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Anastaziya [24]
3 years ago
15

A baseball of mass 0.25 kg is hit at home plate with a speed of 40 m/s. When it lands in a seat in the left-field bleachers a ho

rizontal distance 120 m from home plate, it is moving at 30 m/s. If the ball lands 20 m above the spot where it was hit, how much work is done on it by air resistance?
Physics
1 answer:
Pachacha [2.7K]3 years ago
4 0

Answer:

the work done by air resistance is 38.5 J

Explanation:

given information:

mass of the ball, m = 0.25 kg

initial speed, v_{1} = 40 m/s

final speed, v_{2} = 30 m/s

horizontal distance, x = 120 m

Δh = 20 m

according to conventional energy

W = E_{2} - E_{1}

where

E_{1} is initial energy

E_{2}  is final energy

E = KE + PE

KE is kinetic energy

PE is potential energy

W = E_{2} - E_{1}

   = mgh_{2} + \frac{1}{2} mv_{2} ^{2} - (mgh_{1} + \frac{1}{2} mv_{1} ^{2})

   = mg(h_{2} -h_{1} )+ \frac{1}{2} m(v_{2} ^{2} - v_{1} ^{2})

   = m (gΔh + \frac{1}{2} (v_{2} ^{2} - v_{1} ^{2}))

   = 0.25 ( (9.8) (20) + \frac{1}{2} (30 ^{2} - 40 ^{2}))

   = - 38.5 J

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4 0
3 years ago
The energy from 0.015 moles of octane was used to heat 250 grams of water. The temperature of the water rose from 293.0 K to 371
arsen [322]

Answer : The correct option is, (B) -5448 kJ/mol

Explanation :

First we have to calculate the heat required by water.

q=m\times c\times (T_2-T_1)

where,

q = heat required by water = ?

m = mass of water = 250 g

c = specific heat capacity of water = 4.18J/g.K

T_1 = initial temperature of water = 293.0 K

T_2 = final temperature of water = 371.2 K

Now put all the given values in the above formula, we get:

q=250g\times 4.18J/g.K\times (371.2-293.0)K

q=81719J

Now we have to calculate the enthalpy of combustion of octane.

\Delta H=\frac{q}{n}

where,

\Delta H = enthalpy of combustion of octane = ?

q = heat released = -81719 J

n = moles of octane = 0.015 moles

Now put all the given values in the above formula, we get:

\Delta H=\frac{-81719J}{0.015mole}

\Delta H=-5447933.333J/mol=-5447.9kJ/mol\approx -5448kJ/mol

Therefore, the enthalpy of combustion of octane is -5448 kJ/mol.

5 0
3 years ago
Find the separation of two points on the Moon’s surface that can just be resolved by the 200 in. (= 5.1 m) telescope at Mount Pa
rewona [7]

Answer:

The separation of the 2 points should be 50.0 meters.

Explanation:

According to Rayleigh's scattering criteria the angular separation between 2 points to be resolved equals

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

Applying the given values we get

\theta =1.22\cdot \frac{550\times 10^{-9} }{5.1}=0.1316\times 10^{-6}rads

thus the linear separation equals L=\theta \times Distance

Applying the given values we get

L=0.1316\times 10^{-6} \times 3.8\times 10^{5}\times 10^{3}meters\\\\\therefore L=50.00metes

6 0
3 years ago
A uniform electric field is produced due to the charge distribution inside the closed cylindrical surface. (a) What type of char
DiKsa [7]

Answer with Explanation:

a. Option d is true.

a negatively charged plane parallel to the end faces of the cylinder

b. Radius of cylinder, r=0.66m

Magnitude of electric field, E=300 N/C

We have to find the net flux through the closed surface.

Net electric flux,\phi=-2 EA=-2E(\pi r^2)

\phi=-2\times 300\times (3.14\times (0.66)^2)

\phi=-820.67 Nm^2/C

c.

Net charge,Q=\epsilon_0\times \phi

Where

\epsilon_0=8.85\times 10^{-12}

Q=-820.67\times 8.85\times 10^{-12}

Q=-7.26\times 10^{-9} C

Q=-7.26nC

Where 1nC=10^{-9}C

7 0
3 years ago
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