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sertanlavr [38]

2 years ago

8

How is the density of water different from the densities of most other substances?

1 answer:

Ket [755]2 years ago

4 0

Answer:

In practical terms, density is the weight of a substance for a specific volume. The density of water is roughly 1 gram per milliliter but, this changes with temperature of if there are substances dissolved in it. Ice is less dense than liquid water which is why your ice cubes float in your glass.

Explanation:

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"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

Alja [10]

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

6 0

2 years ago

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Standing still, Bruce, the quarterback, gets tackled by Biff, the 90.0-kg tackle, who is traveling at 7.0 m/s. Upon collision, B

tatyana61 [14]

$\\ \sf\longmapsto \Delta P=P$

$\\ \sf\longmapsto m1v1=m2v2$

$\\ \sf\longmapsto 90(7)=m2(10)$

$\\ \sf\longmapsto 10m2=630$

$\\ \sf\longmapsto m2=\dfrac{630}{10}$

$\\ \sf\longmapsto m2=63kg$

Bruces mass is 63kg

3 0

2 years ago

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There are only 2 GENDERS and it proven with science

VikaD [51]

Answer:

https://young.scot/get-informed/national/gender-identity-terms

Explanation:

6 0

2 years ago

The 3 important steps to a perfect forehand in tennis are : perfect swing, type of racket, and correct finish

Alex17521 [72]

Answer:

False

Explanation:

The 3 important steps to a perfect forehand in tennis are Preparation,Backswing and Swing and contact.

3 0

2 years ago

A car moving at 10.0 m/s encounters a depression in the road that has a circular cross-section with a radius of 30.0 m. What is

Dennis_Churaev [7]

Answer:

F = 789 Newton

Explanation:

Given that,

Speed of the car, v = 10 m/s

Radius of circular path, r = 30 m

Mass of the passenger, m = 60 kg

To find :

The normal force exerted by the seat of the car when the it is at the bottom of the depression.

Solution,

Normal force acting on the car at the bottom of the depression is the sum of centripetal force and its weight.

$N=mg+\dfrac{mv^2}{r}$

$N=m(g+\dfrac{v^2}{r})$

$N=60\times (9.81+\dfrac{(10)^2}{30})$

N = 788.6 Newton

N = 789 Newton

So, the normal force exerted by the seat of the car is 789 Newton.

6 0

2 years ago