All categories

3 years ago

Which of the following is true of changes in momentum?

Physics

1 answer:

soldi70 [24.7K]3 years ago

6 0

Answer: ⇒ Answer is 3

<h2>Explanation: momentum = mass × velocity</h2>

"A small force may produce a large change in momentum by acting on a very massive object".

THEY HAVEN'T GIVEN US THE TIME PERIOD NOR THE DISTANCE TRAVELED. THEREFORE WE CANNOT ACTUALLY DECIDE IF THE FORCE IS KEPT FOR A LONG TIME OR SHORT TIME. ANYWAYS SINCE THE MASS IS GIVEN AS MASSIVE , THE MOMENTUM SHOULD BE DEFINITELY HIGH.

WHY I SAY OTHERS ARE WRONG:

1) For a small force to give a large change in momentum, it should act for a long time interval.

2) By applying a large force for a short time interval, the change of momentum should be large.

3) Correct answer.

4) Acting over a short distance can be the same as acting over a short period of time.Therefore the distance should be large in order for a larger momentum.

I HOPE IT HELPS!

You might be interested in

OSCilloscope shows wave starts cycle minimum 20 units reaches max of 100 units before completing 5 milliseconds what is relation

Vitek1552 [10]

Answer:

$T=0.02\ s$

$f=50\ Hz$

Explanation:

Given:

minimum amplitude at the start of oscillation cycle, $a_0=20\ unit$

the first maximum amplitude after the start of oscillation cycle, $a_{m1}=100\ units$

Time taken to reach from the first minima to the first maxima, $t=5\times 10^{-3}\ s$

As we know that an oscilloscope executes a wave cycle represented by a sine wave. So we can deduce that it has executed one-fourth of the cycle in going from the amplitude of 20 units to 100 units in 0.005 seconds.

<u>So the time taken to complete one cycle of the oscillation:</u>

$T=4\times 0.005$

$T=0.02\ s$ is the time period of the oscillation

<u>We know frequency:</u>

$f=\frac{1}{T}$

$f=\frac{1}{0.02}$

$f=50\ Hz$

6 0

4 years ago

Scenario 2: Use the following information to answer questions 3 and 4:

NemiM [27]

Answer:

A. 52 min

.A. 47 watts

Explanation:

Given that;

jim weighs 75 kg

and he walks 3.3 mph; the objective here is to determine how long must he walk to expend 300 kcal.

Using the following relation to determine the amount of calories burned per minute while walking; we have:

$\dfrac{MET*weight (kg)*3.5}{200}$

here;

MET = energy cost of a physical activity for a period of time

Obtaining the data for walking with a speed of 3.3 mph From the standard chart for MET, At 3.3 mph; we have our desired value to be 4.3

However;

the calories burned in a minute = $\dfrac{4.3*75 (kg)*3.5}{200}$

= 5.644

Therefore, for walking for 52 mins; Jim burns approximately 293.475 kcal which is nearest to 300 kcal.

Given that:

mass m = 75 kg

intensity = 6 kcal/min

The eg ergometer work rate = ??

Applying the formula:

$V_O_2 ( intensity ) = ( \dfrac{W}{m}*1.8)+7$

where ;

$V_O_2 ( intensity ) = \dfrac{1 \ kcal min^{-1}*10^{-3}}{5}$

$V_O_2 ( intensity ) = \dfrac{6*1 \ kcal min^{-1}*10^{-3}}{5}$

$V_O_2 ( intensity ) = 0.0012$

∴ $0.0012 = (\dfrac{W}{75}*1.8)+7 \\ \\ W = \dfrac{0.0012-7}{1.8}*75 \\ \\ W = \dfrac{7*75}{1.8} \\ \\ W = 291.66 \ kg m /min$

Converting to watts;

Since; 6.118kg-m/min is = 1 watt

Then 291.66 kgm /min will be equal to 47.67 watts

≅ 47 watts

3 0

4 years ago

Radiation from the Sun reaching Earth (just outside the atmosphere) has an intensity of 1.4 kW/m2. (a) Assuming that Earth (and

baherus [9]

To solve the problem we will require the concept of Force as a definition of pressure and Area, and the concept of light pressure itself determined by the relationship between intensity and the speed at which light travels. We will match the terms and find the desired force value,

$F = PA$

Here,

P = Pressure

A = Area

Pressure due to the light of the sun will be

$P= \frac{I}{c}$

Here,

I = Intensity

c = Speed velocity

Equation both therms we have that

$F = \frac{IA}{c}$

We have a circular area then

$F = \frac{I(\pi r^2)}{c}$

Replacing with our values (Adding the radius of the Earth)

$F = \frac{(1.4*10^3W/m^2)(\pi (6.32*10^6m)^2)}{3.0*10^8m/s}$

$F = 6.0*10^8N$

Therefore the Force on Earth due to radiation pressure is $6*10^8N$

4 0

3 years ago

Cody hits up food king and uses a scale to weigh the mass of an apple. if the spring potential energy in the scale is .09 j and

Lostsunrise [7]

Answer:

oK so here's <u>what you should do is add .09 and 0.6</u>

Explanation:

7 0

3 years ago

Dee is on a swing in the playground. the chains are 2.5 m long, and the tension in each chain is 450 n when dee is 55 cm above t

Leno4ka [110]

Refer to the diagram shown below.

From the geometry, obtain
x = 2.5 - 0.55 = 1.95 m
cos θ = 1.95/2.5 = 0.78
θ = cos⁻¹ 0.78 = 38.74°

From the free body diagram, the tension in the chain is 450 N.
F is the centripetal force,
W is Dee's weight.

The components of the tension are
Horizontal component = 450 sin(38.74°) = 281.6 N, acting left.
Vertical component = 450 cos(38.74°) = 351.0 N, acting upward.

Answers:
Horizontal: 281.6, acting left.
Vertical: 351.0 N, acting upward.

8 0

4 years ago