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suter [353]
4 years ago
13

Help me please to do those problems...

Physics
1 answer:
Stells [14]3 years ago
3 0

Keeping in mind that:

- Distance is the length of the total path covered by the person, regardless of the directions of the different parts of motion

- Displacement is just the distance in a straight line between the final point and the initial point

Let's apply these concepts to solve the different parts of the problem:

1. 120 m, 80 m west

The total distance is the sum of the length of the different paths:

distance = 100 + 20 = 120 m

To find the displacement, we need to find the distance between the starting point and the ending point. Assuming the starting point as

x = 0

Michelle moved 100 m westward  and 20 m eastward, so the ending point is at

ending point = 100 - 20 = 80 m (westward)

So, the displacement is

displacement = 80 - 0 = 80 m (west)

2. 6400 m, 0

The distance is equal to the length of the track multiplied by the number of laps, so:

distance = 1600 \cdot 4 = 6400 m

The track is circular, and the runner completes exactly 4 laps: it means that at the end of the motion, the runner is at her starting point. So ending point and starting point coincide, and so the displacement

displacement = ending point - starting point = 0

3. 40 m; 10 m forward

The total distance is just the sum of the lengths of the different parts of the motion, so:

distance = 20 + 5 + 10 + 5 = 40 m

The find the displacement, we need to assign signs to every direction:

Forward --> positive along the forward-backward direction

Backard --> negative along the forward-backward direction

Right --> positive direction along the right-left direction

Left --> negative direction along the right-left direction

Along the forward-backward direction, the displacement is:

20 m forward and 10 m backward, so 20 - 10 = 10 (forward)

Along the left right direction, the displacement is:

5 m right and 5 m left, so 5 - 5 = 0

So the net displacement is 10 m (forward)

4. 16 km; 4 km south

Again, t total distance is the sum of the lengths of the different parts of the motion, so:

distance = 2 + 4 + 6 + 4 = 16 km

To find the displacement, we assign signs to every direction:

North --> positive along the north-south direction

south --> negative along the north-south direction

east --> positive direction along the east-west direction

west --> negative direction along the east-west direction

Along the north-south direction, the displacement is:

2 km north and 6 km south, so 2 - 6 = -4 km (4 km south)

Along the east-west direction, the displacement is:

4 km east and 4 km west, so 4 - 4 = 0

So the net displacement is 4 km south

5. 37.7 m; 24 m

The distance covered by the skater is the length of half circumference, so given the radius

r = 12 m

The distance is

distance= \pi r = \pi \cdot (12)=37.7 m

The displacement is the distance in a straight line between the ending position and the starting position. Since the skater ends her motion halfway around the circle, the distance between the initial and final point is equal to the diameter of the circle (two times the radius). So,

displacement = 2r = 2 \cdot 12 = 24 m

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Answer:

B. Technician B

Explanation:

The statement made by Technician A is not right because for an RWD vehicles, its front bearing assembly is not used for sustaining the bearing adjustment. The statement made by the Technician B is right because the main reason for staking the bearing assembly of the wheel of a FWD vehicle is to ensure that the torque is at a specific value. Thus, only Technician B is correct.

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Part D
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Answer:

Which item fits in the left side of the table?Criminal versus Civil Cases Criminal cases Civil cases ? . • Accuser is called the plaintiff • Lower burden of proof Defendant may be found liable • Usually heard by a judge only • Accused not guaranteed an attorney

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3 years ago
A hollow ball with a diameter of 3.81 cm has an average density of 0.0842 g/cm3. What force must be applied to hold it stationar
vfiekz [6]

Answer:

0.26

Explanation:

Given that :

Diameter of ball = 3.81 cm = 3.81/100 = 0.0381 m

Radius (r) = 0.0381 / 2 = 0.01905 m

Average density of ball (Db) = 0.0842 g/cm³ = (0.0842 / 1000)kg / 10^-6 = 0.0842/ 1000 * 10^6 = 84.2kg/m³

Density of water (Dw) = 1000kg/m³

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V = (4/3) * π * 0.01905^3

V = 0.0000289583 m³

Required force = (Dw * V * g) - (Db * V * g)

= (1000 * 0.0000289583 * 9.8) - (84.2 * 0.0000289583 * 9.8)

= 0.259896109172

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A car initially traveling at 27.7 m/s undergoes a constant negative acceleration of magnitude 2.00 m/s2 after its brakes are app
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The car's velocity at time <em>t</em> is given by

v=27.7\dfrac{\rm m}{\rm s}+\left(-2.00\dfrac{\rm m}{\mathrm s^2}\right)t

It comes to a stop when <em>v</em> = 0, which happens when

0=27.7\dfrac{\rm m}{\rm s}+\left(-2.00\dfrac{\rm m}{\mathrm s^2}\right)t\implies t=13.85\,\mathrm s

or after about 13.9 s.

In this time, the car travels a distance <em>x</em> given by

x=\left(27.7\dfrac{\rm m}{\mathrm s}\right)(13.85\,\mathrm s)+\dfrac12\left(-2.00\dfrac{\rm m}{\mathrm s^2}\right)(13.85\,\mathrm s)^2=191.823\,\mathrm m

or about 192 m.

In one complete revolution, each tire covers a distance equal to its circumference,

2\pi(0.340\,\mathrm m)\approx2.13628\,\mathrm m

or about 2.14 m.

This means each tire will complete approximately 192/2.14 ≈ 90 revolutions.

8 0
3 years ago
Tom kicks a soccer ball on a flat, level field with initial speed 20 m/s at an angle 35 degrees above the horizontal. (1) How lo
Arada [10]

Explanation:

Given:

x₀ = 0 m

y₀ = 0 m

v₀ = 20 m/s

θ = 35°

aᵧ = -9.8 m/s²

1) Find t when y = 0.

y = y₀ + v₀ᵧ t + ½ aᵧ t²

0 = 0 + (20 sin 35°) t + ½ (-9.8) t²

0 = t (20 sin 35° - 4.9 t)

t = 0, t = 2.34

The ball stays in the air 2.34 seconds.

2) Find y when vᵧ = 0.

vᵧ² = v₀ᵧ² + 2aᵧ (y - y₀)

0² = (20 sin 35)² + 2(-9.8) (y - 0)

y = 6.71 m

The ball reaches a maximum height of 6.71 meters.

3) Find x when y = 0.

x = x₀ + v₀ₓ t + ½ aₓ t²

x = 0 + (20 cos 35°) (2.34) + ½ (0) (2.3)²

x = 38.4 m

The ball lands 38.4 meters from Tom.

4) Find v when y = 0.

vₓ = aₓ t + v₀ₓ

vₓ = (0) (2.34) + 20 cos 35°

vₓ = 16.4 m/s

vᵧ = aᵧ t + v₀ᵧ

vᵧ = (-9.8) (2.34) + 20 sin 35°

vᵧ = -11.5 m/s

v = √(vₓ² + vᵧ²)

v = √((16.4)² + (-11.5)²)

v = 20 m/s

The ball has a speed of 20 m/s just before it lands.

5 0
3 years ago
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