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3 years ago

9

The temperature of a 50 g sample of aluminum is raised from 20°c to 60°c when 440 cal of heat are added. the specific heat capac

ity of the aluminum is

1 answer:

GaryK [48]3 years ago

4 0

40 percent is how much aluminum is added

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a 282 kg bumper car moving right at 3.50 m/s collides with a 155 kg bumper car moving 1.88 m/s left. afterwards, the 282 kg car

Vaselesa [24]

The momentum of the 155 kg car afterwards is 469.7 kg m/s to the right

Explanation:

We can solve the problem by using the law of conservation of momentum: the total momentum of the system is conserved before and after the collision, so we can write:

$p_i = p_f\\m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$

where:

$m_1 = 282 kg$ is the mass of the bumper car

$u_1 = 3.50 m/s$ is the initial velocity of the bumper car (we take the right as positive direction)

$v_1 = 0.800 m/s$ is the final velocity of the bumper car

$m_2 = 155 kg$ is the mass of the second bumper car

$u_2 = -1.88 m/s$ is the initial velocity of the second car (moving to the left)

$v_2$ is the final velocity of the second car

Solving for $v_2$ ,

$v_2 = \frac{m_1 u_1+m_2 u_2 - m_1 v_1}{m_2}=\frac{(282)(3.50)+(155)(-1.88)-(282)(0.800)}{155}=3.03 m/s$

where the positive sign means the direction is to the right.

And now we can find the momentum of the 155 kg afterwards, which is

$p_2 = m_2 v_2 = (155)(3.03)=469.7 kg m/s$ (to the right)

Learn more about momentum:

brainly.com/question/7973509

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7 0

3 years ago

April stands on a flat surface . If she has a mass of 72 what is the nirmal force acting on her

gayaneshka [121]

The normal force acting on April as she stands on a flat surface is 705.6 N.

Normal force of the girl

Fₙ = mg

where;

Fₙ is the normal force of the girl
m is mass of the girl
g is acceleration due to gravity

Fₙ = 72 x 9.8

Fₙ = 705.6 N

Thus, the normal force acting on April as she stands on a flat surface is 705.6 N.

Learn more about normal force here: brainly.com/question/14486416

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7 0

2 years ago

You open a bank account and deposit $10. Every month you deposit $25 dollars. Let T represent the total money in your account af

Pavlova-9 [17]

The total money that will be present in the bank account, disregarding the percent rate of money increase is the sum of the initial deposit and the all the amounts deposited for m months, 25m. Thus, the answer should be,
T = 10 + 25m

4 0

3 years ago

What is the cell emf for the concentrations given? Express your answer using two significant figures.

alisha [4.7K]

Complete Question

A voltaic cell is constructed with two $Zn^{2+}$ $- Zn$ electrodes. The two cell compartment have $[Zn^{2+}] = 1.6 \ M$ and $[Zn^{2+}] = 2.00*10^{-2} \ M$ respectively.

What is the cell emf for the concentrations given? Express your answer using two significant figures

Answer:

The value is $E = 0.06 V$

Explanation:

Generally from the question we are told that

The concentration of $[Zn^{2+}]$ at the cathode is $[Zn^{2+}]_a = 1.6 \ M$

The concentration of $[Zn^{2+}]$ at the anode is $[Zn^{2+}]_c = 2.00*10^{-2} \ M$

Generally the the cell emf for the concentration is mathematically represented as

$E = E^o - \frac{0.0591}{2} log\frac{[Zn^{2+}]a}{ [Zn^{2+}]c}$

Generally the $E^o$ is the standard emf of a cell, the value is 0 V

So

$E = 0 - \frac{0.0591}{2} * log[\frac{ 2.00*10^{-2}}{1.6} ]$

=> $E = 0.06 V$

4 0

3 years ago

A weather balloon is designed to expand to a maximum radius of 24 m at its working altitude, where the air pressure is 0.030 atm

nlexa [21]

Answer:

<em>Radius at liftoff 8.98 m</em>

Explanation:

At the working altitude;

maximum radius = 24 m

air pressure = 0.030 atm

air temperature = 200 K

At liftoff;

temperature = 349 K

pressure = 1 atm

radius = ?

<em>First, we assume balloon is spherical in nature,</em>

<em>and that the working gas obeys the gas laws.</em>

from the radius, we can find the volume of the balloon at working atmosphere.

Volume of a sphere = $\frac{4}{3} \pi r^{3}$

volume of balloon = $\frac{4}{3}$ x 3.142 x $24^{3}$ = 57913.34 m^3

using the gas equation,

$\frac{P1V1}{T1}$ = $\frac{P2V2}{T2}$

<em>The subscript 1 indicates the properties of the gas at working altitude, and the subscript 2 indicates properties of the gas at liftoff.</em>

imputing values, we have

$\frac{0.03*57913.34}{200}$ = $\frac{1*V2}{349}$

0.03 x 57913.34 x 349 = 200V2

V2 = 606352.67/200 = <em>3031.76 m^3 this is the volume occupied by the gas in the balloon at liftoff.</em>

from the formula volume of a sphere,

V = $\frac{4}{3} \pi r^{3}$ = $\frac{4}{3}$ x 3.142 x $r^{3}$ = 3031.76

4.19 $r^{3}$ = 3031.76

$r^{3}$ = 3031.76/4.19

radius r of the balloon on liftoff = $\sqrt[3]{723.57}$ = <em>8.98 m</em>

4 0

3 years ago