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3 years ago

8

An object of mass 0.9 kg is attached to a massless string of length 3 m, and swung with 3 a tangential velocity of 3 m/s. What i

s the force exerted by the string on the object?
a. 32 N
b. 27 N
C. 5 N
d. 2.7 N

1 answer:

Ganezh [65]3 years ago

5 0

Answer:

d

Explanation:

you have to divide by the square root and then use the factorial to multiply

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In a novel from 1866 the author describes a spaceship that is blasted out of a cannon with a speed of about 11.000 m/s. The spac

Elan Coil [88]

Answer:

$a=0.284\ m/s^2$

Explanation:

Given that,

Initially, the spaceship was at rest, u = 0

Final velocity of the spaceship, v = 11 m/s

Distance accelerated by the spaceship, d = 213 m

We need to find the acceleration experienced by the occupants of the spaceship during the launch. It is a concept based on the equation of kinematics. Using the third equation of motion to find acceleration.

$v^2-u^2=2ad\\\\a=\dfrac{v^2-u^2}{2d}\\\\a=\dfrac{(11)^2-(0)^2}{2\times 213}\\\\a=0.284\ m/s^2$

So, the acceleration experienced by the occupants of the spaceship is $0.284\ m/s^2$ .

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3 years ago

A skier is pulled by a towrope up a frictionless ski slope that makes an angle of 12 with the horizontal. The rope moves paralle

MaRussiya [10]

Answer

given,

angle with horizontal = 12°

constant speed with the slope = 1 m/s

Work on the skier to move 8 m = 900 J

a) Rope moved with constant speed

change in kinetic energy is equal to zero

work done by the force of the rope = 900 J

rate of force of the rope when skier move with

b) v = 1 m/s total time t = 8 s

Power = $\dfrac{work\ done}{time}$

= $\dfrac{900}{8}$

= 112.5 W

c) v = 2 m/s total time t = 8/2 = 4 s

Power = $\dfrac{work\ done}{time}$

= $\dfrac{900}{4}$

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3 years ago

Waves with a higher frequency have a ______ wavelength and _____ energy. *

Helen [10]

Answer:

bsdk........m.....MC...

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3 years ago

How do u work out b)<br> ii)

Yakvenalex [24]

Explanation:

potential difference = current × risstance

see part a point3 current is same

v= IR

v = 8×0.2

v= 1.6

see part a point 5

this potential difference is less than cell

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3 years ago

Male Rana catesbeiana bullfrogs are known for their loud mating call. The call is emitted not by the frog's mouth but by its ear

Sidana [21]

Answer:

The amplitude of the eardrum's oscillation is 6.65×10^-13 m.

Explanation:

Given data:

The sound has a frequency of 262 Hz

The sound level is 84 dB

The air density is 1.21 kg/m^3

The speed of sound is 346 m/s

Solution:

As, Intensity of sound is given by,

I = Io×10^(s/10 db)

I = 2×π^2×ρ×v×f^2×Sm^2

Thus,

Sm = √(Io×10^(s/10 db)) / √( 2×π^2×ρ×v×f^2)

Now, put the values,

Sm = √( 10^-12 × 10^(84/10) ) / √( 2×(3.14)^2×1.21×346×(262)^2 )

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Sm = √0.443×10^-12

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3 years ago