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3 years ago

What is true of the mass and volume of all the sinking objects

Physics

2 answers:

kumpel [21]3 years ago

5 0

Answer:

Idk

Explanation:

algol133 years ago

4 0

The mass of any sinking object, when divided by its volume, results in a number
<span>that is greater than the density of the fluid in which the object is sinking.

THE ABOVE STATEMENT IS TRUE OF THE MASS AND VOLUME OF ALL SINKING OBJECT

We have,
Density = Mass / volume

For objects sinking in a liquid,
Density > Mass / Volume</span>

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Inertia is the property that every material object has that causes objects to resist changes in its state of motion

jonny [76]

True :) good luck on your quiz

3 0

3 years ago

Read 2 more answers

two test cars of equal mass moving towards each other collide on a horiontal frictionless surface. Before the collision, car A h

LiRa [457]

Answer:

4 m/s

Explanation:

m1 = m2 = m

u1 = 20 m/s, u2 = - 12 m/s

Let the speed of composite body is v after the collision.

Use the conservation of momentum

Momentum before collision = momentum after collision

m1 x u1 + m2 x u2 = (m1 + m2) x v

m x 20 - m x 12 = (m + m) x v

20 - 12 = 2 v

8 = 2 v

v = 4 m/s

Thus, the speed of teh composite body is 4 m/s.

4 0

3 years ago

In a tug-of-war game on one campus, 15 students pull on a rope at both ends in an effort to displace the central knot to one sid

mojhsa [17]

Answer:

Net pull = 110 N to the left

Explanation:

Group the different pulls according to the direction (right or left)

2 pull 196 N each to the right

4 pull 98 N each to the left

5 pull 62 N each to the left

3 pull 150 N each to the right

1 pull 250 N to the left

Since positive direction is to the right, the pulls to the left will have a minus (-)

$Net Force= 2(196)+4(-98)+5(-62)+3(150)+1(-250) \\Net Force = -110$

The resulting force is negative, meaning the direction is to the left

6 0

3 years ago

An object is 10 cm from thé mirror, its height is 1 cm and thé focal length is 5 cm. What is thé distance from thé mirror? S1= _

Viefleur [7K]

Note: I assume the mirror is concave, so that its focal length is positive (it is not specified in the text)

1a) We can use the mirror equation to find the distance of the image from the mirror:
$\frac{1}{f}= \frac{1}{p}+ \frac{1}{q}$
where
$f=5 cm$ is the focal length
$p=10 cm$ is the distance of the object from the mirror
$q$ is the distance of the image from the mirror.

Rearranging the equation, we find
$\frac{1}{q}= \frac{1}{f}- \frac{1}{p}= \frac{1}{5}- \frac{1}{10}= \frac{1}{10 cm}$
so, the distance of the image from the mirror is $q=10 cm$ .

1b) The image height is given by the magnification equation:
$\frac{h_i}{h_o}=- \frac{p}{q}$
where $h_i$ is the heigth of the image and $h_o=1 cm$ is the height of the object. By rearranging the equation and using p and q, we find
$h_i=-h_o \frac{p}{q}=-(1 cm) \frac{10 cm}{10 cm}=-1 cm$
and the negative sign means the image is inverted.

2) As before, we can find the distance of the image from the mirror by using the mirror equation:
$\frac{1}{f}= \frac{1}{p}+ \frac{1}{q}$
Rearranging it, we find
$\frac{1}{q}= \frac{1}{f}- \frac{1}{p}= \frac{1}{2}- \frac{1}{10}= \frac{4}{10 cm}$
so, the distance of the image from the mirror is
$q= \frac{10}{4}cm= 2.5 cm$

3) As before, we find the distance of the image from the mirror by using the mirror equation:
$\frac{1}{f}= \frac{1}{p}+ \frac{1}{q}$
Rearranging it, we find
$\frac{1}{q}= \frac{1}{f}- \frac{1}{p}= \frac{1}{2}- \frac{1}{10}= \frac{4}{10 cm}$
so, the distance of the image from the mirror is
$q= \frac{10}{4}cm= 2.5 cm$

And now we can use the magnification equation to find the image height:
$\frac{h_i}{h_o}=- \frac{p}{q}$
Rearranging it, we find
$h_i=-h_o \frac{p}{q}=-(3cm) \frac{10 cm}{2.5 cm}=-12 cm$
and the negative sign means the image is inverted.

5 0

3 years ago

5 What is the angular displacement at the end of the 25-mm-diameter shaft and the linear displacement of point A of Figure P5.5

Anvisha [2.4K]

5 What is the angular displacement at the end of the 25-mm-diameter shaft and the linear displacement of point A of Figure P5.5

<h3>What is displacement ?</h3>

A displacement is a vector in geometry and mechanics that has a length equal to the shortest distance between a point P's initial and final positions. It calculates the length and angle of the net motion, or total motion, in a straight line from the starting point to the destination of the point trajectory. The translation that links the starting point and the ending point can be used to spot a displacement.

The final location xf of a point relative to its beginning position xi, or a relative position (derived from the motion), is another way to express a displacement. The difference between the end and beginning positions can be used to define the equivalent displacement vector

To learn more about displacement from the given link:

brainly.com/question/321442

#SPJ4

3 0

1 year ago