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AleksandrR [38]
3 years ago
10

What is the density of the baseball?

Physics
2 answers:
Georgia [21]3 years ago
7 0

Answer:

Chile the answer is yes

Explanation:

I just took the quiz

Aleksandr [31]3 years ago
5 0

The density of the baseball is 0.875 g/cm^3

Explanation:

The density of an object is given by the equation

d=\frac{m}{V}

where

m is the mass of the object

V is its volume

For the baseball in this problem, we have:

m = 350 g is its mass

V=400 mL = 400 cm^3 is the volume

Substituting into the equation, we find the density:

d=\frac{350}{400}=0.875 g/cm^3

Therefore, the ball's density is outside the acceptable range.

Learn more about density:

brainly.com/question/5055270

brainly.com/question/8441651

#LearnwithBrainly

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mina [271]

Answer:

This link was diagram

Explanation:

https://doubtnut.app.link/FnsNC80Dccb

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3 years ago
A shot-putter accelerates a 7.2 kg shot from rest to 17 m/s . what work did the shot-putter do on the ball?
garri49 [273]
<span>1.0x10^3 Joules The kinetic energy a body has is expressed as the equation E = 0.5 M V^2 where E = Energy M = Mass V = Velocity Since the shot was at rest, the initial energy is 0. Let's calculate the energy that the shot has while in motion E = 0.5 * 7.2 kg * (17 m/s)^2 E = 3.6 kg * 289 m^2/s^2 E = 1040.4 kg*m^2/s^2 E = 1040.4 J So the work performed on the shot was 1040.4 Joules. Rounding the result to 2 significant figures gives 1.0x10^3 Joules</span>
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Which object has more gravitational potential energy? Use PE = m × g × h, where g = 9.8 meters/second2.
Andrew [12]
Gravitational potential energy can be calculated using the formula <span>PE = m × g × h, where g is the gravitational acceleration and is constant hence the energy is dependent directly to mass and the height of the object. Hence more PE is registered when the object is heavier and/or at greater initial height. </span>
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A charge is moving in a magnetic field that points to the left. What direction can the charge move and experience no magnetic fo
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In a hydroelectric power plant, water flows from an elevation of 400 ft to a turbine, where electric power is generated. For an
VashaNatasha [74]

Answer:

V=3.475ft^3/s=3.48ft^3/s

Explanation:

We have here values from SI and English Units. I will convert the units to English Units.

We hace for the power P,

P= 100kW \rightarrow 100kW*(\frac{737.56ft.lbf/s}{1kW})

P= 73.7*10^{3}ft.lbf/s

we have other values such h=400ft and  \gamma= 62.42lb/ft^3 (specific weight of the water), and 0.85 for \eta

We need to figure the flow rate of the water (V) out, that is,

V=\frac{P}{\gamma h \eta_0}

Where \eta_0 is the turbine efficiency, at which is,

\eta_0 = \frac{P}{\gamma Vh}

Replacing,

V=\frac{73.7*10^{3}}{62.42*400*0.85}

V=3.475ft^3/s

With this value (the target of this question) we can also calculate the mass flow rate of the waters,

through the density and the flow rate,

m=\rho V\\m= 3.475*1.94 \\m=6.7415 slugs/s

converting the slugs to lbm, 1slug = 32.174lbm, we have that the mass flow rate of the water is,

m= 217lbm/s

6 0
3 years ago
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