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loris [4]
4 years ago
12

Can an object with constant acceleration reverse its direction of travel? can it reverse its direction twice? in each case, expl

ain and give an example.
Physics
1 answer:
Kay [80]4 years ago
4 0

If an object is moving under constant acceleration then it is possible to reverse its direction once

Like when we throw an object upwards under the action of constant gravity then the object first moves upwards then stops after reaching maximum height and reverse its direction to move back.

now for the next case it is not possible to reverse the direction twice when an object moves under constant acceleration

so when once object reverse it direction then after that it comes in the direction of acceleration and then continue in the same direction for whole motion

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A paper airplane with mass 0.1 kg is flying 1.5 m above the ground with a speed of 2 m/s. what is the total mechanical energy of
Drupady [299]
Mechanical(ME) energy, in physical sciences, is the sum of kinetic energy (KE) and potential energy (PE). Below are the calculation in obtaining the energies,
 
     (1) KE =   0.5mv²     =  0.5(0.1 kg) x (2 m/s)² = 0.2 J
     (2) PE =    md          = (0.1 kg) x (1.5 m)        = 0.15 J
      (3) ME =  KE + PE = 0.2 J + 0.15 J               = 0.35 J

Thus, the mechanical energy is 0.35 Joules. 

3 0
3 years ago
g While hauling a log in the back of a flatbed truck, a driver is pulled over by the state police. Although the log cannot roll
irga5000 [103]

Answer:

The minimum coefficient of static friction required, µ = 0.10

<em>Note. The question is incomplete. The complete question is given below:</em>

<em>While hauling a log in the back of a flatbed truck, a driver is pulled over by the state police. Although the log cannot roll sideways, the police claim that the log could have slid out the back of the truck when accelerating from rest. The driver claims that the truck could not possibly accelerate at the level needed to achieve such an effect. Regardless, the police write a ticket anyway and now the driver court date is approaching.</em>

<em>The log has a mass of m = 929 kg; the truck has a mass of M = 8850 kg. According to the truck manufacturer, the truck can accelerate from 0 to 55 mph in 23.0 seconds, but this does not account for the additional mass of the log. Calculate the minimum coefficient of static friction μs needed to keep the log in the back of the truck.</em>

Explanation:

First, velocity in mph is converted to m/s

1 mph = 0.447 m/s

55 mph ≈ 24.6 m/s

The acceleration of the empty truck is a = v/t = 24.6 / 23 = 1.07 m/s²

Force that can be generated by the truck, F = ma

F = 8850kg * 1.07 m/s² = 9469.5 N

However, with the added mass of the log on it, the acceleration of the truck will become;

a = F / m = 9469.5 N / (8550+929)kg = 0.97 m/s²

Frictional force between the log and the truck = 0.97 m/s² * 929 kg = 901.13 N

Normal reaction on the truck due to the weight of the log, R = mg

R = 929 kg * 9.8m/s² = 9104.2 N

Coefficient of static friction, µ = F/R

µ = 901.13/9104.2

µ = 0.098 ≈ 0.10

Therefore, the minimum static friction required is µ = 0.10

8 0
3 years ago
I need help!!!! I don’t understand physical science at all.
allochka39001 [22]
B is the correct answer
3 0
3 years ago
A) About 9.1<br> B) about 14.1<br> C) about 17.2<br> D) about 18.1 <br><br> Please help me
Mnenie [13.5K]

D. 18.1

K^+H^+G^=180° (sum of int angles of triangle)

K^+30+62=180

K^=88°

GH/sinK=KG/sinH

X/sin88=16/sin62

X*sin62/sin62=16*sin88/sin62

X=18.1

8 0
3 years ago
A large bottle contains 150 L of water, and is open to the atmosphere. If the bottle has a flat bottom with an area of 2 ft, cal
Yuri [45]

Answer:

Total pressure exerted at bottom =  119785.71 N/m^2

Explanation:

given data:

volume of water in bottle = 150 L = 0.35 m^3

Area of bottle = 2 ft^2

density of water = 1000 kg/m

Absolute pressure on bottom of bottle will be sum of atmospheric pressure and pressure due to water

Pressure due to water P = F/A

F, force exerted by water = mg

m, mass of water = density * volume

                             =  1000*0.350 = 350 kg

F  = 350*9.8 = 3430 N

A = 2 ft^2 = 0.1858 m^2  

so, pressure P = 3430/ 0.1858 = 18460.71 N/m^2

Atmospheric pressure

At sea level atmospheric pressure is 101325 Pa

Total pressure exerted at bottom  = 18460.71 + 101325 = 119785.71 N/m^2

Total pressure exerted at bottom =  119785.71 N/m^2

6 0
4 years ago
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