3 years ago

A solid Cute hug sides 0.50m

Physics

1 answer:

irina [24]3 years ago

3 0

Answer:

Explanation:

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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

vovangra [49]

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

A projectile enters a resisting medium at x = 0 with an initial velocity v0 = 910 ft/s and travels 5 in. before coming to rest.

evablogger [386]

Answer:

a = - 1.987 × 10⁶ ft/s²

t = 6.84 × 10⁻⁴ s

Explanation:

v₀ = 910 ft/s

x = 5 in.

relation v = v₀ - k x

v = 0 as body comes to rest

0 = 900 - 5k/12

k = 2184 s⁻¹

acceleration

$\frac{\mathrm{d} v}{\mathrm{d} t} = -k\frac{\mathrm{d} x}{\mathrm{d} t}$

where

(A) a = -k × v

at v= 910 ft/s

a = - 1.987 × 10⁶ ft/s²

(B) at x = 3.9 in.

v = 910 - 3.9(2184)/12

v = 200.2 m/s

$\frac{\mathrm{d} v}{\mathrm{d} t} = -k\frac{\mathrm{d} x}{\mathrm{d} t}$

$\frac{dv}{v} = -kdt$

$\int\limits^{200.2}_{900} {\frac{1}{v} }dv = -k\int\limits^t_0 dt$

$ln(200.2)-ln(900) = -kt$

t = 6.84 × 10⁻⁴ s

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