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3 years ago

A box of books has _______inertia than a box of cotton balls.

2 answers:

4 0

Talking about a quantity of inertia is exactly the same as talking about a quantity of mass. So, if the boxes are anywhere near the same size, then the box of books has <u><em>more</em></u> inertia than the box of cotton balls, because books have more mass than an identical volume of cotton.

natulia [17]3 years ago

3 0

A box of books has more inertia than a box of cotton balls.

<u>Explanation:</u>

Inertia is related to the mass of the object. Inertia is usually defined as the resistance offered by the object in response to the movement or the velocity of the object.

The object will higher mass will have higher inertia as it should offer more resistance to remain in inertia and the object with lesser mas can offer smaller resistance comparatively. So here, the box of books has more inertia than a box of cotton balls.

You might be interested in

A planet orbits a start with the path shown below.

kvv77 [185]

PART a)

As we know that gravitational potential energy is given by the formula

$U = -\frac{Gm_1m_2}{r}$

here we can see that gravitational potential energy inversely varies with the distance

so here when distance from the sun is minimum then magnitude of gravitational potential energy is maximum while since it is given with negative sign so its overall value is minimum at that position

So gravitational potential energy is minimum at the nearest point and maximum at the farthest point

PART b)

Since we know that sum of kinetic energy and potential energy is constant here

so the points of minimum potential energy is the point where kinetic energy is maximum which means speed is maximum

So here speed is maximum at the nearest point

Part C)

since gravitational potential energy inversely varies with distance so it's graph will be like hyperbolic graph with distance

4 0

3 years ago

Suppose a 2.0kg bird is flying at a speed of of 1m/s. It’s kinetic energy would be what?

Ganezh [65]

As we know that the formula of kinetic energy will be

$KE = \frac{1}{2} mv^2$

now here we know that

m = 2 kg

v = 1 m/s

so from the above equation we have

$KE = \frac{1}{2}(2)(1^2)$

$KE = 1 J$

7 0

2 years ago

This is for physical science, need help :/

marusya05 [52]

1). From the frame of reference of a passenger on the airplane looking out of his window, the tree appears to be moving, at roughly 300 miles per hour toward the left of the picture.

2). The SI unit best suited to measuring the height of a building is the meter.

3). 'Displacement' is the straight-line distance and direction from the start-point to the end-point, regardless of the path that was followed to get there.

The ball started out in the child's hand, and it ended up 2 meters away from her in the direction of the wall. So the displacement of the ball from the beginning to the end of the story is: 2 meters toward the wall.

6 0

2 years ago

Free Fall: A rock is thrown directly upward from the edge of a flat roof of a building that is 56.3 meters tall. The rock misses

Slav-nsk [51]

Answer:

v₀₁= 5.525 m / s

Explanation

Freefall Formulas :

The sign of acceleration due to gravity (g) is positive if the object is going down and negative if the object is going up.

vf= v₀+gt

vf²=v₀²+2*g*h

h= v₀t+ (1/2)*g*t²

Where:

h: hight in meters (m)

t : time in seconds (s)

v₀: initial speed in m/s

vf: final speed in m/s

g: acceleration due to gravity in m/s²

Kinematics of the rock from the starting point with vo until it reaches its maximum height:

vf₁= v₀₁-gt₁ :vf₁ =0 to maximum height

0= v₀₁-gt₁

v₀₁ = g*t₁

t₁ =v₀₁ / g Equation (1)

vf₁²= v₀₁²-2*g*h₁ : vf₁ =0 to maximum height

0 = v₀₁²-2*g*h₁

2*g*h₁ = v₀₁²

h₁ = (v₀₁²)/(2g) Equation (2)

Kinematics of the rock when it falls from the maximum height until it touches the floor

h₂= v₀₂t+ (1/2)*g*t₂² v₀₂=vf₁ =0

h₂= 0+ (1/2)*g*t₂²

h₂= (1/2)*g*t₂² Equation (3)

Equation that relates h₁ to h₂

h₂= h₁ + 56.3 , h₁ = (v₀₁²)/(2g)

h₂= (v₀₁²)/(2g) + 56.3 Equation (4)

Equation that relates t₁ to t₂

t₁ + t₂ =4 s

t₂ =4 -t₁

t₂ =4 -(v₀₁/g )

Calculation of v₀₁

We replace equation 4 and equation 5 in equation 3

(v₀₁²)/(2g) + 56.3 = (1/2)*g*(4 -(v₀₁/g ) )²

(v₀₁²)/(2g) + 56.3 = (1/2)*g* (16 - 2*4*(v₀₁/g )+((v₀₁/g )²)

we eliminate (v₀₁²)/(2g) on both sides of the equation

56.3 = (1/2)*g* (16 - 2*4*(v₀₁/g ))

56.3 = 78.4 - 4*v₀₁

4*v₀₁ =78.4-56.3

v₀₁= (78.4-56.3) / ( 4)

v₀₁= 5.525 m / s

7 0

3 years ago

The international Space Station (ISS) orbits the Earth once every 90 mins at an altitude of 409 km. How high would it have to be

Oksi-84 [34.3K]

It would have to be 36,719 Km high in order to be to be in geosynchronous orbit.

To find the answer, we need to know about the third law of Kepler.

<h3>What's the Kepler's third law?</h3>

It states that the square of the time period of orbiting planet or satellite is directly proportional to the cube of the radius of the orbit.
Mathematically, T²∝a³

<h3>What's the radius of geosynchronous orbit, if the time period and altitude of ISS are 90 minutes and 409 km respectively?</h3>

The time period of geosynchronous orbit is 24 hours or 1440 minutes.
As the Earth's radius is 6371 Km, so radius of the ISS orbit= 6371km + 409 km = 6780km.
If T1 and T2 are time period of geosynchronous orbit and ISS orbit respectively, a1 and a2 are radius of geosynchronous orbit and ISS orbit, as per third law of Kepler, (T1/T2)² = (a1/a2)³
a1= (T1/T2)⅔×a2

= (1440/90)⅔×6780

= 43,090 km

Altitude of geosynchronous orbit = 43,090 - 6371= 36,719 km

Thus, we can conclude that the altitude of geosynchronous orbit is 36,719km.

Learn more about the Kepler's third law here:

brainly.com/question/16705471

#SPJ4

6 0

1 year ago