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Llana [10]
3 years ago
11

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!! Joules could be used to measure

Physics
1 answer:
lyudmila [28]3 years ago
4 0

All of the above, work is a measurement of energy transfer, in Joules.

Potential energy = Joules

Kinetic energy = Joules

The key thing here is that anything having to do with just energy or energy transfer is measured in joules.

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car rides on four wheels that are connected to the body of the car by spring assemblies that let the wheels move up and down ove
Tomtit [17]

Answer:

78.4 KN/m

Explanation:

Given

mass of person 'm' =80 kg

car dips about i.e spring stretched 'x'=   1 cm  => 0.01m

acceleration due to gravity 'g'= 9.8 m/s^2

as we know that,in order to find approximate spring constant we use Hooke's Law i.e  F=kx

where,

F = the force needed

x= distance the spring is stretched or compressed beyond its natural length

k= constant of proportionality called the spring constant.

F=kx ---> (since f=mg)

mg=kx

k=(mg)/x

k=(80 x 9.8)/ 0.01

k=78.4x10^3

k=78.4 KN/m

3 0
3 years ago
Read 2 more answers
One of the most important properties of materials in many applications is strength. Two of the qualitative measures of the stren
katrin2010 [14]

To solve this exercise it is necessary to take into account the concepts related to Tensile Strength and Shear Strenght.

In Materials Mechanics, generally the bodies under certain loads are subject to both Tensile and shear strenghts.

By definition we know that the tensile strength is defined as

\sigma = \frac{F}{A}

Where,

\sigma =Tensile strength

F = Tensile Force

A = Cross-sectional Area

In the other hand we have that the shear strength is defined as

\sigma_y = \frac{F_y}{A}

where,

\sigma_y =Shear strength

F_y = Shear Force

A_0 =Parallel Area

PART A) Replacing with our values in the equation of tensile strenght, then

311*10^6 = \frac{F}{(15*10^{-6})(30*10^{-2})}

Resolving for F,

F= 1399.5N

PART B) We need here to apply the shear strength equation, then

\sigma_y = \frac{F_y}{A}

210*10^6 = \frac{F_y}{15*10^{-6}30*10^{-2}}

F_y = 945N

In such a way that the material is more resistant to tensile strength than shear force.

6 0
3 years ago
A man does 4,475 J of work in the process of pushing his 2.50 103 kg truck from rest to a speed of v, over a distance of 26.0 m.
Tcecarenko [31]

Answer:

a) 1.89 m/s  b) 172.1 N

Explanation:

a)

  • Applying the work-energy theorem, if we can neglect the friction between truck and road, the total change in kinetic energy must be equal to the work done by the external forces.
  • This work, is just 4,475 J.
  • So we can write the following equation:

        \Delta K = \frac{1}{2} * m*v^{2} = 4,475 J

  • where m= mass of the truck = 2.5*10³ kg.
  • So, we can find the speed v, as follows:

        v =\sqrt{\frac{2*W}{m}} =\sqrt{\frac{2*4,475J}{2.5e3kg} }  = 1.89 m/s

b)

  • The work done by the man, is just the horizontal force applied, times the displacement produced by the force horizontally:

        W = F*d

  • We can solve for F, as follows:

        F = \frac{W}{d} = \frac{4,475 J}{26.0m} =  172.1 N

4 0
2 years ago
A 392 N wheel comes off a moving truck and rolls without slipping along a highway. At the bottom of a hill it is rotating at 24
alex41 [277]

Answer:

h=12.41m

Explanation:

N=392

r=0.6m

w=24 rad/s

I=0.8*m*r^{2}

So the weight of the wheel is the force N divide on the gravity and also can find momentum of inertia to determine the kinetic energy at motion

N=m*g\\m=\frac{N}{g}\\m=\frac{392N}{9.8\frac{m}{s^{2}}}

m=40kg

moment of inertia

M_{I}=0.8*40.0kg*(0.6m} )^{2}\\M_{I}=11.5 kg*m^{2}

Kinetic energy of the rotation motion

K_{r}=\frac{1}{2}*I*W^{2}\\K_{r}=\frac{1}{2}*11.52kg*m^{2}*(24\frac{rad}{s})^{2}\\K_{r}=3317.76J

Kinetic energy translational

K_{t}=\frac{1}{2}*m*v^{2}\\v=w*r\\v=24rad/s*0.6m=14.4 \frac{m}{s}\\K_{t}=\frac{1}{2}*40kg*(14.4\frac{m}{s})^{2}\\K_{t}=4147.2J

Total kinetic energy  

K=3317.79J+4147.2J\\K=7464.99J

Now the work done by the friction is acting at the motion so the kinetic energy and the work of motion give the potential work so there we can find height

K-W=E_{p}\\7464.99-2600J=m*g*h\\4864.99J=m*g*h\\h=\frac{4864.99J}{m*g}\\h=\frac{4864.99J}{392N}\\h=12.41m

6 0
3 years ago
As part of an exercise program, a woman walks south at a speed of 2.00 m/s for 60.0 minutes. She then turns around and walks nor
Anastaziya [24]

Answer:

Option A

Solution:

As per the question:

The distance covered by the woman in the North direction, d = 3000 m

Time taken to travel in North direction, t = 25.0 min = 1500 s

Velocity of woman in the south direction, v = 2.00 m/s

Time taken in the south direction, t' = 60.0 min = 3600 s

Now,

The distance covered in the south direction, d' = vt' = 2.00\times 3600 = 7200\ m

Now, the total displacement is given by:

D = d' - d = 7200 - 3000 = 4200 m in South

(a) Average velocity of the woman in the whole journey is given by:

v_{avg} = \frac{Total\ displacement}{Total\ time} = \frac{4200}{t + t'}

v_{avg} = \frac{4200}{1500 + 3600} = 0.8235\ m/s ≈ 0.824 m/s South

6 0
3 years ago
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