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2 years ago

Which of the following is not an SI unit. A. pound B. Newton C. Meter D. kilogram

Physics

1 answer:

Hunter-Best [27]2 years ago

3 0

Answer:

pound

Explanation:

Pound is an SAE or imperial unit not SI

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Damping is negligible for a 0.135-kg object hanging from a light, 6.30-N/m spring. A sinusoidal force with an amplitude of 1.70

IceJOKER [234]

Answer:

0.72 Hz minimum frequency

Explanation:

When the damping is negligible,Amplitude is given as

$A = (F/m)/[\sqrt{(\omega ^{^{2}}-\omega _{o}^{2}})^2$

here $\omega _{o}^{2}= k/m$ = (6.30)/(0.135) = 46.67 N/m kg

$F / mA$ = 1.70/(0.135)(0.480) = 26.2 N/m kg

From the above equation , rearranging for ω,

$\omega ^{2}= \omega _{o}^{2}\pm F/m$

⇒ ω² =46.67 ± 26.2 = 72.87 or 20.47

⇒ ω = 8.53 or 4.52 rad/s

Frequency = f

ω=2 π f

⇒ f = ω / 2π = 8.53 /6.28 or 4.52 / 6.28 = 1.36 Hz or 0.72 Hz

The lower frequency is 0.72 Hz and higher is 1.36 Hz

8 0

3 years ago

A toroid has a square cross section with the length of an edge equal to the radius of the inner surface. The ratio of the magnit

Tasya [4]

Answer:

Explanation:

7 0

3 years ago

Read 2 more answers

A 4.75 kg block is sent up a ramp inclined at an angle ????=31.5° from the horizontal. It is given an initial velocity ????0=15.

Colt1911 [192]

Answer:

d = 13.7 m

Explanation:

When block is moving upwards along the inclined plane

then the block is decelerated due to gravity as well as due to friction and speed of the block by which it is projected upwards is given

$v_i = 15 m/s$

deceleration caused to the block due to net force opposite to the motion is given as

$F = - mg sin\theta - \mu mgcos\theta$

$a = \frac{F}{m}$

$a = -g(sin\theta + \mu cos\theta)$

since block is sliding on the inclined plane

so here we can say that the coefficient of the friction must be kinetic friction here

$a = -9.81(sin31.5 + 0.368 cos31.5)$

$a = - 8.2 m/s^2$

now for finding the distance upto which it will stop is given as

$v_f^2 - v_i^2 = 2 a d$

$0 - 15^2 = 2(-8.2) d$

$d = 13.7 m$

8 0

3 years ago

A ball is launched from a 300m cliff and lands 380m away from the cliff in 9 seconds. Calculate the initual speed and the angle

Marysya12 [62]

Answer:

Explanation:

Given

Maximum height H = 300m

Range (horizontal distance) = 380m

Required

Initial speed U and the angle of the ball when it was launched.

Range = U√2H/g

380 = U√2(300)/9.8

380 = U√600/9.8

380 = 7.8246U

U = 380/7.8246

U = 48.57m/s

The initial speed is 48.57m/s

b) Using the formula for calculating time of flight;

T = 2Usin theta/g

9 = 2(48.57)sin theta/9.8

9*9.8 = 97.14sin theta

88.2 = 97.14sin theta

88.2/97.14 = sin theta

sin theta = 0.9079

theta = sin^-1(0.9079)

theta = 65.23°

hence the angle when the ball was launched is 65.23°

6 0

2 years ago

Do you think there are any forces acting on the ball when it is rolling around the inside of the roll of tape? Why or why not?

deff fn [24]

Yes because it is sticky

5 0

2 years ago