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2 years ago

What is the estimated density of the golf ball? Record your answer to the nearest hundreth.

Physics

1 answer:

Hitman42 [59]2 years ago

7 0

The estimated density of the golf ball is 170 kg/m³

What is density?

Density is the ratio as Mass per unit Volume.

In displacement method,

First , we measuring the volume of water displaced by an object which tell us the volume of the object then we will use the physical balance to determine its mass.

Then calculate the density by dividing the mass by the volume.

Here, D = m/V

By displacement method , the estimated volume of golf ball is 100 cm³ and estimated mass is 600 g

Then ,

Density = 100 cm³ / 600 g = 0. 17 g/ cm³ = 170 kg/m³

So the estimated density of golf ball is 170 kg/m³

For more content related to Density visit here;

brainly.com/question/12006917

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A long distance runner sees the finish line and accelerates at a rate in 1.2 m/s2 for

Nuetrik [128]

Answer: he did travel 15 meters.

Explanation:

We have the data:

Acceleration = a = 1.2 m/s^2

Time lapes = 3 seconds

Initial speed = 3.2 m/s.

Then we start writing the acceleration:

a(t) = 1.2 m/s^2

now for the velocity, we integrate over time:

v(t) = (1.2 m/s^2)*t + v0

with v0 = 3.2 m/s

v(t) = (1.2 m/s^2)*t + 3.2 m/s

For the position, we integrate again.

p(t) = (1/2)*(1.2 m/s^2)*t^2 + 3.2m/s*t + p0

Because we want to know the displacementin those 3 seconds ( p(3s) - p(0s)) we can use p0 = 0m

Then the displacement at t = 3s will be equal to p(3s).

p(3s) = (1/2)*(1.2 m/s^2)*(3s)^2 + 3.2m/s*3s = 15m

6 0

3 years ago

If 27 J of work are needed to stretch a spring from 15 cm to 21 cm and 45 J are needed to stretch it from 21 cm to 27 cm, what i

kupik [55]

Answer:

9 cm.

Explanation:

The energy used for stretch the spring from $15 cm$ to $21 cm$ will be , $E_{1}=27J$

The energy used for stretch the spring from $21 cm$ to $27 cm$ will be , $E_{2}=45J$

using the energy of spring formula ,we find that

$27 = \frac{1}{2}K((21-L^{2})-(15-L^{2}))$

$45 = \frac{1}{2}K((27-L^{2})-(21-L^{2}))$

Dividing both the equation will get,

$\frac{3}{5}=\frac{(21-L)^{2}-(15-L)^{2}}{(27-L)^{2}-(21-L)^{2}}\\5((21-L)^{2}-(15-L)^{2})=3((27-L)^{2}-(21-L)^{2})\\3(729 - 54L + L^{2}- 441 + 42L - L^{2} ) = 5(441 - 42L + L^{2} - 225 + 30L - L^{2} )\\3(288 - 12L) = 5(216 - 12L)\\24L = 216\\L = 9 cm$

Therefore, the natural length of the spring is, 9 cm.

4 0

3 years ago

All objects emit what radiation

Eduardwww [97]

If an object is not at Absolute Zero, then it is
absorbing and radiating thermal (heat) energy.

5 0

3 years ago

At what time after being ejected is the boulder moving at a speed 20.7 m/s upward?

Svetlanka [38]

The time after being ejected is the boulder moving at a speed 20.7 m/s upward is 2.0204 s.

<h3>What is the time after being ejected is the boulder moving at a speed 20.7 m/s upward?</h3>

The motion of the boulder is a uniformly accelerated motion, with constant acceleration

a = g = -9.8 $$$m / s^2$

downward (acceleration due to gravity).

By using Suvat equation:

v = u + at

where: v is the velocity at time t

u = 40.0 m/s is the initial velocity

a = g = -9.8 $$$m/s^2$ is the acceleration

To find the time t at which the velocity is v = 20.7 m/s

Therefore,

$$t=\frac{v-u}{a}=\frac{20.7-40}{-9.8}=2.0204 \mathrm{~s}$

The time after being ejected is the boulder moving at a speed 20.7 m/s upward is 2.0204 s.

The complete question is:

A large boulder is ejected vertically upward from a volcano with an initial speed of 40.0 m/s. Ignore air resistance. At what time after being ejected is the boulder moving at 20.7 m/s upward?

To learn more about uniformly accelerated motion refer to:

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4 0

2 years ago

What is the correct answer?

sashaice [31]

Answer:

D) 4

Explanation:

Roots of a polynomial must be factors of the last term.

In this case, the factors of 6 are +1, -1, +2, -2, +3, -3, +6, -6. The only factor that doesn't show up, given the options, is 4. This means that D is the correct answer.

7 0

3 years ago