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katrin2010 [14]

1 year ago

9

What is the magnitude of the acceleration that is produced when the brakes of a 2.4 x 103 kg car apply a 3.2 x 103 N force to st

op the car?
A .75 m/s2

B 1.3 m/s2

C 5.6 x 103 m/s2

D 7.7 x 106 m/s2

1 answer:

andriy [413]1 year ago

4 0

Answer:

B 1.3 m/s2

Explanation:

Newton's second law of motion states that the acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force. This in form of an equation is F = m*a

Since you want to know the acceleration, you have to solve for a

F = m*a

a = F / m

and now you substitute the data

a = 3.2x10^3 N / 2.4x10^3 kg

a = 1.3 m/s^2

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It can be calculated using Boyle's Law. A.

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De Vico Comet orbits the Sun every 74.0 years and has an orbital eccentricity of 0.96. Find the comet's average distance from th

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Answer: The comet's average distance from the sun is 17.6AU

Explanation:

From Kepler's 3rd Law, P^2=a^3

Where P is period in years

and a is length of semi-major axis or the average distance of the comet to the sun.

Given the orbital period to be 74 years

74^2 =a^3

5476 = a^3

Cube root of 5476 =a

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2 years ago

A 15.0-gram lead ball at 25.0°C was heated with 40.5 joules of heat. Given the specific heat of lead is 0.128 J/g∙°C, what is th

mr Goodwill [35]

Answer:

$T=4985.5^{\circ}K$

Explanation:

The equation that relates heat Q with the temperature change $T-T_0$ of a substance of mass <em>m </em>and specific heat <em>c </em>is $Q=mc(T-T_0)$ .

We want to calculate the final temperature <em>T, </em>so we have:

$T=\frac{Q}{mc}+T_0$

Which for our values means (in this case we do not need to convert the mass to Kg since <em>c</em> is given in g also and they cancel out, but we add $273^{\circ}$ to our temperature in $^{\circ}C$ to have it in $^{\circ}K$ as it must be):

$T=\frac{Q}{mc}+T_0=\frac{40.5J}{(15g)(0.128J/g^{\circ}C)}+(298^{\circ}K)=4985.5^{\circ}K$

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1 year ago

Two small aluminum spheres, each of mass 0.0250 kilograms, areseparated by 80.0 centimeters.

sineoko [7]

Answer:

Total number of electrons

$N = 7.25 \times 10^{24}$

electrons removed from each sphere

$N = 5.27 \times 10^{15}$

Fraction of electrons transferred is given as

$f = 7.27 \times 10^{-10}$

Explanation:

As we know that moles is defined as

$n =\frac{mass}{molar mass}$

$n = \frac{0.0250}{0.026982}$

$n = 0.93$

so number of atoms of Al in each sphere is given as

$N = 0.93(6.02 \times 10^{23})$

$N = 5.58 \times 10^{23}$

Now number of electrons in each atom is given as

atomic number = number of electrons in each atom = 13

total number of electrons in each sphere is

$N = 13 \times (5.58 \times 10^{23})$

$N = 7.25 \times 10^{24}$

Also we know that force of attraction between them is given as

$F= \frac{kq_1q_2}{r^2}$

$1.00 \times 10^4 = \frac{(9\times 10^9)q^2}{0.80^2}$

$q = 8.4 \times 10^{-4} C$

now we have

$q = Ne$

$8.4 \times 10^{-4} = N(1.6 \times 10^{-19}$

$N = \frac{(8.4 \times 10^{-4})}{1.6 \times 10^{-19}}$

$N = 5.27 \times 10^{15}$

Fraction of electrons transferred is given as

$f = \frac{5.27 \times 10^{15}}{7.25 \times 10^{24}}$

$f = 7.27 \times 10^{-10}$

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Nana76 [90]

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1 year ago

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