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1 month ago

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A car has a mass of 1600 kg. It is stuck in the snow and is being pulled out by a cable that applies a force of 7560 N due north

. The resistance of the snow and mud also applies a force to the car, which has a magnitude of 7340 N and points due south. What is the acceleration of the car?

1 answer:

Ostrovityanka [3.2K]1 month ago

7 0

Response:

The car's acceleration will be $a=0.1375m/sec^2$

Reasoning:

We are provided with a mass for the ball, m = 1600 kg

The force directed northward is F= 7560 N

The resistance force that counters the car's motion $F_R=7340N$

Thus, the overall force acting on the car $F_{net}=F-F_R=7560-7340=220N$

According to Newton's second law, we understand that F=ma

Therefore $220=1600\times a$

$a=0.1375m/sec^2$

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Answer:

Wnet, in, = 133.33J

Explanation:

Provided that

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Given the heat pump is entirely reversible, the performance coefficient expression is formulated as follows:

According to the first law of thermodynamics,

COP(HP, rev) = 1/(1-TL/TH)

COP(HP, rev) = 1/(1-260/300)

COP(HP, rev) = 1/(1-0.867)

COP(HP, rev) = 1/0.133

COP(HP, rev) = 7.5

The power necessary to operate the heat pump is given by

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<padditionally...><pbased on="" the="" first="" law="" rate="" at="" which="" heat="" is="" extracted="" from="" lower="" temperature="" reservoir="" calculated="" as="">

QL=QH-Wnet, in

QL=1000-133.333

QL=866.67J

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1 month ago

A small cork with an excess charge of +6.0µC is placed 0.12 m from another cork, which carries a charge of -4.3µC.

serg [3582]

A) 16.1 N

The force of electricity acting between the corks can be calculated using Coulomb's law:

$F=k\frac{q_1 q_2}{r^2}$

where

k represents Coulomb's constant

$q_1 = 6.0 \mu C=6.0 \cdot 10^{-6} C$ denotes the charge magnitude on the first cork

$q_2 = 4.3 \mu C = 4.3 \cdot 10^{-6}C$ indicates the charge magnitude on the second cork

r = 0.12 m is the distance separating the corks

By inserting the values into the formula, we arrive at

$F=(9\cdot 10^9 N m^2 C^{-2} )\frac{(6.0\cdot 10^{-6}C)(4.3\cdot 10^{-6} C)}{(0.12 m)^2}=16.1 N$

B) Attractive

<pas per="" coulomb="" law="" the="" orientation="" of="" electric="" force="" between="" two="" charged="" entities="" relies="" on="" their="" charge="" signs.=""><pmore specifically="">

- when both are similarly charged (e.g. positive-positive or negative-negative), the force is repulsive

- when charges are of opposite signs (e.g. positive-negative), the resulting force is attractive

<pin this="" case="" we="" have="">

Cork 1 holds a positive charge

Cork 2 possesses a negative charge

<pthus the="" force="" acting="" between="" them="" is="" attractive.="">

C) $2.69\cdot 10^{13}$

The total charge of the negative cork is

$q_2 = -4.3 \cdot 10^{-6}C$

<pwe understand="" that="" a="" single="" electron="" has="" charge="" of="">

$e=-1.6\cdot 10^{-19}C$

<pthe total="" charge="" of="" the="" negative="" cork="" arises="" from="" having="" n="" extra="" electrons="" so="" we="" can="" express="" it="" as="">

$q_2 = Ne$

<pafter solving="" for="" n="" we="" can="" determine="" the="" count="" of="" excess="" electrons:="">

$N=\frac{q_2}{e}=\frac{-4.3\cdot 10^{-6} C}{-1.6\cdot 10^{-19} C}=2.69\cdot 10^{13}$

D) $3.75\cdot 10^{13}$

The overall charge on the positive cork is

$q_1 = +6.0\cdot 10^{-6}C$

<pthe charge="" of="" a="" single="" electron="" is="" known="" to="" be="">

$e=-1.6\cdot 10^{-19}C$

<pthe total="" charge="" of="" the="" positive="" cork="" results="" from="" n="" excess="" electrons="" which="" can="" be="" depicted="" as="">

$q_1 = -Ne$

<pby calculating="" for="" n="" we="" derive="" the="" number="" of="" electrons="" cork="" has="" lost:="">

$N=-\frac{q_1}{e}=-\frac{+6.0\cdot 10^{-6} C}{-1.6\cdot 10^{-19} C}=3.75\cdot 10^{13}$

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1 month ago

A new roller coaster contains a loop-the-loop in which the car and rider are completely upside down. If the radius of the loop i

Sav [3153]

Respuesta:

11.4 m/s

Explicación:

La fórmula para la aceleración centrípeta es:

$a=\frac{v^2}{R}$

donde, a es la aceleración, v la velocidad alrededor de la circunferencia y R el radio del círculo.

En este problema,

a = g = aceleración debida a la gravedad en la cima = $9.81\ m/s^2$

v = ?

R = 13.2 m

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$9.81=\frac{v^2}{13.2}$

$v^2=9.81\times {13.2}$

v = 11.4 m/s

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2 months ago

What is the least possible initial kinetic energy in the oxygen atom could have and still excite the cesium atom?

inna [3103]

K=E[(m+M)/M] Kmin=4.4

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1 month ago

Modern wind turbines generate electricity from wind power. The large, massive blades have a large moment of inertia and carry a

Yuliya22 [3333]

Answer:

a)106.48 x 10⁵ kg.m²

b)144.97 x 10⁵ kgm² s⁻¹

Explanation:

a)Given

m = 5500 kg

l = 44 m

The moment of inertia for one blade

$I$ = 1/3 x m l²

where m denotes the mass of the blade

l represents the length of each blade.

Substituting the necessary values, the moment of inertia for one blade is

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Total moment of inertia for 3 blades

$I$ = 3 x 35.49 x 10⁵ kg.m²

$I$ = 106.48 x 10⁵ kg.m²

b) The angular momentum 'L' is calculated using

L = $I$ x ω

where,

$I$ = the moment of inertia of the turbine i.e 106.48 x 10⁵ kg.m²

ω= angular velocity =2π f

f represents the frequency of rotation of the blade i.e 13 rpm

f = 13 rpm=>= 13 / 60 revolutions per second

ω = 2π f => 2π x 13 / 60 rad / s

L= $I$ x ω =>106.48 x 10⁵ x 2π x 13 / 60

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19 days ago