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Stells [14]
2 years ago
5

When a baseball is hit, it travels around 65 mps (meters per second). the mass of the baseball is 0.145 kg. what is the kinetic

energy of the baseball?
Physics
1 answer:
nika2105 [10]2 years ago
4 0

Answer:

306J

Explanation:

The formula for kinetic energy is given as ½mv².

Using this, we get our answer by substituting the values into the formula.

½ × 0.145kg × 65m/s

= 306.31J

can be rounded off to 306J

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What is hydrostatic equilibrium in a star? a) The balance between radiation from the surface and the rotation rate b) The expans
Dahasolnce [82]

Answer: d) The balance between the force of gravity directed in and thermal pressure directed

Explanation:

Hydrostatic Equilibrium helps to put in perspective star as self- regulating systems . It makes it plain that the energy generated in the star's hot core, is carried outward towards the cooler surface.

6 0
4 years ago
An unstrained horizontal spring has a length of 0.36 m and a spring constant of 320 N/m. Two small charged objects are attached
lakkis [162]

Answer:

1.been both -ve charged or both +be charged particles

2. 3.52mC

Explanation:

For the charge particle to cause an extension or movement of the string from its unrestrained position they would have been both -ve charged or both +be charged particles that's because like charges repel.

Now the Force sustain by the extended string is

F = Ke;

Where K is the force constant of the string, 320 N/m

e is the extension,0.033 m

F = 320 × 0.033 =10.56N

2.But according to columns law of charge;

F = kQ1 Q2

But Q1=Q2{ since the charge are of the same magnitude}.

Hence F = KQ^2

Where K is columns constant =9×10^9F/m

Hence Q=√F/K

Q= √10.56/9×10^9

=3.52×10^-3C

= 3.52mC

8 0
3 years ago
A major obstacle to developing good human relations is that
OLEGan [10]
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3 0
3 years ago
A hot-air balloonist, rising vertically with a constant velocity of magnitude v = 5.00 m/s , releases a sandbag at an instant wh
lutik1710 [3]
<span>Data:

Initial velocity upward: Vo = 5.00 m/s ,
Initial position: h = 40.0 m above the ground

Type of motion: free fall.

A) Compute the position of the sandbag at a time 1.05 s after its release.

Equation: y = h + Vo*t - g*(t^2) / 2

y = 40.0 m + 5.00 m/s * 1.05s - (9.8 m/s^2) * (1.05 s)^2 / 2 = 39.8 m

B)Compute the velocity of the sandbag at a time 1.05 s after its release.

Equation: Vf = Vo - g*t

=> Vf = 5.00 m/s - (9.8m/s^2) * (1.05 s) = - 5.29 m/s


Negative sign means that the sandbag is going down.

c) How many seconds after its release will the bag strike the ground?

Equation:

y = yo + Vo*t - g*(t^2) / 2

0 = 40.0 + 5.00t - 4.9 t^2

=> 4.9 t^2 - 5t - 40 = 0

Use the quadratic formula and you get: t = 3.41 s
</span>
7 0
3 years ago
Read 2 more answers
ASAP! HELP!
andre [41]

Answer:

2. 1.85

Explanation:

the equations weight * distance / other persons weight = distance from fulcrum. so 66.4*2.35/84.2=1.8532 round down to 1.85

8 0
3 years ago
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