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MArishka [77]
3 years ago
7

If you stand at the edge of a cliff that is 75 m high and throw a rock directly up into the air with a velocity of 20 m/s, at wh

at time will the rock hit the ground? (note: the quadratic formula will give two answers, but only one of them is reasonable.)
Physics
2 answers:
lesya [120]3 years ago
5 0

Answer:

  At time 6.45 seconds rock will hit ground.  

Explanation:

We have equation of motion , s= ut+\frac{1}{2} at^2, s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

 Let us take up direction as positive, so displacement = -75 m, initial velocity = 20 m/s, acceleration = acceleration due to gravity = -9.8m/s^2

 Substituting

    -75= 20*t-\frac{1}{2} *9.8*t^2\\ \\ 4.9t^2-20t-75=0

    t = 6.45 seconds or t =-2.37 seconds

    Since negative time is not possible, t = 6.45 seconds.

    So at time 6.45 seconds rock will hit ground.

My name is Ann [436]3 years ago
4 0

The vertical position of the rock is given by:

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

where

h=75 m is the initial height

v0=20 m/s is the initial vertical velocity

g=9.81 m/s^2 is the acceleration due to gravity

The problem asks to find the time t at which the rock hits the ground, so the time t at which y(t)=0, so the equation above becomes:

75 + 20 t - 4.9t^2 =0

Which has two solutions:

t=-2.37 s

t=6.45 s

The first solution is a negative time so it has no physical meaning, therefore the correct answer is t=6.45 s.


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mixas84 [53]
I believe its the third answer
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2 years ago
You are using a hydrogen discharge tube and high quality red and blue light filters as the light source for a Michelson interfer
boyakko [2]

Answer:

final displacement = +24484.5 nm

Explanation:

The path difference when 158 bright spots were observed with red light (λ1 = 656.3 nm) is given as;

Δr = 2d2 - 2d1 = 150λ1

So, 2d2 - 2d1 = 150λ1

Dividing both sides by 2 to get;

d2 - d1 = 75λ1 - - - - eq1

Where;

d1 = distance between the fixed mirror and the beam splitter

d2 = position of moveable mirror from splitter when 158 bright spots are observed

Now, the path difference between the two waves when 114 bright spots were observed is;

Δr = 2d'2 - 2d1 = 114λ1

2d'2 - 2d1 = 114λ1

Divide both sides by 2 to get;

d'2 - d1 = 57λ1

Where;

d'2 is the new position of the movable mirror from the splitter

Now, the displacement of the moveable mirror is (d2 - d'2). To get this, we will subtract eq2 from eq1.

(d2 - d1) - (d'2 - d1) = 75λ1 - 57λ2

d2 - d1 - d'2 + d1 = 75λ1 - 57λ2

d2 - d'2 = 75λ1 - 57λ2

We are given;

(λ1 = 656.3 nm) and λ2 = 434.0 nm.

Thus;

d2 - d'2 = 75(656.3) - 57(434)

d2 - d'2 = +24484.5 nm

5 0
3 years ago
Can someone answer these
LenKa [72]

Four

Sometimes I think the creators of problems out to drawn and quartered. 60 g does not mean 60 grams. It means 60 * the acceleration due to gravity.

So the question really reads. The acceleration delivered by the air bag is 60 times that of a normal gravitational. This acceleration is delivered to the person where his mass is putting up a whole lot of resistance because he and his 75 kg are moving forward with the impact of the car. The 36 msec. has nothing to do with the problem.

The Force of the Air Bag is mass * a

F_airbag = mass * acceleration = 75 kg * 60 * 9.81 mass * acceleration = 44145 newtons

The answer is 4.41 * 10^4

Answer C

Five

This problem is governed by one formula that you sort of have to get out of your hat -- a piece of magic if you will.

Fg - Bf = m * a

Fg = the Force of gravity

Bf = the braking force

The mass of the rocket is derived from its weight

The acceleration is derived from one of your big 4 equations.

m of the rocket = 75600 / 9.81 = 7706 kg

The acceleration =

vi = 1 km/s = 1000 m/s

vf = 0

t = 2 minute * 60 sec/ min = 120 seconds

a = (vf - vi)/t = (0 - 1000 m/s) / 120 sec

a = - 8.333 m/s^2 The minus sign makes perfect sense. Remember the rocket is slowing down

The net downward force = mass * acceleration = - 7706 kg * - 8.333 m/s^2

The net force = - 64217 N

So going back to the problem's equation we have

Gravitational force - Braking Force = Net Force

Gravitational Force = 75600

Net Force = - 64217

Bracking force = ?

75600 - Bracking force = - 64217  Subtract 75600 from both sides

- Bracking force = - 64217 - 75600

- Braking force = - 139817

Braking force = 139817 N = 1.398 * 10^5 N

Braking Force = 1.4 * 10^5

Answer: Last One.

Six

The first thing you should do is derive a general formula for this problem.

The force pulling both masses down is M*g where g is the acceleration due to gravity.

The formula for this problem is

Mg = (m + M) * a

Now you need to solve for a

a =  [M/(M + m) ] * g

Look what is happening. is a smaller or larger than g? This is a question you should really pay attention to. If it was larger, everyone would have this system in their basement because you'd get more energy output than you put in. Something for nothing is always appealing.

So what's the answer? (I get to ask it. No one posing the question ever should).

A

A is incorrect. M never goes away. The acceleration may get very tiny, but there always is some acceleration.

B must be true. It is just what I finished saying about A

C Who said anything about velocity? It's a red herring. If the velocity became 0 the acceleration would have to turn minus. This answer sounds good, but sounds good doesn't make it right. C is wrong.

D The acceleration does not remain constant no matter what. The answer to A still applies. So D is wrong.

4 0
3 years ago
While rearranging a dorm room, a student does 310 J of work in moving a desk 2.9 m. What was the magnitude of the applied horizo
Viefleur [7K]

Answer:

The horizontal force is 106.89 N.

Explanation:

Given that,

Work done = 310 J

Distance = 2.9 m

We need to calculate the horizontal force

Using formula of work done

W= Fd\cos\theta

Where, \theta=0^{\circ}

W=F\cdot d

Put the value into the formula

310=F\cdot 2.9

F=\dfrac{310}{2.9}

F=106.89\ N

Hence, The horizontal force is 106.89 N.

3 0
3 years ago
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irinina [24]

Answer:

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