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3 years ago

8

A truck accelerating at 0.0083 meters/secondÆ covers a distance of 5.8 × 10Ê meters. If the truck's mass is 7,000 kilograms, wha

t is the work done to reach this distance?

2 answers:

laiz [17]3 years ago

8 0

Answer:

58.1*(5.8*10^4) =3,369,800 Joules. hope it helps

tigry1 [53]3 years ago

3 0

work done = force * distance moved (in direction of the force)

force= mass* acceleration

force=58.1N

58.1*(5.8*10^4)
=3,369,800 J

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The length of an aluminum wire is quadrupled and the radius is doubled. By which factor does the resistance change?

erik [133]

The resistance of a conductive wire is given by:
$R= \frac{\rho L}{A}$
where
$\rho$ is the material resistivity
$L$ is the wire length
$A$ is the cross-sectional area of the wire

The length of the wire is quadrupled, so if we call L the original length and L' the new length, we can write
$L'=4 L$

Similarly, the radius of the wire is doubled (r'=2r), so the new area is
$A'= \pi (r')^2 = \pi (2r)^2 = 4 \pi r^2 = 4A$

And if we substitute into the equation, we find that the new resistance of the wire is
$R'= \frac{\rho L'}{A'}= \frac{\rho (4L)}{4 A'} = \frac{\rho L}{A}=R$
Therefore, R=R': this means that the resistance of the wire did not change.

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3 years ago

You prepare tea with 0.250 kg of water at 85.0 ºC and let it cool down to room temperature (20.0 ºC) before drinking it. Essenti

vladimir2022 [97]

Answer:

232 J/K

Explanation:

The amount of heat gained by the air = the amount of heat lost by the tea.

q_air = -q_tea

q = -mCΔT

q = -(0.250 kg) (4184 J/kg/ºC) (20.0ºC − 85.0ºC)

q = 68,000 J

The change in entropy is:

dS = dQ/T

Since the room temperature is constant (isothermal):

ΔS = ΔQ/T

Plug in values (remember to use absolute temperature):

ΔS = (68,000 J) / (293 K)

ΔS = 232 J/K

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3 years ago

Pumping of groundwater in California most effectively: Select one:

maria [59]

Answer:

A

Explanation:

Ans- A, causes sediments in aquifers that are emptied to compact and the ground above it to sink.

This is one of the major problems for California nowadays. Due to excessive groundwater pumping to satisfy the demand of huge population the aquifers are drying, hence compacting and for that the land above sinks.

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3 years ago

A power supply has an open-circuit voltage of 40.0 V and an internal resistance of 2.00 V. It is used to charge two storage batt

Natali [406]

Complete Question

A power supply has an open-circuit voltage of 40.0 V and an internal resistance of 2.00 $\Omega$ . It is used to charge two storage batteries connected in series, each having an emf of 6.00 V and internal resistance of 0.300 $\Omega$ . If the charging current is to be 4.00 A, (a) what additional resistance should be added in series? At what rate does the internal energy increase in (b) the supply, (c) in the batteries, and (d) in the added series resistance? (e) At what rate does the chemical energy increase in the batteries?

Answer:

a

The additional resistance is $R_z = 4.4 \Omega$

b

The rate at which internal energy increase at the supply is $Z_1 = 32 W$

c

The rate at which internal energy increase in the battery is $Z_1 = 32 W$

d

The rate at which internal energy increase in the added series resistance is $Z_3 = 70.4 W$

e

the increase rate of the chemically energy in the battery is $C = 48 W$

Explanation:

From the question we are told that

The open circuit voltage is $V = 40.0V$

The internal resistance is $R = 2 \Omega$

The emf of each battery is $e = 6.00 V$

The internal resistance of the battery is $r = 0.300V$

The charging current is $I = 4.00 \ A$

Let assume the the additional resistance to to added to the circuit is $R_z$

So this implies that

The total resistance in the circuit is

$R_T = R + 2r +R_z$

Substituting values

$R_T = 2.6 +R_z$

And the difference in potential in the circuit is

$E = V -2e$

=> $E = 40 - (2 * 6)$

$E = 28 V$

Now according to ohm's law

$I = \frac{E}{R_T}$

Substituting values

$4 = \frac{28}{R_z + 2.6}$

Making $R_z$ the subject of the formula

So $R_z = \frac{28 - 10.4}{4}$

$R_z = 4.4 \Omega$

The increase rate of internal energy at the supply is mathematically represented as

$Z_1 = I^2 R$

Substituting values

$Z_1 = 4^2 * 2$

$Z_1 = 32 W$

The increase rate of internal energy at the batteries is mathematically represented as

$Z_2 = I^2 r$

Substituting values

$Z_2 = 4^2 * 2 * 0.3$

$Z_2 = 9.6 \ W$

The increase rate of internal energy at the added series resistance is mathematically represented as

$Z_3 = I^2 R_z$

Substituting values

$Z_3 = 4^2 * 4.4$

$Z_3 = 70.4 W$

Generally the increase rate of the chemically energy in the battery is mathematically represented as

$C = 2 * e * I$

Substituting values

$C = 2 * 6 * 4$

$C = 48 W$

6 0

3 years ago

A 2kg hockey puck is sliding across the ice skating rink at 2 m/s. A player hits the puck so it's velocity increases to 10 m/s.

konstantin123 [22]

The work done on the puck is 96 J

Explanation:

According to the work-energy theorem, the work done on the hockey puck is equal to the change in kinetic energy of the puck.

Mathematically:

$W=K_f -K_i= \frac{1}{2}mv^2-\frac{1}{2}mu^2$

where

$K_f = \frac{1}{2}mv^2$ is the final kinetic energy of the puck, with

m = 2 kg being the mass of the puck

v = 10 m/s is the final speed

$K_i = \frac{1}{2}mu^2$ is the initial kinetic energy of the puck, with

u = 2 m/s being the initial speed of the puck

Substituting numbers into the equation, we find the work done by the player on the puck:

$W=\frac{1}{2}(2)(10)^2 - \frac{1}{2}(2)(2)^2=96 J$

Learn more about work and kinetic energy:

brainly.com/question/6763771

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3 years ago