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Roman55 [17]
3 years ago
14

Which of the following measurment is most accurate? A. 9×10-²m

Physics
1 answer:
sweet-ann [11.9K]3 years ago
6 0

Answer:

B. 900 × 10⁴m

Explanation:

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A force of 200N due South and another force of 300N due East each act on an object simultaneously.
dangina [55]

Answer:

a) 500

b)-500, north west

Explanation:

a) sum of F= F1+F2= 200+300= 500

b) sum of forces=0

so 200+300-500+0

4 0
2 years ago
Suppose you first walk 12.0 m in a direction 20? west of north and then 20.0 m in a direction 40.0? south of west. how far are y
Gnesinka [82]
The representation of this problem is shown in Figure 1. So our goal is to find the vector \overrightarrow{R}. From the figure we know that:

\left | \overrightarrow{A} \right |=12m \\ \\ \left | \overrightarrow{B} \right |=20m \\ \\ \theta_{A}=20^{\circ} \\ \\ \theta_{B}=40^{\circ}

From geometry, we know that:

\overrightarrow{R}=\overrightarrow{A}+\overrightarrow{B}

Then using vector decomposition into components:

For \ A: \\ \\ A_x=-\left | \overrightarrow{A} \right |sin\theta_A=-12sin(20^{\circ})=-4.10 \\ \\ A_y=\left | \overrightarrow{A} \right |cos\theta_A=12cos(20^{\circ})=11.27 \\ \\ \\ For \ B: \\ \\ B_x=-\left | \overrightarrow{B} \right |cos\theta_B=-20cos(40^{\circ})=-15.32 \\ \\ B_y=-\left | \overrightarrow{B} \right |sin\theta_B=-20sin(40^{\circ})=-12.85

Therefore:

R_x=A_x+B_x=-4.10-15.32=-19.42m \\ \\ R_y=A_y+B_y=11.27-12.85=-1.58m

So if you want to find out <span>how far are you from your starting point you need to know the magnitude of the vector \overrightarrow{R}, that is:
</span>
\left | \overrightarrow{R} \right |=&#10;\sqrt{R_x^2+R_y^2}=\sqrt{(-19.42)^2+(-1.58)^2}=\boxed{19.48m}

Finally, let's find the <span>compass direction of a line connecting your starting point to your final position. What we are looking for here is an angle that is shown in Figure 2 which is an angle defined with respect to the positive x-axis. Therefore:

</span>\theta_R=180^{\circ}+tan^{-1}(\frac{\left | R_y \right |}{\left | R_x \right |}) \\ \\ \theta_R=180^{\circ}+tan^{-1}(\frac{1.58}{19.42}) \\ \\ \theta_R=180^{\circ}+4.65^{\circ}=185.85^{\circ}


6 0
3 years ago
A) Show that the surface temperature of a star can be inferred from measurements of blackbodyfluxes at two different frequencies
Roman55 [17]

Answer:

The answers to the questions have been solved in the attachment.

Explanation:

The answers to part a to e are all contained in the attachment. For answer part b, temperature and frequency were assumed to be fixed or constant. V² is directly proportional to T telling us that variation in T gives us a square in the frequency variation. This tells us why it is difficult when both frequencies are on this side of the black body.

3 0
3 years ago
if you have a kinetic energy of 1470 J, and you are 60kg mass and 0 m above the ground, what is you velocity?
laiz [17]

Answer:

The 39.

Explanation:

8 0
2 years ago
Read 2 more answers
If a person is 6.25 m away from a 60.0 w speaker, what is the sound level they are hearing?
vodka [1.7K]

Answer:

110.87 dB

Explanation:

(I got it right on Acellus)

I= P/4(pi)r^2 = 60/4(pi)6.25^2

60/4(pi)6.25^2=0.12223

B=10log(I/Io)

B=10log(0.12223/1*10^-12) = 110.87 dB

111 in sigfigs

4 0
3 years ago
Read 2 more answers
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