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A physics student standing on the edge of a cliff throws a stone vertically downward with an initial speed of 10.0 m/s. The instant before the stone hits the ground below, it is traveling at a speed of 30.0 m/s. If the physics student were to throw the rock horizontally outward from the cliff instead, with the same initial speed of 10.0 m/s, what is the magnitude of the velocity of the stone just before it hits the ground? Ignore any effects of air resistance.

Answers

Answer 1

Answer:

Answer:

vf = 30 m/s : (the magnitude of the velocity of the stone just before it hits the ground)

Explanation:

Because the stone moves with uniformly accelerated movement we apply the following formulas:

vf²=v₀²+2*g*h Formula (1)

Where:

h: displacement in meters (m)

v₀: initial speed in m/s

vf: final speed in m/s

g: acceleration due to gravity in m/s²

Free fall of the stone

Data

v₀ = 10 m/s

vf = 30.0 m/s

g = 9,8 m/s²

We replace data in the formula (1) to calculate h:

vf²=v₀²+2*g*h

(30)² = (10)² + (2)(9.8)*h

(30)²- (10)²= (2)(9.8)*h

h =( (30)²- (10)²) /( 2)(9.8)

h = 40.816 m

Semiparabolic movement of the stone

Data

v₀x = 10 m/s

v₀y = 0 m/s

g = 9.8 m/s²

h= 40.816 m

We replace data in the formula (1) to calculate vfy :

vfy² = v₀y² + 2*g*h

vfy² = 0 + (2)(9.8)( 40.816)

$v_(fy)=√(2*9.8*40.816) = 28.284 (m)/(s)$

$v_(f)=\sqrt{v_(ox)^2+v_(fy)^2}=√((10)^2+(28.284)^2) = 30(m)/(s)$

Answer 2

Answer:

The magnitude of the velocity of the stone just before it hits the ground is 30 m/s.

The given parameters;

initial vertical velocity of the stone, $v_y_0$ = 10 m/s

final vertical velocity of the stone, $v_y_f$ = 30 m/s

The height traveled by the stone before it hits the ground is calculated as;

$v_y_f^2 = v_y_0^2 + 2gh\n\nh = (v_y_f^2- v_y_0^2)/(2g) \n\nh = ((30)^2 - (10)^2)/(2* 9.8) \n\nh = 40.82 \ m$

If the the stone is projected horizontally with initial velocity of 10 m/s;

the initial vertical velocity = 0

Final vertical velocity of the stone is calculated as follow;

$v_y_f^2 = v_y_0^2 + 2gh\n\nv_y_f^2 = 0 + 2* 9.8* 40.82\n\nv_y_f^2 = 800.07\n\nv_y_f = √(800.07) \n\nv_y_f = 28.28 \ m/s$

The horizontal velocity doesn't change.

the final horizontal velocity, $v_x_f$ = initial horizontal velocity = 10 m/s

The resultant of the final velocity of the stone before it hits the ground;

$v _f= √(v_x_f^2 + v_y_f^2) \n\nv_f = √(10^2 + 28.28^2) \n\nv_f= 29.99 \ m/s \approx 30 \ m/s$

Thus, the magnitude of the velocity of the stone just before it hits the ground is 30 m/s.

Learn more here:brainly.com/question/13533552

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Answers

Answer:correct

Explanation: Period T is the reciprocal of frequency (i.e T=1/f)

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A toy rocket is launched straight up by using a spring. The rocket is initially pressed down on the spring so that the spring is compressed by 9 cm. If the spring constant is 1050 N/m and the mass of the rocket is 50 g, how high will the rocket go? You may neglect the effects of air resistance.

Answers

Answer:

Rocket will go to a height of 8.678 m

Explanation:

Mass of the rocket m = 50 gram = 0.05 kg

Spring constant k = 1050 N /m

Spring is stretched to 9 cm

So x = 0.09 m

Work done in stretching the spring

$E=(1)/(2)kx^2=(1)/(2)* 1050* 0.09^2=4.2525J$

From energy conservation this energy will convert into potential energy

Potential energy is equal to $E=mgh$ , here m is mass, g is acceleration due to gravity and h is height

So $0.05* 9.8* h=4.2525$

$h=8.678m$

So rocket will go to a height of 8.678 m

Answer:

8.68 m

Explanation:

compression in spring, x = 9 cm = 0.09 m

Spring constant, K = 1050 N/m

mass of rocket, m = 50 g = 0.05 kg

Let it go upto height h.

Use conservation of energy

Potential energy stored in spring = potential energy of the rocket

$(1)/(2)kx^(2)=mgh$

0.5 x 1050 x 0.09 x 0.09 = 0.05 x 9.8 x h

h = 8.68 m

Thus, the rocket will go upto height 8.68 m.

Someone please help with these 2

Answers

Answer:

Explanation:

The formula that you are working with is F = m*a

Since mass is one part of the formula if you increase the mass, you are going to increase the force.

The second one is much more difficult to answer because it is basically incomplete. This is one way to interpret it. If you start at a certain speed and increase during a known time period then effectively you are defining acceleration which is "a" in the formula.

Without those modifications, there is no answer.

What sound frequency could a human detect

Answers

Answer:

People can hear sounds at frequencies from about 20 Hz to 20,000 Hz,

20 Hz up to 20,000 Hz

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Answers

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