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Travka [436]
3 years ago
8

Bowling balls are roughly the same size, but come in a variety of weights. Given its official radius of roughly 0.110 m, calcula

te the heaviest bowling ball that will float in a fluid of density 1.100.
Physics
1 answer:
velikii [3]3 years ago
5 0

Answer:

6.1328 kg

60.16284 N

Explanation:

r = Radius of ball = 0.11 m

\rho = Density of fluid = 1.1\times 10^3\ kg/m^3 (Assumed)

g = Acceleration due to gravity = 9.81 m/s²

m = Mass of ball

V = Volume of ball = \frac{4}{3}\pi r^3

The weight of the bowling ball will balance the buouyant force

W=F_b\\\Rightarrow mg=V\rho g\\\Rightarrow m=\frac{V\rho g}{g}\\\Rightarrow m=V\rho\\\Rightarrow m=\frac{4}{3}\pi 0.11^3\times 1.1\times 10^3\\\Rightarrow m=6.1328\ kg

The mass of the bowling ball will be 6.1328 kg

Weight will be 6.1328\times 9.81=60.16284\ N

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lidiya [134]
Series,effective resistance =R₁+R₂+R₃...
parallel,effective resistance 1/R=1/R₁ +1/R₂ +1/R₃...
 
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voltage V=9 V 
current I=V/R
           
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3 years ago
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A 500 kg block is attached to a horizontal spring that is at its equilibrium length, and whose force constant is 30 N/m. The blo
m_a_m_a [10]

Answer:

x = 0.396 m

Explanation:

The best way to solve this problem is to divide it into two parts: one for the clash of the putty with the block and another when the system (putty + block) compresses it is   spring

Data the putty has a mass m1 and velocity vo1, the block has a mass m2 .  t's start using the moment to find the system speed.

Let's form a system consisting of putty and block; For this system the forces during the crash are internal and the moment is preserved. Let's write the moment before the crash

    p₀ = m1 v₀₁

Moment after shock

    p_{f} = (m1 + m2) v_{f}

   p₀ = p_{f}

   m1 v₀₁ = (m1 + m2) v_{f}

  v_{f} = v₀₁ m1 / (m1 + m2)

   v_{f}= 4.4 600 / (600 + 500)

  v_{f} = 2.4 m / s

With this speed the putty + block system compresses the spring, let's use energy conservation for this second part, write the mechanical energy before and after compressing the spring

Before compressing the spring

   Em₀ = K = ½ (m1 + m2) v_{f}²

After compressing the spring

   E_{mf} = Ke = ½ k x²

As there is no rubbing the energy is conserved

   Em₀ = E_{mf}

   ½ (m1 + m2) v_{f}² = = ½ k x²

   x = v_{f} √ (k / (m1 + m2))

   x = 2.4 √ (11/3000)

   x = 0.396 m

7 0
3 years ago
A well-thrown ball is caught in a well-padded mitt. If the deceleration of the ball is 2.10×104 m/s2 , and 1.85 ms (1 ms = 10−3
ankoles [38]

Answer:

u = - 38.85 m/s^-1

Explanation:

given data:

acceleration = 2.10*10^4 m/s^2

time = 1.85*10^{-3} s

final velocity = 0 m/s

from equation of motion we have following relation

v = u +at

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0 = u + (21 *1.85)

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negative sign indicate that the ball bounce in opposite directon

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3 years ago
Exercício:
zysi [14]

Answer:

a

Explanation:

a

a

a

a

a

a

a

a

a is the answer

7 0
3 years ago
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lesantik [10]

Answer:

For communication to be communicated between you and your cousin who is in Houston,Texas, when a call is made by you, a request is made to the specific phone, and the telephone tower will get the request from the mobile phone. Then a signal is sent via a transmitter underground, by this the satellite communicates with the local receiver in Houston, that is linked to the local tower over there, the tower would request for your cousin's number and connects the two of you, once a link is set.

Explanation:

From the example stated, what is required for such for a far distance, is a communication satellite link.

When a call is made by you, the a connection request is sent to the specified phone.The telephone tower receives the request from The mobile phone. The local tower(Birmingham,Al) is linked to a ground transmitter by the means of a Fiber optical cable.

A signal is sent to satellite via the ground transmitter.The satellite then set's off the local receiver in (Houston,Texas) which on it's end is connected to the local tower there. This tower then ask for your cousin's mobile for a call that will be incoming, a link is set, once he/she receives the call, from there a conversation can be done.

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3 years ago
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