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3 years ago

Which of the following solutions will have the lowest freezing point?

1 answer:

7 0

C. 1.0 mol/kg sodium phosphide (Na3P)<span>

This is because it will produce 4 ions, and thus has the highest molalitiy. (more particles in solution lowers the freezing point)

</span><span>remember if you ever need anything and you dont want to waste points you can just post it on my pfp or message me the question</span>

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A thin rod of length 1.3 m and mass 250 g is suspended freely from one end. It is pulled to one side and then allowed to swing l

Nutka1998 [239]

(a) 2.42 J

The kinetic energy of a rotating object is given by:

$K=\frac{1}{2}I \omega^2$

where

I is the moment of inertia

$\omega$ is the angular speed

Here we have

$\omega=5.88 rad/s$ at the lowest point of the trajectory

While the moment of inertia of a rod rotating around one end is

$I=\frac{1}{3}ML^2 = \frac{1}{3}(0.250 kg)(1.3 m)^2=0.14 kg m^2$

And substituting in the previous formula, we find the kinetic energy at the lowest position:

$K=\frac{1}{2}(0.14 kg m^2)(5.88 rad/s)^2=2.42 J$

(b) 0.99 m

According to the law of conservation of energy, the total mechanical energy (sum of kinetic energy and potential energy) must be conserved:

$E=K+U$

At the lowest point, we can take the potential energy as zero, so the mechanical energy is just kinetic energy:

$E=K=2.42 J$

At the highest point in the trajectory, the rod is stationary, so the kinetic energy will be zero, and the mechanical energy will simply be equal to the gravitational potential energy:

$E=2.42 J = U = mgh$

where h is the heigth of the centre of mass of the rod with respect to the lowest point of the trajectory. Solving for h, we find

$h=\frac{E}{mg}=\frac{2.42 J}{(0.250 kg)(9.81 m/s^2)}=0.99 m$

5 0

3 years ago

Two construction cranes are each able to lift a maximum load of 20000 N to a height of 250 m. However, one crane can lift that l

Neporo4naja [7]

Answer:

Explanation:

Given

Load $W=20000\ N$

height to which load is raised $h=250\ m$

Another crane take $\frac{1}{6}$ th time to lift the load

Energy required required to lift the Weight

$E=W\times h$

$E=20000\times 250$

$E=5,000,000\ J$

Suppose $P_1$ and $P_2$ is the Power required to lift the weight in t and $\frac{t}{6}$ time

$E=P_1\times t$

$E=P_2\times \frac{t}{6}$

$P_1\times t=P_2\times \frac{t}{6}$

thus

$P_2=6P_1$

Second Crane requires 6 times more power than the slow crane

5 0

3 years ago

A 40.0-$kg$ body is moving in the direction of the positive x axis with a speed of 238 $m/s$ when, owing to an internal explosio

Julli [10]

Answer:

$v_3 = 400\ m/s$

Explanation:

given,

mass of the body = 40 Kg

speed in x-axis = 238 m/s

mass break into three part

m₁ = 7 kg

v₁ = 356 m/s (along the positive y axis)

m₂ = 4.5 kg

v₂ = 357 m/s(along the negative x axis)

m₃ = 40 - (7 + 4.5) = 28.5 Kg

v₃ = ?

using conservation of momentum

MV = m₁v₁ + m₂v₂ + m₃v₃

$(40)(238) \hat{i} = (7)(356) \hat{j} - (4.5)(357) \hat {i} + 28.5 v_3$

$(9520) \hat{i} = 2492 \hat{j} - 1606.5\hat {i} + 28.5 v_3$

$11126.5 \hat{i} - 2492 \hat{j} = 28.5 v_3$

$v_3 =390.40 \hat{i} - 87.44 \hat{j}$

$v_3 = \sqrt{390.40^2 + 87.44^2}$

$v_3 = 400\ m/s$

4 0

3 years ago

What are the sign and magnitude in coulomb's of a point charge that produces a potential of -1.50 V at a distance of 2.00 mm

Charra [1.4K]

Answer:

The sign of the charge is negative

The magnitude of the charge is 3.33 x 10⁻¹³ C

Explanation:

Given;

potential difference, V = -1.5 V

distance of the point charge, r = 2 mm = 2 x 10⁻³ m

The magnitude of the charge is calculated as follows;

$V = \frac{kq}{r} \\\\q = \frac{Vr}{k} \\\\where;\\\\k \ is \ coulomb's \ constant = 9\times 10^9 \ Nm^2/C^2\\\\q = \frac{-1.5 \times 2\times 10^{-3}}{9\times 10^9 } \\\\q = -3.33 \times 10^{-13} \ C\\\\Magnitude \ of \ the\ charge, q = 3.33 \times 10^{-13} \ C$