2. What is the speed of section A? Section B? Section C?

When you calculate the SLOPE of a line segment, what does the SLOPE represent? (Choose all that apply) the Distance traveled the

dolphi86 [110]

Answer:

Please find the answer in the explanation

Explanation:

When you calculate the SLOPE of a line segment, what does the SLOPE represent? (Choose all that apply) the Distance traveled the Displacement the Velocity the Acceleration None of the above

The slope of any time graph can not give you distance or displacement except for position - time graph.

When you plot either distance or displacement against time, that is, distance time graph or displacement time graph, you can get speed or velocity as the slope of the line segment.

You can only acceleration as a slope in a line of best fit if velocity is plotted against time. That is, in a velocity time graph.

5 0

3 years ago

The electrons in the beam of a television tube have an energy of 19.0 keV. The tube is oriented so that the electrons move horiz

igomit [66]

Answer:

The direction is due south

Explanation:

From the question we are told that

The energy of the electron is $E = 19.0keV = 19.0 *10^3 eV$

The earths magnetic field is $B = 42.3 \muT = 42.3 *10^{-6} T$

Generally the force on the electron is perpendicular to the velocity of the elecrton and the magnetic field and this is mathematically reresented as

$\= F = q (\= v * \=B)$

On the first uploaded image is an illustration of the movement of the electron

Looking at the diagram we can see that in terms of direction the magnetic force is

$\= F =q(\=v * \= B)= -( -\r i * - \r k)$

$= -(- (\r i * \r k))$

generally i cross k = -j

so the equation above becomes

$\= F = -(-(- \r j))$

$= - \r j$

This show that the direction is towards the south

3 0

3 years ago

5/6 When switched on, the grinding machine accelerates from rest to its operating speed of 3450 rev/min in 6 seconds. When switc

ludmilkaskok [199]

Answer:

Δθ₁ = 172.5 rev

Δθ₁h = 43.1 rev

Δθ₂ = 920 rev

Δθ₂h = 690 rev

Explanation:

Assuming uniform angular acceleration, we can use the following kinematic equation in order to find the total angle rotated during the acceleration process, from rest to its operating speed:

$\Delta \theta = \frac{1}{2} *\alpha *(\Delta t)^{2} (1)$

Now, we need first to find the value of the angular acceleration, that we can get from the following expression:

$\omega_{f1} = \omega_{o} + \alpha * \Delta t (2)$

Since the machine starts from rest, ω₀ = 0.
We know the value of ωf₁ (the operating speed) in rev/min.
Due to the time is expressed in seconds, it is suitable to convert rev/min to rev/sec, as follows:

$3450 \frac{rev}{min} * \frac{1 min}{60s} = 57.5 rev/sec (3)$

Replacing by the givens in (2):

$57.5 rev/sec = 0 + \alpha * 6 s (4)$

Solving for α:

$\alpha = \frac{\omega_{f1}}{\Delta t} = \frac{57.5 rev/sec}{6 sec} = 9.6 rev/sec2 (5)$

Replacing (5) and Δt in (1), we get:

$\Delta \theta_{1} = \frac{1}{2} *\alpha *(\Delta t)^{2} = \frac{1}{2} * 6.9 rev/sec2* 36 sec2 = 172.5 rev (6)$

in order to get the number of revolutions during the first half of this period, we need just to replace Δt in (6) by Δt/2, as follows:

$\Delta \theta_{1h} = \frac{1}{2} *\alpha *(\Delta t/2)^{2} = \frac{1}{2} * 6.9 rev/sec2* 9 sec2 = 43.2 rev (7)$

In order to get the number of revolutions rotated during the deceleration period, assuming constant deceleration, we can use the following kinematic equation:

$\Delta \theta = \omega_{o} * \Delta t + \frac{1}{2} *\alpha *(\Delta t)^{2} (8)$

First of all, we need to find the value of the angular acceleration during the second period.
We can use again (2) replacing by the givens:
ωf =0 (the machine finally comes to an stop)
ω₀ = ωf₁ = 57.5 rev/sec
Δt = 32 s

$0 = 57.5 rev/sec + \alpha * 32 s (9)$

Solving for α in (9), we get:

$\alpha_{2} =- \frac{\omega_{f1}}{\Delta t} = \frac{-57.5 rev/sec}{32 sec} = -1.8 rev/sec2 (10)$

Now, we can replace the values of ω₀, Δt and α₂ in (8), as follows:

$\Delta \theta_{2} = (57.5 rev/sec*32) s -\frac{1}{2} * 1.8 rev/sec2\alpha *(32s)^{2} = 920 rev (11)$

In order to get finally the number of revolutions rotated during the first half of the second period, we need just to replace 32 s by 16 s, as follows:
$\Delta \theta_{2h} = (57.5 rev/sec*16 s) -\frac{1}{2} * 1.8 rev/sec2\alpha *(16s)^{2} = 690 rev (12)$

7 0

2 years ago

4

12345 [234]

Answer:

charcoal is the correct answer

hope it helps , pls mark me as brainliest

6 0

3 years ago

A 0.810 kg ball falls 2.5m. How much work does the force of gravity do on the ball?

lisabon 2012 [21]

Answer:

W = 19.845 J

Explanation:

Work is defined as W = Fdcos $\theta$ , where F is the force exerted and d is the distance. Because the direction the ball is falling is the same direction as the force itself, $\theta$ = 0 deg, and since cos(0) = 1, this equation is equivalent to W = Fd. In this case, the force exerted is the weight force, which is equivalent to m * g. Substituting you get:

W = mgd = 0.810 kg * 9.8 m/s^2 * 2.5m

W = 19.845 J

6 0

2 years ago