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3 years ago

Help. I need to find the degree using the sine law.

Physics

1 answer:

Marina CMI [18]3 years ago

4 0

Use pythagorean theorem
$a ^{2} + b {}^{2} = c {}^{2}$
to find the opposite side, which is 7.3

so then you can just use inverse sinA=7.3/10 which equals 46.9 degrees

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The first stage in the GAS model of stress is

Vladimir [108]

The first stage in the Gas model of stress is alarm and mobilization. So the correct option in regards to the given question is option “d”. Hans Selye is the person that evolved this model and he has explained this model in complete details. He has broken down his model into three stages. The first stage involves alarm and mobilization. The second stage includes resistance. The third and the final stage include the exhaustion stage. These are the stages that an organism goes through to restore back the balance when stress is exerted from outside.

8 0

3 years ago

an athlete in a hammer-throw event swings a 7.0-kilogram hammer in a horizontal circle at a constant speed of 12 meters per seco

Semenov [28]

Answer:

ac = 72 m/s²

Fc = 504 N

Explanation:

We can find the centripetal acceleration of the hammer by using the following formula:

$a_c = \frac{v^2}{r}$

where,

ac = centripetal acceleration = ?

v = constant speed = 12 m/s

r = radius = 2 m

Therefore,

$a_c = \frac{(12\ m/s)^2}{2\ m}$

ac = 72 m/s²

Now, the centripetal force applied by the athlete on the hammer will be:

$F_c = ma_c\\F_c = (7\ kg)(72\ m/s^2)$

Fc = 504 N

6 0

2 years ago

Uniformly charged ring with 180 nC/m and radius R= 58 cm. Find the magnitude of the electric field in KN/C at a point P on the a

raketka [301]

Answer:

3.135 kN/C

Explanation:

The electric field on the axis of a charged ring with radius R and distance z from the axis is E = qz/{4πε₀[√(z² + R²)]³}

Given that R = 58 cm = 0.58 m, z = 116 cm = 1.16m, q = total charge on ring = λl where λ = charge density on ring = 180 nC/m = 180 × 10⁻⁹ C/m and l = length of ring = 2πR. So q = λl = λ2πR = 180 × 10⁻⁹ C/m × 2π(0.58 m) = 208.8π × 10⁻⁹ C and ε₀ = permittivity of free space = 8.854 × 10⁻¹² F/m

So, E = qz/{4πε₀[√(z² + R²)]³}

E = 208.8π × 10⁻⁹ C × 1.16 m/{4π8.854 × 10⁻¹² F/m[√((1.16 m)² + (0.58 m)²)]³}

E = 242.208 × 10⁻⁹ Cm/{35.416 × 10⁻¹² F/m[√(1.3456 m² + 0.3364 m²)]³}

E = 242.208 × 10⁻⁹ Cm/35.416 × 10⁻¹² F/m[√(1.682 m²)]³}

E = 6.839 × 10³ Cm²/[1.297 m]³F

E = 6.839 × 10³ Cm²/2.182 m³F

E = 3.135 × 10³ V/m

E = 3.135 × 10³ N/C

E = 3.135 kN/C

3 0

3 years ago

A 1.90-m-long barbell has a 25.0 kg weight on its left end and a 37.0 kg weight on its right end. if you ignore the weight of th