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3 years ago

A skater increases her velocity from 2.0 m/s to 10.0 m/s in 3.0 seconds. What is the skater's acceleration?

Physics

1 answer:

geniusboy [140]3 years ago

8 0

As change is velocity is

Δv=10 m s−1−2 m s−1=8 m s−1

and time taken is 3 seconds so acceleration is

acceleration=change in velocitychange in time

a=8 m s−13 s=2.7 m s−2

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Two loudspeakers emit sound waves along the x-axis. A listener in front of both speakers hears a maximum sound intensity when sp

grandymaker [24]

Answer:

frequency of the sound = f = 1,030.3 Hz

phase difference = Φ = 229.09°

Explanation:

Step 1: Given data:

Xini = 0.540m

Xfin = 0.870m

v = 340m/s

Step 2: frequency of the sound (f)

f = v / λ

λ = Xfin - Xini = 0.870 - 0.540 = 0.33

f = 340 / 0.33

f = 1,030.3 Hz

Step 3: phase difference

phase difference = Φ

Φ = (2π/λ)*(Xini - λ) = (2π/0.33)* (0.540-0.33) = 19.04*0.21 = 3.9984

Φ = 3.9984 rad * (360°/2π rad)

Φ = 229.09°

Hope this helps!

5 0

3 years ago

A 58 kg skier is going down a 35 degree slope. The areaof each

maxonik [38]

To solve this problem we will use a free body diagram that allows us to determine the Normal Force.

In general, the normal force would be equivalent to

$N = mgcos\theta$

Since the skier is standing on two skis, his weight will be divide by two

$N' = \frac{mgcos\theta}{2}$

Pressure is given as the force applied in a given area, that is

$P = \frac{F}{A}$

Replacing F with N'

$P = \frac{N'}{A}$

$P = \frac{\frac{mgcos\theta}{2}}{A}$

Our values are given as,

$m = 58kg$

$g = 9.8m/s^2$

$\theta = 35\°$

$A = 0.3m^2$

Replacing we have that

$P = \frac{\frac{(58)(9.8)cos(35)}{2}}{0.3}$

$P = 776.01Pa$

Therefore the pressure exerted by each ski on the snow is 776.01Pa

6 0

3 years ago

An astronaut holds a rock 100 m above surface of Planet X. The rock is then thrown upwards with a sleek of 15m/s. The rock reach

Gelneren [198K]

Answer: $5 m/s^{2}$

Explanation:

This problem is related to vertical motion, and the equation that models it is:

$y=y_{o}+V_{o}sin\theta t-\frac{1}{2}gt^{2}$ (1)

Where:

$y=0m$ is the rock's final height

$y_{o}=100 m$ is the rock's initial height

$V_{o}=15 m/s$ is the rock's initial velocity

$\theta=90\°$ is the angle at which the rock was thrown (directly upwards)

$t=10 s$ is the time

$g$ is the acceleration due gravity in Planet X

Isolating $g$ and taking into account $sin(90\°)=1$ :

$g=(-\frac{2}{t^{2}})(y-y_{o}-V_{o}t)$ (2)

$g=(-\frac{2}{(10 s)^{2}})(0 m-100 m-(15 m/s)(10 s))$ (3)

$g=5 m/s^{2}$ (4) This is the acceleration due gravity in Planet X

5 0

2 years ago

Explain what happens to the energy of a rock on the edge of a cliff as it falls from the

dolphi86 [110]

Answer:

A rock sitting on the edge of a cliff. If the rock falls, the potential energy will be converted to kinetic energy, as the rock will be moving. The potential energy decreases as the kinetic energy increases. The potential energy decreases as the kinetic energy decreases.

3 0

3 years ago

1 point

elixir [45]

Answer:Squinting is often a sign of vision

problems. The result of squinting the

eye will result in a force being aplied

that

Explanation:

7 0

3 years ago