All categories

2 years ago

What is the kinetic energy of a 0.15 kg baseball that is moving with a velocity of 50 m/s?

1 answer:

7 0

Using the equation:

$ke \: = \frac{1}{2} m{v}^{2}$
$ke = \frac{1}{2} (0.15) {(50)}^{2}$
$ke = 188 \: (3sf)$

Ans: 188 J

You might be interested in

A student observes that it is hard to hear music underwater in a pool. They state that the sound is always muffled. They

s344n2d4d5 [400]

Answer:

FALSE

Explanation:

The answer is false.

The speed of the sound in water is faster when compared to the speed of sound in air. This is because, the particles in air is loosely packed and are far from each other as compared to water or liquid.

The water particles are close to each other than air particles, so water particles are able to transmit the vibrations of the sound faster than the air particles.

Therefore sound waves travels faster in water than in air.

5 0

2 years ago

Why does the biosphere not have distinct boundaries?

Alexeev081 [22]

Answer:

living things are found in air, water, and soil.

Explanation:

7 0

3 years ago

A 20 μF capacitor initially charged to 30 μC is discharged through a 1.5 kΩ resistor. Part A How long does it take to reduce the

Natasha_Volkova [10]

Answer:

it will take 36.12 ms to reduce the capacitor's charge to 10 μC

Explanation:

Qi= C×V

then:

Vi = Q/C = 30μ/20μ = 1.5 volts

and:

Vf = Q/C = 10μ/20μ = 0.5 volts

then:

v = v₀e^(–t/τ)

v₀ is the initial voltage on the cap

v is the voltage after time t

R is resistance in ohms,

C is capacitance in farads

t is time in seconds

RC = τ = time constant

τ = 20µ x 1.5k = 30 ms

v = v₀e^(t/τ)

0.5 = 1.5e^(t/30ms)

e^(t/30ms) = 10/3

t/30ms = 1.20397

t = (30ms)(1.20397) = 36.12 ms

Therefore, it will take 36.12 ms to reduce the capacitor's charge to 10 μC.

7 0

3 years ago

Un puente de acero de 100 m de largo a 8° C aumenta su temperatura a 24°C ¿Cuánto medirá su longitud? Valor del coeficiente de d

BartSMP [9]

La longitud final del puente de acero es 100.018 metros.

Asumamos que la dilatación térmica experimentada por el puente de acero es pequeña, de modo que podemos emplear la siguiente aproximación lineal para determinar la longitud final del puente de acero ( $L$ ), en metros:

$L = L_{o}\cdot [1+\alpha\cdot (T_{f}-T_{o})]$ (1)

Donde:

$L_{o}$ - Longitud inicial del puente, en metros.
$\alpha$ - Coeficiente de dilatación, sin unidad.
$T_{o}$ - Temperatura inicial, en grados Celsius.
$T_{f}$ - Temperatura final, en grados Celsius.

Si tenemos que $L_{o} = 100\,m$ , $\alpha = 11.5\times 10^{-6}$ , $T_{o} = 8\,^{\circ}C$ y $T_{f} = 24\,^{\circ}C$ , entonces la longitud final del puente de acero es: