Does this graph represent constant or changing speed? How do you know? Find the average speed

Compare the energy consumption of two commonly used items in the household. Calculate the energy used by a 1.40 kW toaster oven,

Nikolay [14]

Explanation:

It is given that,

Power consumed by toaster oven, $P=1.4\ kW$

Time taken, t = 6 minutes = 0.1 hour

Energy used by toaster, $W_{toaster}=P\times t$

$W_{toaster}=1.4\ kW\times 0.1\ h=0.14\ kWh$

Similarly,

Power consumed by toaster oven, $P=11\ W$

Time taken, t = 9 minutes = 0.15 hour

Energy used by toaster, $W_{toaster}=P\times t$

$W_{toaster}=11\ W\times 0.15\ h=1.65\ Wh=0.0016\ kWh$

Hence, this is the required solution.

3 0

3 years ago

Intelligence tests that are given to participants in groups

White raven [17]

a. are typically paper-and-pencil measures.

Similar with psychological tests, mostly structured personality tests.

Psychological tests comes two ways: 
The structure psychological tests or, objectives tests and unstructured psychological tests or, also called projective tests. By what you are referring the responder strongly asserts a projective tests which in definition comes with an unambiguous stimuli or no paper test just drawings and images. If what the responder’s suggesting is correct you are referring to the Rorschach projective tests, these tests are a figure symmetrically placed in an inkblot that lets you visualize or create a mental picture out of it, and makes you describe what you in see much detail as you can.

4 0

3 years ago

An electron moves at 0.130 c as shown in the figure (Figure 1). There are points: A, B, C, and D 2.10 μm from the electron.

Olegator [25]

Hi there!

We can use Biot-Savart's Law for a moving particle:
$B= \frac{\mu_0 }{4\pi}\frac{q\vec{v}\times \vec{r}}{r^2 }$

B = Magnetic field strength (T)
v = velocity of electron (0.130c = 3.9 × 10⁷ m/s)

q = charge of particle (1.6 × 10⁻¹⁹ C)

μ₀ = Permeability of free space (4π × 10⁻⁷ Tm/A)

r = distance from particle (2.10 μm)

There is a cross product between the velocity vector and the radius vector (not a quantity, but specifies a direction). We can write this as:

$B= \frac{\mu_0 }{4\pi}\frac{q\vec{v} \vec{r}sin\theta}{r^2 }$

Where 'θ' is the angle between the velocity and radius vectors.

a)
To find the angle between the velocity and radius vector, we find the complementary angle:

θ = 90° - 60° = 30°

Plugging 'θ' into the equation along with our other values:

$B= \frac{\mu_0 }{4\pi}\frac{q\vec{v} \vec{r}sin\theta}{r^2 }\\\\B= \frac{(4\pi *10^{-7})}{4\pi}\frac{(1.6*10^{-19})(3.9*10^{7}) \vec{r}sin(30)}{(2.1*10^{-5})^2 }$

$B = \boxed{7.07 *10^{-10} T}$

b)
Repeat the same process. The angle between the velocity and radius vector is 150°, and its sine value is the same as that of sin(30°). So, the particle's produced field will be the same as that of part A.

c)

In this instance, the radius vector and the velocity vector are perpendicular so

'θ' = 90°.

$B= \frac{(4\pi *10^{-7})}{4\pi}\frac{(1.6*10^{-19})(3.9*10^{7}) \vec{r}sin(90)}{(2.1*10^{-5})^2 } = \boxed{1.415 * 10^{-9}T}$