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Gennadij [26K]
4 years ago
12

A pitcher hurls a 0.25kg ball around a vertical circular path of radius 0.6 m, applying a tangential force of 30N, before releas

ing it at the bottom of the circle (underhand pitch). If the speed of the ball at the top of the circle was been 15 m/s, what will be the speed just after it's released?
Physics
1 answer:
Sever21 [200]4 years ago
3 0
Thank you for posting your question here at brainly. Below is the solution:

Ke up top = 1/2*.25 *225 
<span>gain of Pot energy = .25*9.81*1.2 </span>
<span>work input = (1/4)(2 pi *.6)*30 </span>

<span>so </span>
<span>sum of those 3 energies = </span>
<span>(1/2)(.25)v^2</span>
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An incident light ray hits a plane surface at 20 degrees (hint: remember where the angle is measured from). What is the angle be
Juli2301 [7.4K]

Answer:

40^{\circ}

Explanation:

The incident angle is 20^{\circ} so the reflected ray will also be 20^{\circ}.

The normal divides the angle between the incident and reflected ray. So, the angle between them would be

\text{Incident angle}+\text{Reflected angle}=20+20=40^{\circ}

So, the angle between the incident and reflected rays is 40^{\circ}.

8 0
3 years ago
A certain radioactive nuclide decays with a disintegration constant of 0.0178 h-1.
umka21 [38]

Explanation:

Given that,

The disintegration constant of the nuclide, \lambda=0.0178\ h^{-1}

(a) The half life of this nuclide is given by :

t_{1/2}=\dfrac{ln(2)}{\lambda}

t_{1/2}=\dfrac{ln(2)}{0.0178}

t_{1/2}=38.94\ h

(b) The decay equation of any radioactive nuclide is given by :

N=N_oe^{-\lambda t}

\dfrac{N}{N_o}=e^{-\lambda t}

Number of remaining sample in 4.44 half lives is :

t_{1/2}=4.44\times 38.94

t_{1/2}=172.89\ h^{-1}

So, \dfrac{N}{N_o}=e^{-0.0178\times 172.89}

\dfrac{N}{N_o}=0.046

(c) Number of remaining sample in 14.6 days is :

t_{1/2}=14.6\times 24

t_{1/2}=350.4\ h^{-1}

So, \dfrac{N}{N_o}=e^{-0.0178\times 350.4}

\dfrac{N}{N_o}=0.0019

Hence, this is the required solution.

4 0
4 years ago
(50 POINT REWARD) PHYSICS, IMAGE IS ATTACHED BELOW (TRIG/VECTORS)
daser333 [38]

Answer:

Explanation:

The east vector is 10cos45 = 7.07 km

The north vector is 10sin45 - 10 = -2.93 km

The displacement magnitude is

d = √(7.07² + 2.93²) = 7.65... = 8 km

θ = arctan(-2.93/7.07) = -22.51... = 23° south of east.

6 0
3 years ago
A new Cat6 cable run and keystone jack were installed for a workstation that was moved from the sales department to the shipping
blagie [28]

Answer:

The cable run exceeds the specifications for Ethernet over twisted pair.

Explanation:

When using an Ethernet network, the network's router also serves as a bridge to the Internet. The router connects to the modem, which carries the Internet signal, sending and receiving data packet requests and routing them to the proper computers on the network. Ethernet is a way of connecting computers together in a local area network or LAN. It has been the most widely used method of linking computers together in LAN s since the 1990 s. The basic idea of its design is that multiple computers have access to it and can send data at any time.

4 0
3 years ago
The filament of a certain lamp has a resistance that increases linearly with temperature. When a constant voltage is switched on
alekssr [168]

Answer:

The change in temperature is \Delta T  = 1795 K

Explanation:

From the question we  are told that

   The temperature coefficient is  \alpha  =  4 * 10^{-3 }\  k^{-1 }

The resistance of the filament is mathematically represented as

           R  =  R_o [1 + \alpha  \Delta T]

Where R_o is the initial resistance

Making the change in temperature the subject of the formula

     \Delta T = \frac{1}{\alpha } [\frac{R}{R_o} - 1 ]

Now from ohm law

           I = \frac{V}{R}

This implies that current varies inversely with current so

           \frac{R}{R_o} = \frac{I_o}{I}

Substituting this we have

       \Delta T  = \frac{1}{\alpha } [\frac{I_o}{I} - 1 ]

From the question we are told that

    I  = \frac{I_o}{8}

Substituting this we have

   \Delta T  = \frac{1}{\alpha } [\frac{I_o}{\frac{I_o}{8} } - 1 ]

=>     \Delta T  = \frac{1}{3.9 * 10^{-3}} (8 -1 )

        \Delta T  = 1795 K

6 0
4 years ago
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