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2 years ago

12

the diameter of the wheels on your car ( including the tires) is 25 inches. you are going to drive 250 miles today. each of your

wheels is goingnto turn by an angle of

1 answer:

timofeeve [1]2 years ago

3 0

So the answer is 324.

Explanation
Because I took that test and I got right if u don’t believe me u can brainy it

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Plz, help the rubber ball be dropped from the top of a ladder. It bounces on the same spot on the ground several times to a less

Vladimir [108]

Answer:

What do you mean?

Explanation:

6 0

2 years ago

A 0.750kg block is attached to a spring with spring constant 13.5N/m . While the block is sitting at rest, a student hits it wit

vazorg [7]

Answer:

A)A=0.075 m

B)v= 0.21 m/s

Explanation:

Given that

m = 0.75 kg

K= 13.5 N

The natural frequency of the block given as

$\omega =\sqrt{\dfrac{K}{m}}$

The maximum speed v given as

$v=\omega A$

A=Amplitude

$v=\sqrt{\dfrac{K}{m}}\times A$

$0.32=\sqrt{\dfrac{13.5}{0.75}}\times A$

A=0.075 m

A= 0.75 cm

The speed at distance x

$v=\omega \sqrt{A^2-x^2}$

$v=\sqrt{\dfrac{K}{m}}\times \sqrt{A^2-x^2}$

$v=\sqrt{\dfrac{13.5}{0.75}}\times \sqrt{0.075^2-(0.075\times 0.75)^2}$

v= 0.21 m/s

5 0

3 years ago

A source of sound is kept in a jar in a vacuum. Air is slowly introduced in to the jar. What happens to the sound coming out of

timama [110]

<span>The loudness of the sound increases gradually as the air is slowly introduced in to the jar. This is because sound needs a physical medium and in a vacuum there is none. The air provides that medium and as it is introduced, the transfer of sound energy increases</span>

3 0

3 years ago

Which statement best describes the acceleration of a baseball?

insens350 [35]

Answer:

C the baseball began at rest and rolls at a rate of 14.7 m/s after 1.5 seconds

3 0

2 years ago

A rubber-wheeled 50 kg cart rolls down a 35 degree concrete incline. Coefficient of rolling friction between rubber and concrete

lapo4ka [179]

Answer:

(a) 5.62 m/s²

(b) 5.46 m/s²

Explanation:

The given values are:

mass,

m = 50 kg

angle

= 35°

(a)

If friction neglected,

⇒ $F_x=mgSin \theta=ma$

⇒ $a_x=gSin \theta$

$=9.8 \ Sin35^{\circ}$

$=5.62 \ m/s^2$

(b)

If friction present,

⇒ $F_x=mgCos \theta$

⇒ $F_x=mgSin \theta-\mu_r mgCos \theta$

⇒ $a=gSin \theta-\mu_rgCos \theta$

$=9.8 \ Sin35^{\circ}-0.02\times 9.8 \ Cos30^{\circ}$

$=5.46 \ m/s^2$

8 0

3 years ago