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4 years ago

What is the bump on the lateral side just proximal to the 5th metatarsal?

1 answer:

6 0

The description refers to the styloid process or tuberosity. This styloid process is seen in the 5th metatarsal and this area shows the fifth metatarsal's growth plate. This bone is being attached by strong tendon known as the <span>Peroneus brevis tendon.</span>

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The minimum distance required to stop a car moving at 35.6 mi/h is 40.8 ft. What is the minimum stopping distance, in ft, for th

maria [59]

Answer:

Minimum stopping distance = 164.69 ft

Explanation:

Speed of car = 35.6 mi/hr = 15.82 m/s

Stopping distance = 40.8 ft = 12.44 m

We have equation of motion

v² = u² + 2as

0²=15.82²+ 2 x a x 12.44

a = -10.06 m/s²

Now wee need to find minimum stopping distance, in ft, for the same car moving at 71.5 mi/h.

Speed of car = 71.5 mi/h = 31.78 m/s

We have

v² = u² + 2as

0² = 31.78² - 2 x 10.06 x s

s = 50.2 m = 164.69 ft

Minimum stopping distance = 164.69 ft

5 0

3 years ago

You are sitting on a Ferris wheel moving at a constant speed of 8 m/s. When you are at the bottom of the rotation, what is the n

Reptile [31]

Answer:

Explanation:

Formula for calculating normal force is expressed as:

$F = ma\\since \ a = \frac{v^2}{r} \\F = \frac{mv^2}{r}$

m is the mass of the body = 50kg

v is the velocity of the Ferris wheel = 8m/s

r is the radius of the Ferris wheel

Substitute into the formula:

$F = \frac{50 \times 8}{10}\\ F = \frac{400}{10}\\ F = 40N\\$

Hence the normal force of the seat is 40N

3 0

3 years ago

A simple harmonic oscillator consists of a spring of constant k = 100 N/m and a block of mass 2.0 kg. When t = 1.0 s, the positi

ludmilkaskok [199]

Answer:

a) 0,18 m b) 0,18 m c) 0 m/s

Explanation:

5 0

3 years ago

A 2060 kg railroad flatcar, which can move with negligible friction, is motionless next to a platform. A 241 kg sumo wrestler ru

vampirchik [111]

Answer

given,

mass of railroad = 2060 Kg

mass of sumo wrestler = 241 Kg

speed = 5.3 m/s

a) using conservation of momentum

m v = (m + M) V

241 x 5.3 = (241 + 2060)V

$V = \dfrac{1277.3}{2301}$

V = 0.556 m/s

b) Let v be the relative speed

Again using conservation of momentum

241 x 5.3 =241 x (5.3+v) + 2060 v

241 x 5.3 =241 x 5.3+241 x v + 2060 v

2301 v = 0

v = 0 m/s

c) the speed of the flatcar if he then turns and run sat 5.3 m/s relative to the flat car opposite his original direction

241 x 5.3 =241 x (v - 5.3) + 2060 v

241 x 5.3 = - 241 x 5.3+241 x v + 2060 v

2301 v = 2554.6

v = 1.11 m/s

3 0

4 years ago

Linda has a cup of coffee sitting in the cup holder in her car. She has to suddenly slam on her brakes and stop. The cup remaine

wolverine [178]

<h3>Answer:</h3><h2>The Newton's first law of motion will be applied in this situation.</h2>
<h3>Explanation:</h3>

As mention in the question, the car was moving then every object inside the car was also in the motion state with the same velocity as the car had. when the brake applied on the car, the brake only stopped the motion of the car than the objects inside it. According to the first law of motion, due to inertia, the body keeps its state intact until an external force applies to change its state. Because of inertia, the coffee inside the cup was in moving state with the car. When brake applied the coffee spilled out of the car to maintain its state and spread all around. If the cup would keep out of the cup holder, It will move also forward on the applied brake.

8 0

3 years ago