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2 years ago

The position within a baseball team that is responsible for catching balls in the outfield is called the catcher ? true or false

Physics

2 answers:

Andreyy892 years ago

7 0

The answer is false....

aksik [14]2 years ago

4 0

Answer:

False

Explanation:

the catcher, who is located just behind the batter and part of whose job it is to catch any ball near home plate;

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A steel ball of mass 0.500 kg is fastened to a cord that is 70.0 cm long and fixed at the far end. The ball is then released whe

Liula [17]

Answer:

a) v₁fin = 3.7059 m/s (→)

b) v₂fin = 1.0588 m/s (→)

Explanation:

a) Given

m₁ = 0.5 Kg

L = 70 cm = 0.7 m

v₁in = 0 m/s ⇒ Kin = 0 J

v₁fin = ?

hin = L = 0.7 m

hfin = 0 m ⇒ Ufin = 0 J

The speed of the ball before the collision can be obtained as follows

Einitial = Efinal

⇒ Kin + Uin = Kfin + Ufin

⇒ 0 + m*g*hin = 0.5*m*v₁fin² + 0

⇒ v₁fin = √(2*g*hin) = √(2*(9.81 m/s²)*(0.70 m))

⇒ v₁fin = 3.7059 m/s (→)

b) Given

m₁ = 0.5 Kg

m₂ = 3.0 Kg

v₁ = 3.7059 m/s (→)

v₂ = 0 m/s

v₂fin = ?

The speed of the block just after the collision can be obtained using the equation

v₂fin = 2*m₁*v₁ / (m₁ + m₂)

⇒ v₂fin = (2*0.5 Kg*3.7059 m/s) / (0.5 Kg + 3.0 Kg)

⇒ v₂fin = 1.0588 m/s (→)

7 0

3 years ago

Sitting in a chairlift, Rebecca has a gravitational potential energy of 5,997.6

stira [4]

Answer:

B) 12 m

Explanation:

Gravitational potential energy is:

PE = mgh

Given PE = 5997.6 J, and m = 51 kg:

5997.6 J = (51 kg) (9.8 m/s²) h

h = 12 m

8 0

2 years ago

Can anyone plz help me

juin [17]

Answer:

D,B,C,A,C

Explanation:

I believe that is the correct answers but it is unclear. I don't think the key for the second last question would let the current flowing so the bulb would be off.

8 0

2 years ago

Before colliding, the momentum of block A is -100 kg*m/, and block B is -150 kg*m/s. After, block A has a momentum -200 kg*m/s.

rjkz [21]

Answer:

Momentum of block B after collision = $-50\ kg\ ms^{-1}$

Explanation:

Given

Before collision:

Momentum of block A = $p_{A1}$ = $-100\ kg\ ms^{-1}$

Momentum of block B = $p_{B1}$ = $-150\ kg\ ms^{-1}$

After collision:

Momentum of block A = $p_{A2}$ = $-200\ kg\ ms^{-1}$

Applying law of conservation of momentum to find momentum of block B after collision $p_{B2}$ .

$p_{A1}+p_{B1}=p_{A2}+p_{B2}$

Plugging in the given values and simplifying.

$-100-150=-200+p_{B2}$

$-250=-200+p_{B2}$

Adding 200 to both sides.

$200-250=-200+p_{B2}+200$

$-50=p_{B2}$

∴ $p_{B2}=-50\ kg\ ms^{-1}$

Momentum of block B after collision = $-50\ kg\ ms^{-1}$

6 0

2 years ago

If it takes 20 seconds to push the table across the room.What is the power ?

olchik [2.2K]

Answer: Its applied force and friction, The friction is the drag and the applied force is you pushing

3 0

2 years ago