All categories

3 years ago

The ____ button on the rehearsal toolbar clears the slide time box and resets the timer to 0:00:00.

1 answer:

6 0

The Rehearsal toolbar is used in Microsoft Power Point. It has buttons that allow the user to start, pause and repeat viewing the slides in the slide show. The buttons are: next, pause recording, slide time, repeat and elapsed time. The slide time button <span>on the rehearsal toolbar clears the slide time box and resets the timer to 0:00:00.</span>

You might be interested in

Water is considered to be

Natalka [10]

Answer: the thing that brings you life- h2o- idrk the question but if this helps your welcome ^-^have a good day

Explanation:

3 0

3 years ago

calculate minimum number of incident photons per area and of the minimum dose needed to visualize an object of 1 mm squared usin

Lynna [10]

Answer:

minimum number of photon is 4.05 × $10^{7}$

Explanation:

given data

energy = 50 keV = 50 × $10^{3}$ eV = 50 × $10^{3}$ × 1.602× $10^{-19}$ J

thickness = 10^-3

contrast = 1%

to find out

number of incident photons

solution

we know here equation that is

E = n × h × ν .......................1

put here all these value

50 × $10^{3}$ = n × 6.6× $10^{-34}$ × c/ 1× $10^{-3}$

50 × $10^{3}$ × 1.602× $10^{-19}$ = n × 6.6× $10^{-34}$ ×( 3 × $10^{8}$ / 1× $10^{-3}$ )

solve it and find n

n = 4.05 × $10^{7}$

so here minimum number of photon is 4.05 × $10^{7}$

3 0

3 years ago

A +2.00nc point charge is at the origin, and a second -5.00nc point charge is on the x-axis at x = 0.800m find the magnitude of

Damm [24]

You have to reduce 2.00 an5.00 I order to use the×that=0.800

3 0

3 years ago

Mohs scale is used in describing which characteristic?

AleksandrR [38]

Mohs scale of mineral hardness is a qualitative scale that characterizes the scratch resistance of various minerals with the ablity of harder material to scratch softer material

7 0

3 years ago

Read 2 more answers

A 0.25 kg ball is suspended from a light 0.65 m string as shown. The string makes an angle of 31° with the vertical. Let U = 0 w

steposvetlana [31]

Explanation:

a) The height of the ball h with respect to the reference line is

$h = L - L\cos{31°} = L(1 - \cos{31°})$

so its initial gravitational potential energy $U_0$ is

$U = mgh = mgL(1 - \cos{31°})$

$\:\:\:\:\:=(0.25\:\text{kg})(9.8\:\text{m/s}^2)(0.65\:\text{m})(1 - \cos{31})$

$\:\:\:\:\:=0.23\:\text{J}$

b) To find the speed of the ball at the reference point, let's use the conservation law of energy:

$\Delta{K} + \Delta{U} = 0 \Rightarrow K_0 + U_0 = K + U$

We know that the initial kinetic energy $K_0,$ as well as its final gravitational potential energy $U$ are zero so we can write the conservation law as

$mgL(1 - \cos{31°}) = \frac{1}{2}mv^2$

Note that the mass gets cancelled out and then we solve for the velocity v as

$v = \sqrt{2gL(1 - \cos{31°})}$

$\:\:\:\:\:= \sqrt{2(9.8\:\text{m/s}^2)(0.65\:\text{m})(1 - \cos{31°})}$

$\:\:\:\:\:= 1.3\:\text{m/s}$

5 0

3 years ago

Read 2 more answers