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3 years ago

Which of these experiments tests a chemical property of an object??

2 answers:

8 0

B. shining a bright light on the objects and testing for decomposition 

In explanation, chemical property is a characteristic of a certain substance came from an outcome due to chemical change or reaction. In the situation above, more specifically toxicity is involved in the chemical property/change. Hence, when the object is tested for decomposition. Like for an example of decomposition simply in metals, rusting. Rusting a process of degeneration of metals. Here it works the same. Toxicity is how much damage did a certain entity do to the object.

Ne4ueva [31]3 years ago

5 0

B is the correct answer

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A 425 g block is released from rest at height h0 above a vertical spring with spring constant k = 460 N/m and negligible mass. T

Morgarella [4.7K]

Answer:

(a) = +5.38m (b) = -5.38m (c) = 1.246m (d) = +0.3771m.

Explanation:

Initially the spring is at equilibrium,

Work done by all forces = change in kinetic energy

Work = ∇K.E

Work = Kf -Ki =0

Since the work done = 0 since the body is at rest.

W(spring) + W(gravity) = 0

W(spring) = -W(gravity)

Work done by the block on the spring = W(block/spring)

W(block/spring) + W(spring) = 0

W(spring) = -∫kx.dx

W(spring) = ½k(X²i - X²f) ; Xi =0, Xf = 15.3cm = 0.153m

W(spring) = -½* 460 * (0.153)²

W(spring) = - 5.38NM

Work done by block on spring = + 5.38NM

(b). Workdone by spring on the block = -5.38NM.

Note: This is so because the displacement of the force is in the opposite direction to the previous one since they counter each other to maintain equilibrium.

(C). W(spring) +W(gravity) = 0

½kx² + mg(h + x) = 0

-5.83 + mg(h + 0.153) =0

5.83 = 0.425*9.8 (h + 0.153)

5.83 = 4.165(h + 0.153)

H = 1.399 - 0.153

H = 1.246m

(D).

If the release height was 6ho

H = 6* 1.246m = 7.476m

W(spring) = W(gravity)

½kx² = mg(7.476 + x)

Note: At maximum compression, the blocks would be at rest.

½Kx² = mg(h + x)

½ * 460 * x² = 0.425 * 9.8 * (7.476 + x)

230x² = 4.165 (7.476 + x)

230x² = 31.137 + 4.165x

230x² - 4.165x - 31.137 = 0

Solving the quadratic equation ( i would suggest you use formula method for easy navigation of the variables)

X = + 0.3771m or -0.3589m

But we can't have a negative compression value,

X = + 0.3771m

7 0

3 years ago

Read 2 more answers

If the volume is held constant, what happens to the pressure of a gas as temperature is decreased? Explain.

Lady bird [3.3K]

Answer:Decreases

Explanation:

Given

Volume is held constant that is it is a isochoric process.

We know that

PV=nRT

as n,V& R are constant therefore only variables are

P & T

so $\frac{P_1}{T_1}=\frac{P_2}{T_2}$

$\frac{P_1}{P_2}=\frac{T_1}{T_2}$

As $T_1$ is decreasing therefore Pressure must also decrease so that ratio remains constant.

6 0

3 years ago

Una barra de plata de 335.2 g con una temperatura de 100 ºC se introduce un calorímetro de aluminio de 60 g de masa que contiene

sdas [7]

Respuesta:

0,0560 cal / gºC.

Explicación:

Cantidad de calor; (Q)

Q = mcΔt; Δt = t2 - t1

m = masa, c = capacidad calorífica específica; Δt = cambio de temperatura

c de agua = 1 cal / gºC

c de aluminio = 0,22 cal / gºC

QTotal = Q de agua + Q de aluminio

Q de agua = 450 * 1 * (26 - 23) = 1350 cal

Q de aluminio = 60 * 0.22 * (26 - 23) = 39.6 cal

QTotal = 1350 + 39,6 = 1389,6 cal

Calor perdido = calor ganado

QTotal = calor perdido

- 1389,6 = 335,2 * c * (26 - 100)

-1389,6 = −24804,8 * c

c = 1389,6 / 24804,8

c = 0,056021 cal / gºC.

Capacidad calorífica específica de la plata = 0,0560 cal / gºC.

8 0

3 years ago

M84, M87, and NGC 4258 all have accretion disks around their central black holes for which the rotational velocities have been m

givi [52]

Answer:

For M84:

M = 590.7 * 10³⁶ kg

For M87:

M = 2307.46 * 10³⁶ kg

Explanation:

1 parsec, pc = 3.08 * 10¹⁶ m

The equation of the orbit speed can be used to calculate the doppler velocity:

$v = \sqrt{\frac{GM}{r} }$

making m the subject of the formula in the equation above to calculate the mass of the black hole:

$M = \frac{v^{2} r}{G}$ .............(1)

For M84:

r = 8 pc = 8 * 3.08 * 10¹⁶

r = 24.64 * 10¹⁶ m

v = 400 km/s = 4 * 10⁵ m/s

G = 6.674 * 10⁻¹¹ m³/kgs²

Substituting these values into equation (1)

$M = \frac{( 4*10^{5}) ^{2} *24.64* 10^{16} }{6.674 * 10^{-11} }$

M = 590.7 * 10³⁶ kg

For M87:

r = 20 pc = 20 * 3.08 * 10¹⁶

r = 61.6* 10¹⁶ m

v = 500 km/s = 5 * 10⁵ m/s

G = 6.674 * 10⁻¹¹ m³/kgs²

Substituting these values into equation (1)

$M = \frac{( 5*10^{5}) ^{2} *61.6* 10^{16} }{6.674 * 10^{-11} }$

M = 2307.46 * 10³⁶ kg

The mass of the black hole in the galaxies is measured using the doppler shift.

The assumption made is that the intrinsic velocity dispersion is needed to match the line widths that are observed.

3 0

3 years ago

In this football play the quarterback keeps the ball and goes straight ahead, behind the blocking of the center and the guards.

oksano4ka [1.4K]

Answer:

Explanation:

3 0

3 years ago