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2 years ago

9

If v = 5.00 meters/second and makes an angle of 60° with the negative direction of the y–axis, calculate all possible values of

vx.

1 answer:

Goshia [24]2 years ago

8 0

Vx = + 4.33 m/s. Hope this helps

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A charge q1 of -5.00 x 10^-9 C and a charge q2 of -2.00x 10^-9 C are separated by a distance of 40.0 cm. Find the equilibrium po

Blababa [14]

The magnitude of charge on a proton and electron is the same, 1.602 x 10-19 C. Protons are +, and electrons -.

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2 years ago

The position of an object in simple harmonic motion is defined by the function y = (0.50 m) sin (πt/2). Determine the maximum sp

Gnoma [55]

The maximum speed of the object under simple harmonic motion is 0.786 m/s.

The given parameters:

Position of the particle, y = 0.5m sin(πt/2)

<h3>Wave equation for simple harmonic motion;</h3>

y = A sin(ωt + Ф)

where;

A is the amplitude = 0.5 m
ω is the angular speed = π/2

The maximum speed of the object is calculated as follows;

$V_{max} = A \omega\\\\V_{max} = 0.5 \times \frac{\pi}{2} = \frac{\pi}{4} \ m/s = 0.786 \ m/s$

Thus, the maximum speed of the object under simple harmonic motion is 0.786 m/s.

Learn more about simple harmonic motion here: brainly.com/question/17315536

5 0

1 year ago

The alpha line in the balmer series of the hydrogen spectrum consists of light having a wavelength of 6.56. calculate the freque

guajiro [1.7K]

The alpha line in the Balmer series is the transition from n=3 to n=2 and with the wavelength of λ=656 nm = 6.56*10^-7 m. To get the frequency we need the formula: v=λ*f where v is the speed of light, λ is the wavelength and f is the frequency, or c=λ*f. c=3*10^8 m/s. To get the frequency: f=c/λ. Now we input the numbers: f=(3*10^8)/(6.56*10^-7)=4.57*10^14 Hz. So the frequency of the light from alpha line is f= 4.57*10^14 Hz.

5 0

3 years ago

A 53.0-kg skater is traveling due east at a speed of 2.90 m/s. A 72.0-kg skater is moving due south at a speed of 6.20 m/s. They

soldier1979 [14.2K]

Answer:

a. $\thta=71^{\circ}$

b. $v_f=3.78 m/s$

Explanation:

We are given that

$m_1=53 kg$

$v_1=2.9 m/s$

$m_2= 72 kg$

$v_2=6.2 m/s$

a.We have to find the angle

$\theta=tan^{-1}{\frac{m_2v_2}{m_1v_1}=\frac{72(6.2)}{53(2.9)}=71^{\circ}$

$\theta=71^{\circ}$

b. We have to find the speed $v_f$

According to law of conservation of momentum

$m_1v_1=(m_1+m_2)v_fcos\theta$

$53(2.9)=(53+72)v_fcos 71^{\circ}=40.7 v_f$

$v_f=\frac{153.7}{40.7}=3.78 m/s$

3 0

2 years ago

PLEASE HELP ME WITH THIS ONE QUESTION

Ipatiy [6.2K]

k = $\dfrac{ (\dfrac{h}{ \lambda} )^{2} }{2m}$

k = (6.626×10-¹⁹/590 × 10-⁹ )^{2} /2 × 1.673 × 10-²⁷

k = (1.12 × 10-³⁰)^2/3.346×10-²⁷

k = 1.25 × 10-⁶⁰ /3.346×10-²⁷

k = 0

ldk why, my answer is coming this :(

4 0

2 years ago