Ask question

Ask question

All categories

2 years ago

5

You are given two objects of identical size, one made of aluminum and the other of lead. You hang each object separately from a

spring balance. Because lead is denser than aluminum, the lead object weighs more. Now you weigh each object while it is submerged in water. Will the difference between the weights of the aluminum object and the lead object in water be greater than, less than, or the same as it was when the objects were weighed in air? greater than

1 answer:

ollegr [7]2 years ago

7 0

Answer:

Explanation:

Volume of lead object = volume of aluminium object = V

mass of lead object > mass of aluminium object

When both the objects immersed in water, the buoyant force acting on both the objects.

Buoyant force = Volume immersed x density of water x gravity

As the volume of both the objects is same, so the buoyant force acting on both the objects is same.

So, weight in air of lead object is more than the weight in air of aluminium object.

You might be interested in

Kind of mixture that has the same mixture throughout

Ainat [17]

It's a homogeneous mixture
<span />

8 0

2 years ago

Read 2 more answers

The magnetic field or force seems to be associated with the lineup of____________ within the magnet.

Nastasia [14]

The magnetic field or force seems to be associated with the lineup of electrons withim the magnet

6 0

2 years ago

Read 2 more answers

(a) What is the potential between two points situated 10 cm and 20 cm from a 3.0-μC point charge? (b) To what location should th

julia-pushkina [17]

Answer:

(a) 135 kV

(b) The charge chould be moved to infinity

Explanation:

(a)

The potential at a distance of <em>r</em> from a point charge, <em>Q</em>, is given by

$V = -\dfrac{kQ}{r}$

where $k = 9\times 10^9 \text{ F/m}$

Difference in potential between the points is

$kQ\left[-\dfrac{1}{0.2\text{ m}} -\left( -\dfrac{1}{0.1\text{ m}}\right)\right] = \dfrac{kQ}{0.2\text{ m}} = \dfrac{9\times10^9\text{ F/m}\times3\times10^{-6}\text{ C}}{0.2\text{ m}}$

$PD = 135\times 10^3\text{ V} = 135\text{ kV}$

(b)

If this potential difference is increased by a factor of 2, then the new pd = 135 kV × 2 = 270 kV. Let the distance of the new location be <em>x</em>.

$270\times10^3 = kQ\left[-\dfrac{1}{x}-\left(-\dfrac{1}{0.1\text{ m}}\right)\right]$

$10 - \dfrac{1}{x} = \dfrac{270000}{9\times10^9\times3\times10^{-6}} = 10$

$\dfrac{1}{x} = 0$

$x = \infty$

The charge chould be moved to infinity

7 0

3 years ago

The question is in the picture

Sedbober [7]

Answer:

e) 120m/s

Explanation:

When the ball reaches its highest point, its velocity becomes zero, meaning

$v_0-gt = 0$ .

where $v_0$ is the initial velocity.

Solving for $t$ we get

$t = \dfrac{v_0}{g}$

which is the time it takes the ball to reach the highest point.

Now, after the ball has reached its highest point, it turns around and falls downwards. After time $t_0$ since it had reached the highest point, the ball has traveled downwards and the velocity $v_f$ it has gained is

$v_f = gt_0$ ,

and we are told that this is twice the initial velocity $v_0$ ; therefore,

$v_f = 2v_0 = gt_0$

which gives

$t_0 = \dfrac{2v_0}{g}.$

Thus, the total time taken to reach velocity $2v_0$ is

$t_{tot} = t+t_0 = \dfrac{v_0}{g}+\dfrac{2v_0}{g}$

$t_{tot} = \dfrac{3v_0}{g}.$

This $t_{tot}$ , we are told, is 36 seconds; therefore,

$36= \dfrac{3v_0}{g},$

and solving for $v_0$ we get:

$v_0 = \dfrac{36g}{3}$

$v_0 = \dfrac{36s(10m/s^2)}{3}$

$\boxed{v_0 = 120m/s}$

which from the options given is choice e.

7 0

3 years ago

Digiron [165]

It would be the temperature of the steel decreases and the temperature of the water increases

7 0

2 years ago

Read 2 more answers