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2 years ago

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A yo-yo can be thought of as a solid cylinder of mass m and radius r that has a light string wrapped around its circumference (s

ee below). One end of the string is held fixed in space. If the cylinder falls as the string unwinds without slipping, what is the acceleration of the cylinder? (Enter the magnitude. Use the following as necessary: g, m, and r.)

1 answer:

lbvjy [14]2 years ago

4 0

Answer:

6.5 m/s^2

Explanation:

The net force acting on the yo-yo is

F_net = mg-T

ma=mg-T

now T= mg-ma

net torque acting on the yo-yo is

τ_net = Iα

I= moment of inertia (= 0.5 mr^2 )

α = angular acceleration

τ_net = 0.5mr^2(a/r)

Tr= 0.5mr^2(a/r)

(mg-ma)r=0.5mr^2(a/r)

a(1/2+1)=g

a= 2g/3

a= 2×9.8/3 = 6.5 m/s^2

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A steel playground slide is 5.25 m long and is raised 2.75 m on one end. A 45.0 kg child slides down from the top starting at re

Temka [501]

Answer:

$F=32.24N$

Explanation:

From the question we are told that:

Height $h= 2.75 m$

Length $l = 5.25 m$

Mass $m=45kg$

Final speed $v_f=6.81$

Generally the equation for Potential Energy P.E is mathematically given by

$P.E=mgh$

Therefore

Initial potential energy

$P.E_1=45*9.8*2.75 \\\\P.E_1= 1212.75 J$

Generally the equation for Kinetic Energy K.E is mathematically given by

$K.E=0.5mv^2$

Therefore

Final kinetic energy

$K.E_2= 1/2*45*6.81*6.81 \\\\K.E_2= 1043.46J$

Generally the equation for Work_done is mathematically given by

$W=P.E_1-K.E_2\\\\W=169.3$

Therefore

$F=\frac{W}{d}\\\\F=\frac{169.3}{5.25}$

$F=32.24N$

8 0

2 years ago

A 225-kg object and a 525-kg object are separated by 3.80 m. (a) Find the magnitude of the net gravitational force exerted by th

pav-90 [236]

Answer: $F_{net}=3.383\times 10^{-7}\ N$

Explanation:

Given

Mass of first object $m_1=225\ kg$

Mass of second object $m_2=525\ kg$

Distance between them $d=3.8\ m$

$m_3=61\ kg$ object is placed between them

So force exerted by $m_1$ on $m_3$

$F_{13}=\frac{Gm_1m_3}{1.9^2}$

$F_{13}=\frac{6.674\times 10^{-11}(225\times 61)}{1.9^2}$

$F_{13}=2.5374141274×10^{−7}\ N$

Force exerted by $m_2\ on\ m_3$

$F_{23}=\frac{Gm_2m_3}{1.9^2}$

$F_{23}=\frac{6.674\times 10^{-11}(525\times 61)}{1.9^2}$

$F_{23}=5.920632964\times 10^{-7}\ N$

So net force on $m_3$ is

$F_{net}=F_{23}-F_{13}$

$F_{net}=5.920632964\times 10^{-7}-2.5374141274\times 10^{-7}$

$F_{net}=3.383\times 10^{-7}\ N$

i.e. net force is towards $m_2$

(b)For net force to be zero on $m_3$ , suppose

So force exerted by $m_1$ and $m_2$ must be equal

$F_{13}=F_{23}$

$\Rightarrow \frac{Gm_1m_3}{x^2}=\frac{Gm_2m_3}{(3.8-x)^2}$

$\Rightarrow \frac{m_1}{x^2}=\frac{m_2}{(3.8-x)^2}$

$\Rightarrow (\frac{3.8-x}{x})^2=\frac{m_2}{m_1}$

$\Rightarrow \frac{3.8-x}{x}=\sqrt{\frac{525}{225}}$

$\Rightarrow 3.8-x=1.52752x$

$\Rightarrow 3.8=2.52x$

$\Rightarrow x=1.507\ m$

4 0

2 years ago

You are watching an archery tournament when you start wondering how fast an arrow is shot from the bow. Remembering your physics

olga nikolaevna [1]

Answer:112.82 m/s

Explanation:

Given

range of arrow=68 m

$Angle=3^{\circ}$

as the arrow travels it acquire a vertical velocity $v_y$

$v_y=u+at$

$v_y=0+9.81\times t$ -------1

Range is given by

R=ut

where u=initial velocity

$68=u\times t$

$t=\frac{68}{u}$

substitute the value of t in eqn 1

$v_y=9.81\times \frac{68}{u}$

$v_y\times u=9.81\times 68=667.08$ --------2

and $tan(3)=\frac{v_y}{u}$

$v_y=utan(3)=0.0524u$

substitute it in 2

$0.0524 u^2=667.08$

$u^2=12,728.644$

u=112.82 m/s

6 0

3 years ago

) A satellite of mass m has an orbital period T when it is in a circular orbit of radius R around the earth. If the satellite in

Mrrafil [7]

Answer:

A) T.

Explanation:

Kepler's third law states that the orbital period (T) of a satellite is related with the radius (R) and the mass of the object (M) it orbits:

$T=\frac{2\pi R^{\frac{3}{2}}}{\sqrt{GM}}$

So the orbital period is independent of the mass of the satellite, that means no matter the mass every satellite at a radius R around the earth have an orbital period A.

4 0

3 years ago

A rigid tank contains nitrogen gas at 227 °C and 100 kPa gage. The gas is heated until the gage pressure reads 250 kPa. If the a

aleksley [76]

Answer:

T₂ =602 °C

Explanation:

Given that

T₁ = 227°C =227+273 K

T₁ =500 k

Gauge pressure at condition 1 given = 100 KPa

The absolute pressure at condition 1 will be

P₁ = 100 + 100 KPa

P₁ =200 KPa

Gauge pressure at condition 2 given = 250 KPa

The absolute pressure at condition 2 will be

P₂ = 250 + 100 KPa

P₂ =350 KPa

The temperature at condition 2 = T₂

We know that

$\dfrac{T_2}{T_1}=\dfrac{P_2}{P_1}\\T_2=T_1\times \dfrac{P_2}{P_1}\\T_2=500\times \dfrac{350}{200}\ K\\$

T₂ = 875 K

T₂ =875- 273 °C

T₂ =602 °C

5 0

3 years ago