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2 years ago

The weight of a 630 g piece of ham is _____.

1 answer:

8 0

The weight of an object is given by:
$F=mg$
where
F is the weight
m is the mass of the object
$g=9.81 m/s^2$ is the gravitational acceleration

The piece of ham in this problem has a mass of
m=630 g = 0.63 kg
therefore its weight is
$F=mg=(0.63 kg)(9.81 m/s^2)=6.18 N$

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Two speedboats are traveling at the same speed relative to the water in opposite directions in a moving river. An observer on th

DENIUS [597]

Answer:

a) vboat = 5.95 m/s b) vriver= 1.05 m/s

Explanation:

a) As observed from the shore, the speed of the boats can be expressed as the vector sum, of the boat speed relative to the water and the river speed relative to the shore, as follows:

vb₁s = vb₁w + vrs

In one case, the boat is moving in the same direction as the water:

vb₁s = vb₁w + vrs = 7.0 m/s (1)

For the other boat, it is clear that is moving in an opposite direction:

vb₂s = vb₂w - vrs = 4.9 m/s (2)

As we know that vb₁w = vb₂w, adding both sides, we can remove the river speed from the equation, as follows:

vb₁w = vb₂w = $\frac{7.0 m/s + 4.9 m/s}{2} =5.95 m/s$

b) Replacing this value in (1) and solving for vriver, we have:

vriver = 7.0 m/s - 5.95 m/s = 1.05 m/s

(we could have arrived to the same result subtracting both sides in (1), and (2))

3 0

2 years ago

An ideal monatomic gas initially has a temperature of T and a pressure of p. It is to expand from volume V1 to volume V2. If the

yawa3891 [41]

Answer:

Isothermal : P2 = ( P1V1 / V2 ) , work-done $pdv = nRT * In( \frac{V2}{v1} )$

Adiabatic : : P2 = $\frac{P1V1^{\frac{5}{3} } }{V2^{\frac{5}{3} } }$ , work-done =

W = $(3/2)nR(T1V1^(2/3)/(V2^(2/3)) - T1)$

Explanation:

initial temperature : T

Pressure : P

initial volume : V1

Final volume : V2

A) If expansion was isothermal calculate final pressure and work-done

we use the gas laws

= PIVI = P2V2

Hence : P2 = ( P1V1 / V2 )

work-done :

$pdv = nRT * In( \frac{V2}{v1} )$

B) If the expansion was Adiabatic show the Final pressure and work-done

final pressure

$P1V1^y = P2V2^y$

where y = 5/3

hence : P2 = $\frac{P1V1^{\frac{5}{3} } }{V2^{\frac{5}{3} } }$

Work-done

W = $(3/2)nR(T1V1^(2/3)/(V2^(2/3)) - T1)$

Where $T2 = T1V1^(2/3)/V2^(2/3)$

3 0

2 years ago

The food calorie, equal to 4186J , is a measure of how much energy is released when food is metabolized by the body. A certain b

vovangra [49]

<h2>The hiker will go up to 850 m on the hill</h2>

Explanation:

The total energy gained by the hiker = 140 x 4186 J

This energy is consumed in the potential energy acquired , while climbing up the hill.

The potential energy P.E = mass of hiker x acceleration due to gravity x height

Thus

140 x 4186 = 69 x 10 x h

or h = $\frac{4186x140}{69x10}$ = 850 m

If the 20% of the total energy is used

the height h₀ = $\frac{0.2x4186x140}{69x10}$ = 170 m

5 0

2 years ago

An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The el

Tresset [83]

Complete Question

An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The electric field in the wire changes with time as E(t)=0.0004t2−0.0001t+0.0004 newtons per coulomb, where time is measured in seconds.

I = 1.2 A at time 5 secs.

Find the charge Q passing through a cross-section of the conductor between time 0 seconds and time 5 seconds.

Answer:

The charge is $Q =2.094 C$

Explanation:

From the question we are told that

The diameter of the wire is $d = 0.205cm = 0.00205 \ m$

The radius of the wire is $r = \frac{0.00205}{2} = 0.001025 \ m$

The resistivity of aluminum is $2.75*10^{-8} \ ohm-meters.$

The electric field change is mathematically defied as

$E (t) = 0.0004t^2 - 0.0001 +0.0004$

Generally the charge is mathematically represented as

$Q = \int\limits^{t}_{0} {\frac{A}{\rho} E(t) } \, dt$

Where A is the area which is mathematically represented as

$A = \pi r^2 = (3.142 * (0.001025^2)) = 3.30*10^{-6} \ m^2$

$\frac{A}{\rho} = \frac{3.3 *10^{-6}}{2.75 *10^{-8}} = 120.03 \ m / \Omega$

Therefore

$Q = 120 \int\limits^{t}_{0} { E(t) } \, dt$

substituting values

$Q = 120 \int\limits^{t}_{0} { [ 0.0004t^2 - 0.0001t +0.0004] } \, dt$

$Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | t} \atop {0}} \right.$

From the question we are told that t = 5 sec

$Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | 5} \atop {0}} \right.$

$Q = 120 [ \frac{0.0004(5)^3 }{3} - \frac{0.0001 (5)^2}{2} +0.0004(5)] }$

$Q =2.094 C$

5 0

2 years ago

Dane is holding an 8 kilogram box 2 metres above the ground. How much energy is in the box's gravitational potential energy stor

9966 [12]

156.8 Joules of energy is in the box's gravitational potential energy store

<u>Explanation</u>:

<em>Given:</em>

Mass of the box Dane is holding = 8 Kilograms

Height at which Dane is holding the box above the ground= 2 metres

Gravitational potential energy in the box=?

<em>Solution:</em>

gravitational potential energy is the work done per mass on a object to move that object from one fixed location to to another location against gravity.Its unit is joules or J

Thus Gravitational potential energy is represented as,

$PE_g=mgh$