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2 years ago

A dog is 60m away while moving at constant velocity of 10m/s towards you. Where is the dog after 4 seconds?

Physics

2 answers:

Kitty [74]2 years ago

6 0

The dog is 20m away.
if the dog moves 10m/s, multiply 10 by 4, subtract 40 from 60. 20

evablogger [386]2 years ago

4 0

20m away

the dog was 60m away from. you subtract 40m since it is 10m/s x 4 seconds

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A 0.600-mm diameter wire stretches 0.500% of its length when it is stretched with a tension of 20.0 n. what is the young's modul

Rashid [163]

The Young modulus is given by:
$E= \frac{F /A}{\Delta L / L_0}$
where
F is the force applied
$L_0$ is the initial length of the wire
$A$ is the cross-sectional area of the wire
$\Delta L$ is the stretch of the wire

The wire in the problem stretches by $0.5%$ of its length, this means
$\frac{\Delta L}{L_0} = 0.005$

We can also calculate the area of the wire; its radius is in fact half the diameter:
$r= \frac{d}{2}= \frac{0.600 mm}{2}=0.300 mm=0.3 \cdot 10^{-3} m$
and so the area is
$A=\pi r^2 = \pi (0.3 \cdot 10^{-3} m)^2 = 2.83 \cdot 10^{-7} m^2$

We know the force applied to the wire, F=20 N, so now we have everything to calculate the Young modulus:

$E= \frac{F/A}{\Delta L / L_0} = \frac{20 N/(2.83 \cdot 10^{-7} m^2)}{0.005}=1.42 \cdot 10^{10} N/m^2$

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2 years ago

A charge is moving in a magnetic field that points to the left. What direction can the charge move and experience no magnetic fo

tiny-mole [99]

The answer is left and right.

4 0

2 years ago

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A 0.200-kg mass is attached to the end of a spring with a spring constant of 11 N/m. The mass is first examined (t = 0) when the

inn [45]

Answer:

a) x (t) = 0.3187 cos (7.416 t + 1.008) , b) v = -2,363 sin (7,416 t + 1,008)

c) a = - 17.52 cos (7.416t + 1.008)

Explanation:

The spring mass system creates a harmonic oscillator that is described by the equation

x = Acos (wt + φ)

Where is the amplitude, w the angular velocity and fi the phase

a) Let's reduce the SI system

x = 17.0 cm (1 m / 100 cm) = 0.170 m

The angular velocity is given by

w = √ (k / m)

w = √ 11 / 0.200

w = 7.416 rad / s

Let's look for the terms of the equation with the data for time zero (t = 0 s)

0.170 = A cos φ

Body speed can be obtained by derivatives

v = dx / dt

v = -A w sin (wt + φ)

2.0 = -A 7.416 sin φ

Let's write the two equations

0.170 = A cos φ

2.0 / 7.416 = -A sin φ

Let's divide those equations

tan φ= 2.0 / (7.416 0.170)

φ= tan⁻¹ (1,586)

φ= 1.008 rad

We calculate A

A = 0.170 / cos φ

A = 0.170 / cos 1.008

A = 0.3187 m

With these values we write the equation of motion

x (t) = 0.3187 cos (7.416 t + 1.008)

b) the speed can be found by derivatives

v = dx / dt

v = - 0.3187 7.416 sin (7.416 t +1.008)

v = -2,363 sin (7,416 t + 1,008)

c) the acceleration we look for conserved

a = dv / dt

a = -2,363 7,416 cos (7,416 t + 1,008)

a = - 17.52 cos (7.416t + 1.008)

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Explanation:

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v-u=2m/s

v-u is the change is velocity.

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Which of the following ways is usable energy lost?

tamaranim1 [39]

A. Friction or all of the above

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