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Tamiku [17]
2 years ago
6

A baseball player hits a baseball with a bat. The mass of the ball is 0.25 kg. The ball accelerates at 200 m/s2. What is the for

ce, to the nearest N, of the bat on the baseball?
Physics
1 answer:
Brums [2.3K]2 years ago
6 0

Answer:

50N

Explanation:

Force (N) = mass (kg) × acceleration (m/s²)

0.25kg times 200m/s² = 50N

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Within the theory of G relativity what, exactly, is meant by " the speed of light WITHIN A VACUUM" ? & what does that have t
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seconds \times  \frac{meters}{seconds} =meters
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3 0
3 years ago
Consider two wheels with fixed hubs. The hub cannot move, but the wheel can rotate about it. The hubs are fixed to a stationary
kykrilka [37]

Answer:

Magnitude of force on wheel B is 4 N

Explanation:

Given that

I=mr^2

For wheel A

m= 1 kg

d= 1 m,r= 0.5 m

F=1 N

We know that

T= F x r

T=1 x 0.5 N.m

T= 0.5 N.m

T= I α

Where I is the moment of inertia and α is the angular acceleration

I_A=1 \times 0.5^2\ kg.m^2

I_A=0.25\ kg.m^2

T= I α

0.5= 0.25 α

\alpha = 2\ rad/s^2

For Wheel B

m= 1 kg

d= 2 m,r=1 m

I_B=1 \times 1^2\ kg.m^2

I_B=1 \ kg.m^2

Given that angular acceleration is same for both the wheel

\alpha = 2\ rad/s^2

T= I α

T= 1 x 2

T= 2 N.m

Lets force on wheel is F then

T = F x r

2 = F x 1

So F= 2 N

Magnitude of force on wheel B is 2 N

3 0
3 years ago
2.5 g of helium at an initial temperature of 300 K interacts thermally with 9.0 g of oxygen at an initial temperature of 620 K .
muminat

Answer:

Explanation:

2.5 g of He = 2.5 / 4  mole

= .625 moles

9 g of oxygen = 9/32

= .28 mole of oxygen

C_p of He = 3/2 R

C_p of O₂ = 5/2 R

A ) Initial thermal energy of He = 3/2 n R T

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= 2340 J

Initial thermal energy of O₂ = 5/2 n R T

= 2.5 x .28 x 8.32 x 620

= 3610.88 J

B ) If T be the equilibrium temperature after mixing

gain of heat by helium

= n C_p Δ T

= .625 x 3/2 R x ( T - 300 )

Loss of heat by oxygen

n C_p Δ T

= .28 x 5/2 R x ( 620 - T )

Loss of heat = gain of heat

.625 x 3/2 R x ( T - 300 ) = .28 x 5/2 R x ( 620 - T )

1.875 T- 562.5 = 868- 1.4 T

3.275 T = 1430,5

T = 436.8 K

Thermal energy of He

= 1.5 x .625 x 8.32 x 436.8

= 3407 J

thermal energy of O₂

= 2.5 x .28  x 8.32 x 436.8

= 2543.92 J

C )

Heat energy transferred

=  .28 x 5/2 R x ( 620 - T )

=  .28 x 5/2 x  8.32 x ( 620 - 436.8 )

1066.95 J

Heat will flow from O₂ to He

Final temperature is 436.8 K

7 0
3 years ago
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