What is the answer to these questions?

The specific heat of aluminum is 0.90 J/gC . How much heat is given off when 25 grams of aluminum is cooled from 55 C to 25 C?

sp2606 [1]

Answer:

2. How many joules of heat are needed to raise the temperature of 10.0 g of aluminum from 22°C to 55°C, if the specific heat of aluminum is 0.90 J/gºC? c=0.90J/g. 9 (2 sigs.)

Explanation:

8 0

3 years ago

The propeller of a World War II fighter plane is 2.30 m in diameter. (a) What is its angular velocity in radians per second if i

Sergio [31]

Answer with Explanation:

We are given that

Diameter of fighter plane=2.3 m

Radius= $r=\frac{d}{2}=\frac{2.3}{2}=1.15 m$

a.We have to find the angular velocity in radians per second if it spins=1200 rev/min

Frequency= $\frac{1200}{60}=20 Hz$

1 minute=60 seconds

Angular velocity= $\omega=2\pi f$

Angular velocity= $2\times \frac{22}{7}\times 20=125.7 rad/s$

b.We have to find the linear speed of its tip at this angular velocity if the plane is stationary on the tarmac.

$v=r\omega=1.15\times 125.7=144.56 m/s$

c.Centripetal acceleration= $\omega^2 r=(125.7)^2(1.15)=18170.56 m/s^2$

Centripetal acceleration== $\frac{18170.56\times g}{9.81}=1852.25 g m/s^2$

7 0

4 years ago

5. Each of five satellites makes a circular orbit about an object that is much more massive than any of the satellites. The mass

Vikentia [17]

The options of the question are missing. I have attached it.

Answer:

Satellite in option B will have the greatest speed.

Explanation:

From kepplers third law, we know that

V² = GM/R

Thus, v = √(GM/R)

Where;

v is velocity

G is gravitational constant

M is mass

R is radius

Looking at the options, let's start from the first one;

Option A

Here, mass = (1/2)m and radius = R

So, v = √(GM/R) thus v = √(G(m/2)/R) = √(Gm/2R)

Option B

Here, mass = m and radius = (1/2)R

Thus v = √(Gm/(R/2)) = √(2Gm/R)

Option C

Here, mass = m and radius = R

Thus v = √(Gm/R)

Option D

Here, mass = m and radius = 2R

Thus v = √(Gm/(2R))

Now, inspecting all the options, it's clear that option B will have the greatest velocity because it's numerator is the biggest and will in turn lead to higher velocity.

4 0

4 years ago

How much work is done (by a battery, generator, or some other source of potential difference) in moving Avogadro's number of ele

luda_lava [24]

Answer:

W = 1.5 x 10⁶ J = 1.5 MJ

Explanation:

First, we calculate the potential difference between the given 2 points. So, we have:

V₁ = Electric Potential at Initial Position = 6.7 V

V₂ = Electric Potential at Final Position = - 8.9 V

Therefore,

ΔV = Potential Difference = V₂ - V₁ = -8.9 V - 6.7 V = - 15.6 V

Since, we use magnitude in calculation only. Therefore,

ΔV = 15.6 V

Now, we calculate total charge:

Total Charge = q = (No. of Electrons)(Charge on 1 Electron)

where,

No. of Electrons = Avagadro's No. = 6.022 x 10²³

Charge on 1 electron = 1.6 x 10⁻¹⁹ C

Therefore,

q = (6.022 x 10²³)(1.6 x 10⁻¹⁹ C)

q = 96352 C

Now, from the definition of potential difference, we know that it is equal to the worked done on a unit charge moving it between the two points of different potentials:

ΔV = W/q

W = (ΔV )(q)

where,

W = work done = ?

W = (15.6 V)(96352 C)

W = 1.5 x 10⁶ J = 1.5 MJ

7 0

3 years ago

The half-life of Iodine-131 is 8.0252 days. If 14.2 grams of I-131 is released in Japan and takes 31.8 days to travel across the

MakcuM [25]

Answer:

Explanation:

Half-life problems are modeled as exponential equations. The half-life formula is $P=P_o\left (\dfrac{1}{2} \right)^{\frac{t}{k}}$ where $P_o$ is the initial amount, $k$ is the length of the half-life, $t$ is the amount of time that has elapsed since the initial measurement was taken, and $P$ is the amount that remains at time $t$ .

$P=14.2\left (\dfrac{1}{2} \right)^{\frac{t}{8.0252}}$

Deriving the half-life formula

If one forgets the half-life formula, one can derive an equivalent equation by recalling the basic an exponential equation, $y=a b^{t}$ , where $t$ is still the amount of time, and $y$ is the amount remaining at time $t$ . The constants a and b can be solved for as follows:

Knowing that amount initially is 14.2g, we let this be time zero:

$y=a b^{t}$

$(14.2)=ab^{(0)}$

$14.2=a *1$

$14.2=a$

So, $a=14.2$ , which represents out initial amount of the substance, and our equation becomes: $y=14.2 b^{t}$

Knowing that the "half-life" is 8.0252 days (note that the unit here is "days", so times for all future uses of this equation must be in "days"), we know that the amount remaining after that time will be one-half of what we started with:

$\left(\frac{1}{2} *14.2 \right)=14.2 b^{(8.0252)}$

$\dfrac{7.1}{14.2}=\dfrac{14.2 b^{8.0252}}{14.2}$

$0.5=b^{8.0252}$

$\sqrt[8.0252]{\frac{1}{2}}=\sqrt[8.0252]{b^{8.0252}}$

$\sqrt[8.0252]{\frac{1}{2}}=b$

Recalling exponent properties, one could find that $\left ( \frac{1}{2} \right )^{\frac{1}{8.0252}}=b$ , which will give the equation identical to the half-life formula. However, recalling this trivia about exponent properties is not necessary to solve this problem. One can just evaluate the radical in a calculator:

$b=0.9172535661...$

Using this decimal approximation has advantages (don't have to remember the half-life formula & don't have to remember as many exponent properties), but one minor disadvantage (need to keep more decimal places to reduce rounding error).

So, our general equation derived from the basic exponential function is:

$y=14.2* (0.9172535661)^t$ or $y=14.2*(0.5)^{\frac{t}{8.0252}}$ where y represents the amount remaining at time t.