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Citrus2011 [14]
3 years ago
12

If we have two objects with the same mass but different densities (Lets assume object 1 is denser,therefore volume is lower rela

tive to object 2).Assume both objects have a lower density than water.Now if we put both in water,both will float.Since they float, buoynacy force is equal to the weight of the object,so buoynacy force in 1 and 2 should be equal .However if we use F(boyancy)=pvg--> then becuase the submerged v is less for the denser object and p fluid is the same,then we can conclude that F bouynacy are not the same for both objects. True or false?

Physics
1 answer:
o-na [289]3 years ago
7 0

take the volume that is submerged. That's the volume of water displaced.

but the volume of submerged is different in those two

so buoyancy force is different in those two

weights are the same

probably. since the densities* are different.

In question it says wights are the same and diffferent volumes

so it seems that the one with more density should have a lower buyonacy force

but you said that when an object floats the buoyancy force equals weight,so since both objects have the same weight,then buoynacy force should be equal in those two


The more dense object will float with a greater percentage of its volume immersed, not less.

2) If they have the same MASS, the more dense one will have less VOLUME

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Highest common factor of 12r and 10
Scilla [17]

I think the number 2, not sure

3 0
2 years ago
Find the average velocity for the time interval beginning when t = 4 with duration 1 seconds, 0.5 seconds, and 0.05 seconds.
s344n2d4d5 [400]

This question is incomplete, the complete question is;

A student dropped a textbook from the top floor of his dorm and it fell according to the formula s(t) = -16t² + 8√t, where t is the time in seconds and s(t) is the distance in feet from the top of the building.

(a) Write a formula for the average velocity of the ball for t near 4.

(b) Find the average velocity for the time interval beginning when t = 4 with duration 1 seconds, 0.5 seconds, and 0.05 seconds

(c) What is your estimate for the instantaneous velocity of the ball at t = 4

Answer:

a)

Average velocity, (Vavg)  of the ball for t near 4.

Vavg = [s(4) - s(0)] / (4 - 0)

Where s(4) = -16 × 4² + 8 × √4= - 240 m

s(0) = -16 × 0 + 8 * 0 = 0

b)

duration = 1 sec

Vavg = [s(5) - s(4)] / (5 - 4)

s(5) = -16 × 52 + 8 × √5 = - 382 m

s(4) = -16 × 42 + 8  √4 = - 240 m

Vavg = (-382 - (-240)) / (5 - 4)

Vavg = - 142.1 m/s

duration = 0.5 sec

Vavg = [s(4.5) - s(4)] / (4.5 - 4)

s(4.5) = -16 × 4.52 + 8 × √4.5 = - 307 m

s(4) = -16 × 42 + 8 × √4 = - 240 m

Vavg = (-307 - (-240)) / (4.5 - 4)

Vavg= - 134.1 m/s

duration = 0.05 sec

Vavg = [s(4.05) - s(4)] / (4.05 - 4)

s(4.05) = -16 × 4.052 + 8 × √4.05 = - 246 m

s(4) = -16 × 42 + 8 × √4 = - 240 m

Vavg = (-246 - (-240)) / (4.05 - 4)

Vavg= - 126.8 m/s

c)

Instantaneous velocity, v = ds/dt

= - 16 × 2 × t + 8 ×× (0.5 / √t )

= - 32 × t + 4/√t

ds/dt at t = 4 is,

v = - 32 × 4 + 4 / √4

= - 126 m/s

5 0
3 years ago
A helicopter pulls upward by means of a rope on a 250 kg crate to lift it UNIFORMLY. What is the net force on the crate?
Cloud [144]

Answer:

The net force = 0

Explanation:

The given information includes;

The mass of the crate = 250 kg

The way the helicopter lifts the crate = Uniformly (constant rate (speed), no acceleration)

In order to pull the crate upwards, the helicopter has to provide a force equivalent to the weight of the crate keeping the helicopter on the ground.

The weight of the crate = The mass of the crate × The acceleration due gravity acting on the crate

The weight of the crate, F_w↓ = 250 kg × 9.81 m/s² = 2,452.5 N

The force the helicopter should provide to just lift the crate, F_{(helicopter)}↑ = The weight of the crate = 2,452.5 N

The net force, F_{(net)} = F_{(helicopter)}↑ - F_w↓ = 2,452.5 N - 2,452.5 N = 0

The net force = 0.

3 0
3 years ago
When a bow is stretched back and an arrow is shot, what type of energy conversion has occurred?
Lelechka [254]
D. mechanical to elastic
6 0
3 years ago
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jenyasd209 [6]

Answer:

G. It will take twice as long.

Explanation:

Let's call v the original speed of the plane and d the distance between Dallas and Pensacola. The time the plane originally takes to complete the flight is

t=\frac{d}{v}

In this problem, we are told that the plane encounters wind moving at half of its speed: \frac{v}{2}, in the opposite direction. This means that the new speed of the plane is

v'=v-\frac{v}{2}=\frac{v}{2}

And so, the time the plane takes now to complete the flight is

t'=\frac{d}{v/2}=2\frac{d}{v}=2t

So, the plane takes twice the time as before.

4 0
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