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zzz [600]
3 years ago
8

A 55.0-kg lead ball is dropped from the Leaning Tower of Pisa. The tower is 55.0 m high. What is the speed of the ball after it

has traveled 4.20 m downward
Physics
1 answer:
PolarNik [594]3 years ago
4 0

Answer:

The speed of the ball is 9.07 m/s.

Explanation:

Given that,

Mass of the lead ball, m = 55 kg

Height of the tower, h = 55 m

We need to find the speed of the ball it has traveled 4.20 m downward, x = 4.2 meters

The initial speed of the ball will be 0 as it was at rest initially. Let v is the speed of the ball after it has traveled 4.20 m downward. It is a case of equation of motion such that :

v^2-u^2=2ax

v^2=2ax

Here, a = g

v^2=2\times 9.8\times 4.2

v = 9.07 m/s

So, the speed of the ball is 9.07 m/s. Therefore, this is the required solution of given condition.

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A 15-kg block at rest on a horizontal frictionless surface is attached to a very light ideal spring of force constant 450 N/m. T
den301095 [7]

Answer:

0.266 m

Explanation:

Assuming the lump of patty is 3 Kg then applying the principal of conservation of linear momentum,

P= mv where p is momentum, m is mass and v is the speed of an object. In this case

m_pv_p=v_c(m_p+m_b) where sunscripts p and b represent putty and block respectively, c is common velocity.

Substituting the given values then

3*8=v(15+3)

V=24/18=1.33 m/s

The resultant kinetic energy is transferred to spring hence we apply the law of conservation of energy

0.5(m_p+m_b)v_c^{2}=0.5kx^{2} where k is spring constant and x is the compression of spring. Substituting the given values then

(3+15)*1.33^{2}=450*x^{2}\\x\approx 0.266 m

7 0
3 years ago
A series of optical telescopes produced an image that has a resolution of about 0.00350 arc second.
Mila [183]

Answer:

The smallest diameter is D =122 \ m

Explanation:

From the question we are told that

       The resolution of the telescope is \theta  =  0.00350 \ arc \ second

           The wavelength is  \lambda = 1.70 \mu m = 1.70 *10^{-6} \ m

From the question we are told that

        1 arc \ sec = \frac{1}{3600^o}

So      0.00350 \ arc \ second = x

Therefore

             x =  0.00350  *  \frac{1}{3600 }

              x = ( 9.722*10^{-7} )^o

Now  1^o  =  \frac{\pi}{180}

   So  (9.722*10^{-7})^o =  \theta

  =>    \theta  =  (9.722*10^{-7}) * \frac{\pi}{180}

           \theta  =  1.69*10^{-8} rad

The smallest diameter is mathematically represented  as

          D = \frac{1.22 \lambda }{\theta  }

substituting values

           D = \frac{1.22 * 1.7 *10^{-6}} {1.69 *10^{-8}  }

           D =122 \ m

   

6 0
3 years ago
Which of these is a type of matter that rarely interacts with other matter and is created in some nuclear fusion reactions withi
marysya [2.9K]

Answer:

the answer is helium you already know that because im in college on my way to the military next week bless me

Explanation:

7 0
3 years ago
The breaking car had 10,000 J of kinetic energy before breaking after breaking it had 2000 J of kinetic energy. How much thermal
WINSTONCH [101]

Answer:

8000J

Explanation:

The kinetic energy of the car lost during breaking are converted to thermal energy and are gained by the brakes.

Kinetic energy loss by car = thermal energy gained by brakes.

∆K.E = ∆T.E ....1

The Kinetic energy loss by car can be expressed as;

∆K.E = K.E1 - K.E2

Initial K.E = K.E1 = 10000J

Final K.E = K.E2 = 2000J

∆K.E= 10000J - 2000J = 8000J

From equation 1,

∆K.E = ∆T.E

∆T.E = 8,000J

thermal energy gain by brakes = 8,000J

8 0
3 years ago
Several light bulbs, each of resistance 1.5 Ω, are connected in a series across a 120 V source of emf. If the current through th
Leni [432]
<h3><u>Answer;</u></h3>

40 light bulbs

<h3><u>Explanation</u>;</h3>

The total resistance of components or bulbs in series is given as the sum of resistance of all the components.

Thus; if there are bulbs in series each with a resistance of 1.5 Ω, the the total resistance will be; 1.5nΩ

From the ohms law;

V = IR , where V is the voltage, I is the current and R is the resistor.

Thus; R = V/i

         R = 120/2

            = 60 Ω

But, there are n bulbs each with 1.5 Ω; thus there are;

n = 60/1.5

<u>  = 40 Bulbs </u>

7 0
3 years ago
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