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Ne4ueva [31]
2 years ago
14

a pitcher threw a baseball straight up at 35.8 meters per second. what is the ball's velocity after 2.50 seconds?

Physics
1 answer:
Andrei [34K]2 years ago
5 0
Ok the velocity of an object in free fall is given by the equation :

v=v0-gt, where v0 is the original velocity, g is the gravitational constant (9.8 m/s^2) and t is the time. 

so, we substitute values into this equation. v=35.8-9.8*2.5; v=11.3 m/s
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A lead fishing weight of mass 0.2 kg is tied to a fishing line that is 0.5 m long. the weight is then whirled in a vertical circ
Mariana [72]
In the movement of the weight in vertical circle, using momentum balance, the largest tension is at the bottom of the circle. This is represented by: 

<span>F = T - m g </span>
<span>T = F + m g 
</span>F (centripetal) = mv^2/r
<span>= m v^2 / r + m g </span>

<span>m v^2 / r = T - m g </span>
<span>T= 0.5m * 100kgm/s^2 / 0.2kg - 9.81m/s^2 * 0.5m </span>
<span>T= 245 m^2/s^2 </span>


7 0
2 years ago
A 6.0-cm-diameter, 11-cm-long cylinder contains 100 mg of oxygen (O2) at a pressure less than 1 atm. The cap on one end of the c
butalik [34]

Answer:

The temperature of the gas is 1197.02 K

Explanation:

From ideal gas law;

PV = nRT

Where;

P is the pressure of the gas

V is the volume of the gas

R is ideal gas constant = 8.314 L.kPa/mol.K

T is the temperature of the gas

n is the number of moles of gas

Volume of the gas in the cylindrical container = πr²h

Given;

r = 6/2 = 3 cm = 0.03 m

h = 11 cm = 0.11 m

V = π × (0.03)² × 0.11 = 3.11 × 10⁻⁴ m³ = 0.311 L

number of moles of oxygen gas = Reacting mass / molar mass

=\frac{0.1}{32} = 0.003125, moles

T = \frac{PV}{nR} = \frac{100X0.311}{0.003125X8.314} =1197.02K

Therefore, the temperature of the gas is 1197.02 K

6 0
2 years ago
A 0.500 kg aluminum pan on a stove is used to heat 0.250 liters of water from 20.0ºC to 80.0ºC. (a) How much heatis required? Wh
Jet001 [13]

Answer:

heat used to rise temperature pan =  30.1%

heat used to rise temperature water =  69.9%

Explanation:

Given data

mass of water = 0.250 liter = 0.250 kg

aluminum pan mass = 0.500 kg

initial temperature = 20.0ºC

final temperature =  80.0ºC

to find out

heat used to rise temperature of  pan and water

solution

we find here heat transferred to the water that is

heat transferred to the water = mass of water × specific heat of water × change in temperature    ...........1

specific heat of water is 4186 J/kgºC

so

heat transferred to the water = 0.250 × 4186 × (80-20) kJ

heat transferred to the water = 62.8 kJ

and

heat transferred to the aluminum that is

heat transferred to the aluminum = mass of aluminum × specific heat of aluminum × change in temperature    ...........2

here specific heat of aluminum is 900 J/kgºC

heat transferred to the aluminum = 0.500 × 900 × (80-20) kJ

heat transferred to the aluminum = 27 kJ

so

total heat = 62.8 + 27 = 89.8 kJ

so

heat used to rise temperature pan = 27/89.8 ×100% = 30.1%

heat used to rise temperature water = 62.8 / 89.8 ×100% = 69.9%

8 0
2 years ago
The pull of the moon on Earth's tidal bulge is causing _____. the earth to gradually rotate faster the earth to slowly expand in
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2 years ago
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A pendulum is used in a large clock. The pendulum has a mass of 2kg. If the pendulum is moving at a speed of 2.9 m/s when it rea
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2 years ago
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