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adoni [48]
4 years ago
15

A 0.80 kg object tied to the end of a 2.0 m string swings as a pendulum. At the lowest point of its swing, the object has a kine

tic energy of 10 J. Determine the speed of the object at the instant when the string makes an angle of 50o with the vertical.
Physics
1 answer:
devlian [24]4 years ago
6 0

Answer:

3.32 m/s

Explanation:

From the law of conservation of energy, the sum of mechanical and kinetic energy should be equal to the 10 J given. Potential energy is given by mgh where m is mass, g is acceleration due to gravity and h is the height. For this case, h= l(1-cos\theta) and l is string length, given as 2 m, \theta is given as 50 degrees. Kinetic energy is given by 0.5mv^{2} and it is this velocity that is unknown.

10J=0.5\times 0.8kg\times v^{2}+ 0.8kg\times 9.81\times 2m(1-cos 50^{\circ})\\v\approx 3.32 m/s

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1. Cold front

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6 0
4 years ago
A machine part has the shape of a solid uniform sphere of mass 250 g and a diameter of 4.30 cm. It is spinning about a frictionl
zysi [14]

Answer:\alpha =9.302\ rad/s^2

Explanation:

Given

mass of sphere m=250\ gm

diameter of sphere d=4.30\ cm

radius r=\frac{4.30}{2}\ cm

f=0.0200\ N

friction will provide resisting torque so

f\times r=I\times \alpha

where I=\text{moment of Inertia}

f=\text{friction force}

\alpha =\text{angular acceleration}

I=\frac{2}{5}mr^2

0.02\times r=\frac{2}{5}mr^2\times \alpha

\alpha =\frac{5}{2r}\times f

\alpha =\frac{5}{2}\times \frac{2}{4.3\times 10^{-2}}\times 0.02

\alpha =9.302\ rad/s^2

(b)time taken to decrease its rotational speed by 21\ rad/s

t=\dfrac{\Delta \omega }{\alpha }

t=\dfrac{21}{9.302}

t=2.25\ s

6 0
3 years ago
Which statements about water are true? Choose more than one answer.
insens350 [35]

water is not found in the periodic table.

water has a lot of empty space.

I think that's it I know those to are true.

4 0
3 years ago
The energy levels of one‑electron ions are given by the equation En=(−2.18×10−18 J)(Z^2/n^2) where Z is atomic number and n is t
Alexxandr [17]

Answer:

\lambda=4.86*10^{-7}m

Explanation:

Using the given equation, we calculate the energy associated with the excited state n_i=8 and n_f=4

E_n=\frac{-2.18*10^-18J(Z^2)}{n^2}

Helium has an atomic number (Z) equal to 2, for n=8:

E_8=\frac{-2.18*10^{-18}J(2^2)}{8^2}\\E_8=-1.36*10{-19}J

For n=4:

E_4=\frac{-2.18*10^{-18}J(2^2)}{4^2}\\E_4=-5.45*10{-19}J

When an electron jumps from an energy level with greater energy E_i to one with lower energy E_f the wavelength of the emitted photon is given by:

\lambda=\frac{hc}{E_i-E_f}

h is the Planck constant and c the speed of light in vaccum. So, we have:

\lambda=\frac{hc}{E_8-E_4}\\\lambda=\frac{6.63*10^{-34}J \cdot s(3*10^8\frac{m}{s})}{-1.36*10{-19}J-(-5.45*10{-19}J)}\\\lambda=4.86*10^{-7}m

5 0
3 years ago
What are the conditions required for a rigid body to be in translational equilibrium?
kolezko [41]

Answer:

Explanation:

The condition for translation equilibrium is that is that the net force acting on the body must be zero.

The sum all the external forces acting on the body in horizontal as well as vertical direction must be zero.

∑Fₓ=0  and ∑Fy=0

now if the above two condition are satisfied the rigid body is said to be in  translational equilibrium.

God bless... hope this help to clear your doubt.

5 0
3 years ago
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