All categories

3 years ago

How heavy is an object that displaces 400N of water in a pool?

1 answer:

7 0

All we can say is that the object's volume is about 41 liters. That's the same as the volume of water displaced.

We can't say anything about the object's weight. There is no direct connection between the weight of the object and the weight of the water it displaces.

You might be interested in

If an object is placed 10cm in front of a converging lens that has a focal length of 15cm. What are the properties of the image?

Contact [7]

Answer:

Enlarged [Size]

Virtual and Erect [Nature]

On the same side of the lens as the object [Position]

Explanation:

7 0

3 years ago

I will give brainliest) According to Newton's second law of motion, when an object is acted on by an unbalanced force, how will

evablogger [386]

Newton's first law of motion has been frequently stated throughout this lesson. Anobject at rest stays at rest and an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced force.

4 0

3 years ago

Read 2 more answers

What is the likely identity of a metal if a sample has a mass of 63.5 g when measured in air and an apparent mass of 60.2 g when

Gre4nikov [31]

Answer:

Gold

Explanation:

Given:

Mass of sample = 63.5 g

Mass of water = 60.2 g

Find:

Object

Computation:

Mass of water displaced = 63.5 g - 60.2 g

Mass of water displaced = 3.3 g

So, volume in water = 3.3 cm³

Density = Mass / Volume

Density = 63.5 g / 3.3

Density = 19.24

So,

Object ,must be gold.

7 0

2 years ago

Consider a metal single crystal oriented such that the normal to the slip plane and the slip direction are at angles of 60 and 3

Scilla [17]

Explanation:

Given that,

Angle by the normal to the slip α= 60°

Angle by the slip direction with the tensile axis β= 35°

Shear stress = 6.2 MPa

Applied stress = 12 MPa

We need to calculate the shear stress applied at the slip plane

Using formula of shear stress

$\tau=\sigma\cos\alpha\cos\beta$

Put the value into the formula

$\tau=12\cos60\times\cos35$

$\tau=4.91\ MPa$

Since, the shear stress applied at the slip plane is less than the critical resolved shear stress

So, The crystal will not yield.

Now, We need to calculate the applied stress necessary for the crystal to yield

Using formula of stress

$\sigma=\dfrac{\tau_{c}}{\cos\alpha\cos\beta}$

Put the value into the formula

$\sigma=\dfrac{6.2}{\cos60\cos35}$

$\sigma=15.13\ MPa$

Hence, This is the required solution.

3 0

3 years ago

A 40.0 kg beam is attached to a wall with a hi.nge and its far end is supported by a cable. The angle between the beam and the c

GalinKa [24]

The magnitude of the force that the beam exerts on the hi.nge will be,261.12N.

To find the answer, we need to know about the tension.

<h3>How to find the magnitude of the force that the beam exerts on the hi.nge?</h3>

Let's draw the free body diagram of the system using the given data.
From the diagram, we have to find the magnitude of the force that the beam exerts on the hi.nge.
For that, it is given that the horizontal component of force is equal to the 86.62N, which is same as that of the horizontal component of normal reaction that exerts by the beam on the hi.nge.

$N_x=86.62N$

We have to find the vertical component of normal reaction that exerts by the beam on the hi.nge. For this, we have to equate the total force in the vertical direction.

$N_y=F_V=mg-Tsin59\\$