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2 years ago

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Superman is off to visit Lois lane. He sees he's late and increases his flying speed. I know he defies the laws of physics anywa

y, but let's pretend, okay? If our lead-phobic super hero accelerates uniformly from 1210 m/s to 1600 m/s in 0.40 seconds, then what is his rate of acceleration

1 answer:

mario62 [17]2 years ago

7 0

Answer:

975 m/s^2

Explanation:

The formula for rate of acceleration is

(Final velocity - intial velocity)/ time taken

If you plug in your data you will get 975 m/s^2

(1600 m/s -1210 m/s )/.4 s = 975 m^2

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A puck of mass m = 0.085 kg is moving in a circle on a horizontal frictionless surface. It is held in its path by a massless str

Rina8888 [55]

Answer:

T = 11.93 N

Explanation:

Newton's second law to the puck in the circular path

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in radial direccion (N)

m : puck mass (kg)

a : radial acceleration of the puck (m/s²)

Data:

m = 0.085 kg

L = 0.72 m = R : radium of the circular path (m)

θ= one revolution = 2Π rad

t= 0.45 s

Angular speed of the puck

ω = θ/t

ω = 1 rev/0.45 s = (2π/0.45) rad/s

ω = 13.96 rad /s

Radial acceleration or centripetal

a = ω²*R

a = (13.96) ²* (0.72)

a = 140.3 m/s²

Magnitude of the tension in the string (T)

We apply the Formula (1)

∑F = m*a

T = (0.085 kg )* (140.3 m/s² )

T = 11.93 N

4 0

2 years ago

A trebuchet was a hurling machine built to attack the walls of a castle under siege. A large stone could be hurled against a wal

Studentka2010 [4]

(a) 18.9 m/s

The motion of the stone consists of two independent motions:

- A horizontal motion at constant speed

- A vertical motion with constant acceleration ( $g=9.8 m/s^2$ ) downward

We can calculate the components of the initial velocity of the stone as it is launched from the ground:

$u_x = v_0 cos \theta = (25.0)(cos 41.0^{\circ})=18.9 m/s\\u_y = v_0 sin \theta = (25.0)(sin 41.0^{\circ})=16.4 m/s$

The horizontal velocity remains constant, while the vertical velocity changes due to the acceleration along the vertical direction.

When the stone reaches the top of its parabolic path, the vertical velocity has became zero (because it is changing direction): so the speed of the stone is simply equal to the horizontal velocity, therefore

$v=18.9 m/s$

(b) 22.2 m/s

We can solve this part by analyzing the vertical motion only first. In fact, the vertical velocity at any height h during the motion is given by

$v_y^2 - u_y^2 = 2ah$ (1)

where

$u_y = 16.4 m/s$ is the initial vertical velocity

$v_y$ is the vertical velocity at height h

$a=g=-9.8 m/s^2$ is the acceleration due to gravity (negative because it is downward)

At the top of the parabolic path, $v_y = 0$ , so we can use the equation to find the maximum height

$h_{max} = \frac{-u_y^2}{2a}=\frac{-(16.4)^2}{2(-9.8)}=13.7 m$

So, at half of the maximum height,

$h = \frac{13.7}{2}=6.9 m$

And so we can use again eq(1) to find the vertical velocity at h = 6.9 m:

$v_y = \sqrt{u_y^2 + 2ah}=\sqrt{(16.4)^2+2(-9.8)(6.9)}=11.6 m/s$

And so, the speed of the stone at half of the maximum height is

$v=\sqrt{v_x^2+v_y^2}=\sqrt{18.9^2+11.6^2}=22.2 m/s$

(c) 17.4% faster

We said that the speed at the top of the trajectory (part a) is

$v_1 = 18.9 m/s$

while the speed at half of the maximum height (part b) is

$v_2 = 22.2 m/s$

So the difference is

$\Delta v = v_2 - v_2 = 22.2 - 18.9 = 3.3 m/s$

And so, in percentage,

$\frac{\Delta v}{v_1} \cdot 100 = \frac{3.3}{18.9}\cdot 100=17.4\%$

So, the stone in part (b) is moving 17.4% faster than in part (a).

4 0

2 years ago

Which statement about electrons and atomic orbitals is NOT true?

balandron [24]

<em>An electron has the same amount of energy in all orbitals is not true</em>

$\boxed{\boxed{\bold{Further~explanation}}}$

In an atom there are levels of energy in the skin and sub skin.

This energy level is expressed in the form of electron configurations.

Writing electron configurations starts from the lowest to the highest sub-shell energy level.

So electrons that occupy the orbitals in the lowest sub-skin have the lowest energy level

In the principle of Pauli's prohibition it was stated that there are no two electrons in one atom that can have the same four quantum numbers.

So suppose that there are two electrons occupying one orbital can have the same main quantum number (n), azimuth (l) and magnetic (m), then the last quantum number that is the quantum spin number (s) must be different.

So that the two electrons are different from just the quantum spin number, even though the other quantum numbers are the same.

So in one orbital only a maximum of 2 electrons is occupied, because if there is a third electron, this third electron will have the same quantum spin number as the previous electron

The electron cloud is a visual representation of the location of electrons in an atom.

Orbital is the place around the nucleus where electrons may be found

Electron clouds show the state of electrons in their orbitals

So electron clouds can show the condition of all orbitals in an atom

The lowest energy level of an electron occupies a sub-skin of 1s which has only one orbital

Charging electrons in the sub skin uses the following sequence:

1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d¹⁰, 4p⁶, 5s², 4d¹⁰, 5p⁶, 6s², etc.

Statement about electrons and atomic orbitals is not true is An electron has the same amount of energy in all orbitals

the electron configuration for barium (Ba) in noble-gas notation brainly.com/question/11147367

the formation of a bond.

brainly.com/question/11311275

quantum number

brainly.com/question/2292596

Keywords: the electron configuration, orbitals, atoms, energy, skin, sub skin, electron clouds

8 0

2 years ago

Read 2 more answers

A rocket exhausts fuel with a velocity of 1500m/s, relative to the rocket. It starts from rest in outer space with fuel comprisi

babunello [35]

Answer:

v= 2413.5 m/s

Explanation:

maximum change of speed of rocket

=(initial exhaust velocity)×ln [(initialmass/finalmass)]

let initial mass= m

final mass = m-m(4/5) = m/5

[since the 80% of mass which is fuel is exhausted]

V-0 = 1500 ln (1/0.2)

V= 1500×1.609 = 2413.5 m/s

therefore, its exhaust speed v= 2413.5 m/s

4 0

2 years ago

Read 2 more answers

While ice skating, you unintentionally crash into a person. Your mass is 60 kg, and you are traveling east at 8.0 m/s with respe

kaheart [24]

Answer:

6.18 m/s

Explanation:

Roller skate collision

The final direction of the system (me=M + person=P) velocity vector is at an angle; Ф, to the direction running south to north. Apply the component form of the impulse-momentum equation, firstly;

x-axis component form (+x east);

$P_{Miy}$ + $p_{Piy}$ + $j_{y}$ = $P_{Mfy}$ + $P_{pfy}$

$m_{Mu_{Miy}+ m_{pu_{piy}}+0=(m_{M}+m_{p})V_{f} sin$ Ф

60 ·8 + 0 = (60 + 80) $V_{f}sin$ Ф

480 = 140 $V_{f} sin$ Ф................. (I)

y-axis component form (+y north);

$P_{Mix}$ + $p_{Pix}$ + $j_{x}$ = $P_{Mfx}$ + $P_{pfx}$

$m_{Mu_{Mix}+ m_{pu_{pix}}+0=(m_{M}+m_{p})V_{f} cos$ Ф

0 + 80.9 = (60 + 80) $V_{f}cos$ Ф

720= 140 $V_{f}cos$ Ф

140Vf= $\frac{720}{cos}$ Ф......................................(2)

Substituting (2) into (1) to give the angle;

480 = 720tan Ф

Ф = arctan(0.67) =33.69°.......................(3)

Evaluating (1) with (3) gives the velocity magnitude

480 = 140Vfsin 33.69°

Vf=6.18 m/s

note 1:

This angle corresponds to a direction; 90° - 33.69° = 56.31° north of east.

7 0

2 years ago