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3 years ago

8

a 0.15 kg baseball has a momentum of 0.78 kg*m/s just before it lands on the ground. What was the ball's speed just before landi

ng?

1 answer:

vovangra [49]3 years ago

4 0

5.2m/s

Explanation:

Given parameters:

Mass of baseball = 0.15kg

Momentum of baseball = 0.78kgm/s

Unknown:

Speed of baseball = ?

Solution:

The momentum of the baseball is a function of the product of the mass and velocity. It is a vector quantity:

Momentum = mass x velocity

Since the speed of the ball is unknown:

Velocity = $\frac{momentum }{mass}$

= $\frac{0.78}{0.15}$

= 5.2m/s

The speed of the baseball before it lands is 5.2m/s

Learn more:

Momentum brainly.com/question/9484203

#learnwithBrainly

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3 years ago

Calculate the force that the 4kg block exerts on the 10kg block

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2 years ago

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2 years ago

Read 2 more answers

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 <em>final</em> del puente de acero es 100.018 metros.

Asumamos que la dilatación <em>térmica</em> experimentada por el puente de acero es <em>pequeña</em>, de modo que podemos emplear la siguiente aproximación <em>lineal</em> para determinar la longitud <em>final</em> 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:

$L = (100\,m)\cdot [1+(11.5\times 10^{-6})\cdot (24\,^{\circ}C - 8\,^{\circ}C)]$

$L = 100.018\,m$

La longitud <em>final</em> del puente de acero es 100.018 metros.

Para aprender más sobre dilatación térmica, invitamos cordialmente a ver esta pregunta verificada: brainly.com/question/24953416

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2 years ago

A diffraction grating with 600 lines/mmlines/mm is illuminated with light of wavelength 510 nmnm. A very wide viewing screen is

Ksenya-84 [330]

Answer:

A.2.95 m

B.7

Explanation:

We are given that

Diffraction grating=600 lines/mm

d= $\frac{1 mm}{600}=\frac{1\times 10^{-3} m}{600}=1.67\times 10^{-6} m$

Wavelength of light, $\lambda=510 nm=510\times 10^{-9} m$

l=4.6 m

A.We have to find the distance between the two m=1 bright fringes

$sin\theta=\frac{m\lambda}{d}$

For first bright fringe, =1

$sin\theta=\frac{1\times 510\times 10^{-9}}{1.67\times 10^{-6}}=0.305$

$\theta=sin^{-1}(0.305)=17.76^{\circ}$

The distance between two m=1 fringes

$x=2ltan\theta=2\times 4.6 tan(17.76^{\circ})=2.95 m$

Hence, the distance between two m=1 fringes=2.95 m

B.For maximum number of fringes,

$sin\theta=1$

$sin\theta=\frac{m\lambda}{d}$

Substitute the values

$1=\frac{m\times 510\times 10^{-9}}{1.67\times 10^{-6}}$

$m=\frac{1.67\times 10^{-6}}{510\times 10^{-9}}=3.3\approx 3$

Maximum number of bright fringes on the scree= $2m+1=2(3)+1=7$

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3 years ago