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snow_tiger [21]
2 years ago
15

Question 7 (2 points)

Physics
1 answer:
DiKsa [7]2 years ago
6 0
This has to do with independent and dependent variables. The independent variable is what affects the dependant variable, so the question is, does a students mark on a test affect their hours of sleep or does their hours of sleep affect their test results? Which one makes more sense? I would say the latter.
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A 6.0-kilogram cart initially traveling at 4.0 meters per second east accelerates uniformly at 0.50 meter per second squared eas
VMariaS [17]

Answer: 5.5m/s

Explanation:

vf=vi+at

vf= 4.0m/s + (0.50m/s^2)(3.0s)

3 0
2 years ago
Can concave lens be used to make hand lens .why?​
mario62 [17]

Answer:

No concave lens cannot be used to make hand lens because in hand Lens we use convex lens so as to converge the rays of the light After refraction and helps to produce a magnified image of an object.

Hope it will help you :)❤

8 0
3 years ago
Find the radius Rrigel of the star Rigel, the bright blue star in the constellation Orion that radiates energy at a rate of 2.7×
Zinaida [17]

Answer:

r = 1.61 x 10^{11} m

Explanation:

energy radiated (H) = 2.7 x 10^31 W

surface temperature (T)  = 11,000 k

assuming ε = 1 and taking σ = 5.67 x 10^{-8} W/m^{2}.K^{4}

we can find the radius of the star from the equation below

H = A x  ε x σ x T^{4}              

where area (A) = 4 x π x r^{2} (assuming it is a sphere)

therefore  the equation becomes

H = 4 x π x r^{2} x  ε x σ x T^{4}  

2.7 x 10^31  = 4 x π x r^{2} x  1 x 5.67 x 10^{-8}  x (11,000)^{4}

r = \sqrt{\frac{2.7 x 10^31}{4 x π x 1 x 5.67 x 10^{-8}  x (11,000)^{4}} }

r = 1.61 x 10^{11} m

       

4 0
3 years ago
A light, flexible cable is wrapped around a solid cylinder with mass 3.3 kg and a radius of 0.8 meters. The cylinder rotates on
kari74 [83]

Answer:

9.16rad/s^2

Explanation:

We are given that

Mass,m_1=3.3 kg

Radius,r=0.8 m

m_2=4.9 kg

Height,h=2.9 m

We have to find the angular acceleration of the cylinder.

According to question

4.9g-T=4.9a

Tr=I\alpha

Where

\alpha=\frac{a}{r}

Tr=\frac{1}{2}m_1ra

T=\frac{1}{2}m_1a=\frac{1}{2}(3.3)a

Substitute the value

4.9g-\frac{1}{2}(3.3a)=4.9a

4.9\times 9.8=4.9a+\frac{3.3a}{2}

Where g=9.8 m/s^2

48.02=a(4.9+1.65)=6.55a

a=\frac{48.02}{6.55}=7.33m/s^2

Angular acceleration,\alpha=\frac{a}{r}=\frac{7.33}{0.8}=9.16rad/s^2

7 0
3 years ago
An industrial laser is used to burn a hole through a piece of metal. The average intensity of the light is W/m². What is the rms
inn [45]

I think your question should be:

An industrial laser is used to burn a hole through a piece of metal. The average intensity of the light is

S = 1.23*10^9 W/m^2

What is the rms value of (a) the electric field and

(b) the magnetic field in the electromagnetic wave emitted by the laser

Answer:

a) 6.81*10^5 N/c

b) 2.27*10^3 T

Explanation:

To find the RMS value of the electric field, let's use the formula:

E_r_m_s = sqrt*(S / CE_o)

Where

C = 3.00 * 10^-^8 m/s;

E_o = 8.85*10^-^1^2 C^2/N.m^2;

S = 1.23*10^9 W/m^2

Therefore

E_r_m_s = sqrt*{(1.239*10^9W/m^2) / [(3.00*10^8m/s)*(8.85*10^-^1^2C^2/N.m^2)]}

E_r_m_s= 6.81 *10^5N/c

b) to find the magnetic field in the electromagnetic wave emitted by the laser we use:

B_r_m_s = E_r_m_s / C;

= 6.81*10^5 N/c / 3*10^8m/s;

B_r_m_s = 2.27*10^3 T

8 0
3 years ago
Read 2 more answers
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