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2 years ago

Calculate the tangential speed of a yo-yo twirled at the end of a 4 meter long string at 2 revolutions per second.

Physics

1 answer:

Oliga [24]2 years ago

8 0

Answer: 50 m/s

Explanation: speed v = 2· pi·n·r = 2· 3.14· 2 s^-1· 4 m

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2. (1) A piece of rubber is 50 cm long when a weight of

Harrizon [31]

Answer

By F = -kx {-ve just indicating the sign of the force}

=>35 = k x (85-50) x 10^-2

=>k = 100 N/m

Again by F = -kx

8 0

2 years ago

A clam of mass 0.12 kg dropped by a seagull takes 3.0 s to hit the ground. [Neglect friction.]

jarptica [38.1K]

-- The acceleration of gravity is 9.8 m/s².
So if there's no air resistance, the speed of a falling object
always increases by 9.8 m/s for every second it falls.

   Speed = (original speed) + (gravity x falling time)

-- If it has no vertical speed when it started, then at the end
of 3 seconds, its speed is

   = (0)    + (9.8 m/s² x 3 sec)

   Velocity = 29.4 m/s downward .

6 0

2 years ago

A force in the +x -direction with magnitude F(x)=18.0N−(0.530N/m)x is applied to a 7.90 kg box that is sitting on the horizontal

dsp73

Answer:

$v\approx 8.570\,\frac{m}{s}$

Explanation:

The equation of equlibrium for the box is:

$\Sigma F_{x} = 18\,N-(0.530\,\frac{N}{m} )\cdot x = (7.90\,kg)\cdot a$

The formula for the acceleration, given in $\frac{m}{s^{2}}$ , is:

$a = \frac{18\,N-(0.530\,\frac{N}{m} )\cdot x}{7.90\,kg}$

Velocity can be derived from the following definition of acceleration:

$a = v\cdot \frac{dv}{dx}$

$v\, dv = a\, dx$

$\frac{1}{2}\cdot v^{2} = \int\limits^{17\,m}_{0\,m} {\frac{18\,N-(0.530\,\frac{N}{m} )\cdot x}{7.90\,kg} } \, dx$

$\frac{1}{2}\cdot v^{2} =\frac{18\,N}{7.90\,kg} \int\limits^{17\,m}_{0\,m}\, dx - \frac{0.530\,\frac{N}{m} }{7.90\,kg} \int\limits^{17\,m}_{0\,m} {x} \, dx$

$\frac{1}{2}\cdot v^{2} = (2.278\,\frac{m}{s^{2}})\cdot x |_{0\,m}^{27\,m}-(0.034\,\frac{1}{s^{2}})\cdot x^{2}|_{0\,m}^{27\,m}$

$v =\sqrt{2\cdot[(2.278\,\frac{m}{s^{2}})\cdot x |_{0\,m}^{27\,m}-(0.034\,\frac{1}{s^{2}})\cdot x^{2}|_{0\,m}^{27\,m}] }$

The speed after the box has travelled 17 meters is:

$v\approx 8.570\,\frac{m}{s}$

3 0

2 years ago

A satellite that goes around the earth once every 24 hours iscalled a geosynchronous satellite. If a geosynchronoussatellite is

lesantik [10]

Answer:

35870474.30504 m

Explanation:

r = Distance from the surface

T = Time period = 24 h

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

m = Mass of the Earth = 5.98 × 10²⁴ kg

Radius of Earth = $6.38\times 10^6\ m$

The gravitational force will balance the centripetal force

$\dfrac{GMm}{R^2}=m\dfrac{v^2}{R}\\\Rightarrow v=\sqrt{\dfrac{GM}{R}}$

$T=\dfrac{2\pi r}{v}\\\Rightarrow T=\dfrac{2\pi r}{\sqrt{\dfrac{GM}{r}}}$

From Kepler's law we have relation

$T^2=\dfrac{4\pi^2r^3}{GM}\\\Rightarrow r^3=\dfrac{T^2GM}{4\pi^2}\\\Rightarrow r=\left(\dfrac{(24\times 3600)^2\times 6.67\times 10^{-11}\times 5.98\times 10^{24}}{4\pi^2}\right)^{\dfrac{1}{3}}\\\Rightarrow r=42250474.30504\ m$

Distance from the center of the Earth would be

$42250474.30504-6.38\times 10^6=\mathbf{35870474.30504\ m}$

8 0

2 years ago

What is the approximate efficiency of the engine?

Elodia [21]

Answer:91% I think

Explanation:

8 0

2 years ago

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