What is Shortening melting

A cat’s crinkle ball toy of mass 15g is thrown straight up with an initial speed of 3m/s . Assume in this problem that air drag

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Answer:

(a)0.0675 J

(b)0.0675 J

(c)0.0675 J

(d)0.0675 J

(e)-0.0675 J

(f)0.459 m

Explanation:

15g = 0.015 kg

(a) Kinetic energy as it leaves the hand

$E_k = \frac{mv^2}{2} = \frac{0.015*3^2}{2} = 0.0675 J$

(b) By the law of energy conservation, the work done by gravitational energy as it rises to its peak is the same as the kinetic energy as the ball leave the hand, which is 0.0675 J

(c) The change in potential energy would also be the same as 0.0675J in accordance with conservation law of energy.

(d) The gravitational energy at peak point would also be the same as 0.0675J

(e) In this case as the reference point is reversed, we would have to negate the original potential energy. So the potential energy as the ball leaves hand is -0.0675J

(f) Since at the maximum height the ball has potential energy of 0.0675J. This means:

mgh = 0.0675

0.015*9.81h = 0.0675

h = 0.459 m

The ball would reach a maximum height of 0.459 m

4 0

3 years ago

Consider an elevator with a table and a book on top of the table. The mass of the table is 10kg and the mass of the book is 2kg.

Step2247 [10]

Newton's second law allows us to find the force of the block on the table is 126 N

Newton's second law says that the net force is proportional to the product of the mass and the acceleration of the body

Σ F = m a

Where the bold letters indicate vectors, m is the mass and the acceleration of the body

A free body diagram is a diagram where the forces are represented without the details of the bodies, in the attached we can see a free body diagram of the system.

Let's start by finding the acceleration of the elevator with kinematics

v = v₀ + a t

a = $\frac{v-v_o}{a}$

Where v and v₀ are the current and initial velocity, respectively, at acceleration and t is the time

a = $\frac{8-1}{2}$

a = 3.5 m / s²

Let's write Newton's second law for each body

The book

N₂ - W₂ + N₁ = m a

Table

N₁ - W₁ - W₂ = M a

W₁ = Mg

W₂ = mg

N₁ = (M + m) g + M a

N₁ = (10 + 2) 9.8 + 10 3.5

N₁ = 152.6 N

This is the reaction of the earth to the support of the block and the table

N₂ = ma + m g - N₁

N₂ = m ( a +g) - N₁

N₂ = 2 (3.5 + 9.8) - 152.6

N₂ = 26.6 - 152.6

N₂ = -126 N

The negative sign indicates that the direction is opposite to the one assigned, this is the action of the block on the table.

In conclusion using Newton's second law we can find the forces of the block on the table is 126 N

Learn more here: brainly.com/question/19860811

4 0

3 years ago

Se lanza verticalmente una esfera con una rapidez de 30m/se. Determinar la rapidez de la esfera a una altura de 40m (g=10m/s2)

sergeinik [125]

$v^2-{v_0}^2=2a(x-x_0)$

dónde $v$ es la velocidad de la esfera, $v_0$ es suya velocidad inicial, $a=-g$ la aceleración debida a la gravedad, $x$ la posición, y $x_0$ la posición inicial. Tomamos $x_0=0\,\mathrm m$ a referirse a la posición de la esfera en el momento que la esfera fue lanzada.

Entonces

$v^2-\left(30\,\dfrac{\mathrm m}{\mathrm s}\right)^2=2\left(-10\,\dfrac{\mathrm m}{\mathrm s^2}\right)(40\,\mathrm m)$

$\implies v^2=100\,\dfrac{\mathrm m^2}{\mathrm s^2}\implies v=\pm10\,\dfrac{\mathrm m}{\mathrm s}$

Esto nos dice que la esfera alcanza una altura de 40 m en dos momentos - una vez hacia arriba y una vez hacia abajo. Sin embargo, independientemente del signo de la velocidad, sabemos que suya magnitud es 10 m/s, y así tenemos una rapidez de 10 m/s también en ambos momentos.

4 0

3 years ago

Magnetic pole reversal worksheet

Likurg_2 [28]

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8 0

3 years ago

Calculate the wavelength of a neutron that has a velocity of 200. cm/s. (the mass of a neutron = 1.675 × 10–27 kg and h = 6.63 ×

Licemer1 [7]

<span>To find the wavelength of a neutron can be calculated by using the formula Wavelength=h/m x v Where h is planck's constant m=mass of neutron v= velocity of the particle By substituting the given values Wavelength= 6.63 × 10–34 j s / 1.675 × 10–27 kg x 2 m/s^-1 Wavelength of a neutron=1.979 x 10^-7 m</span>

6 0

3 years ago

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