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irga5000 [103]
3 years ago
14

a child pulls on a string that is attached to a car. if the child does 80.2 J of work while pulling the car 25.0 m, with what fo

rce is the child pulling?
Physics
1 answer:
12345 [234]3 years ago
3 0

Answer:

F = 3.20 N

Explanation:

Given:

Work done by child = 80.2 j

Distance that the car moves = 25.0 m

We need to find the force acting on the car.

Solution:

Using work done formula as.

W = F\times d

Where:

W = Work done by any object.

F = Force (push or pull)

d = distance that the object moves.

Substitute W = 80.2\ J\ and\ d =25.0\ m in work done formula.

80.2 = F\times 25

F=\frac{80.2}{25}

F = 3.20 N

Therefore, force acting on the car F = 3.20 N

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A horse of mass 242 kg pulls a cart of mass 224 kg. The acceleration of gravity is 9.8 m/s 2 . What is the largest acceleration
Rufina [12.5K]

To solve this problem it is necessary to apply the concepts related to Newton's second Law and the force of friction. According to Newton, the Force is defined as

F = ma

Where,

m= Mass

a = Acceleration

At the same time the frictional force can be defined as,

F_f = \mu N

Where,

\mu = Frictional coefficient

N = Normal force (mass*gravity)

Our values are given as,

m_h = 242 kg\\m_c = 224 kg\\\mu = 0.894\\

By condition of Balance the friction force must be equal to the total net force, that is to say

F_{net} = F_f

m_{total}a = \mu m_hg

(m_h+m_c)a = \mu*m_h*g

Re-arrange to find acceleration,

a= \frac{\mu*m_h*g}{(m_h+m_c)}

a = \frac{0.894*242*9.8}{(242+224)}

a = 4.54 m/s^2

Therefore the acceleration the horse can give is 4.54m/s^2

3 0
3 years ago
Where c is a constant that depends on the initial gas pressure behind the projectile. The initial position of the projectile is
Ronch [10]

Complete Question

A gas gun uses high pressure gas tp accelerate projectile through the gun barrel.

If the acceleration of the projective is : a = c/s m/s​2

Where c is a constant that depends on the initial gas pressure behind the projectile. The initial position of the projectile is s= 1.5m and the projectile is initially at rest. The projectile accelerates until it reaches the end of the barrel at s=3m. What is the value of the constant c such that the projectile leaves the barrel with velocity of 200m/s?

Answer:

The value of the constant is  c = 28853.78 \ m^2 /s^2

Explanation:

From the question we are told that

         The acceleration is  a =  \frac{c}{s}\   m/s^2

         The  initial position of the projectile is s= 1.5m

         The final position of the projectile is s_f =  3 \ m

          The velocity is  v = 200 \ m/s

     Generally  time  =  \frac{ds}{dv}

   and  acceleration is a =  \frac{v}{time }

so

            a = v  \frac{dv}{ds}

 =>        vdv  =  a ds

             vdv  = \frac{c}{s}  ds

integrating both sides

           \int\limits^a_b  vdv  = \int\limits^c_d \frac{c}{s}  ds

Now for the limit

          a =  200 m/s

             b = 0 m/s  

         c = s= 3 m

          d =s_f= 1.5 m

So we have  

           \int\limits^{200}_{0}  vdv  = \int\limits^{3}_{1.5} \frac{c}{s}  ds

              [\frac{v^2}{2} ] \left | 200} \atop {0}} \right.  = c [ln s]\left | 3} \atop {1.5}} \right.

            \frac{200^2}{2}  =  c ln[\frac{3}{1.5} ]

=>           c = \frac{20000}{0.69315}

              c = 28853.78 \ m^2 /s^2

     

5 0
3 years ago
Each line (unit) on the horizontal axis represents: 5 years 10 years 20 years all of the above
zlopas [31]
The correct answer is 10 years
6 0
3 years ago
What is 25 degrees Celsius equal to in Fahrenheit? I need to know how to do it to, so give me a formula, and it would help if yo
andreev551 [17]
C= 5/9*(F - 32) so F=C*9/5 + 32

F= 25*9/5 +32 = 77 Fahrenheit
5 0
3 years ago
What is 20°C in Fahrenheit? A. 68°F B. 62°F C. 100°F D. 36°F
creativ13 [48]

A. 68*F
hope this helps!
7 0
3 years ago
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