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2 years ago

Which of the following classifications of star temperature is hottest? K O B M

Physics

2 answers:

nikdorinn [45]2 years ago

6 0

O is the correct answer!

Harrizon [31]2 years ago

5 0

The hottest would be the O type and the coolest is M

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Two forces,one of 12 N and another of 24 N,act on a body in such a way that they make an angle of 90degree with each other.Find

jok3333 [9.3K]

Answer:

26.83 N.

Explanation:

If the angle between two vector is 90°, to get the resultant, we use Pythagoras theorem.

a² = b²+c²......................... Equation 1

Where a = R = Resultant, b = 12 N, c = 24 N.

Substitute these values into equation 1

R² = 12²+24²

R² = 144+576

R² = 720

√R² = √720

R = 26.83 N.

Hence, the result of the two force is 26.83 N.

4 0

2 years ago

I need help in my physics class and show me how it’s done

Korolek [52]

If we have the angle and magnitude of a vector A we can find its Cartesian components using the following formula

$A_x = |A|cos(\alpha)\\\\A_y = |A|sin(\alpha)$

Where | A | is the magnitude of the vector and $\alpha$ is the angle that it forms with the x axis in the opposite direction to the hands of the clock.

In this problem we know the value of Ax and Ay and we need the angle $\alpha$ .

Vector A is in the 4th quadrant

So:

$A_x = 6\\\\A_y = -6.5$

So:

$|A| = \sqrt{6^2 + (-6.5)^2}\\\\|A| = 8.846$

So:

$Ay = -6.5 = 8.846cos(\alpha)\\\\sin(\alpha) = \frac{-6.5}{8.846}\\\\sin(\alpha) = -0.7348\\\\\alpha = sin^{- 1}(- 0.7348)$

$\alpha$ = -47.28 ° +360° = 313 °

$\alpha$ = 313 °

Option 4.

4 0

3 years ago

A 1369.4 kg car is traveling at 28.9 m/s when the driver takes his foot off the gas pedal. It takes 5.1 s for the car to slow do

Darya [45]

Answer:

F = 2389.603 N

Explanation:

Given:

Mass m = 1,369.4 kg

Initial velocity u = 28.9 m/s

Final velocity v = 20 m/s

Time t = 5.1 s

Find:

Net force

Computation:

a = (v - u)/t

a = (20 - 28.9)/5.1

a = -1.745 m/s²

F = ma

F = (1369.4)(1.745)

F = 2389.603 N

7 0

2 years ago

A 0.50 kg toy is attached to the end of a 1.0 m very light string. The toy is whirled in a horizontal circular path on a frictio

xenn [34]

Answer:

The maximum speed will be 26.475 m/sec

Explanation:

We have given mass of the toy m = 0.50 kg

radius of the light string r = 1 m

Tension on the string T = 350 N

We have to find the maximum speed without breaking the string

For without breaking the string tension must be equal to the centripetal force

So $T=\frac{mv^2}{r}$

So $350=\frac{0.5\times v^2}{1}$

$v^2=700$

v = 26.475 m /sec

So the maximum speed will be 26.475 m/sec

6 0

2 years ago

A bird flies at a speed of 2.3 m/s if it has 14 j of kinetic energy what is the mass

Sidana [21]

Kinetic Energy = 1/2 * mv²

Kinetic Energy = 14 J, v = 2.3 m/s , m = ?

14 = 1/2 * m* 2.3²

14 = 0.5*m*2.3*2.3

m = 14 / (0.5*2.3*2.3)

m = 5.29 kg.

Mass = 5.29 kg.

7 0

3 years ago

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