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11 days ago

A 8.00g sample of substance (substance, molar mass = 152.0 g/mol) was combusted in a bomb calorimeter with a heat capacity of 6.

21 kJ/°C. If the temperature rose from 25.0°C to 75.0°C, determine the value of ΔH°comb for substance.

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1. Determina el momento que produce una fuerza de 7 N tangente a una rueda de un metro de diámetro, sabiendo que el punto de apl

Sav [3153]

The torque resulting from a force is expressed as τ= F r into the blade. The force's moment is mathematically represented as τ = F x r, where the bold terms signify vectors. We can express this in terms of magnitude as τ = F r sin θ. In our scenario, since the force is tangential to the wheel, the angle between F and the radius is 90º, with sin 90 = 1. Hence, τ= F r. The torque's direction can be determined using the right-hand rule, where fingers curling in accordance with the torque direction from force to radius, with the thumb indicating the torque's direction. For a clockwise rotation, the fingers curl in that direction, and the thumb points inward toward the blade, indicating the direction of the torque.

5 0

2 months ago

Radiation emitted from human skin reaches its peak at λ = 940 µm. (a) What is the frequency of this radiation? (b) What type of

inna [3103]

Answer:

a) The frequency is 3.191x10^11.

b) The type of radiation is microwave radiation.

c) The energy of one quantum of this radiation amounts to 1.32x10^-3ev.

Explanation:

For further details and calculations, please refer to the image below.

8 0

3 months ago

A 2 000-kg sailboat experiences an eastward force of 3 000 N by the ocean tide and a wind force against its sails with a magnitu

ValentinkaMS [3465]

Respuesta:

La magnitud de la aceleración resultante es 2.2 $m/s^2$

Explicación:

La masa (m) del velero es 2000 kg

La fuerza que actúa sobre el velero debido a la marea del océano es $F_1$ = 3000N

Hacia el este significa que se da a lo largo de la dirección positiva del eje x

Entonces $F_{1x}$ = 3000N y $F_{1y}$ = 0

La fuerza del viento que actúa sobre el velero es $F_2$ 6000N dirigida hacia el noroeste, lo que significa a un ángulo de 45 grados sobre el eje negativo x

Luego

$F_{2x}$ = -(6000N) cos 45 grados = -4242.6 N

$F_{2y}$ = (6000N) cos 45 grados = 4242.6 N

Por lo tanto, la fuerza neta que actúa sobre el velero en la dirección x es

$F_x = F_{1x}+ F_{2x}$

= - 3000 N + 4242.6 N

= - 3000 N +4242.6 N

= 1242.6N

La fuerza neta que actúa sobre el velero en la dirección y es

= 0+ 4242.6N

= 4242.6N $F_y = F_{1y}+ F_{2y}$ La magnitud de la fuerza resultante =

Usando el teorema de Pitágoras de 1243 N y 4243 N

4420.8 N $\sqrt{(1242.6)^2 + (4242.6)^2$ F = ma

$\sqrt{(1544054.76) + (17999654.8)}$

$\sqrt{(19543709.5)^2}$

= 2.2

4 0

2 months ago

An 800-N billboard worker stands on a 4.0-m scaffold weighing 500 N and supported by vertical ropes at each end. How far would t

Maru [3345]

Answer:

2.5 m

Explanation:

Billboard worker's weight = 800 N

Number of ropes = 2

Length of scaffold = 4 m

Weight of scaffold = 500 N

Tension present in rope = 550 N

The total torques will be

$-800(4-x)-500\times 2+550\times 4=0\\\Rightarrow -800(4-x)=500\times 2-550\times 4\\\Rightarrow -800(4-x)=-1200\\\Rightarrow -x=\dfrac{1200}{800}-4\\\Rightarrow -x=-2.5\\\Rightarrow x=2.5\ m$

The worker is positioned at 2.5 m

7 0

3 months ago