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2 years ago

5

for a certain chemical reaction, the reactants contain 52 kj of potential energy and the products contain 32 kj how much energy

is absorbed or released by the reaction. A.84 kj is released B.20 kj is released C. 20 kj is absorbed D.84 kj is absorbed

2 answers:

Luden [163]2 years ago

7 0

B, is the answer. (20 kj is released)

mojhsa [17]2 years ago

4 0

Answer:

B. 20 kj is released

Explanation:

A.P.E.X. CST

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A nonconducting sphere has radius R = 1.29 cm and uniformly distributed charge q = +3.83 fC. Take the electric potential at the

zalisa [80]

Answer:

a) -2.516 × 10⁻⁴ V

b) -1.33 × 10⁻³ V

Explanation:

The electric field inside the sphere can be expressed as:

$E= \frac{kqr}{R^3}$

The potential at a distance can be represented as:

V(r) - V(0) = $-\int\limits^r_0 {\frac{kqr}{R^3} } \, dr^2$

V(r) - V(0) = $[\frac{qr^2}{8 \pi E_0R^3 }]$ ₀

V(r) = $-[\frac{qr^2}{8 \pi E_0R^3 }]$ ₀

Given that:

q = +3.83 fc = 3.83 × 10⁻¹⁵ C

r = 0.56 cm

= 0.56 × 10⁻² m

R = 1.29 cm

= 1.29 × 10⁻² m

E₀ = 8.85 × 10⁻¹² F/m

Substituting our values; we have:

$V(r)$ $= -\frac{(3.83*10^{-15}C)(0.560*10^{-2}m)^2}{8 \pi (8.85*10^{-12}F/m)(1.29*10^{-2}m)^3}$

$V(r)$ = -2.15 × 10⁻⁴ V

The difference between the radial distance and center can be expressed as:

V(r) - V(0) = $-\int\limits^R_0 {\frac{kqr}{R^3} } \, dr^2$

V(r) - V(0) = $[\frac{qr^2}{8 \pi E_0R^3 }]^R$

V(r) = $-\frac{qR^2}{8 \pi E_0R^3 }$

V(r) = $-\frac{q}{8 \pi E_0R }$

V(r) $= -\frac{(3.83*10^{-15}C)}{8 \pi (8.85*10^{-12}F/m)(1.29*10^{-2}m)}$

V(r) = -0.00133

V(r) = - 1.33 × 10⁻³ V

8 0

3 years ago

If 3600 j of work is done in 3.0 s what is the power<br>0.00083W<br>1200W<br>3600W<br>11000W

Viktor [21]

Answer:

1200 W

Explanation:

Power is given by the ratio between work done and time taken:

$P=\frac{W}{t}$

where W is the work done and t the time taken.

In this problem, W = 3600 J and t = 3.0 s. Therefore, the power in this exercise is

$P=\frac{3600 J}{3.0 s}=1200 W$

5 0

3 years ago

Read 2 more answers

A bowling ball that has a radius of 11.0 cm and a mass of 5.00 kg rolls without slipping on a level lane at 2.80 rad/s.

NemiM [27]

Answer:

$\dfrac{K_t}{K_r}=\dfrac{5}{2}$

Explanation:

Given that,

Mass of the bowling ball, m = 5 kg

Radius of the ball, r = 11 cm = 0.11 m

Angular velocity with which the ball rolls, $\omega=2.8\ rad/s$

To find,

The ratio of the translational kinetic energy to the rotational kinetic energy of the bowling ball.

Solution,

The translational kinetic energy of the ball is :

$K_t=\dfrac{1}{2}mv^2$

$K_t=\dfrac{1}{2}m(r\omega)^2$

$K_t=\dfrac{1}{2}\times 5\times (0.11\times 2.8)^2$

The rotational kinetic energy of the ball is :

$K_r=\dfrac{1}{2}I \omega^2$

$K_r=\dfrac{1}{2}\times \dfrac{2}{5}mr^2\times \omega^2$

$K_r=\dfrac{1}{2}\times \dfrac{2}{5}\times 5\times (0.11)^2\times (2.8)^2$

Ratio of translational to the rotational kinetic energy as :

$\dfrac{K_t}{K_r}=\dfrac{5}{2}$

So, the ratio of the translational kinetic energy to the rotational kinetic energy of the bowling ball is 5:2

4 0

3 years ago

Climatology is a subspecialty of what?

m_a_m_a [10]

Climatology is a subspecialty of Meterology.

3 0

3 years ago

Read 2 more answers

The law of conservation of momentum states that the total momentum of interacting objects does not change . This means the total

pickupchik [31]

Answer:

The momentum of an object is equal to the product of its mass and its velocity.

Explanation:

Consider an object of mass $m$ travelling at a velocity $\vec{v}$ . The momentum $\vec{p}$ of this object would be:

$\vec{p} = m \cdot \vec{v}$ .

For the law of conservation of momentum, consider two objects: object $\rm a$ and object $\rm b$ . Assume that these two objects collided with each other.

Let $m_{\rm a}$ and $m_{\rm b}$ denote the mass of the two objects.
Let $\vec{v}_{\rm a}(\text{initial})$ and $\vec{v}_{\rm b}(\text{initial})$ denote the velocity of the two object right before the interaction.
Let $\vec{v}_{\rm a}(\text{final})$ and $\vec{v}_{\rm b}(\text{final})$ denote the velocity of the two objects right after the interaction.

The momentum of the two objects right before the collision would be $m_{\rm a}\cdot \vec{v}_{\rm a}(\text{initial})$ and $m_{\rm b}\cdot \vec{v}_{\rm b}(\text{initial})$ , respectively.
The momentum of the two objects right after the collision would be $m_{\rm a}\cdot \vec{v}_{\rm a}(\text{final})$ and $m_{\rm b}\cdot \vec{v}_{\rm b}(\text{final})$ , respectively.

The sum of the momentum of the two objects would be:

$m_{\rm a}\cdot \vec{v}_{\rm a}(\text{initial}) + m_{\rm b}\cdot \vec{v}_{\rm b}(\text{initial})$ right before the collision, and
$m_{\rm a}\cdot \vec{v}_{\rm a}(\text{final}) + m_{\rm b}\cdot \vec{v}_{\rm b}(\text{final})$ right after the collision.

Assume that the system of these two objects is isolated. By the law of conservation of momentum, the sum of the momentum of these two objects should be the same before and after the collision. That is:

$m_{\rm a}\cdot \vec{v}_{\rm a}(\text{initial}) + m_{\rm b}\cdot \vec{v}_{\rm b}(\text{initial}) = m_{\rm a}\cdot \vec{v}_{\rm a}(\text{final}) + m_{\rm b}\cdot \vec{v}_{\rm b}(\text{final})$ .

4 0

3 years ago