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2 years ago

Which of the following statements describes the law of inertia?

Physics

1 answer:

Drupady [299]2 years ago

8 0

Answer:

B. objects remain in the same state of motion unless a force acts on them

Explanation:

Sana nakatulong

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How can you use graphs to calculate the displacement of an object?

tekilochka [14]

Answer:

c. find the slope of the velocity time graph

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3 years ago

To practice Problem-Solving Strategy 15.1 Mechanical Waves. Waves on a string are described by the following general equation y(

DerKrebs [107]

Answer:

0.0549 m

Explanation:

Given that

equation y(x,t)=Acos(kx−ωt)

speed v = 8.5 m/s

amplitude A = 5.5*10^−2 m

wavelength λ = 0.5 m

transverse displacement = ?

v = angular frequency / wave number

and

wave number = 2π/ λ

wave number = 2 * 3.142 / 0.5

wave number = 12.568

angular frequency = v k

angular frequency = 8.5 * 12.568

angular frequency = 106.828 rad/sec ~= 107 rad/sec

equation y(x,t)=Acos(kx−ωt)

y(x,t)= 5.5*10^−2 cos(12.568 x−107t)

when x =0 and and t = 0

maximum y(x,t)= 5.5*10^−2 cos(12.568 (0) − 107 (0))

maximum y(x,t)= 5.5*10^−2 m

and when x = x = 1.52 m and t = 0.150 s

y(x,t)= 5.5*10^−2 cos(12.568 (1.52) −107(0.150) )

y(x,t)= 5.5*10^−2 × (0.9986)

y(x,t) = 0.0549 m

so the transverse displacement is 0.0549 m

5 0

3 years ago

A 1.50 x 109 kg railroad car moving at 7.00 m/s to the north collides with

barxatty [35]

GIVE ME YOUR MONEY

Explanation:

8 0

3 years ago

Help me please I will give you a crown!!!!

grigory [225]

Answer:

ID:6207328365

p.a.s.s:qwerty

o.n.l.y f.o.r g.i.r.l.s.

8 0

3 years ago

Read 2 more answers

The 8 kg block is then released and accelerates to the right, toward the 2 kg block. The surface is rough and the coefficient of

natita [175]

Answer:

3.258 m/s

Explanation:

k = Spring constant = 263 N/m (Assumed, as it is not given)

x = Displacement of spring = 0.7 m (Assumed, as it is not given)

$\mu$ = Coefficient of friction = 0.4

Energy stored in spring is given by

$U=\dfrac{1}{2}kx^2\\\Rightarrow U=\dfrac{1}{2}\times 263\times 0.7^2\\\Rightarrow U=64.435\ J$

As the energy in the system is conserved we have

$\dfrac{1}{2}mv^2=U-\mu mgx\\\Rightarrow v=\sqrt{2\dfrac{U-\mu mgx}{m}}\\\Rightarrow v=\sqrt{2\dfrac{64.435-0.4\times 8\times 9.81\times 0.7}{8}}\\\Rightarrow v=3.258\ m/s$

The speed of the 8 kg block just before collision is 3.258 m/s

7 0

3 years ago