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2 years ago

14

A 1.2 KG rubber ball is being thrown in the air if the ball is traveling at 2.0 M/S when it is 3.0 M off the ground what is the

total energy

1 answer:

Vitek1552 [10]2 years ago

6 0

Answer:

37.7 J

Hope this helps! (see pictures)

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A 0.29 kg particle moves in an xy plane according to x(t) = - 19 + 1 t - 3 t3 and y(t) = 20 + 7 t - 9 t2, with x and y in meters

Artist 52 [7]

Answer:

Part a)

$F = 7.76 N$

Part b)

$\theta = -137.7 degree$

Part c)

$\theta = -127.7 degree$

Explanation:

As we know that acceleration is rate of change in velocity of the object

So here we know that

$x = -19 + t - 3t^3$

$y = 20 + 7t - 9t^2$

Part a)

differentiate x and y two times with respect to time to find the acceleration

$a_x = \frac{d^2}{dt^2}(-19 + t - 3t^3)$

$a_x = \frac{d}{dt}(0 +1 - 9t^2)$

$a_x = -18t$

$a_y = \frac{d^2}{dt^2}(20 + 7t - 9t^2)$

$a_y = \frac{d}{dt}(0 +7 - 18t)$

$a_y = -18$

Now the acceleration of the object is given as

$\vec a = (-18t)\hat i + (-18)\hat j$

at t= 1.1 s we have

$\vec a = -19.8 \hat i - 18 \hat j$

now the net force of the object is given as

$\vec F = m\vec a$

$\vec F = (0.29 kg)(-19.8 \hat i - 18 \hat j)$

$\vec F = -5.74 \hat i - 5.22 \hat j$

now magnitude of the force will be

$F = \sqrt{5.74^2 + 5.22^2} = 7.76 N$

Part b)

Direction of the force is given as

$tan\theta = \frac{F_y}{F_x}$

$tan\theta = \frac{-5.22}{-5.74}$

$\theta = -137.7 degree$

Part c)

For velocity of the particle we have

$v_x = \frac{dx}[dt}$

$v_x = (0 +1 - 9t^2)$

$v_y = \frac{dy}{dt}$

$v_y = (0 +7 - 18t)$

now at t = 1.1 s

$\vec v = -9.89\hat i - 12.8 \hat j$

now the direction of the velocity is given as

$\theta = tan^{-1}(\frac{v_y}{v_x})$

$\theta = tan^{-1}(\frac{-12.8}{-9.89})$

$\theta = -127.7 degree$

7 0

3 years ago

Honeybees accumulate charge as they fly, and they transfer charge to the flowers they visit. Honeybees are able to sense electri

Vilka [71]

Answer:

ΔE> E_minimo

We see that the field difference between these two flowers is greater than the minimum field, so the bee knows if it has been recently visited, so the answer is if it can detect the difference

Explanation:

For this exercise let's use the electric field expression

E = k q / r²

where k is the Coulomb constant that is equal to 9 109 N m² /C², q the charge and r the distance to the point of interest positive test charge, in this case the distance to the bee

let's calculate the field for each charge

Q = 24 pC = 24 10⁻¹² C

E₁ = 9 10⁹ 24 10⁻¹² / 0.20²

E₁ = 5.4 N / C

Q = 32 pC = 32 10⁻¹² C

E₂ = 9 10⁹ 32 10⁻¹² / 0.2²

E₂ = 7.2 N / C

let's find the difference between these two fields

ΔE = E₂ -E₁

ΔE = 7.2 - 5.4

ΔE = 1.8 N / C

the minimum detection field is

E_minimum = 0.77 N / C

ΔE> E_minimo

We see that the field difference between these two flowers is greater than the minimum field, so the bee knows if it has been recently visited, so the answer is if it can detect the difference

8 0

2 years ago

How do you know what state of matter the particles are in?

lianna [129]

Answer: The amount of energy consisted in the molecules determines the state of matter.

Explanation:

7 0

2 years ago

The scale model of an airplane is 1:130, where 1 inch on the model is equal to 130 inches on the actual airplane. Which statemen

Delicious77 [7]

Answer:

The \: parts \: of \: the \: model \\ airplane \: are \: in \: the \\ same \: proportions \: as \: the \\ actual \: airplane.

3 0

2 years ago

If your heart is beating at 76. 0 beats per minute, what is the frequency of your heart's oscillations in hertz?

nikitadnepr [17]

If the heart is beating at 76. 0 beats per minute, the Frequency of heart's oscillations in hertz is 1.25 Hertz.

The Frequency is defined as the number of oscillations completed in 1s.

Our heart beats continuously giving rise to the frequency of the heart.

Measuring frequency would be possible if we measure the time in seconds.

75 beats per minute means that 75 beats occur in 60s.

Frequency = Number of beats / Total time taken

Frequency = 75 / 60

Frequency = 1.25 Hz

Hence, number of beats per second is 1.25 Beats.

Beats per second or Cycles per second is same as the frequency of oscillation called as Hertz (Hz)

Hence, frequency of the heart oscillation is 1.25 Hertz (Hz).

Learn more about Frequency here, brainly.com/question/14316711

#SPJ4

5 0

1 year ago