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3 years ago

13

If you measured all the energy related to motion and all the stored energy in the particles of a substance, which would you be m

easuring? heat capacity heat thermal energy

1 answer:

Nady [450]3 years ago

3 0

If you measured all the energy related to motion and all the stored energy in the particles of a substance, you would be measuring the thermal energy of the particles. If there is movement of the particles, they are also releasing energy in the form of heat.

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Which of the following shows kinetic energy being converted into potential energy?

Komok [63]

B) a rock being tossed high into the air

3 0

3 years ago

Issac and Blaise decide to race. They both start at the same position at the same time. Issac runs at 2m/s but decides to take a

FromTheMoon [43]

Let the Blaise runs for time "t" to complete the race

so the total distance he moved is given by

$d_1 = 1* t$

Now Issac runs for time t = "t - 2*60"

because it took rest for 2 minutes

$d_2 = 2*(t - 120)$

now it is given that Blaise wins by 10 m distance

$d_1 - d_2 = 10$

$1* t - 2*(t - 120) = 10$

$t - 2t + 240 = 10$

$t = 230 s$

now the distance moved by Blaise is given by

$d_1 = 1*230 = 230 m$

6 0

2 years ago

A man walks along a straight path at a speed of 4 ft/s. A searchlight is located on the ground 6 ft from the path and is kept fo

BARSIC [14]

We are given that,

$\frac{dx}{dt} = 4ft/s$

We need to find $\frac{d\theta}{dt}$ when $x=8ft$

The equation that relates x and $\theta$ can be written as,

$\frac{x}{6} tan\theta$

$x = 6tan\theta$

Differentiating each side with respect to t, we get,

$\frac{dx}{dt} = \frac{dx}{d\theta} \cdot \frac{d\theta}{dt}$

$\frac{dx}{dt} = (6sec^2\theta)\cdot \frac{d\theta}{dt}$

$\frac{d\theta}{dt} = \frac{1}{6sec^2\theta} \cdot \frac{dx}{dt}$

Replacing the value of the velocity

$\frac{d\theta}{dt} = \frac{1}{6} cos^2\theta (4)^2$

$\frac{d\theta}{dt} = \frac{8}{3} cos^2\theta$

The value of $cos \theta$ could be found if we know the length of the beam. With this value the equation can be approximated to the relationship between the sides of the triangle that is being formed in order to obtain the numerical value. If this relation is known for the value of x = 6ft, the mathematical relation is obtained. I will add a numerical example (although the answer would end in the previous point) If the length of the beam was 10, then we would have to

$cos\theta = \frac{6}{10}$

$\frac{d\theta}{dt} = \frac{8}{3} (\frac{6}{10})^2$

$\frac{d\theta}{dt} = \frac{24}{25}$

Search light is rotating at a rate of 0.96rad/s

4 0

2 years ago

A net force of 60 N north acts on an object with a mass of 30 kg. Use Newton's second law of

earnstyle [38]

Answer:

Explanation:

F = ma. For us, this looks like

60 = 30a and

a = 2 m/s/s

If the force goes up to, say, 90, then

90 = 30a and

a = 3...if the force goes up, the acceleration also goes up.

If the mass goes up to say, 60, and the force stays the same, then

60 = 60a and

a = 1...if the mass goes up, the acceleration goes down.

7 0

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