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Lelechka [254]
3 years ago
6

Net force is the sum of all the forces acting on an object. If a spring balance pulls on a body with a force of 10 N, and fricti

on acts on the body in the opposite direction with a force of 1 N, the net force would be 9 N in the direction of the spring balance (10 N – 1 N = 9 N).
What is the net force acting on the object when the spring balance pulls the rope with a force of 25 N and friction acts on the body with a force of 20N?
Physics
1 answer:
julsineya [31]3 years ago
3 0

Answer:

5N

Explanation:

(25 N - 20 N = 5 N)

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2. Ms. Evelyn lined up her students at recess on the soccer field. She told each student to kick a soccer ball as hard as they c
mash [69]

Answer

A. Yes, because they did several tests with the soccer balls and recorded their observations. B. Yes, because every student kicked

Explanation:

5 0
2 years ago
What are the basic rules of basketball? Are the rules different for men’s versus women’s competition?
Vlad [161]

The 5 main rules are two teams of five players each try to score by shooting a ball through a hoop elevated 10 feet above the ground. The game is played on a rectangular floor called the court, and there is a hoop at each end. The rules are basically the same but the women's ball is one inch smaller than the mens ball.

4 0
3 years ago
Read 2 more answers
During a testing process, a worker in a factory mounts a bicycle wheel on a stationary stand and applies a tangential resistive
olganol [36]

Answer:

The force is     F_c  =  789.03 \  N    

Explanation:

From the question we are told that

   The tangential  resistive force is F_t  =   115 \ N

   The mass of the wheel is  m  = 1.80 kg

  The diameter of the wheel is  d =  50.0 cm  = 0.5 \ m

   The diameter of the sprocket is  d_c  =  8.50 \ cm =0.085 \ m

  The angular acceleration considered is  \alpha  =  4.30\ rad/s^2

Generally the radius of the wheel is

       r = \frac{d}{2}

=>     r = \frac{0.5}{2}

=>     r = 0.25 \ m

Generally the radius of the sprocket is

       r_c = \frac{d_c}{2}

=>     r_c = \frac{0.085}{2}

=>     r_c = 0.0425 \ m

Generally the moment of inertia of the wheel is mathematically represented as

      I  =  m  *  r^2

=>    I  =  1.80  *  0.25^2

=>    I  = 1.1125 \ kg \cdot m^2

Generally the torque experienced by the wheel due to the forces acting on it  is mathematically represented as

      \tau =  F_c *  r_c  -  F_t  * r

Here  F_c is the force acting on the sprocket

So  

      \tau =  F_c *  0.0425 - 115  * 0.25

       \tau = 0.0425F_c  -  28.75

Generally the torques that will cause the wheel to move with \alpha  =  4.30\ rad/s^2 is mathematically represented as

       \tau  =  I  * \alpha

So

        0.0425F_c  -  28.75  =   I  * \alpha

        0.0425F_c  -  28.75  =   1.1125  *4.30    

       0.0425F_c  -  28.75  =   1.1125  *4.30    

        F_c  =  789.03 \  N    

5 0
3 years ago
Detailed measurements of the disk and central bulge region of our Galaxy suggest our Milky Way is a:Select one:A. quasar. B. bar
chubhunter [2.5K]

I believe the answer is:

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4 0
3 years ago
Determine the specific heat of a certain metal if a 450 gram sample of it loses 34 500 Joules of heat as its temperature starts
slamgirl [31]

Answer:

c = 0.4356 J/gK

Explanation:

Given the following data;

Mass = 450 grams

Initial temperature, T1 = 150°C

Final temperature, T2 = 53°C

Quantity of heat = 34500 Joules

To find the specific heat capacity of the metal;

Heat capacity is given by the formula;

Q = mcdt

Where;

Q represents the heat capacity or quantity of heat.

m represents the mass of an object.

c represents the specific heat capacity of water.

dt represents the change in temperature.

dt = T2 - T1

dt = 53 - 150

dt = -97°C

Converting the temperature in Celsius to Kelvin, we have;

dt = 273 + (-97) = 176 Kelvin

Making c the subject of formula, we have;

c = \frac {Q}{mdt}

Substituting into the equation, we have;

c = \frac {34500}{450*176}

c = \frac {34500}{79200}

c = 0.4356 J/gK

6 0
3 years ago
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