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Natalija [7]
2 years ago
12

The fastest pitched baseball was clocked at 47 m/s. Assume that the pitcher exerted his force (assumed to be horizontal and cons

tant) over a distance of 1.0 m, and a baseball has a mass of 145 g.(a) What force did he produce on the ball during this record-setting pitch? (b) Draw free-body diagrams of the ball during the pitch and just after it left the pitcherâs hand.
Physics
1 answer:
arsen [322]2 years ago
4 0

Answer:

Explanation:

F ×1 = 0.5×0.145×47×47

F = 160.15 N

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Three wires are made of copper having circular cross sections. Wire 1 has a length l and radius r. Wire 2 has a length l and rad
Alex73 [517]

Explanation:

Below is an attachment containing the solution.

4 0
2 years ago
Evaporation of sweat requires energy and thus take excess heat away from the body. Some of the water that you drink may eventual
kotegsom [21]

Answer:

The amount of heat required is H_t =  1.37 *10^{6} \ J

Explanation:

From the question we are told that

The mass of water is m_w  =  20 \ ounce = 20 * 28.3495 = 5.7 *10^2 g

The temperature of the water before drinking is T_w  =  3.8 ^oC

The temperature of the body is T_b  =  36.6^oC

Generally the amount of heat required to move the water from its former temperature to the body temperature is

H=  m_w  *  c_w * \Delta T

Here c_w is the specific heat of water with value c_w = 4.18 J/g^oC

So

H=   5.7 *10^2 * 4.18 * (36.6 - 3.8)

=> H= 7.8 *10^{4} \  J

Generally the no of mole of sweat present mass of water is

n = \frac{m_w}{Z_s}

Here Z_w is the molar mass of sweat with value

Z_w =  18.015 g/mol

=> n = \frac{5.7 *10^2}{18.015}

=> n = 31.6 \  moles

Generally the heat required to vaporize the number of moles of the sweat is mathematically represented as

H_v  =  n  *  L_v

Here L_v is the latent heat of vaporization with value L_v  = 7 *10^{3} J/mol

=> H_v  =  31.6 * 7 *10^{3}

=> H_v  = 1.29 *10^{6} \  J

Generally the overall amount of heat energy required is

H_t =  H +  H_v

=> H_t =  7.8 *10^{4} +  1.29 *10^{6}

=> H_t =  1.37 *10^{6} \ J

4 0
2 years ago
Which factors could be potential sources of error in the experiment? check all that apply.
Vadim26 [7]

(A)energy lost in the lever due to friction

(C) visual estimation of height of the beanbag

(E)position of the fulcrum for the lever affecting transfer of energy

6 0
2 years ago
Read 2 more answers
In general it is best to conceptualize vectors as arrows in space, and then to make calculations with them using their component
Dimas [21]

(1) A - B

(2) B - C

(3) - A + B - C

(4) 3A - 2C

(5) - 2A + 3B - C

(6) 2A - 3 (B - C)

Answer:

(1)  (3,-5,-4)

(2) (-5, 4, 0)

(3) (-6, 4, 3)

(4) (-3, -2, -11)

(5) (-11, 14, 8)

(6) (17, -12, -6)

Explanation:

A⃗ =(1,0,−3)

B⃗ =(−2,5,1)

C⃗ =(3,1,1)

Vector additions and subtraction are done on a component by component basis, that is, only data from component î can be added to or subtracted from another Vector's component î. And so on for components j and k.

1) (A - B) = (1,0,−3) - (−2,5,1) = (1-(-2), 0-5, -3-1) = (3,-5,-4)

2)  (B - C) = (−2,5,1) - (3,1,1) = (-2-3, 5-1, 1-1) = (-5, 4, 0)

3) -A + B - C = -(1,0,−3) + (−2,5,1) - (3,1,1) = (-1-2-3, 0+5-1, 3+1-1) = (-6, 4, 3)

4) 3A - 2C = 3(1,0,−3) - 2(3,1,1) = (3,0,-9) - (6,2,2) = (3-6, 0-2, -9-2) = (-3, -2, -11)

5) -2A + 3B - C = -2(1,0,−3) + 3(−2,5,1) - (3,1,1) = (-2,0,6) + (-6,15,3) - (3,1,1) = (-2-6-3, 0+15-1, 6+3-1) = (-11, 14, 8)

6) 2A - 3 (B - C) = 2(1,0,−3) - 3[(−2,5,1) - (3,1,1)] = (2,0,-6) - 3(-5,4,0) = (2+15, 0-12, -6-0) = (17, -12, -6)

3 0
2 years ago
Suppose you are myopic (nearsighted). You can clearly focus on objects that are as far away as 52.5 cm away. You can clearly foc
Lilit [14]

Answer:

Explanation:

Image of distant object will be made at far point or at 52.5 so

object distance u = infinity

image distance v = - 52.5 cm

focal length required = f

Lens formula

1 / v - 1 / u = 1 / f

1 /  - 52.5 - 0 = 1 / f

f =  -52.5 cm

= -.525 m

Power P = 1 / f = -  1 / .525

= -  1.90

now , for eye with glass we shall find new near point .

v = ?

u = - 17.2 cm

f = -  52.5 cm

1 / v - 1 / u = 1 / f

  1 / v + 1 / 17.2 = -  1 / 52.5

1 / v  = - 1 / 17.2 -    1 / 52.5

= - .05813 -  .019

= - .07713

u = - 12.96 cm

so new near point will be 12.96 cm

5 0
2 years ago
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