What is the speed of a wave that has a wavelength of 0.4 m and a frequency of 10 Hz

How much heat is required to convert 5.53 g of ice at -12.0 ∘C to water at 24.0 ∘C? (The heat capacity of ice is 2.09 J/(g⋅∘C),

fredd [130]

Answer:

2.55 × 10³ J =2.55 kJ

Explanation:

Specific heat capacity of ice = 37.8 J / mol °C

Specific heat capacity of water = 76.0 J/ mol °C

Ice at -12 °C is converted to ice at 0 °C by absorbing heat Q₁

Ice at 0°C melts to water at 0 °C. Let Heat absorbed during this phase change be Q₂ .

Let heat absorbed to raise the temperature of water from 0 C to 24°C be Q₃ .

Total heat = Q = Q₁ + Q₂ + Q₃

Q₁ = (37.8 j/mol C )(5.53 g /18.01532 g/ mol )( 0-(-12)) = 139.23749 j

Q₂ =(5.53 g/18.01532 g H₂O / mol ) (6.02 x10³ j) = 1847.905 j

Q₃ = (76 j/mol C) ( (5.53 g/18.01532 g H₂O / mol )(24-0) = 559.8968 j

Total Heat required = Q = 139.23749 j + 1847.905 j + 559.8968 j

= 2547.039 j = 2.55 × 10³ J =2.55 kJ

5 0

4 years ago

As a quality test of ball bearings, you drop bearings (small metal balls), with zero initial velocity, from a height of 1.94 m i

11Alexandr11 [23.1K]

Answer:

The average acceleration of the bearings is $0.77\times10^{3}\ m/s^2$

Explanation:

Given that,

Height = 1.94 m

Bounced height = 1.48 m

Time interval $t=14.86\times10^{-3}\ s$

Velocity of the ball bearing just before hitting the steel plate

We need to calculate the velocity

Using conservation of energy

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

Put the value into the formula

$9.8\times1.94=\dfrac{1}{2}\times v^2$

$v=\sqrt{2\times9.8\times1.94}$

$v=6.166\ m/s$

Negative as it is directed downwards

After bounce back,

We need to calculate the velocity

Using conservation of energy

$mgh=\dfrac{1}{2}mv^2'$

Put the value into the formula

$9.8\times1.48=\dfrac{1}{2}\times v^2'$

$v'=\sqrt{2\times9.8\times1.48}$

$v'=5.38\ m/s$

We need to calculate the average acceleration of the bearings while they are in contact with the plate

Using formula of acceleration

$a=\dfrac{v-v'}{t}$

Put the value into the formula

$a=\dfrac{5.38-(-6.166)}{14.86\times10^{-3}}$

$a=776.98\ m/s^2$

$a=0.77\times10^{3}\ m/s^2$

Hence,The average acceleration of the bearings is $0.77\times10^{3}\ m/s^2$

6 0

3 years ago

Masses A and B rest on very light pistons that enclose a fluid.There is no friction between the pistons and the cylinders they f

RSB [31]

Answer:

D)Not enough information

Explanation:

According to Pascal's principle, the pressure exerted on the two pistons is equal:

$p_A = p_B$

Pressure is given by the ratio between force F and area A, so we can write

$\frac{F_A}{A_A}=\frac{F_B}{A_B}$

The force exerted on each piston is just equal to the weight of the corresponding mass: $F=W=mg$ , where m is the mass and g is the gravitational acceleration. So the equation becomes

$\frac{m_A g}{A_A}=\frac{m_B g}{A_B}$

Now we can rewrite the mass as the product of volume, V, times density, d:

$\frac{V_A d_A g}{A_A}=\frac{V_B d_B g}{A_B}$

We also know that $A_B = 2.0 m^2\\A_A = 1.0 m^2$

So we can further re-arrange the equation (and simplify g as well):

$\frac{V_A d_A}{1}=\frac{V_B d_B}{2}$

$\frac{d_A}{d_B}=\frac{V_B}{2V_A}$

We are also told that block B has bigger volume than block A: $V_B > V_A$ . However, this information is not enough to allow us to say if the fraction on the right is greater than 1 or smaller than 1: therefore, we cannot conclude anything about the densities of the two objects.