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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?

1 answer:

julsineya [31]3 years ago

3 0

Answer:

5N

Explanation:

(25 N - 20 N = 5 N)

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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

$0.0425F_c - 28.75 = I * \alpha$

$0.0425F_c - 28.75 = 1.1125 *4.30$

$0.0425F_c - 28.75 = 1.1125 *4.30$

$F_c = 789.03 \ N$

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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

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