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3 years ago

Select the correct answer.

Physics

1 answer:

Zinaida [17]3 years ago

8 0

I would say 2 Hours, because 45 miles an hour, so 45 mph x 2 h = 90 miles

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Maurinko [17]

The answer is C. I hope this helps you.

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3 years ago

The uniform bar of mass m and length l is balanced in the vertical position when the horizontal force p is applied to the roller

WARRIOR [948]

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7 0

3 years ago

Electrical wire with a diameter of .5 cm is wound on a spool with a radius of 30 cm and a height of 24 cm.

kow [346]

Answer:

a) # lap = 301.59 rad , b) L = 90.48 m

Explanation:

a) Let's use a direct proportions rule (rule of three). If one turn of the wire covers 0.05 cm, how many turns do you need to cover 24 cm

# turns = 1 turn (24 cm / 0.5 cm)

# laps = 48 laps

Let's reduce to radians

# laps = 48 laps (2 round / 1 round)

# lap = 301.59 rad

b) Each lap gives a length equal to the length of the circle

L₀ = 2π R

L = # turns L₀

L = # turns 2π R

L = 48 2π 30

L = 9047.79 cm

L = 90.48 m

6 0

3 years ago

Could I get help on this question please . My parents won’t help me /:

vovikov84 [41]

Answer:

Tarzan will be moving at 7.4 m/s.

Explanation:

From the question given above, the following data were obtained:

Height (h) of cliff = 2.8 m

Initial velocity (u) = 0 m/s

Final velocity (v) =?

NOTE: Acceleration due to gravity (g) = 9.8 m/s²

Finally, we shall determine how fast (i.e final velocity) Tarzan will be moving at the bottom. This can be obtained as follow:

v² = u² + 2gh

v² = 0² + (2 × 9.8 × 2.8)

v² = 0 + 54.88

v² = 54.88

Take the square root of both side

v = √54.88

v = 7.4 m/s

Therefore, Tarzan will be moving at 7.4 m/s at the bottom.

3 0

2 years ago

In a mail-sorting facility, a 2.50-kg package slides down an inclined plane that makes an angle of 20.0° with the horizontal. Th

lawyer [7]

Answer:

The coefficient of kinetic friction is 0.382.

Explanation:

Given:

Angle of inclination is, $\theta=20.0$ °

Mass of package is, $m=2.50\ kg$

Initial speed of package is, $u=2.00\ m/s$

Final speed of the package at the bottom is, $v=0\ m/s$

Distance of travel along the incline is, $d=12.0\ m$

Acceleration due to gravity is, $g=9.8\ m/s^2$

Let the coefficient of kinetic friction be $\mu$ .

Now, the frictional force will be acting along the incline but in the direction opposite to the direction of motion.

So, the net acceleration acting on the package will be up the incline and is equal to:

$a=\mu g\cos\theta-g\sin\theta$ ----------------- 1

Now, using equation of motion, we have:

$v^2-u^2=2ad\\\\0-(2.00)^2=2a(12.0)$

Solving for 'a', we get:

$-4.00=24.0a\\\\a=-\frac{4}{24}=-\frac{1}{6}\ m/s^2$

Now, plug in the value of 'a' in equation (1). This gives,

$\mu g\cos\theta-g\sin\theta=\frac{1}{6}$ ( Neglecting negative sign)

Plug in all the given values and solve for $\mu$ . This gives,

$9.8(-sin(20)+\mu cos(20))=\frac{1}{6}\\\\-0.342+\mu\times 0.94=0.017\\\\0.94\mu=0.342+0.017\\\\0.94\mu=0.359\\\\\mu=\frac{0.359}{0.94}=0.382$

Therefore, the coefficient of kinetic friction is 0.382.

5 0

3 years ago