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3 years ago

A tiger leaps horizotally from a 15.5 m high rock with a speed of 7.5 m/s. How

Physics

1 answer:

aniked [119]3 years ago

5 0

The tiger lands 13.4 m from the base of the rock

Explanation:

The motion of the tiger is a projectile motion, therefore it consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction

- A uniformly accelerated motion, with constant acceleration (acceleration of gravity) in the downward direction

We start by analyzing the vertical motion, to find the time of flight of the tiger. Using the following suvat equation:

$s=u t+\frac{1}{2}at^2$

where, taking downward as positive direction, we have:

s = 15.5 m is the vertical displacement of the tiger (the height of the rock)

$u=0$ is the initial vertical velocity

t is the time of the fall

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

Solving for t, we find :

$t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(15.5)}{9.8}}=1.78 s$

Therefore the tiger takes 1.78 s to land on the ground.

Now we analyze the horizontal motion: the tiger moves horizontally with a constant velocity of

$v_x = 7.5 m/s$

So, the distance it covers in a time t is given by:

$d=v_x t$

and by substituting t = 1.78 s, we find the horizontal distance covered by the tiger:

$d=(7.5)(1.78)=13.4 m$

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

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While entering a freeway, a car accelerates from rest at a rate of 2.40 m/s2 for 12.0 s. (a) Draw a sketch of the situation. (b)

ArbitrLikvidat [17]

Answer:

a) See attached picture, b) We know the initial velocity = 0, initial position=0, time=12.0s, acceleration= $2.40m/s^{2}$ , c) the car travels 172.8m in those 12 seconds, d) The car's final velocity is 28.8m/s

Explanation:

a) In order to draw a sketch of the situation, I must include the data I know, the data I would like to know and a drawing of the car including the direction of the movement and its acceleration, just like in the attached picture.

b) From the information given by the problem I know:

initial velocity =0

acceleration = $2.40m/s^{2}$

time = 12.0 s

initial position = 0

unknown:

displacement.

in order to choose the appropriate equation, I must take the knowns and the unknown and look for a formula I can use to solve for the unknown. I know the initial velocity, initial position, time, acceleration and I want to find out the displacement. The formula that contains all this data is the following:

$x=x_{0}+V_{x0}t+\frac{1}{2}a_{x}t^{2}$

Once I got the equation I need to find the displacement, I can plug the known values in, like this:

$x=0+0(12s)+\frac{1}{2}(2.40\frac{m}{s^{2}} )(12s)^{2}$

after cancelling the pertinent units, I get that my answer will be given in meters. So I get:

$x=\frac{1}{2} (2.40\frac{m}{s^{2}} )(12s)^{2}$

which solves to:

$x=172.8m$

So the displacement of the car in 12 seconds is 172.8m, which makes sense taking into account that it will be accelerating for 12 seconds and each second its velocity will increase by 2.4m/s.

d) So, like the previous part of the problem, I know the initial position of the car, the time it travels, the initial velocity and its acceleration. Now I also know what its final position is, so we have more than enough information to find this answer out.

I need to find the final velocity, so I need to use an equation that will use some or all of the known data and the unknown. In order to solve this problem, I can use the following equation:

$a=\frac{V_{f}-V_{0} }{t}$

Next, since I need to find the final velocity, I can solve the equation just for that, I can start by multiplying both sides by t so I get:

$at=V_{f}-V_{0}$

and finally I can add $V_{0}$ to both sides so I get:

$V_{f}=at+V_{0}$

and now I can proceed and substitute the known values:

$V_{f}=at+V_{0}$

$V_{f}=(2.40\frac{m}{s^{2}}} (12s)+0$

which solves to:

$V_{f}=28.8m/s$

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What type of lever has the fulcrum between the resistance arm and the effort arm?

crimeas [40]

I’m not really sure but I think it’s D type 1 lever

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At a picnic, there is a contest in which hoses are used to shoot water at a beach ball from three different directions. As a res

hram777 [196]

Answer:

F₃ = 122.88 N

θ₃ = 20.63°

Explanation:

First we find the components of F₁:

For x-component:

F₁ₓ = F₁ Cos θ₁

F₁ₓ = (50 N) Cos 60°

F₁ₓ = 25 N

For y-component:

F₁y = F₁ Sin θ₁

F₁y = (50 N) Sin 60°

F₁y = 43.3 N

Now, for F₂. As, F₂ acts along x-axis. Therefore, its y-component will be zero and its x-xomponent will be equal to the magnitude of force itself:

F₂ₓ = F₂ = 90 N

F₂y = 0 N

Now, for the resultant force on ball to be zero, the sum of x-components of the forces and the sum of the y-component of the forces must also be equal to zero:

F₁ₓ + F₂ₓ + F₃ₓ = 0 N

25 N + 90 N + F₃ₓ = 0 N

F₃ₓ = - 115 N

for y-components:

F₁y + F₂y + F₃y = 0 N

43.3 N + 0 N + F₃y = 0 N

F₃y = - 43.3 N

Now, the magnitude of F₃ can be found as:

F₃ = √F₃ₓ² + F₃y²

F₃ = √[(- 115 N)² + (- 43.3 N)²]

and the direction is given as:

θ₃ = tan⁻¹(F₃y/F₃ₓ) = tan⁻¹(-43.3 N/-115 N)

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How do i do a Wavelength and frequency problem

KengaRu [80]

Answer:

wavelength = v/f or wavelength equals to velocity over frequency

frequency= v/w or velocity over wavelength

frequency= 1/p or one over period or time

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