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2 years ago

A car possesses 20,000 units of momentum. what would be the car's new momentum if ... its velocity was doubled?

Physics

1 answer:

pochemuha2 years ago

7 0

10,000 units of momentum.
p=mv
20,000=m(2v)
10,000=mv

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Coffee is poured at a uniform rate at 20 cm3 / s into a cup whose inside is shaped like a truncated cone. if the upper and lower

Vladimir [108]

Let the cup is filled to height h after some time

now the total volume of coffee filled in the cup is given as

$\frac{2}{y} = \frac{4}{6+y}$

$2y = 6 + y$

$y = 6 cm$

now volume of the coffee will be

$V = \frac{1}{3}\pi r^2(y + 6) - \frac{1}{3}\pi 2^2 (6)$

here we know that

$\frac{r}{y+6} = \frac{2}{6}$

$r = \frac{y+6}{3}$

$V = \frac{1}{3}\pi (\frac{y+6}{3})^2(y+6) - \frac{1}{3}\pi 2^2(6)$

now we know that volume flow rate is given as

$Q = \frac{dV}{dt}$

$20 cm^3/s = \frac{1}{3}\pi (\frac{1}{9})(3(y+6)^2)\frac{dy}{dt}$

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1 year ago

. A spring has a length of 0.200 m when a 0.300-kg mass hangs from it, and a length of 0.750 m when a 1.95-kg mass hangs from it

kap26 [50]

Answer:

29.4 N/m

0.1

Explanation:

a) From the restoring Force we know that :

F_r = —k*x

the gravitational force :

F_g=mg

Where:

F_r is the restoring force .

F_g is the gravitational force

g is the acceleration of gravity

k is the constant force

xi , x2 are the displacement made by the two masses.

Givens:

m1 = 1.29 kg

m2 = 0.3 kg

x1 = -0.75 m

x2 = -0.2 m

g = 9.8 m/s^2

Plugging known information to get :

F_r =F_g

-k*x1 + k*x2=m1*g-m2*g

k=29.4 N/m

b) To get the unloaded length 1:

l=x1-(F_1/k)

Givens:

m1 = 1.95kg , x1 = —0.75m

Plugging known infromation to get :

l= x1 — (F_1/k)

= 0.1

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