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Dima020 [189]
3 years ago
14

"A block of metal weighs 40 N in air and 30 N in water. What is the buoyant force on the block due to the water? The density of

water is 1000 kg/m3."
Physics
2 answers:
Alja [10]3 years ago
6 0

Answer:

buoyant force on the block due to the water= 10 N

Explanation:

We know that

buoyant force(F_B) on a block= weight of the block in air (actual weight) - weight of block in water.

Given:

A block of metal weighs 40 N in air and 30 N in water.

F_B =  40-30= 10 N

therefore,  buoyant force on the block due to the water= 10 N

Yuliya22 [10]3 years ago
3 0

Answer:

10 N

Explanation:

We are given that

Weight of block of metal in air=40 N

Actual weight of object=Weight of object in air

Actual weight of block of metal=40 N

Apparent weight of block of metal=Weight of block in water

Weight of block in water=30 N

Apparent weight of block=30 N

Density of water=1000 kg /cubic meter

We have to find the buoyant force on the block due to water.

Buoyant force=Actual weight-apparent weight

Buoyant force on the block due to water=40-30=10 N

Hence, the buoyant force on the block due to the  water=10 N

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I believe the answer is the second one.
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3 years ago
A sample of chlorine has two naturally occurring isotopes. The isotope Cl-35 (mass 35.0 amu) makes up 75.8% of the sample, and t
irinina [24]

Answer:

M_{av}=35.521amu

Explanation:

As in any sample you will have 75.8% of Cl-35 iosotopes and 24.3% of Cl-37 iosotopes you can get the average atomic mass as:

M_{av}=(35amu*75.8+37amu*24.3)/100=35.521amu

4 0
3 years ago
The note created by a flute will increase the speed of sound increases. When a marching band goes outside on a cold day, what wo
alexgriva [62]

A).

It would decrease because the speed of sound and temperature are proportional.

4 0
3 years ago
Read 2 more answers
A 50.-kilogram rock rolls off the edge of a cliff. if it is traveling at a speed of 24.2 m/s when it hits the ground, what is th
ElenaW [278]

The correct answer to the question is : 29.88 m.

EXPLANATION :

As per the question, the mass of the rock m = 50 Kg.

The rock is rolling off the edges of the cliff.

The final velocity of the rock when it hits the ground v = 24 .2 m/s.

Let the height of the cliff is h.

The potential energy gained by the rock at the top of the cliff = mgh.

Here, g is known as acceleration due to gravity, and g = 9.8\ m/s^2

When the rock rolls off the edge of the cliff, the potential energy is converted into kinetic energy.

When the rock hits the ground, whole of its potential energy is converted into its kinetic energy.

The kinetic energy of the rock when it touches the ground is given as -

                Kinetic energy K.E = \frac{1}{2}mv^2.

From above we know that -

   Kinetic energy at the bottom of the cliff = potential energy at a height h

                 \frac{1}{2}mv^2=\ mgh

                ⇒ v^2=\ 2gh

                ⇒ h=\ \frac{v^2}{2g}

                ⇒ h=\ \frac{(24.2)^2}{2\times 9.8}

                ⇒ h=\ 29.88\ m

Hence, the height of the cliff is 29.88 m

             


5 0
3 years ago
Two long, straight wires both carry current to the right, are parallel, and are 25 cm apart. Wire one carries a current of 2.0 A
Artemon [7]

Answer:

x = 7.14 meters

Explanation:

It is given that,

Current in wire 1, I_1=2\ A

Current in wire 2, I_2=5\ A  

Distance between parallel wires, r = 25 cm

Let at P point the net magnetic field equal to 0. The magnetic field at a point midway between the is given by :

B=\dfrac{\mu_oI}{2\pi r}

Let the distance is x from wire 1. So,

\dfrac{I_1}{r}=\dfrac{I_2}{(0.25-r)}

\dfrac{2}{r}=\dfrac{5}{(25-r)}

x=\dfrac{50}{7}\ m

x = 7.14 meters

So, the magnetic field will be 0 at a distance of 7.14 meters from wire 1. Hence, this is the required solution.

6 0
3 years ago
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