All categories

3 years ago

How much heat is required to completely boil away 0.200 kg of water (ΔHv = 2.26 × 103 kJ/kg)?

2 answers:

alexandr1967 [171]3 years ago

4 0

We will use the formula:
Q = ml
where Q is the heat, m is the mass of substance, and l is the latent heat of vaporization of the substance.

Q = 0.2 x 2.26 x 10³
Q = 452 kJ

ratelena [41]3 years ago

3 0

<span>The heat required to completely boil away 0.200 kg of water is 4.52 × 102 kJ.
</span>

You might be interested in

Which of the following statements about Gaussʹs law are correct? (There may be more than one correct choice.) Question 3 options

irina1246 [14]

Answer:

If a Gaussian surface is completely inside an electrostatic conductor, the electric field must always be zero at all points on that surface.

Explanation:

Option A is incorrect because, given this case, it is easier to calculate the field.

Option B is incorrect because, in a situation where the surface is placed inside a uniform field, option B is violated

Option C is also incorrect because it is possible to be a field from outside charges, but there will be an absence of net flux through the surface from these.

Hence, option D is the correct answer. "If a Gaussian surface is completely inside an electrostatic conductor, the electric field must always be zero at all points on that surface."

3 0

2 years ago

What type on msd causes inflammation of the tendons that correct bones to muscles

svp [43]

I believe it would be Tendonitis

8 0

3 years ago

In a laboratory, you determine that the density of a certain solid is 5.23× 10 −6 kg/m m 3 . convert this density into kilograms

tamaranim1 [39]

Density is given as

$\rho = 5.23 * 10^{-6} kg/mm^3$

now we have to convert this density into $kg/m^3$

now we have

$1 m = 1000 mm$

$\rho = 5.23 * 10^{-6} \frac{kg}{(1*10^-3 m)^3}$

$\rho = 5.23 * 10^{-6} \frac{1*10^9 kg}{m^3}$

$\rho = 5.23 * 10^3 kg/m^3$

$\rho = 5230 kg/m^3$

8 0

3 years ago

Which expression allows you to determine the mechanical advantage of an inclined plane?

Leya [2.2K]

Answer:

Explanation:

The mechanical advantage of an inclined plane can be determined using different variables. In this case, the geometry of the setup is relevant. The advantage is proportional to the length of the plane, and inversely proportional to the height: it is the ratio (length) / (height) of the plane. For example, given a desired, fixed height, a long inclined plane gives you a bigger mechanical advantage than a short inclined plane. In this example, pushing an object up the long plane will require a smaller force, than it would on the short plane.

Strictly speaking, (D) would also "allow you to determine the mechanical advantage" because you could simply invert the ratio listed under (D). However, (C) is the best, direct, answer.

4 0

2 years ago

Two manned satellites approach one another at a relative velocity of v=0.190 m/s, intending to dock. The first has a mass of m1=

drek231 [11]

Answer:

Their final relative velocity is 0.190 m/s

Explanation:

The relative velocity of the satellites, v = 0.190 m/s

The mass of the first satellite, m₁ = 4.00 × 10³ kg

The mass of the second satellite, m₂ = 7.50 × 10³ kg

Given that the satellites have elastic collision, we have;

$v_2 = \dfrac{2 \cdot m_1}{m_1 + m_2} \cdot u_1 - \dfrac{m_1 - m_2}{m_1 + m_2} \cdot u_2$

$v_2 = \dfrac{ m_1 - m_2}{m_1 + m_2} \cdot u_1 + \dfrac{2 \cdot m_2}{m_1 + m_2} \cdot u_2$

Given that the initial velocities are equal in magnitude, we have;

u₁ = u₂ = v/2

u₁ = u₂ = 0.190 m/s/2 = 0.095 m/s

v₁ and v₂ = The final velocities of the satellites

We get;

$v_1 = \dfrac{2 \times 4.0 \times 10^3}{4.0 \times 10^3 + 7.50 \times 10^3} \times 0.095 - \dfrac{4.0 \times 10^3- 7.50\times 10^3}{4.0 \times 10^3+ 7.50\times 10^3} \times 0.095 = 0.095$

$v_2 = \dfrac{ 4.0 \times 10^3 - 7.50\times 10^3}{4.0 \times 10^3 + 7.50 \times 10^3} \times 0.095 + \dfrac{2 \times 7.50\times 10^3}{4.0 \times 10^3+ 7.50\times 10^3} \times 0.095 = 0.095$

The final relative velocity of the satellite, $v_f$ = v₁ + v₂

∴ $v_f$ = 0.095 + 0.095 = 0.190

The final relative velocity of the satellite, $v_f$ = 0.190 m/s

4 0

2 years ago