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3 years ago

6

a 74.9 kg person sits at rest on an icy pond holding a physics book. he throws the physics book west at 8.25 m/s and he recoils

at 0.215 m/s what is the mass of the physics book

1 answer:

kifflom [539]3 years ago

3 0

Answer:

1.95 kg

Explanation:

Momentum is conserved.

m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂

0 = (74.9) (-0.215) + m (8.25)

m = 1.95

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A wagon wheel consists of 8 spokes of uniform diameter, each of mass ms and length L cm. The outer ring has a mass mring. What i

just olya [345]

Answer:

The moment of inertial of the wheel, $I = 8(\frac{1}{3}M_sL^2 ) + M_rL^2$

Explanation:

Given;

8 spokes of uniform diameter

mass of each spoke, = $M_s$

length of each spoke, = L

mass of outer ring, = $M_r$

The moment of inertial of the wheel will be calculated as;

$I = 8I_{spoke} + I_{ring}$

where;

$I_{spoke$ is the moment of inertia of each spoke

$I_{ring$ is the moment of inertia of the rim

Moment of inertia of each spoke $=\frac{1}{3}M_sL^2$

Moment of inertial of the wheel

$I = 8(\frac{1}{3}M_sL^2 ) + M_rL^2$

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2 years ago

Calculate the current flowing if a charge of 36 kilocoulombs flows in 1 hour.

kupik [55]

Answer:

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Explanation:

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3 years ago

Titus drives his jetski a distance of 1000 meters in 7.045 seconds. How fast was he moving in meters per second? How fast was he

Ksenya-84 [330]

Answer:

a) v = 141.9 m/s

b) v = 317.4 miles/h

Explanation:

a) How fast was he moving in meters per second?

$v = \frac{d}{t} = \frac{1000 m}{7.045 s} = 141.9 m/s$

Hence, the jet ski is moving at 141.9 meters per second.

b) How fast was he moving in miles per hour?

$v = \frac{d}{t} = \frac{1000 m}{7.045 s} = 141.9 m/s*\frac{3600 s}{1 h}*\frac{1 mile}{1609.34 m} = 317.4 miles/h$

Therefore, the jet ski is moving at 317.4 miles per hour.

I hope it helps you!

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3 years ago

What does it mean to do science

Basile [38]

Answer:

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2 years ago

The drawing shows an adiabatically isolated cylinder that is divided initially into two identical parts by an adiabatic partitio

Sveta_85 [38]

Answer:

temperature on left side is 1.48 times the temperature on right

Explanation:

GIVEN DATA:

$\gamma = 5/3$

T1 = 525 K

T2 = 275 K

We know that

$P_1 = \frac{nRT_1}{v}$

$P_2 = \frac{nrT_2}{v}$

n and v remain same at both side. so we have

$\frac{P_1}{P_2} = \frac{T_1}{T_2} = \frac{525}{275} = \frac{21}{11}$

$P_1 = \frac{21}{11} P_2$ ..............1

let final pressure is P and temp $T_1 {f} and T_2 {f}$

$P_1^{1-\gamma} T_1^{\gamma} = P^{1 - \gamma}T_1 {f}^{\gamma}$

$P_1^{-2/3} T_1^{5/3} = P^{-2/3} T_1 {f}^{5/3}$ ..................2

similarly

$P_2^{-2/3} T_2^{5/3} = P^{-2/3} T_2 {f}^{5/3}$ .............3

divide 2 equation by 3rd equation

$\frac{21}{11}^{-2/3} \frac{21}{11}^{5/3} = [\frac{T_1 {f}}{T_2 {f}}]^{5/3}$

$T_1 {f} = 1.48 T_2 {f}$

thus, temperature on left side is 1.48 times the temperature on right

6 0

3 years ago