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2 years ago

14

Paula has walked in a straight line, 30.5° north of west, for 1650 meters. How far south and east should she walks to return to

her original location

1 answer:

olchik [2.2K]2 years ago

8 0

The answer is A.
Sy = 1650 x sin30.5 = 837.4 m toward south
Sx = 1650 x cos30.5 = 1421.7 m toward east

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True or false, only a radioactive isotope will have a half-life

maks197457 [2]

False not radioactive isotope will have a half-life

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3 years ago

A toy plane is flying in a horizontal circle by being attached to a 0.75 meter string. The plane has a mass of 101.7 grams and m

nadezda [96]

Answer:

F = 2,894 N

Explanation:

For this exercise let's use Newton's second law

F = m a

The acceleration is centripetal

a = v² / r

Angular and linear variables are related.

v = w r

Let's replace

F = m w² r

The radius r and the length of the rope is related

cos is = r / L

r = L cos tea

Let's replace

F = m w² L cos θ

Let's reduce the magnitudes to the SI system

m = 101.7 g (1 kg / 1000g) = 0.1017 kg

θ = 5 rev (2π rad / rev) = 31,416 rad

w = θ / t

w = 31.416 / 5.1

w = 6.16 rad / s

F = 0.1017 6.16² 0.75 cos θ

F = 2,894 cos θ

The maximum value of F is for θ equal to zero

F = 2,894 N

7 0

3 years ago

The total energy of a block—spring system is 0.18 J. The amplitude is 14.0 cm and the maximum speed is 1.25 m/s. Find: (a) the m

algol13

a) The mass is 0.23 kg

b) The spring constant is 1.25 N/m

c) The frequency is 1.42 Hz

d) The speed of the block is 1.08 m/s

Explanation:

a)

We can find the mass of the block by applying the law of conservation of energy: in fact, the total mechanical energy of the system (which is sum of elastic potential energy, PE, and kinetic energy, KE) is constant:

$E=PE+KE=const.$

The potential energy is given by

$PE=\frac{1}{2}kx^2$

where k is the spring constant and x is the displacement. When the block is crossing the position of equilibrium, x = 0, so all the energy is kinetic energy:

$E=KE_{max}=\frac{1}{2}mv_{max}^2$ (1)

where

m is the mass of the block

$v_{max}=1.25 m/s$ is the maximum speed

We also know that the total energy is

$E=0.18 J$

Re-arranging eq.(1), we can find the mass:

$m=\frac{2E}{v_{max}^2}=\frac{2(0.18)}{(1.25)^2}=0.23 kg$

b)

The maximum speed in a spring-mass system is also given by

$v_{max} =\sqrt{\frac{k}{m}} A$

where

k is the spring constant

m is the mass

A is the amplitude

Here we have:

$v_{max}=1.25 m/s$ is the maximum speed

m = 0.23 kg is the mass

A = 14.0 cm = 0.14 m is the amplitude

Solving for k, we find the spring constant

$k=\frac{v_{max}^2}{A^2}m=\frac{1.25^2}{0.14^2}(0.23)=18.3 N/m$

c)

The frequency in a spring-mass system is given by

$f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}$

where

k is the spring constant

m is the mass

In this problem, we have:

k = 18.3 N/m is the spring constant (found in part b)

m = 0.23 kg is the mass (found in part a)

Substituting and solving for f, we find the frequency of the system:

$f=\frac{1}{2\pi}\sqrt{\frac{18.3}{0.23}}=1.42 Hz$

d)

We can solve this part by using the law of conservation of energy; in fact, we have

$E=PE+KE=\frac{1}{2}kx^2 + \frac{1}{2}mv^2$

Where v is the speed of the system when the displacement is equal to x.

We know that the total energy of the system is

E = 0.18 J

Also we know that

k = 18.3 N/m is the spring constant

m = 0.23 kg is the mass

Substituting

x = 7.00 cm = 0.07 m

We can solve the equation to find the corresponding speed v:

$v=\sqrt{\frac{2E-kx^2}{m}}=\sqrt{\frac{2(0.18)-(18.3)(0.07)^2}{0.23}}=1.08 m/s$

#LearnwithBrainly

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2 years ago

Range of motion is the distance an object can travel when separated from another object. Please select the best answer from the

Lady_Fox [76]

Based on the given statement above, the correct answer would be FALSE. It is not true that range of motion is the distance an object can travel when separated from another object because range of motion or ROM is the distance--linear or angular--<span>that a movable object may normally travel while properly ATTACHED (not separated) to another. Hope this answer helps.</span>

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3 years ago

If a particle undergoes shm with amplitude 0.21 m, what is the total distance it travels in one period?

Ymorist [56]

If a particle undergoes simple harmonic motion with an amplitude of 0.21 meters, this means that the maximum displacement of the particle from its resting position is 0.21. For one period, it traveled from its starting position which is twice the amplitude and then back to its original position which is another distance that is twice the amplitude as well. Therefore, the total distance it traveled is 2*amplitude + 2*amplitude = 2*0.21 + 2*0.21 = 0.42 + 0.42 = 0.84 meters.

7 0

3 years ago