The nucleus of an atom has all of the following characteristics except that it

A man makes a 27.0 km trip in 16 minutes. (a.) How far was the trip in miles? (b.) If the speed limit was 55 miles per hour, wa

Maru [1056]

Answer:

(a) 16.777 miles

(b) Yes, he exceeded the speed limit

Explanation:

(a)

We need to perform the necessary calculations to convert kilometers to miles:

$27km*\frac{1000m}{1km} *\frac{1mi}{1609.34m} =16.77706389mi$

Thus, the distance of the trip in miles is:

$d=16.77706389mi$

(b)

Next, we will compute the man's speed during the journey:

$v=\frac{d}{t}$

Before that, we must convert minutes to hours:

$16min*\frac{1h}{60min} =2.666666667h$

The resulting speed is:

$v=\frac{16.77706389mi}{2.666666667h} =62.91398959\frac{mi}{h}$

Consequently:

$62.91398959\frac{mi}{h}>55\frac{mi}{h}$

Thus, it can be concluded that the driver was speeding

8 0

17 hours ago

A container is filled with an ideal diatomic gas to a pressure and volume of P1 and V1, respectively. The gas is then warmed in

Maru [1056]

Answer:

Explanation:

The transitions occur as follows:

P₁ V₁ changes to 3P₁, V₁ (with constant volume) — first phase.

Subsequently, 3P₁,V₁ transitions to 3P₁, 5V₁ (with constant pressure) — second phase.

During the initial phase, the temperature must be escalated by a factor of 3. Therefore, if the starting temperature is T₁, then the ending temperature will be 3 T₁.

P₁V₁ = n R T₁, where n represents the number of moles of gas.

Thus, nRT₁ = P₁V₁.

The heat added at constant volume is given by n Cv (3T₁ - T₁),

= n x 5/3 R X 2T₁ (noting that for diatomic gas, Cv = 5/3 R).

= 10/3 x nRT₁

= 10/3 x P₁V₁.

In the second phase, the temperature must rise 5 times. Thus, if the initial temperature is 3T₁, then the final temperature will be 15 T₁.

The heat added at constant pressure in this scenario becomes:

= n Cp (15T₁ - 3T₁)

= n x 7/3 R X 12T₁ (for diatomic gases, Cp = 7/3 R).

= 28 x nRT₁

= 28 P₁V₁.

6 0

12 days ago

The same physics student jumps off the back of her Laser again, but this time the Laser is

Yuliya22 [1153]

a) The student's speed after jumping is 1.07 m/s

b) The final speed of the laser is 10.4 m/s

Explanation:

a)

This issue can be approached through the momentum conservation principle: In the absence of external forces, the combined momentum of the student and the laser must remain unchanged. Hence, we can express:

$p_i = p_f\\0=mv+MV$

where:

The initial momentum is zero

m = 42 kg signifies the mass of the laser

v = 1.5 m/s is the laser's final velocity

M = 59 kg is the mass of the student

V denotes the student's final velocity

Solving this for V, we can determine the student's speed:

$V=-\frac{mv}{M}=-\frac{(42)(1.5)}{59}=-1.07 m/s$

Thus, the student's final speed calculates to 1.07 m/s.

b)

Here, both the laser and the student have a combined speed of 3.1 m/s prior to the student's jump; thus, the initial momentum isn't zero.

<pSo, we formulate the equation of momentum conservation as:

$(m+M)u=mv+MV$

where:

m = 42 kg denotes the mass of the laser

M = 59 kg is the student’s mass

u = 3.1 m/s is their starting velocity

V = -2.1 m/s indicates the student's speed post-jump (she jumps backward)

v signifies the laser's final speed

When we resolve for v, we have:

$v=\frac{(m+M)u-MV}{m}=\frac{(42+59)(3.1)-(59)(-2.1)}{42}=10.4 m/s$

Learn more about momentum:

3 0

12 days ago

Read 2 more answers

Approximately 1.000 g each of four gasses H2, Ne, Ar, and Kr are placed in a sealed container all under1.5 atm of pressure. Assu

serg [1198]

Answer:

The partial pressure of H2 is 0.375 atm.

The partial pressure of Ne also stands at 0.375 atm.

Explanation:

Mass of H2 = 1 g

Mass of Ne = 1 g

Mass of Ar = 1 g

Mass of Kr = 1 g

Overall mass of the gas mixture totals 4 g.

Pressure in the sealed container is 1.5 atm.

Calculating the partial pressure for H2 yields: (mass of H2/total mass of gas mixture) × pressure of sealed container = 1/4 × 1.5 = 0.375 atm.

Calculating the partial pressure for Ne similarly gives: (mass of Ne/total mass of gas mixture) × pressure of sealed container = 1/4 × 1.5 = 0.375 atm.

7 0

11 days ago

The gas tank of Dave’s car has a capacity of 12 gallons. The tank was 38 full before Dave filled it to capacity. It cost him $2.

Sav [1105]

Answer:

$ 18.75

Explanation:

Given:

Dave's car has a capacity of 12 gallons.

If the tank is at 3/8 full prior to filling:

The cost of gasoline per gallon is $2.50.

The volume Dave needs to fill: