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aliya0001 [1]
3 years ago
12

A baseball pitcher throws a ball with a speed of 41 m/s. Estimate the average acceleration of the ball during the throwing motio

n. In throwing the baseball, the pitcher accelerates the ball through a displacement of about 3.5 m, from behind the body to the point where it is released.
Physics
1 answer:
dusya [7]3 years ago
8 0
<span>When t=0, v=0, d=0 When t=tf, v=41m/s, d=3.5m We have 2 formulas – the ones corresponding to uniformly accelerated linear movement: vf=a*t+vo d=(1/2)*a*t^2+vo*t Let’s put the data in the formulas: 41m/s=a*t+0=a*t 3.5m=(1/2)*a*t^2+0*t=1/2*a*t^2 You can use a variety of methods to find t and a. I will choose substitution. t=(41m/s)/a 3.5m=(1/2)*a*((41m/s)/a)^2=(1/2)*a*(41m/s)^2/a^2=(1/2)*(41m/s)^2/a a=(1/2)*(41m/s)^2/(3.5m)=(1/2)*41^2(m^2/s^2)/(3.5m) a=41^2(m/s^2)/( 2*3.5)=240m/s^2</span>
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How much heat must be added to make a 5g substance with a specific heat of 2 J/gC that has its temperature go up 10 degrees? Q =
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Answer:

100 Joule

Explanation:

Amount of heat in agiven body is given by Q = m•C•ΔT

where m is the mass of the body

c is the specific heat capacity of body. It is the amount of heat stored in 1 unit weight of body which raises raises the temperature of body by 1 unit of temperature.

ΔT is the change in the temperature of body

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coming back to problem

m = 5g

C = 2J/gC

since, it is given that temperature of body increases by 10 degrees, thus

ΔT = 10 degrees

Using the formula for heat as given

Q = m•C•ΔT

Q = 5* 2 * 10  Joule= 100 Joule

Thus, 100 joule heat must be added to  a  5g substance with a specific heat of 2 J/gC to raise its temperature go up by 10 degrees.

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igor_vitrenko [27]

First, balance the reaction:

_ KClO₃   ==>   _ KCl + _ O₂

As is, there are 3 O's on the left and 2 O's on the right, so there needs to be a 2:3 ratio of KClO₃ to O₂. Then there are 2 K's and 2 Cl's among the reactants, so we have a 1:1 ratio of KClO₃ to KCl :

2 KClO₃   ==>   2 KCl + 3 O₂

Since we start with a known quantity of O₂, let's divide each coefficient by 3.

2/3 KClO₃   ==>   2/3 KCl + O₂

Next, look up the molar masses of each element involved:

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• Cl: 35.453 g/mol

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Convert 10 g of O₂ to moles:

(10 g) / (31.998 g/mol) ≈ 0.31252 mol

The balanced reaction shows that we need 2/3 mol KClO₃ for every mole of O₂. So to produce 10 g of O₂, we need

(2/3 (mol KClO₃)/(mol O₂)) × (0.31252 mol O₂) ≈ 0.20835 mol KClO₃

KClO₃ has a total molar mass of about 122.549 g/mol. Then the reaction requires a mass of

(0.20835 mol) × (122.549 g/mol) ≈ 25.532 g

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