What happens to a day of light that slows down when it hits a new medium at an angle

19,792,000,000 in scientific notation will have how many significant figures

xz_007 [3.2K]

Answer:

= 1.9792 × 10^10

Significant Figures= 5

Explanation:

Look at the attachment below

Hope this helps (:

8 0

4 years ago

Find the object's speeds v1, v2, and v3 at times t1=2.0s, t2=4.0s, and t3=13s.

Burka [1]

Since this is a distance/time graph, the speed at any time is the slope
of the part of the graph that's directly over that time on the x-axis.

At time t1 = 2.0 s
That's in the middle of the first segment of the graph,
that extends from zero to 3 seconds.
Its slope is 7/3 . v1 = 7/3 m/s .

At time t2 = 4.0 s
That's in the middle of the horizontal part of the graph
that runs from 3 to 6 seconds.
Its slope is zero.
v2 = zero .

At time t3 = 13 s.
That's in the middle of the part of the graph that's sloping down,
between 11 and 16 seconds.
Its slope is -3/5 . v3 = -0.6 m/s .

7 0

3 years ago

Read 2 more answers

A new restaurant is interested in determining the best time-temperature combination for roasting a five-pound cut of lamb. The t

Leto [7]

Answer:

C

Explanation:

(c) The two cuts that are being roasted for each time-temperature combination are an example of replication.

In the question it is given that From 10 identical cuts of lamb, 2 are randomly selected to roast using each of the time-temperature combinations in the same oven. Here it is an act of copying the exact sahpe size of the lamb in all cuts, which is nothing but replication. Moreover, this replication can help in proper comparision.

7 0

3 years ago

1. What is the kinetic energy of a 1.75 kg ball travelling at a speed of 54 m/s?

Over [174]

Answer:

We conclude that the kinetic energy of a 1.75 kg ball traveling at a speed of 54 m/s is 2551.5 J.

Explanation:

Given

Mass m = 1.75 kg

Velocity v = 54 m/s

To determine

Kinetic Energy (K.E) = ?

We know that a body can possess energy due to its movement — Kinetic Energy.

Kinetic Energy (K.E) can be determined using the formula

$K.E=\frac{1}{2}mv^2$

where

m is the mass (kg)

v is the velocity (m/s)

K.E is the Kinetic Energy (J)

now substituting m = 1.75, and v = 54 in the formula

$K.E=\frac{1}{2}mv^2$

$K.E=\frac{1}{2}\left(1.75\right)\left(54\right)^2$

$K.E=1458\times 1.75$

$K.E=2551.5$ J

Therefore, the kinetic energy of a 1.75 kg ball traveling at a speed of 54 m/s is 2551.5 J.

7 0

3 years ago

Consider an oblique shock wave with a wave angle of 35 o . Upstream of the wave, the static pressure and temperature are 2,000 l

ziro4ka [17]

Answer:

The pressure is 6570 lbf/ft²

The temperature is 766 ⁰R

The velocity is 2746.7 ft/s

deflection angle behind the wave is 17.56⁰

Explanation:

Speed of air at initial condition:

$a_1 = \sqrt{\gamma RT } = \sqrt{1.4* 1716*520 } = 1117.70 \ ft/s$

γ is the ratio of specific heat, R is the universal gas constant, and T is the initial temperature.

initial mach number

$M_1 = \frac{v_1}{a_1} = \frac{3355}{1117.7} = 3$

then, $M_n = M_1sin \beta = 3sin(35) = 1.721$

based on the values obtained, read off the following from table;

P₂/P₁ = 3.285

T₂/T₁ = 1.473

Mₙ₂ = 0.6355

Thus;

P₂ = 3.285P₁ = 3.285(2000) = 6570 lbf/ft²

T₂ = 1.473T₁ = 1.473(520⁰R) = 766 ⁰R

Again; to determine the velocity and deflection angle, first we calculate the mach number.

$M_t_1 = M_1cos \beta = 3 cos(35) = 2.458$

$w_2 = a_1M_t_1 = 2.458(1117.70) = 2746.7 \ ft/s$

$a_2 = \sqrt{\gamma RT_2} = \sqrt{1.4*1718*766} = 1357.34 \ ft/s$

$v_2 = a_2M_n_2 = 1357.34(0.6355) = 862.59 \ ft/s$

$Tan(\beta -\theta) = \frac{v_2}{w_2} = \frac{862.59}{2746.7} \\\\Tan(\beta -\theta) = 0.314\\\\\beta -\theta= 17.44\\\\\theta = \beta - 17.44 = 17.56^o$

6 0

3 years ago