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3 years ago

Hurry. Answer quick!

Physics

1 answer:

Luda [366]3 years ago

3 0

Answer:

1) The change in momentum of the ball is 8 kgm/s

2) The total momentum of the bodies is 20 kgm/s

Explanation:

Given;

mass of the ball, m = 2 kg

initial velocity of the ball, v₁ = 2 m/s

final velocity of the ball, v₂ = -2 m/s (the ball rebounds)

The change in momentum of the ball is given by;

ΔP = Δmv

ΔP = mv₁ - mv₂

ΔP = m(v₁ - v₂)

ΔP = 2(2 - (-2))

ΔP = 2(2 + 2)

ΔP = 8 kgm/s

Given;

mass of the first body, m₁ = 5kg

mass of the second body, m₂ = 5kg

speed of the first body, v₁ = 2 m/s

speed of the second body, v₂ = 2 m/s

The total momentum of the bodies is given by

P = m₁v₁ + m₂v₂

P = (5 x 2) + (5 x 2)

P = 20 kgm/s

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Answer:

-384.22N

Explanation:

From Coulomb's law;

F= Kq1q2/r^2

Where;

K= constant of Coulomb's law = 9 ×10^9 Nm^2C-2

q1 and q2 = magnitudes of the both charges

r= distance of separation

F= 9 ×10^9 × −7.97×10^−6 × 6.91×10^−6/(0.0359)^2

F= -495.65 × 10^-3/ 1.29 × 10^-3

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A motor is operating at 3450 RPM and is producing a torque of 32 lb-in. What is the output horsepower of the motor?

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3 years ago

A baseball is thrown directly upward from ground level with a velocity of +15 m/s. What are the two times when the ball is 10 m

erastovalidia [21]

Answer:

time is 0.5660 s

and time is - 3.62431 s

Explanation:

velocity u = 15 m/s

height s = 10 m

acceleration due to gravity g = –9.8 m/s²

to find out

time

solution

we will apply here distance equation that is

s = ut - 1/2× gt² ...........1

here put all these value and get time t

here s is height and g is -9.8

s = ut - 1/2× gt²

10 = 15t - 1/2× (-9.8)t²

10 = 15t + 4.9t²

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8 0

3 years ago

A cast-iron flywheel has a rim whose OD is 1 m and whose ID is 0.8 m. The flywheel weight is to be such that an energy fluctuati

grin007 [14]

Answer:

A.Coefficient of speed fluctuation of the flywheel = 0.222

B. The width <em>(thickness)</em> of the rim should be 0.131 m (131 mm)

Explanation:

Coefficient of speed fluctuation $(C_{s})$ $= \frac{N_{2}-N_{1}}{N}$

$N_{1}$ = minimum speed = 200 rpm

$N_{2}$ = maximum speed = 250 rpm

$N$ = average speed $= \frac{N_{2}+N_{1}}{2} = \frac{250+200}{2} = 225 rpm$

∴ $Cs = \frac{250-200}{225}=0.222$

Hence the coefficient of speed fluctuation of the flywheel = 0.222

The moment of Inertia , $I=\frac{E_{2}-E_{1}}{C_{s}\times\omega^{2}}$

Where

$E_{2}-E_{1}=$ energy fluctuation of flywheel = 6.75 J

$\omega^{2}=$ angular velocity of flywheel = $\frac{2\pi N}{60} = \frac{2\pi \times 225}{60}= 23.56 rad/sec$

$C_{s}=$ coefficient of speed fluctuation of the flywheel = 0.222

Hence,

$I=\frac{6.75\times10^{3}}{0.222\times(23.56)^{2}}=54.78 Nms^{2}$

Similarly,

I = $\frac{m}{8}\times(d_{o}^{2}- d_{i}^{2})$

From the moment of Inertia, we can get the weight of the flywheel as

$m=\frac{8I}{(d_{o}^{2}+ d_{i}^{2})}= \frac{8\times 54.78}{(1^{2}+0.8^{2})}=267.22kg$

From this weight, we will be able to calculate the volume of the flywheel and hence, estimate the thickness. But to do this, we need to know its density first. this can be got from standard tables.

Specific weight of cast iron = $70.6KN/m^{3}$ ( from standard material property table)

density of cast iron , $\rho =\frac{70.6\times 10^{3}}{9.81}= 7,197kg/m^{3}$

Volume of cast iron flywheel = $\frac{m}{\rho}= \frac{267.22}{7197}= 0.03713 m^{3}$

similarly, the volume of the flywheel can also be obtained through the formula :

$V= \frac{\pi t(d_{o}^{2}-d_{i}^{2})}{4}$

we can easily estimate the thickness of the flywheel from here by solving for t as shown below

$0.03713=\frac{\pi t(1^{2}-0.8^{2})}{4}$

$0.03713=0.2827t$

$t=\frac{0.03713}{0.2827}$

$\therefore t= 0.131m \approx 131mm$

The width of the rim = 131 mm

8 0

3 years ago