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3 years ago

13

A skier, starting from rest, slides down an icy frictionless 80o incline whose vertical height is 105 m. How fast are they going

when the reach the bottom?

1 answer:

leonid [27]3 years ago

4 0

Answer:

nadaáaaaaaaaaaaaaa aaaaaaaaaaaaaaa

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A projectile is fired with an initial speed of 51.2 m/s at an angle of 44.5 degrees above the horizontal on a long flat firing r

irinina [24]

Use the projectile motion equations

H = v^2 x sin^2(θ) ÷ 2g

t = 2 x v x sinθ ÷ g

R = v^2 x sin2θ ÷ g

3 0

3 years ago

A stone was dropped off a cliff and hit the ground with a speed of 88 ft/s . What is the height of the cliff? (Use 32 ft/s 2 for

bulgar [2K]

To solve this problem we will apply the linear motion kinematic equations, which describe the change in velocity, depending on the acceleration and the distance traveled, that is,

$v_f^2 = v_i^2 +2ah$

Where,

$v_f$ = Final Velocity

$v_i$ = Initial Velocity

a = Acceleration

h = height

Our values are given as,

$v_f = 88 ft/s\\v_i = 0 ft /s\\a = 32 ft/s^2\\$

Replacing we have,

$vf^2 = vi^2 + 2*a*h$

$88^2 = 0 + 2*32*h$

$h= 121 ft$

Therefore the height of the cliff is 121ft

5 0

3 years ago

Please need help on this thank you

lys-0071 [83]

I am pretty sure it is B....

6 0

3 years ago

Will Mark Brainliest if Correct PLZ!!!!! A bullet is shot at some angle above the horizontal at an initial velocity of 87m/s on

qaws [65]

Answer:

≅50°

Explanation:

We have a bullet flying through the air with only gravity pulling it down, so let's use one of our kinematic equations:

Δx=V₀t+at²/2

And since we're using Δx, V₀ should really be the initial velocity in the x-direction. So:

Δx=(V₀cosθ)t+at²/2

Now luckily we are given everything we need to solve (or you found the info before posting here):

Δx=760 m
V₀=87 m/s
t=13.6 s
a=g=-9.8 m/s²; however, at 760 m, the acceleration of the bullet is 0 because it has already hit the ground at this point!

With that we can plug the values in to get:

$760=(87)(cos\theta )(13.6)+\frac{(0)(13.6^{2}) }{2}$

$760=(1183.2)(cos\theta)$

$cos\theta=\frac{760}{1183.2}$

$\theta=cos^{-1}(\frac{760}{1183.2})\approx50^{o}$

3 0

3 years ago

An upward force of 32.6 N is applied via a string to lift a ball with a mass of 2.8 kg. (a) What is the gravitational force acti

Igoryamba

Answer:

a) Fg = -27.4 N

b) Fnet = 5.2 N

c) a = 1.9 m/s2

Explanation:

a)

There are two forces acting on the ball, one directed upward (assuming this direction as positive, along the y-axis) which is the tension on the string (lifting force), and another aimed downward, which is the attractive force due to gravity.
Applying the Newton's Universal Law of Gravitation to a mass close to the surface of the Earth (in this case the ball), we can take the acceleration due to gravity like a constant, that we call by convention g, equal to -9.8 m/s2.
So, we can write the following expression for Fg:

$F_{g} = m*g = 2.8 kg*(-9.8m/s2) = -27.4 N (1)$

b)

The net force on the ball, will be just the difference between the lifting force (32.6 N) and the force due to gravity, Fg:

$F_{net} = T -F_{g} = 32.6 N - 27.4 N = 5.2 N (2)$

c)

According Newton's 2nd Law, the acceleration caused by a net force on a point mass (we can take the ball as one) is given by the following expression:

$a = \frac{F_{net} }{m} = \frac{5.2N}{2.8kg} = 1.9 m/s2 (3)$

3 0

3 years ago