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3 years ago

A bus moving with acceleration 2m/s^2. What does it mean

Physics

2 answers:

den301095 [7]3 years ago

7 0

Answer:

a bus moving with acceleration of 2 meter per second square

densk [106]3 years ago

4 0

Answer:

It means that it's velocity is 2m/s per s

Means it travels 2m in 1st second and 4m in 2nd sec

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A rock is thrown off a cliff at an angle of 53° with respect to the horizontal. The cliff is 100 m high. The initial speed of th

Nadusha1986 [10]

(a) 129.3 m

The motion of the rock is a projectile motion, consisting of two indipendent motions along the x- direction and the y-direction. In particular, the motion along the x- (horizontal) direction is a uniform motion with constant speed, while the motion along the y- (vertical) direction is an accelerated motion with constant acceleration $g=-9.8 m/s^2$ downward.

The maximum height of the rock is reached when the vertical component of the velocity becomes zero. The vertical velocity at time t is given by

$v(t) = v_0 sin \theta +gt$

where

$v_0 = 30 m/s$ is the initial velocity of the rock

$\theta=53^{\circ}$ is the angle

t is the time

Requiring $v(t)=0$ , we find the time at which the heigth is maximum:

$0=v_0 sin \theta + gt\\t=\frac{-v_0 sin \theta}{g}=-\frac{(30)(sin 53^{\circ})}{(-9.8)}=2.44 s$

The heigth of the rock at time t is given by

$y(t) = h+(v_0 sin\theta) t + \frac{1}{2}gt^2$

Where h=100 m is the initial heigth. Substituting t = 2.44 s, we find the maximum height of the rock:

$y=100+(30)(sin 53^{\circ})(2.44)+\frac{1}{2}(-9.8)(2.44)^2=129.3 m$

(b) 44.1 m

For this part of the problem, we just need to consider the horizontal motion of the rock. The horizontal displacement of the rock at time t is given by

$x(t) = (v_0 cos \theta) t$

where

$v_0 cos \theta$ is the horizontal component of the velocity, which remains constant during the entire motion

t is the time

If we substitute

t = 2.44 s

Which is the time at which the rock reaches the maximum height, we find how far the rock has moved at that time:

$x=(30)(cos 53^{\circ})(2.44)=44.1 m$

(c) 7.58 s

For this part, we need to consider the vertical motion again.

We said that the vertical position of the rock at time t is

$y(t) = h+(v_0 sin\theta) t + \frac{1}{2}gt^2$

By substituting

y(t)=0

We find the time t at which the rock reaches the heigth y=0, so the time at which the rock reaches the ground:

$0=100+(30)(sin 53^{\circ})t+\frac{1}{2}(-9.8)t^2\\0=100+23.96t-4.9t^2$

which gives two solutions:

t = -2.69 s (negative, we discard it)

t = 7.58 s --> this is our solution

(d) 136.8 m

The range of the rock can be simply calculated by calculating the horizontal distance travelled by the rock when it reaches the ground, so when

t = 7.58 s

Since the horizontal position of the rock is given by

$x(t) = (v_0 cos \theta) t$

Substituting

$v_0 = 30 m/s\\\theta=53^{\circ}$

and t = 7.58 s we find:

$x=(30)(cos 53^{\circ})(7.58)=136.8 m$

(e) (36.1 m, 128.3 m), (72.2 m, 117.4 m), (108.3 m, 67.4 m)

Using the equations of motions along the two directions:

$x(t) = (v_0 cos \theta) t$

$y(t) = h+(v_0 sin\theta) t + \frac{1}{2}gt^2$

And substituting the different times, we find:

$x(2.0 s)=(30)(cos 53^{\circ})(2.0)=36.1 m$

$y(2.0 s)= 100+(23.96)(2.0)-4.9(2.0)^2=128.3 m$

$x(4.0 s)=(30)(cos 53^{\circ})(4.0)=72.2 m$

$y(4.0 s)= 100+(23.96)(4.0)-4.9(4.0)^2=117.4 m$

$x(6.0 s)=(30)(cos 53^{\circ})(6.0)=108.3 m$

$y(6.0 s)= 100+(23.96)(6.0)-4.9(6.0)^2=67.4 m$

3 0

3 years ago

Why is it important for a scientist to be open-minded but skeptical?

insens350 [35]

It’s important for a scientist to be open-minded because they need to be able to understand or creative anything that was never thought about. It’s also important for a scientist to be skeptical because some ideas or products that they hear about can have false factors. I hope this helped!

5 0

3 years ago

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What would be Kelly's weight be in Newton's if her mass is 70 kilograms?

RSB [31]

It would be the first option, A

because u use the equation: W = m*g

8 0

3 years ago

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a student compared two samples of matter. he recorded the results in the chart. Sample A: Appearance: shiny and yellow. Sample B

Vladimir [108]

Answer:

The answer are B

Explanation:the two samples haves yellow color :)

6 0

3 years ago

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DetermiOne of the lines in the Balmer series of the hydrogen atom emission spectrum is at 397 nm. It results from a transition f

larisa86 [58]

Answer:

a)n =7 , b) n = 5

Explanation:

The energy levels of the hydrogen atom is described by the Bohr model

$E_{n}$ = - 13.606 1/n²

This equation energy is given in elector volts and n is an integer

A transition occurs when the electro sees from a superior to a lower state

E₀ - $E_{n}$ = -13.606 (1/ $n_{f}$ ² - 1/n₀²)

Let's apply this expression

n₀ = 2

Let's look for the energy of the different levels and subtract it

n₀ $E_{n}$ (eV)

1 -13,606

2 -3.4015

3 - 1.5118

4 -0.850375

5 -0.54424

6 -0.3779

7 -0.2777

The wavelength of the transition is 397 nm = 397 10⁻⁹ m

The speed of light is related to wavelength and frequency

c = λ f

The Planck equation gives the energy of a transition

E = h f

E = h c /λ

Let's calculate

E = 6.63 10⁻³⁴ 3 10⁸/397 10⁻⁹

E = 5.01 10⁻¹⁹ J

Let's reduce to eV

E = 5.01 10⁻¹⁹ J (1 eV / 1.6 10⁻¹⁹ J)

E = 3.1313 eV

Let's examine the possible transitions from the initial level ni = 2

ΔE = $E_{2}$ - $E_{n}$ = -3.1313

En = 3.13 -3.4015

$E_{n}$ = 0.2702 eV

When examining the table we see that the level that this energy has is the level of n = 7

Part B the transition is in the infrared

The frequency is 74 10¹² Hz

We use the Planck equation

E = h f

E = 6.63 10⁻³⁴ 74 10¹²

E = 4.9062 10⁻²⁰ J

E = 4.9062 10⁻²⁰ / 1.6 10⁻¹⁹

E = 0.3066 eV

We look for the level with the energy difference

ΔE = E₄- $E_{n}$ = 0.3066

$E_{n}$ = 0.3066 - 0.85037

$E_{n}$ = -0.54376 eV

When examining the table this energy has the level n = 5, therefore from this level the transition occurs

6 0

4 years ago