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Dima020 [189]
3 years ago
15

Lonnie pitches a baseball of mass 0.20 kg. The ball arrives at home plate with a speed of 40 m/s and is batted straight back to

Lonnie with a return speed of 60 m/s. If the bat is in contact with the ball for 0.050 s what is the impulse experienced by the ball? A. 360 N.s B. 20 N.s C. 400 N.s D. 9.0 N.s
Physics
1 answer:
suter [353]3 years ago
4 0

Answer:

B) 20N.s is the correct answer

Explanation:

The formula for the impulse is given as:

Impulse = change in momentum

Impulse = mass × change in speed

Impulse = m × ΔV

Given:

initial speed  = 40m/s

Final speed = -60 m/s (Since the the ball will now move in the opposite direction after hitting the bat, the speed is negative)

mass = 0.20 kg

Thus, we have

Impulse = 0.20 × (40m/s - (-60)m/s)

Impulse = 0.20 × 100 = 20 kg-m/s or 20 N.s

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Respuesta:

0,0560 cal / gºC.

Explicación:

Cantidad de calor; (Q)

Q = mcΔt; Δt = t2 - t1

m = masa, c = capacidad calorífica específica; Δt = cambio de temperatura

c de agua = 1 cal / gºC

c de aluminio = 0,22 cal / gºC

QTotal = Q de agua + Q de aluminio

Q de agua = 450 * 1 * (26 - 23) = 1350 cal

Q de aluminio = 60 * 0.22 * (26 - 23) = 39.6 cal

QTotal = 1350 + 39,6 = 1389,6 cal

Calor perdido = calor ganado

QTotal = calor perdido

- 1389,6 = 335,2 * c * (26 - 100)

-1389,6 = −24804,8 * c

c = 1389,6 / 24804,8

c = 0,056021 cal / gºC.

Capacidad calorífica específica de la plata = 0,0560 cal / gºC.

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3 years ago
What's the name of the compound Kl?
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2 years ago
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What is the similarities of a furnace and the sun?
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2 years ago
A 3.00 kg object is moving in the XY plane, with its x and y coordinates given by x = 5t³ !1 and y = 3t ² + 2, where x and y are
Hatshy [7]

Answer:

The net force acting on this object is 180.89 N.

Explanation:

Given that,

Mass = 3.00 kg

Coordinate of position of x= 5t^3+1

Coordinate of position of y=3t^2+2

Time = 2.00 s

We need to calculate the acceleration

a = \dfrac{d^2x}{dt^2}

For x coordinates

x=5t^3+1

On differentiate w.r.to t

\dfrac{dx}{dt}=15t^2+0

On differentiate again w.r.to t

\dfrac{d^2x}{dt^2}=30t

The acceleration in x axis at 2 sec

a = 60i

For y coordinates

y=3t^2+2

On differentiate w.r.to t

\dfrac{dy}{dt}=6t+0

On differentiate again w.r.to t

\dfrac{d^2y}{dt^2}=6

The acceleration in y axis at 2 sec

a = 6j

The acceleration is

a=60i+6j

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F = 3.00\times(60i+6j)

F=180i+18j

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|F|=\sqrt{(180)^2+(18)^2}

|F|=180.89\ N

Hence, The net force acting on this object is 180.89 N.

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3 years ago
Ram has power of 550 watt. What does it mean?
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