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vampirchik [111]

2 years ago

10

A ski hill is angled 20° above the horizontal. If a skier starts from rest at the top of the hill, how fast will she be travelin

g after she has traveled 62 m 62 ⁢ m down the slope? Assume that friction is negligible.

1 answer:

MaRussiya [10]2 years ago

6 0

Answer:

20 m/s

Explanation:

First, we calculate for the acceleration down the incline using $a = gsin\theta$ .

a = (9.80)sin(20°)

a = 3.35

Then, we use one of the kinematics equations, ${v_{f}}^{2} = {v_{i}}^{2} + 2 a\Delta s$ .

$v_{i}$ = 0 m/s

a = 3.35

$\Delta s$ = 62 m

$v_{f} = \sqrt{2(3.35)(62)}$

$v_{f} =20.381 = 20 m/s$

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lbvjy [14]

Answer:

Impulse = 322.5[kg*m/s], the answer is D

Explanation:

This method it is based on the principle of momentum and the amount of movement; and used to solve problems involving strength, mass, speed and time.

If units of the SI are used, the magnitude of the impulse of a force is expressed in N * s. however, when remembering the definition of the newton.

$N*S=(kg*m/s^{2} )*s = kg*m/s$

Now replacing the values on the following equation that express the definition of impulse

$Impulse = Force * Time\\\\Impulse = 215 * 1.5 = 322.5 [kg*m/s]$

4 0

2 years ago

Read 2 more answers

A steel piano wire, of length 1.150 m and mass 4.80 g is stretched under a tension of 580.0 N.

kaheart [24]

A steel piano wire, of length 1.150 m and mass of 4.80 g is stretched under a tension of 580.0 N.the speed of transverse waves on the wire would be 372.77 m/s

<h3>What is a sound wave?</h3>

It is a particular variety of mechanical waves made up of the disruption brought on by the movements of the energy. In an elastic medium like the air, a sound wave travels through compression and rarefaction.

For calculating the wave velocity of the sound waves generated from the piano can be calculated by the formula

V= √F/μ

where v is the wave velocity of the wave travel on the string

F is the tension in the string of piano

μ is the mass per unit length of the string

As given in question a steel piano wire, of length 1.150 m and mass of 4.80 g is stretched under a tension of 580.0 N.

The μ is the mass per unit length of the string would be

μ = 4.80/(1.150×1000)

μ = 0.0041739 kg/m

By substituting the respective values of the tension on the string and the density(mass per unit length) in the above formula of the wave velocity

V= √F/μ

V=√(580/0.0041739)

V = 372.77 m/s

Thus, the speed of transverse waves on the wire comes out to be 372.77 m/s

Learn more about sound waves from here

brainly.com/question/11797560

#SPJ1

6 0

1 year ago

PLEASE HELP!!! GIVING BRAINLIEST!! ill also answer questions that you have posted if you answer this correctly!!!! (40pts)

melisa1 [442]

D.)downward toward the center of earth

7 0

2 years ago

Firemen are shooting a stream of water at a burning building. A high-pressure hose shoots out the water with a speed of 26.0 m/s

alekssr [168]

Answer:

a) θ = 58.3º

b) vfh = 13.7 m/s

c) g = -9.8 m/s2

d) h = 22.2 m

e) vfb = 15.5 m/s

Explanation:

a)

Assuming that gravity is the only influence that causes an acceleration to the water, due to it is always downward, since both directions are independent each other, in the horizontal direction, the water moves at a constant speed.
Since the velocity vector has a magnitude of 26.0 m/s, we can find its horizontal component as follows:
vₓ₀ = v * cos θ (1)
where θ is the angle between the water and the horizontal axis (which we define as the x-axis, being positive to the right).
Applying the definition of average velocity, taking the end of the hose like the origin, and making t₀ = 0, we can write the following expression:

$x_{f} = v_{ox} * t = v_{o} * cos \theta * t (2)$

Replacing by the givens of xf = 41.0m, t = 3.00 s, and v=26.0 m/s, we can solve for the angle of elevation θ, as follows:

$cos \theta = \frac{x_{f} }{v*t} = \frac{41.0m}{26.0m/s*3.00s} = 0.526 (3)$

⇒θ = cos⁻¹ (0.526) = 58.3º (4)

b)

At the highest point in its trajectory, just before starting to fall, the vertical component of the velocity is just zero.
Since the horizontal component keeps constant during all the journey, we can conclude that the speed at this point is just v₀ₓ, that we can find easily from (1) replacing by the values of v and cos θ, as follows:
vₓ₀ = v * cos θ = 26.0 m/s * 0.526 = 13.7 m/s. (5)

c)

At any point in the trajectory, the only acceleration present is due to the action of gravity, which accepted value is -9.8 m/s2 (taking the upward direction on the vertical y-axis as positive)

d)

Since we know the time when the water strikes the building, it will be the same for the vertical movement, so, we can use the kinematic equation for vertical displacement, as follows:

$\Delta y = v_{oy} * t - \frac{1}{2} *g*t^{2} (6)$

Our only unknown remains v₀y, which can be obtained in the same way than the horizontal component:
v₀y = v * sin θ = 26.0 m/s * 0.85 = 22.1 m/s (7)
Replacing (7) in (6), we get:

$\Delta y = 22.1 m/s* 3.0s - \frac{1}{2} *9.8m/s2*(3.00s)^{2} = 22.2 m (8)$

e)

When the water hits the building the velocity vector, has two components, the horizontal vₓ and the vertical vy.
The horizontal component, since it keeps constant, is just v₀x:
v₀ₓ = 13.7 m/s
The vertical component can be found applying the definition of acceleration (g in this case), solving for the final velocity, as follows:

$v_{fy} = v_{oy} - g*t (9)$

Replacing by the time t (a given), g, and v₀y from (7), we can solve (9) as follows:

$v_{fy} = 22.1 m/s - 9.8m/s2*3.00s = -7.3 m/s (10)$

Since we know the values of both components (perpendicular each other), we can find the magnitude of the velocity vector (the speed, i.e. how fast is it moving), applying the Pythagorean Theorem to v₀ₓ and v₀y, as follows:

$v_{f} = \sqrt{(13.7m/s)^{2} +(-7.3m/s)^{2}} = 15.5 m/s (11)$

3 0

2 years ago

A child is riding a merry-go-round that has an instantaneous angular speed of 12 rpm. If a constant friction torque of 12.5 Nm i

sammy [17]

Answer:

$-0.25 rad/s^2$

Explanation:

The equivalent of Newton's second law for rotational motions is:

$\tau = I \alpha$

where

$\tau$ is the net torque applied to the object

I is the moment of inertia

$\alpha$ is the angular acceleration

In this problem we have:

$\tau = -12.5 Nm$ (net torque, with a negative sign since it is a friction torque, so it acts in the opposite direction as the motion)

$I=50.0 kg m^2$ is the moment of inertia

Solving for $\alpha$ , we find the angular acceleration:

$\alpha = \frac{\tau}{I}=\frac{-12.5 Nm}{50.0 kg m^2}=-0.25 rad/s^2$

3 0

2 years ago