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tino4ka555 [31]

3 years ago

14

What is the average speed if an object travels 200 km in 5 hours

1 answer:

melamori03 [73]3 years ago

8 0

Answer:

40km per hour

Explanation:

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2 years ago

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A motorcycle that is slowing down uniformly. The motorcycle covers 1 ????m=1000 m in 80 sec⁡. The motorcycle then covers the nex

Lynna [10]

Answer:

Part a)

acceleration = -0.042 m/s/s

Part b)

initial speed = 14.17 m/s

final speed = 5.77 m/s

Explanation:

Part a)

Let the initial velocity of the motorcycle is

$v_i = v_o$

now at the end of 80 s let the speed is

$v_f = v_1$

after another 120 s let the speed will be

$v_f' = v_2$

now we know that

$d = \frac{v_i + v_f}{2} (t)$

$d = \frac{v_o + v_1}{2}(80)$

$1000 = 40(v_o + v_1)$

also we know that

$v_1 - v_o = a(80)$

also we have

$1000 = \frac{v_1 + v_2}{2}(120)$

$1000 = 60(v_1 + v_2)$

now we can say

$(v_2 + v_1) - (v_o + v_1) = \frac{50}{3} - \frac{50}{2}$

also we know

$v_2 - v_o = a(120 + 80)$

$-8.33 = 200 a$

$a = -0.042 m/s^2$

Part b)

now we have

$v_1 + v_o = 25$

$v_1 - v_o = (-0.042)(80)$

$v_1 = 10.83 m/s$

so the starting velocity of the trip is

$v_o = 25 - 10.83 = 14.17 m/s$

now speed after t = 200 s is given as

$v_2 = v_o + at$

$v_2 = 14.17 - (0.042)(200)$

$v_2 = 5.77 m/s$

5 0

3 years ago

A mass weighing 24 pounds, attached to the end of a spring, stretches it 4 inches. Initially, the mass is released from rest fro

svetoff [14.1K]

Answer:

The equation of motion is $x(t)=-$ $\frac{1}{3} cos4\sqrt{6t}$

Explanation:

Lets calculate

The weight attached to the spring is 24 pounds

Acceleration due to gravity is $32ft/s^2$

Assume x , is spring stretched length is ,4 inches

Converting the length inches into feet $x=\frac{4}{12} =\frac{1}{3}feet$

The weight (W=mg) is balanced by restoring force ks at equilibrium position

mg=kx

$W=kx$ ⇒ $k=\frac{W}{x}$

The spring constant , $k=\frac{24}{1/3}$

= 72

If the mass is displaced from its equilibrium position by an amount x, then the differential equation is

$m\frac{d^2x}{dt} +kx=0$

$\frac{3}{4} \frac{d^2x}{dt} +72x=0$

$\frac{d^2x}{dt} +96x=0$

Auxiliary equation is, $m^2+96=0$

$m=\sqrt{-96}$

= $\frac{+}{} i4\sqrt{6}$

Thus , the solution is $x(t)=c_1cos4\sqrt{6t}+c_2sin4\sqrt{6t}$

$x'(t)=-4\sqrt{6c_1} sin4\sqrt{6t}+c_2$ $4\sqrt{6}$ $cos4\sqrt{6t}$

The mass is released from the rest x'(0) = 0

$=-4\sqrt{6c_1} sin4\sqrt{6(0)}+c_2$ $4\sqrt{6}$ $cos4\sqrt{6(0)}$ =0

$c_2$ $4\sqrt{6} =0$

$c_2=0$

Therefore , $x(t)=c_1$ $cos 4\sqrt{6t}$

Since , the mass is released from the rest from 4 inches

$x(0)= -4$ inches

$c_1 cos 4\sqrt{6(0)} =-\frac{4}{12}$ feet

$c_1=-\frac{1}{3}$ feet

Therefore , the equation of motion is $-\frac{1}{3} cos4\sqrt{6t}$

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2 years ago

A sound wave traveling through dry air has a frequency of 15 Hz, a

Korolek [52]

Answer:

Option B

Explanation:

Speed of a wave is denoted by:

v=fλ

where f is the frequency which is unchanged 15Hz and λ is the new wavelength which is 28m

v=fλ

$v=15(28)\\v=420m/s$

3 0

3 years ago

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