * h/v_0^2)$

b) $h = v_0^2/ g$

c) $h = v_0^2/ g$

So we have to remember some famous equations like the position and velocity of an object moving in a constant, like this:

$y = y_0 + v_0*t + 1/2 * a * t^2\nv = v_0 + a * t$

where:

y = height at time t

y0 = initial height

v0 = initial velocity

a = acceleration

t = time

v = velocity

a) When the balls collide, h1 = h2. Then,

$h_1 = h_2\nv_0 * t - 1/2 g * t^2 = h - 1/2 * g * t^2\nv_0 * t = h\nt = h / v_0$

Replacing in the equation of the height of the first ball:

$h_1 = v_0 * h/v_0 - 1/2g * h^2/v_0^2\nh_1 = h - 1/2 g * h^2/ v_0^2\nh_1 = h*(1 - 1/2 g * h/v_0^2)$

b) that the balls collide at t = h/v0. Then:

$h/ v_0 = -v_0/-g\nh = v_0^2/ g$

c) The value of h for which the collision occurs at the instant when the first ball is at its highest point is the maximum value,Then:

$h = v_0^2/ g$

See more about velocity at brainly.com/question/862972

A ball of mass 0.7 kg flies through the air at low speed, so that air resistance is negligible. (a) What is the net force acting on the ball while it is in motionWhich components of the ball's momentum will be changed by this force? What happens to the x component of the ball's momentum during its flight? What happens to the y component of the ball's momentum during its flight? It decreases. What happens to the z component of the ball's momentum during its flight?

Answers

A) Net Force is -6.86N

B) The y component of momentum.

C) The x component of momentum should remain the same.

D)The y component of momentum decreases.

E)The z component of momentum should remain constant.

The following information should be considered:

(A)

The net force should be

= -9.8 (0.7)

= -6.86N

(B)

Due to the net force is on the y-axis, so only the vertical component of the momentum should be changed because to the force.

(C)

Because there is no resistance of air, the ball should be in projectilemotion problems, this represents hat the x component of the velocity remains constant, also does the mass.

The y component of momentum reduced, this is due to gravity reduced the y component of the velocity.

E)Because there is no z component of the force there is no change in the z component of the momentum.

Learn more: brainly.com/question/9122916?referrer=searchResults

Final answer:

With negligible air resistance and low speed, the only significant net force on a 0.7 kg ball is gravity, affecting the ball's y component of momentum. The x component remains constant, and z component changes are not discussed without additional forces.

Explanation:

When a ball of mass 0.7 kg flies through the air at low speed with air resistance negligible, the net force acting on the ball while it is in motion is primarily due to gravity, which will be impacting the y component of the ball's momentum. The x component of the ball's momentum remains unchanged because no horizontal force is applied, while the y component changes due to gravity, and the z component would only change if there were forces acting in a direction out of the horizontal plane, which are not mentioned in the scenario. As for the Earth-ball system, momentum is conserved in the vertical direction because the system experiences no net external vertical force.

Proposed Exercise - Circular MovementConsider four pulleys connected by correals as illustrated in the figure below. One motor moves the A pulley with angular acceleration A= 20 rad/s^2/. If pulley A is initially moving with angular acceleration A= 40 rad/s^2, determine the angular speed of pulleys B and C after three seconds. Consider that the belts do not slide

Answers

Answer:

ωB = 300 rad/s

ωC = 600 rad/s

Explanation:

The linear velocity of the belt is the same at pulley A as it is at pulley D.

vA = vD

ωA rA = ωD rD

ωD = (rA / rD) ωA

Pulley B has the same angular velocity as pulley D.

ωB = ωD

The linear velocity of the belt is the same at pulley B as it is at pulley C.

vB = vC

ωB rB = ωC rC

ωC = (rB / rC) ωB

Given:

ω₀A = 40 rad/s

αA = 20 rad/s²

t = 3 s

Find: ωA

ω = αt + ω₀

ωA = (20 rad/s²) (3 s) + 40 rad/s

ωA = 100 rad/s

ωD = (rA / rD) ωA = (75 mm / 25 mm) (100 rad/s) = 300 rad/s

ωB = ωD = 300 rad/s

ωC = (rB / rC) ωB = (100 mm / 50 mm) (300 rad/s) = 600 rad/s

URGENT!!!!!! Assume that a wire has 1.5 ohms of resistance. If the wire is connected to two batteries with a total voltage of 3.0 V, how much current will flow through the wire? 3.0 amps 2.3 amps 2.0 amps 1.0 amps

Answers

Answer:

2.0 amps

Explanation:

Current is the ratio of voltage to resistance:

I = V/R = (3.0)/(1.5) = 2.0

The current in the wire is 2.0 amps.

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