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3 years ago

7

Claire lost weight thanks to the support and encouragement of her friends. They helped her stick to a healthy diet and pushed he

r to engage in physical exercise every day. Identify the FaceTime that helped her lose weight

1 answer:

atroni [7]3 years ago

6 0

Answer:

hello the options to your question is missing below are the options

culture

heredity

physical environment

social environment

answer : social environment

Explanation:

Claire ability to lose weight due to the encouragement and support of her friends shows that she was ability to achieve her goal due to her social circle ( friends ). therefore the face time that helped her lose weight is her social environment and not culture nor any other option as mentioned above.

social environments help/play a vital role in individual's life a lot as long as every human being is a social being

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In A. C. motor capacitor is used

natita [175]

Answer:

d. to decrease D. C.

Explanation:

d. to decrease D. C.

4 0

3 years ago

A worker wants to load a 12 kg crate into a truck by sliding the crate up a straight ramp which is 2.5 m long and which makes an

olga2289 [7]

Answer:

a) The magnitude of the friction force is 55.851 newtons, b) The speed of the crate when it reaches the bottom of the ramp is 2.526 meters per second.

Explanation:

a) This situation can be modelled by the Principle of Energy Conservation and the Work-Energy Theorem, where friction represents the only non-conservative force exerting on the crate in motion. Let consider the bottom of the straight ramp as the zero point. The energy equation for the crate is:

$U_{g,1}+K_{1} = U_{g,2}+K_{2}+ W_{fr}$

Where:

$U_{g,1}$ , $U_{g,2}$ - Initial and final gravitational potential energy, measured in joules.

$K_{1}$ , $K_{2}$ - Initial and final translational kinetic energy, measured in joules.

$W_{fr}$ - Work losses due to friction, measured in joules.

By applying the defintions of translational kinetic and gravitational potential energies and work, this expression is now expanded:

$m\cdot g \cdot y_{1} + \frac{1}{2}\cdot m\cdot v_{1}^{2} = m\cdot g \cdot y_{2} + \frac{1}{2}\cdot m\cdot v_{2}^{2} + \mu_{k}\cdot m \cdot g \cdot \cos \theta$

Where:

$m$ - Mass of the crate, measured in kilograms.

$g$ - Gravitational acceleration, measured in meters per square second.

$y_{1}$ , $y_{2}$ - Initial and final height of the crate, measured in meters.

$v_{1}$ , $v_{2}$ - Initial and final speeds of the crate, measured in meters per second.

$\mu_{k}$ - Kinetic coefficient of friction, dimensionless.

$\theta$ - Ramp inclination, measured in sexagesimal degrees.

The equation is now simplified and the coefficient of friction is consequently cleared:

$y_{1}-y_{2}+\frac{1}{2\cdot g}\cdot (v_{1}^{2}-v_{2}^{2}) = \mu_{k}\cdot \cos \theta$

$\mu_{k} = \frac{1}{\cos \theta} \cdot \left[y_{1}-y_{2}+\frac{1}{2\cdot g}\cdot (v_{1}^{2}-v_{2}^{2}) \right]$

The final height of the crate is:

$y_{2} = (1.6\,m)\cdot \sin 30^{\circ}$

$y_{2} = 0.8\,m$

If $\theta = 30^{\circ}$ , $y_{1} = 0\,m$ , $y_{2} = 0.8\,m$ , $g = 9.807\,\frac{m}{s^{2}}$ , $v_{1} = 5\,\frac{m}{s}$ and $v_{2} = 0\,\frac{m}{s}$ , the coefficient of friction is:

$\mu_{k} = \frac{1}{\cos 30^{\circ}}\cdot \left\{0\,m-0.8\,m+\frac{1}{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)}\cdot \left[\left(5\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}\right] \right\}$

$\mu_{k} \approx 0.548$

Then, the magnitude of the friction force is:

$f =\mu_{k}\cdot m\cdot g \cdot \cos \theta$

If $\mu_{k} \approx 0.548$ , $m = 12\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ and $\theta = 30^{\circ}$ , the magnitude of the force of friction is:

$f = (0.548)\cdot (12\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot \cos 30^{\circ}$

$f = 55.851\,N$

The magnitude of the force of friction is 55.851 newtons.

b) The energy equation of the situation is:

$m\cdot g \cdot y_{1} + \frac{1}{2}\cdot m\cdot v_{1}^{2} = m\cdot g \cdot y_{2} + \frac{1}{2}\cdot m\cdot v_{2}^{2} + \mu_{k}\cdot m \cdot g \cdot \cos \theta$

$y_{1}+\frac{1}{2\cdot g}\cdot v_{1}^{2} =y_{2} + \frac{1}{2\cdot g}\cdot v_{2}^{2} + \mu_{k}\cdot \cos \theta$

Now, the final speed is cleared:

$y_{1}-y_{2}+ \frac{1}{2\cdot g}\cdot v_{1}^{2} -\mu_{k}\cdot \cos \theta= \frac{1}{2\cdot g}\cdot v_{2}^{2}$

$2\cdot g \cdot (y_{1}-y_{2}-\mu_{k}\cdot \cos \theta) + v_{1}^{2} = v_{2}^{2}$

$v_{2} = \sqrt{2\cdot g \cdot (y_{1}-y_{2}-\mu_{k}\cdot \cos \theta)+v_{1}^{2}}$

Given that $g = 9.807\,\frac{m}{s^{2}}$ , $y_{1} = 0.8\,m$ , $y_{2} = 0\,m$ , $\mu_{k} \approx 0.548$ , $\theta = 30^{\circ}$ and $v_{1} = 0\,\frac{m}{s}$ , the speed of the crate at the bottom of the ramp is:

$v_{2}=\sqrt{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot [0.8\,m-0\,m-(0.548)\cdot \cos 30^{\circ}]+\left(0\,\frac{m}{s} \right)^{2}}$

$v_{2}\approx 2.526\,\frac{m}{s}$

The speed of the crate when it reaches the bottom of the ramp is 2.526 meters per second.

7 0

3 years ago

A car, with an initial velocity of 22 m/s west, travels for 345 seconds. If it has a final velocity of 44 m/s west what is its’

cupoosta [38]

Answer: 0.064 m/s²

Explanation:

Initial velocity = 22m/s

Final velocity = 44m/s

Time taken = 345 seconds

Acceleration = (Final velocity - Initial velocity) / Time taken

= (44 - 22) / 345

= 22/345

= 0.0637681

= 0.064 m/s²

Therefore, the acceleration is 0.064 m/s².

4 0

3 years ago

Find the resultant of these two forces : 2.00 x 10^2 N due east and 4.00 x 10^2 N 30.0 degree north of west

alexandr1967 [171]

If two vectors are represented both in magnitude and direction by the two adjacent sides of a triangle taken in an order, then their resultant is given by the third side taken in the reverse order.

Resultant vector :

R = sqrt of A^2 + B^2 + 2ABcosθ

given :

A = 200 units

B = 300 units

θ = 30 degrees

∴ R = sqrt of 200^2 + 300^2 + 2(200)(300)cos35

the answer is 477.805656

7 0

3 years ago

Read 2 more answers

Which of the following are non-contact forces: friction, electrostatic force, magnetic force, gravity?

oee [108]

Magnetic force and Gravity are non-contact forces among friction, electrostatic force, magnetic force and gravity.

4 0

4 years ago

Read 2 more answers