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frutty [35]
3 years ago
15

A launched hopper reach to 1.20 m maximum height. How much is it’s launch velocity?

Physics
1 answer:
garri49 [273]3 years ago
4 0

The launch velocity is 4.8 m/s

Explanation:

We can solve this problem by applying the law of conservation of energy. In fact, the mechanical energy of the hopper (equal to the sum of the potential energy + the kinetic energy) is conserved. So we can write:

U_i +K_i = U_f + K_f

where:

U_i is the initial potential energy, at the bottom

K_i is the initial kinetic energy, at the bottom

U_f is the final potential energy, at the top

K_f is the final kinetic energy, at the top

We can rewrite the equation as:

mgh_i + \frac{1}{2}mu^2 = mgh_f + \frac{1}{2}mv^2

where:

m is the mass of the hopper

g=9.8 m/s^2 is the acceleration of gravity

h_i = 0 is the initial height

u is the launch speed of the hopper

h_f = 1.20 m is the maximum altitude reached by the hopper

v = 0 is the final speed (which is zero when the hopper reaches the maximum height)

Solving the equation for u, we find the launch speed of the hopper:

u=\sqrt{2gh_g}=\sqrt{2(9.8)(1.20)}=4.8 m/s

Learn more about kinetic energy and potential energy:

brainly.com/question/6536722

brainly.com/question/1198647  

brainly.com/question/10770261  

#LearnwithBrainly

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All known frequencies of the visible spectrum are..
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Explanation:

b.  \: light

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3 years ago
Which colors allowed the most electricity to be generated
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Yellow and red hope that helped
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3 years ago
Please help Need good grade
ivann1987 [24]

Answer:

The other angle is 120°.

Explanation:

Given that,

Angle = 60

Speed = 5.0

We need to calculate the  range

Using formula of range

R=\dfrac{v^2\sin(2\theta)}{g}...(I)

The range for the other angle is

R=\dfrac{v^2\sin(2(\alpha-\theta))}{g}....(II)

Here, distance and speed are same

On comparing both range

\dfrac{v^2\sin(2\theta)}{g}=\dfrac{v^2\sin(2(\alpha-\theta))}{g}

\sin(2\theta)=\sin(2\times(\alpha-\theta))

\sin120=\sin2(\alpha-60)

120=2\alpha-120

\alpha=\dfrac{120+120}{2}

\alpha=120^{\circ}

Hence, The other angle is 120°

6 0
3 years ago
A cyclist traveling at constant speed of 12m/s when he passes a stationary bus.The bus starts moving just as the cyclist passes
Bogdan [553]

Answer:

A.) 8 seconds

B.) 16 seconds

C.) 48 m

Explanation:

Given that a cyclist traveling at constant speed of 12 m/s

and the bus accelerates uniformly at 1.5ms²

A.) The bus has the following parameters

Acceleration a = 1.5 m/s^2

Initial velocity U = 0. Since the bus is starting from rest.

Final velocity V = 12 m/s

Use equation one of linear motion.

V = U + at

Substitute V, U and a into the formula

12 = 0 + 1.5t

1.5t = 12

t = 12/1.5

t = 8 seconds

Therefore, the bus reach the same speed as the cyclist at 8 seconds.

B.) For the cyclist moving at constant speed, acceleration a = 0. Using second equation of motion

h = Ut + 1/2at^2

Since a = 0, the equation is reduced to:

h = Ut.

Also, for the bus,

h = Ut + 1/2at^2

Equate the two equations since the h is the same

Ut = Ut + 1/2at^2

Substitute all the parameters into the formula

12t = 0 + 1/2 × 1.5t^2

12t = 0.75t^2

0.75t = 12

t = 12/0.75

t = 16 seconds

Therefore, the bus takes 16 seconds to catch the cyclist

C.) Use third equation of linear motion.

V^2 = U^2 + 2as

Where s = distance

Substitute V, U and a into the formula

12^2 = 0 + 2 × 1.5 S

144 = 3S

S = 144/3

S = 48 m

8 0
3 years ago
The y component of a vector is 36, and the angle between the vector and the x axis is 27 what is the magnitude of the vector
xz_007 [3.2K]

Answer:

Magnitude of Vector = 79.3

Explanation:

When a vector is resolved into its rectangular components, it forms two vector components. These components  are named as x-component and y-component, they are calculated by the following formulae:

x-component of vector = (Magnitude of Vector)(Cos θ)

y-component of vector = (Magnitude of Vector)(Sin θ)

where,

θ = angle of the vector with x-axis = 27°

Therefore, using the values in the equation of y-component, we get:

36 = (Magnitude of Vector)(Sin 27°)

Magnitude of Vector = 36/Sin 27°

<u>Magnitude of Vector = 79.3</u>

3 0
2 years ago
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