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2 years ago

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A freezer has a coefficient of performance of 4.1. You place 0.45 kg of water at 16°C in the freezer, which maintains its temper

ature of -18°C. In this problem you can take the specific heat of water to be 4190 J/kg/K, the specific heat of ice to be 2100 J/kg/K, and the latent heat of fusion for water to be 3.34 ×105 J/kg. How much additional energy, in joules, does the freezer use to cool the water to ice at -18°C?

1 answer:

velikii [3]2 years ago

5 0

Ohh this is quite a lot umm i’ll be back hold on

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A 1300 kg car moving at 20 m/s and a 900 kg car moving at 15 m/s in precisely oppositedirections participate in a head-on crash.

miskamm [114]

Given

Car 1

m1 = 1300 kg

v1 = 20 m/s

m2 = 900 kg

v2 = -15 m/s

(Negative sign shows that direction of car 2 is opposite to car 1)

Procedure

As per the conservation of linear momentum, "The total momentum of the system before the collision must be equal to the total momentum after the collision". And this applies to the perfectly inelastic collision as well. Then the expression is,

$\begin{gathered} m_1v_1+m_2v_2=(m_1+m_2)v \\ v=\frac{m_1v_1+m_2v_2}{m_1+m_2} \\ v=\frac{1300\operatorname{kg}\cdot20m/s-900\operatorname{kg}\cdot15m/s}{1300\operatorname{kg}+900\operatorname{kg}} \\ v=5.681m/s \end{gathered}$

Thus, we can conclude that the speed and direction of the cars after the impact is 5.68 m/s towards the first car.

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1 year ago

VMariaS [17]

Answer:

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3 years ago

A short current element dl⃗ = (0.500 mm)j^ carries a current of 5.40 A in the same direction as dl⃗ . Point P is located at r⃗ =

SashulF [63]

Answer:

The magnetic field along x axis is

$B_{x}=1.670\times10^{-10}\ T$

The magnetic field along y axis is zero.

The magnetic field along z axis is

$B_{z}=3.484\times10^{-10}\ T$

Explanation:

Given that,

Length of the current element $dl=(0.5\times10^{-3})j$

Current in y direction = 5.40 A

Point P located at $\vec{r}=(-0.730)i+(0.390)k$

The distance is

$|\vec{r}|=\sqrt{(0.730)^2+(0.390)^2}$

$|\vec{r}|=0.827\ m$

We need to calculate the magnetic field

Using Biot-savart law

$B=\dfrac{\mu_{0}}{4\pi}\timesI\times\dfrac{\vec{dl}\times\vec{r}}{|\vec{r}|^3}$

Put the value into the formula

$B=10^{-7}\times5.40\times\dfrac{(0.5\times10^{-3})\times(-0.730)i+(0.390)k}{(0.827)^3}$

We need to calculate the value of $\vec{dl}\times\vec{r}$

$\vec{dl}\times\vec{r}=(0.5\times10^{-3})\times(-0.730)i+(0.390)k$

$\vec{dl}\times\vec{r}=i(0.350\times0.5\times10^{-3}-0)+k(0+0.730\times0.5\times10^{-3})$

$\vec{dl}\times\vec{r}=0.000175i+0.000365k$

Put the value into the formula of magnetic field

$B=10^{-7}\times5.40\times\dfrac{(0.000175i+0.000365k)}{(0.827)^3}$

$B=1.670\times10^{-10}i+3.484\times10^{-10}k$

Hence, The magnetic field along x axis is

$B_{x}=1.670\times10^{-10}\ T$

The magnetic field along y axis is zero.

The magnetic field along z axis is

$B_{z}=3.484\times10^{-10}\ T$

7 0

3 years ago

Calculate the kinetic energy in joules of a 1,500 kg car that is moving at a speed of 42 km/h

Rzqust [24]

Data:
KE (Kinetic Energy) = ? (Joule)
m (mass) = 1500 Kg
v (speed) = 42 Km/h
converting to m/s (42 / 3.6), we have: v (speed) = 11.6 m/s

Formula:
$K_{E} = \frac{1}{2} m*v^2$

Solving:
$K_{E} = \frac{1}{2} m*v^2$
$K_{E} = \frac{1}{2} *1500*(11.6)^2$
$K_{E} = \frac{1}{2} *1500*134.56$
$K_{E} = \frac{201840}{2}$
$\boxed{\boxed{K_{E} = 100920\:Joule}}\end{array}}\qquad\quad\checkmark$

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2 years ago

Two horizontal metal plates, each 10.0 cm square, are aligned 1.00 cm apart with one above the other. They are given equal-magni

Ber [7]

Answer:

a) motion is PARABOLIC, b) positive particle is accelerated towards the negative plate, c) x = 6.19 10⁹ m

Explanation:

This exercise looks at the motion of a positively charged particle in an electric field.

a) Since the field is vertical the acceleration in this direction is

F = m a

the electric force is

F = q E

we substitute

q E = m a

a = qE / m

the mass of the particle is m = 2.00 10-16 kg

a = 1.6 10⁻¹⁹ 2.02 10³ / 2.00 10⁻¹⁶ kg

a = 1,616 m / s²

on the x-axis there are no relationships because there are no forces.

Since the particle has velocities in both axes, its motion is PARABOLIC,

b) the positive particle is accelerated towards the negative plate,

The field is descending, for which the event is down

c) where hit the particle on the x-axis

they indicate that the particle leaves the center of the negative plate, for which we will fix our reference system at this point.

Let's find the components of the initial velocity.

sin θ = v_{oy} / v

cos θ = v₀ₓ / v

v_{oy} = v₀ sin θ

v₀ₓ = v₀ cos θ

v_{oy) = 1.02 10⁵ sin 37 = 0.6139 10⁵ m / s

v₀ₓ = 1.02 10⁵ cos 37 = 0.8146 10⁵ m / s

Let's find the time it takes to hit the negative plate

y = y₀I + v_{oy} t + ½ a and t2

in this case the positions are y = y₀ = 0 and the accelerations

a = - 1,616m/s2,

we substitute

0 = 0 + v_{oy} t - ½ a_y t²

v_{oy}= ½ a_y t

t = 2v_{oy} / a_y

let's calculate

t = 2 0.6139 10⁵ / 1.616

t = 7.597 10⁴s

in this time the particle travels a horizontal distance

x = v₀ₓ t

x = 0.8145 10⁵ 7.597 10⁴

x = 6.19 10⁹ m

the particle falls off the plate

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2 years ago