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iren [92.7K]
3 years ago
8

Two vectors, X and Y, form a right angle. Vector X is 48 inches long and vector Y is 14 inches long. The length of the

Physics
1 answer:
Andrews [41]3 years ago
3 0

Answer:

The length of the resultant vector is 50 inches

Explanation:

Use the Pythagorean theorem to find the answer, since the addition of these two perpendicular vectors will have a magnitude (length) equal to the hypotenuse of the right angle triangle formed by the two:

|Resultant|\,=\,\sqrt{x^2+y^2} \\|Resultant|\,=\,\sqrt{48^2+14^2}\\|Resultant|\,=\,\sqrt{2304+196}\\|Resultant|\,=\,\sqrt{2500}\\|Resultant|\,=\,50\,in  

The length of the resultant vector is 50 inches

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sun was created through the hypothesis

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4 years ago
A mass is attached to a vertical spring, which then goes into oscillation. At the high point of the oscillation, the spring is i
andrew-mc [135]

Answer:

0.34 sec

Explanation:

Low point of spring ( length of stretched spring ) = 5.8 cm

midpoint of spring = 5.8 / 2 = 2.9 cm

Determine the oscillation period

at equilibrum condition

Kx = Mg

g= 9.8 m/s^2

x = 2.9 * 10^-2 m

k / m = 9.8 / ( 2.9 * 10^-2 ) =  337.93

note : w = \sqrt{k/m}   = \sqrt{337.93} = 18.38 rad/sec

Period of oscillation =  2\pi  / w

                                  = 0.34 sec

8 0
3 years ago
Peg P is driven by the forked link OA along the path described by r = eu, where r is in meters. When u = p4 rad, the link has an
8_murik_8 [283]

Answer:

The transverse component of acceleration is 26.32 m/s^2 where as radial the component of acceleration is 8.77 m/s^2

Explanation:

As per the given data

u=π/4 rad

ω=u'=2 rad/s

α=u''=4 rad/s

r=e^u

So the transverse component of acceleration are given as

a_{\theta}=(ru''+2r'u')\\

Here

r=e^u\\r=e^{\pi/4}\\r=2.1932 m

r'=e^u.u'\\r'=2.1932 \times 2\\r'=4.3864 m

So

a_{\theta}=(ru''+2r'u')\\a_{\theta}=(2.1932\times 4+2\times 4.3864 \times 2)\\a_{\theta}=26.32 m/s\\

The transverse component of acceleration is 26.32 m/s^2

The radial component is given as

a_r=r''-r\theta'^2

Here

r''=e^u.u'^2+e^u u''\\r''=2.1932 \times (2)^2+2.1932\times 4\\r''=17.5456 m

So

a_r=r''-ru'^2\\a_r=17.5456-2.1932\times (2)^2\\a_r=8.7728 m/s^2

The radial component of acceleration is 8.77 m/s^2

6 0
3 years ago
• 2. A 70 kg man on a 100 kg boat throws a ball. The boat moves backwards 5 meters in 10 seconds. What is
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Answer:

0.5 m/sec

Explanation:

v=S÷t

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v=0.5 m/sec

6 0
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