Please someone help, I’m very confused and it’s due soon, thanks

Answers

Answer 1

Answer:

Answer:

1 s
19.6 m
2 s
0.8 m/s^2
28 m/s
79 m/s
0.37 s
26 m/s
242 m/s
19,930 m

Explanation:

In physics, many of the relationships between speed, distance, and acceleration are tied up in the equations for potential and kinetic energy. For an object of mass M* at height h in a gravity field with acceleration g, the potential energy is

PE = Mgh

At velocity v, the kinetic energy of the object is ...

KE = 1/2Mv^2

When an object is dropped or launched from rest, the height and velocity are related by the fact that kinetic energy gets translated to potential energy, or vice versa. This gives rise to ...

PE = KE

Mgh = (1/2)Mv^2

The mass (M) can be factored out of this, so we have ...

2gh = v^2

This can be solved for height:

h = v^2/(2g) . . . . [eq1]

or for velocity:

v = √(2gh) . . . . [eq2]

When acceleration is constant, as assumed here, the velocity changes linearly (to/from 0). So, over the time of travel, the average velocity is half the final velocity. That is,

t = 2h/v

Depending on whether you start with h or with v, this resolves to two more equations:

t = 2(v^2/(2g))/v = v/g . . . . [eq3]

t = 2h/(√(2gh)) = √(4h^2/(2gh)) = √(2h/g) . . . . [eq4]

The last of these can be rearranged to give distance as a function of time:

h = gt^2/2 . . . . [eq5]

or acceleration as a function of time and distance:

g = 2h/t^2 . . . . [eq6]

These 6 equations can be used to solve the problems posed. Just "plug and chug." For problems in Earth's gravity, we use g=9.8 m/s^2. (You may want to keep these equations handy. Be aware of the assumptions they make.)

_____

* M is used for mass in these equations so as not to get confused with m, which is used for meters.

_____

1) Use [eq4]: t = √(2·6 m/(9.8 m/s^2)) ≈ 1.107 s ≈ 1 s

2) Use [eq5]: h = (9.8 m/s^2)(2 s)^2/2 = 19.6 m

3) Use [eq4]: t = √(25 m/(4.9 m/s^2)) ≈ 2.259 s ≈ 2 s

4) Use [eq6]: g = 2(10 m)/(5 s)^2 = 0.8 m/s^2

5) Use [eq2]: v = √(2·9.8 m/s^2·40 m) = 28 m/s

6) Use [eq2]: v = √(2·9.8 m/s^2·321 m) ≈ 79.32 m/s ≈ 79 m/s

7) Using equation [eq3], we will find the time until Tina reaches her maximum height. Her actual off-the-ground total time is double this value. Using [eq3]: t = v/g = (1.8 m/s)/(9.8 m/s^2) = 9/49 s. Tina is in the air for double this time:

2(9/49 s) ≈ 0.37 s

8) Use [eq2]: v = √(2·9.8 m/s^2·33.5 m) ≈ 25.624 m/s ≈ 26 m/s

9) Use [eq2]: v = √(2·9.8·3000) m/s ≈ 242.49 m/s ≈ 242 m/s

(Note: the terminal velocity in air is a lot lower than this for an object like a house.)

10) Use [eq1]: h = (625 m/s)^2/(2·9.8 m/s^2) ≈ 19,930 m

_____

Additional comment

Since all these questions make use of the same equation development, I have elected to answer them. Your questions are more likely to be answered if you restrict your posts to 3 or fewer questions each.

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Find an expression for the center of mass of a solid hemisphere, given as the distance R from the center of the flat part of the hemisphere. Express your answer in terms of R. Express the coefficients using three significant figures.

Answers

Answer:

z_c = ⅜R

Explanation:

If we assume that the hemisphere has uniform density, we can express the centre of mass as;

z_c = (ρ/M)∫∫∫ z•dV

We know that density(ρ) = mass(M)/volume(V)

Thus, Vρ = M

And volume of hemisphere = 2πr³/3

Thus;

2Vρπr³/3 = M

So;

z_c = (ρ/(2Vρπr³/3))∫∫∫ z•dV

Where r = R in this case.

ρ will cancel out to give;

z_c = (3/(2πr³))∫∫∫_V (z•dV)

In spherical coordinates,

r is radius

Φ = angle between the point and the z − axis

θ = azimuthal angle

Therefore, the integral becomes what it is in the attached image.

I've completed the explanation as well in the attachment.

OFFERING 60 POINTS IF YOU CAN SHOW THE WORK!!!!A 1000 kg roller coaster begins on a 10 m tall hill with an initial velocity of 6m/s and travels down before traveling up a second hill. As the coaster moves from its initial height to its lowest position, 1700J of energy is transformed to thermal energy by friction.

Answers

Answer; 10.6 i think

Explanation:

(a) At the top of the hill, the coaster has total energy (potential and kinetic)

E = (1000 kg) g (10 m) + 1/2 (1000 kg) (6 m/s)² = 116,000 J

As it reaches its lowest position, its potential energy is converted to kinetic energy, and some is lost to friction, making its speed v such that

1/2 (1000 kg) v ² = 116,000 J - 1700 J = 114,300 J

===> v ≈ 15.2 m/s

If no energy is lost to friction as the coaster makes its way up the second hill, all of its kinetic energy would be converted to potential energy at the maximum possible height H.

1/2 (1000 kg) (15.2 m/s)² = (1000 kg) gH

===> H ≈ 11.7 m

(b) At the top of the second hill with minimum height h, and with maximum speed 4.6 m/s, the coaster has energy

E = P + K = (1000 kg) gh + 1/2 (1000 kg) (4.6 m/s)²

Assuming friction isn't a factor again, the energy here should match the energy at the lowest point in part (a), 114,300 J.

(1000 kg) g h + 1/2 (1000 kg) (4.6 m/s)² = 114,300 J

===> h ≈ 10.6 m

A rocket with a mass of 62,000 kg (including fuel) is burning fuel at the rate of 150 kg/s and the speed of the exhaust gases is 6,000 m/s. If the rocket is fired vertically upward from the surface of the Earth, determine its height after 744 kg of its total fuel load has been consumed. Since the mass of fuel consumed is small compared to the total mass of the rocket, you can consider the mass of the rocket to be constant for the time interval of interest.

Answers

Answer:

h≅ 58 m

Explanation:

GIVEN:

mass of rocket M= 62,000 kg

fuel consumption rate = 150 kg/s

velocity of exhaust gases v= 6000 m/s

Now thrust = rate of fuel consumption×velocity of exhaust gases

=6000 × 150 = 900000 N

now to need calculate time t = amount of fuel consumed÷ rate

= 744/150= 4.96 sec

applying newton's law

M×a= thrust - Mg

62000 a=900000- 62000×9.8

acceleration a= 4.71 m/s^2

its height after 744 kg of its total fuel load has been consumed

$h= (1)/(2)at^2$

$h= (1)/(2)4.71*4.96^2$

h= 58.012 m

h≅ 58 m

The normal is a line perpendicular to the reflecting surface at the point of incidence.

Answers

Answer:

True

Explanation:

The normal line is defined as the line which is perpendicular to the reflecting surface at the point where the incident ray meet with the reflecting surface.

The angle of incident is defined as the angle which is subtended by the incident ray with respect to the normal ray by consider the normal ray as the base line and angle is measured from the point where incident ray is incident on the reflecting surface of the mirror.

Similarly reflecting ray can be defined as the ray which is reflected after the incident of a ray and the angle subtended by the reflecting ray is measure with respect to normal ray by considering normal ray as a base line.

Therefore, the normal ray is the perpendicular line to the reflecting surface at the point of incidence.

A student goes skateboarding a few times a week what is the dependent variable

Answers

The dependent variable in this scenario is the outcome or result that you are trying to measure or analyze based on the student's skateboarding activity.

Since the student goes skateboarding a few times a week, the dependent variable could be any aspect related to their skateboarding experience or its effects.

Examples of possible dependent variables could include:

1. Improvement in skateboarding skills (e.g., measured by tricks learned, levels of proficiency).

2. Physical fitness (e.g., measured by changes in endurance, strength, or flexibility).

3. Time spent skateboarding per session.

4. The number of skateboarding injuries or accidents.

5. Overall enjoyment or satisfaction with skateboarding.

6. Changes in stress levels or mood before and after skateboarding sessions.

7. Social interactions and friendships formed through skateboarding.

The specific dependent variable would depend on the research question or hypothesis you are investigating in relation to the student's skateboarding activity.

An earthquake 45 km from a city produces P and S waves that travel outward at 5000 m/s and 3000 m/s, respectively. Once city residents feel the shaking of the P wave, how much time do they have before the S wave arrives in seconds?