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3 years ago

What is the difference between mass and momentum?

Physics

1 answer:

VLD [36.1K]3 years ago

6 0

Answer & Explanation:

Definition:

Mass informs us about how much matter is present in a body. It is the measure of the inertia of a body. It can be measured by measuring the force of gravity acting on it on earth.

On the other hand, Momentum is the quantity of motion of the body. It depends depends upon velocity and the direction of the motion of the body.

Unit:

The SI unit of mass is kilogram (kg) while the SI unit of momentum is kg.m.s-1.

Formula:

Mass = Momentum / Velocity

Momentum = Mass * Velocity

or p = mv

Type:

Mass is scalar quantity while Momentum is a vector quantity.

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A racecar accelerates uniformly from 18.5 mil to 46.1 m/s in 2.47 seconds.

laiz [17]

Answer:

The acceleration of the racecar is $\mathbf{11.17~m/s^2}$

Explanation:

Uniformly Accelerated Motion

It's a type of motion in which the velocity of an object changes by an equal amount in every equal period of time.

Following the definition above, the acceleration is defined as:

$\displaystyle a=\frac{v_f-v_o}{t}$

Where a is the constant acceleration, vo the initial speed, vf the final speed, and t the time.

The racecar goes from vo=18.5 m/s to vf=46.1 m/s in t=2.47 seconds, thus the acceleration is:

$\displaystyle a=\frac{46.1-18.5}{2.47}$

$\displaystyle a=\frac{27.6}{2.47}$

$a = 11.17~m/s^2$

The acceleration of the racecar is $\mathbf{11.17~m/s^2}$

5 0

3 years ago

Which of the following is an example of a push?

Alex777 [14]

Throwing a baseball.

7 0

3 years ago

Read 2 more answers

Unless indicated otherwise, assume the speed of sound in air to be v = 344 m/s. A stationary police car emits a sound of frequen

bixtya [17]

Answer:

a) The velocity of the car is 7.02 m/s and the car is approaching to the police car as the frequency of the police car is increasing.

b) The frequency is 1404.08 Hz

Explanation:

If the police car is a stationary source, the frequency is:

$f_{a} =(\frac{v+v_{c} }{v} )f_{s}$ (eq. 1)

fs = frequency of police car = 1200 Hz

fa = frequency of moving car as listener

v = speed of sound of air

vc = speed of moving car

If the police car is a stationary observer, the frequency is:

$f_{L} =f_{a} (\frac{v}{v-v_{c} } )=(\frac{v+v_{c} }{v-v_{c} } )f_{s}$ (eq. 2)

Now,

fL = frequecy police car receives

fs = frequency police car as observer

a) The velocity of car is from eq. 2:

$1250=1200(\frac{v+v_{c} }{v-v_{c} } )\\1250(v-v_{c} )=1200(v+v_{c} )\\v_{c} =\frac{50*344}{2450} =7.02m/s$

b) Substitute eq. 1 in eq. 2:

$f_{L} =(\frac{v+v_{p} }{v-v_{c} } )(\frac{v+v_{c} }{v-v_{p} } )f_{s} =(\frac{344+20}{344-7.02} )(\frac{344+7.02}{344-20} )*1200=1404.08Hz$

7 0

4 years ago

Traveling 221 miles from Boston Back Bay Station to NYC Penn Station takes 3 hours

Nostrana [21]

Answer:

Approximately $116\; \text{miles}$ for the train from Boston to NYC Penn Station.

Approximately $105\; \text{miles}$ for the train from NYC Penn Station to Boston.

Explanation:

Convert minutes to hours:

$\begin{aligned}t(\text{BOS $\to$ NYC}) &= 3\; {\text{hour}} + 40\; \text{minute} \times \frac{1\; {\text{hour}}}{60\; \text{minute}} \\ &=\left(3 + \frac{2}{3}\right)\; \text{hour}\\ &= \frac{11}{3}\; \text{hour} \end{aligned}$ .

$\begin{aligned}t(\text{NYC $\to$ BOS}) &= 4\; {\text{hour}} + 5\; \text{minute} \times \frac{1\; {\text{hour}}}{60\; \text{minute}} \\ &= \frac{49}{15}\; \text{hour} \end{aligned}$ .

Calculate average speed of each train:

$\begin{aligned}v(\text{BOS $\to$ NYC}) &= \frac{s}{t}\\ &= \frac{221\; \text{mile}}{\displaystyle \frac{11}{3}\; \text{hour}} \\ &= \frac{663}{11}\; \text{mile} \cdot \text{hour}^{-1}\end{aligned}$ .

$\begin{aligned}v(\text{NYC $\to$ BOS}) &= \frac{s}{t}\\ &= \frac{221\; \text{mile}}{\displaystyle \frac{49}{15}\; \text{hour}} \\ &= \frac{2652}{49}\; \text{mile} \cdot \text{hour}^{-1}\end{aligned}$

Assume that it takes a time period of $t$ for the trains to pass by each other after departure. Distance each train travelled would be:

$s(\text{NYC $\to$ BOS}) = v(\text{NYC $\to$ BOS})\, t$ .

$s(\text{BOS $\to$ NYC}) = v(\text{BOS $\to$ NYC})\, t$ .

Since the trains have just passed by each other, the sum of the two distances should be equal to the distance between the stations:

$v(\text{NYC $\to$ BOS})\, t + v(\text{BOS $\to$ NYC})\, t = 221\; \text{mile}$ .

Rearrange and solve for $t$ :

$(v(\text{NYC $\to$ BOS}) + v(\text{BOS $\to$ NYC}))\, t = 221\; \text{mile}$ .

$\begin{aligned}t &= \frac{221\; \text{mile}}{v(\text{NYC $\to$ BOS}) + v(\text{BOS $\to$ NYC})} \\ &= \frac{221\; \text{mile}}{\displaystyle \frac{663}{11}\; \text{mile} \cdot \text{hour}^{-1} + \frac{2652}{49}\; \text{mile} \cdot \text{hour}^{-1}} \\ &= \frac{539}{279}\; \text{hour}\end{aligned}$ .

Distance each train travelled in $t = (539 / 279)\; \text{hour}$ :

$\begin{aligned}s(\text{BOS $\to$ NYC}) &= v\, t \\ &= \frac{663}{11}\; \text{mile} \cdot \text{hour}^{-1} \times \frac{539}{279}\; \text{hour} \\ &\approx 116\; \text{mile}\end{aligned}$ .

$\begin{aligned}s(\text{NYC $\to$ BOS}) &= v\, t \\ &= \frac{2652}{49}\; \text{mile} \cdot \text{hour}^{-1} \times \frac{539}{279}\; \text{hour} \\ &\approx 105\; \text{mile} \end{aligned}$ .

8 0

2 years ago

Why are objects that fall near Earth’s surface rarely in free fall?

bulgar [2K]

Objects that fall near Earth’s surface are rarely in free fall.

"Free fall" is the situation where the ONLY force on an object is
the force of gravity, and nothing else.

Objects near Earth's surface are almost always surrounded by air.
If they are falling, then the air is exerting forces on them, and they
are not in "free fall".

4 0

3 years ago

Read 2 more answers