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Norma-Jean [14]

4 years ago

5

Object A had a density of 10 g/cm3 and a mass of 9 g. Object B had a mass of 4 g. Which object has a greater volume and by how m

uch?

1 answer:

svlad2 [7]4 years ago

8 0

The volume of Object-A is

(9 g) / (10 g/cm³) = 0.9 cm³

We have no way to find the volume of Object-B from its mass,
because we don't know its density. So we can't compare their
volumes.

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HELP!!! Why is plasma considered the most common state of matter in the universe?

olasank [31]

Clue : There is plasma in sun

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3 years ago

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A Meteorite Strikes On October 9, 1992, a 27-pound meteorite struck a car in Peekskill, NY, leaving a dent 22 cm deep in the tru

vagabundo [1.1K]

Answer:

Acceleration of the meteorite, $a=-38409.09\ m/s^2$

Explanation:

It is given that,

A Meteorite after striking struck a car, v = 0

Initial speed of the Meteorite, u = 130 m/s

Distance covered by Meteorite, s = 22 cm = 0.22 m

We need to find the magnitude of its deceleration. It can be calculated using the third equation of motion as :

$a=\dfrac{v^2-u^2}{2s}$

$a=\dfrac{-(130)^2}{2\times 0.22}$

$a=-38409.09\ m/s^2$

So, the deceleration of the Meteorite is $-38409.09\ m/s^2$ . Hence, this is the required solution.

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3 years ago

A 40-kg box is being pushed along a horizontal smooth surface. The pushing force is 15 n directed at an angle of 15Â° below the

Kobotan [32]

Answer:

Acceleration of the crate is 0.362 m/s^2.

Explanation:

Given:

Mass of the box, m = 40 kg

Applied force, F = 15 N

Angle at which the force is applied, $(\theta)$ = 15°

We have to find the magnitude of the acceleration.

Let the acceleration be "a".

FBD is attached with where we can see the horizontal and vertical component of force.

⇒ $F_x=Fcos(\theta)$ and ⇒ $F_y=Fsin (\theta)$

⇒ $F_x=15cos(15)$ ⇒ $F_y=15sin (15)$

⇒ Applying concept of forces.

⇒ $\sum F_x=F_n_e_t =F-f$

⇒ $F_n_e_t =F-f$

⇒ $ma =F-f$ <em> ...Newtons second law Fnet = ma</em>

⇒ $a =\frac{F-f}{m}$

⇒ Plugging the values.

⇒ $a =\frac{15cos(15)-0}{40}$ <em>...f is the friction which is zero here.</em>

⇒ $a =\frac{14.48}{40}$

⇒ $a=0.362\ ms^-^2$

Magnitude of the acceleration of the crate is 0.362 m/s^2.

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3 years ago

If the swimmer starts at rest, slides without friction, and descends through a vertical height of 2.41 m

AveGali [126]

Answer:

6.88 m/s

Explanation:

The Conservation of Energy states that:

Initial Kinetic Energy + Initial Potential Energy = Final Kinetic Energy + Final Potential Energy

So we can write

$mgh_{i}+\frac{1}{2}mv_{i} ^{2}=mgh_{f}+\frac{1}{2}mv_{f} ^{2}$

We can cancel the common factor of $m$ which leaves us with

$gh_{i}+\frac{1}{2}v_{i} ^{2}=gh_{f}+\frac{1}{2}v_{f} ^{2}$

Lets solve for $v_f$

$gh_{i}+\frac{v_{i} ^{2}}{2}=gh_{f}+\frac{v_{f} ^{2}}{2}$

Subtract $gh_f$ from both sides of the equation.

$gh_{i}+\frac{v_{i} ^{2}}{2}-gh_{f}=\frac{v_{f} ^{2}}{2}$

Multiply both sides of the equation by 2.

$2(gh_{i}+\frac{v_{i} ^{2}}{2}-gh_{f})={v_{f} ^{2}$

Simplify the left side.

Apply the distributive property.

$2(gh_{i})+2\frac{v_{i} ^{2}}{2}+2(-gh_{f})={v_{f} ^{2}$

Cancel the common factor of 2.

$2gh_{i}+v_{i} ^{2}-2gh_{f}={v_{f} ^{2}$

Take the square root of both sides of the equation to eliminate the exponent on the right side.

${v_{f}=\sqrt{2gh_{i}+v_{i} ^{2}-2gh_{f}}$

We are given $g,v_{i},h_{i},h_{f}$ .

We can now solve for the final velocity.

${v_{f}=\sqrt{(2*9.81*2.41)+(0^{2})-(2*9.81*0)$

Anything multiplied by 0 is 0.

${v_{f}=\sqrt{2*9.81*2.41$

${v_{f}=\sqrt{47.2842$

$v_f=6.88$

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1 year ago

Let n represent the number of similar resistors in a circuit. Each resistor has a resistance of r. When the resistors are connec

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4 years ago