HELP PLEASE ASAP!!!<br> I CAN'T FAIL PHYSICS<br> xoxo thank you!

A generator converts mechanical energy into _____________________ energy.

erica [24]

Electrical energy is your answer.

6 0

2 years ago

An experiment is conducted such that an applied force is exerted on a 5kg object as it travels across a horizontal surface in wh

Katen [24]

If an experiment is conducted such that an applied force is exerted on an object, a student could use the graph to determine the net work done on the object.

The graph of the net force exerted on the object as a function of the object’s distance traveled is attached below.

A student could use the graph to determine the net work done on the object by Calculating the area bound by the line of best fit and the horizontal axis from 0m to 5m

For more information on work done, visit

brainly.com/subject/physics

5 0

2 years ago

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Find the net work W done on the particle by the external forces during the motion of the particle in terms of the initial and fi

steposvetlana [31]

Answer:

$W=K_f-K_i$

Explanation:

The work done on a particle by external forces is defined as:

$W=\int\limits^{r_f}_{r_i} {F\cdot dr} \,$

According to Newton's second law $F=ma$ . Thus:

$W=\int\limits^{r_f}_{r_i}{ma\cdot dr} \,\\$

Acceleration is defined as the derivative of the speed with respect to time:

$W=m\int\limits^{r_f}_{r_i}{\frac{dv}{dt}\cdot dr} \,\\\\W=m\int\limits^{r_f}_{r_i}{dv \cdot \frac{dr}{dt}} \,$

Speed is defined as the derivative of the position with respect to time:

$W=m\int\limits^{v_f}_{v_i} v \cdot dv \,$

Kinetic energy is defined as $K=\frac{mv^2}{2}$ :

$W=m\frac{v_f^2}{2}-m\frac{v_i^2}{2}\\W=K_f-K_i$

3 0

2 years ago

Which of the following statements is true?

xenn [34]

Correct answer choice is :

C) The freezing and melting temperatures of a substance are the same.

Explanation:

Fluids have a particular temperature at which they convert into solids, identified as their freezing point. In theory, the melting point of a solid should be the same as the freezing point of the liquid. In practice, small variations among these measures can be seen. The freezing point of a matter is the same as that substance's melting point. At this distinct temperature, the substance can exist as either a solid or a liquid. At temperatures below the freezing/ melting point, the substance is a solid.

4 0

2 years ago

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You pull straight up on the string of a yo-yo with a force 0.35 N, and while your hand is moving up a distance 0.16 m, the yo-yo

jarptica [38.1K]

Answer:

a) 0.138J

b) 3.58m/S

c) (1.52J)(I)

Explanation:

a) to find the increase in the translational kinetic energy you can use the relation

$\Delta E_k=W=W_g-W_p$

where Wp is the work done by the person and Wg is the work done by the gravitational force

By replacing Wp=Fh1 and Wg=mgh2, being h1 the distance of the motion of the hand and h2 the distance of the yo-yo, m is the mass of the yo-yo, then you obtain:

$Wp=(0.35N)(0.16m)=0.056J\\\\Wg=(0.062kg)(9.8\frac{m}{s^2})(0.32m)=0.19J\\\\\Delta E_k=W=0.19J-0.056J=0.138J$

the change in the translational kinetic energy is 0.138J

b) the new speed of the yo-yo is obtained by using the previous result and the formula for the kinetic energy of an object:

$\Delta E_k=\frac{1}{2}mv_f^2-\frac{1}{2}mv_o^2$

where vf is the final speed, vo is the initial speed. By doing vf the subject of the formula and replacing you get:

$v_f=\sqrt{\frac{2}{m}}\sqrt{\Delta E_k+(1/2)mv_o^2}\\\\v_f=\sqrt{\frac{2}{0.062kg}}\sqrt{0.138J+1/2(0.062kg)(2.9m/s)^2}=3.58\frac{m}{s}$

the new speed is 3.58m/s

c) in this case what you can compute is the quotient between the initial rotational energy and the final rotational energy

$\frac{E_{fr}}{E_{fr}}=\frac{1/2I\omega_f^2}{1/2I\omega_o^2}=\frac{\omega_f^2}{\omega_o^2}\\\\\omega_f=\frac{v_f}{r}\\\\\omega_o=\frac{v_o}{r}\\\\\frac{E_{fr}}{E_{fr}}=\frac{v_f^2}{v_o^2}=\frac{(3.58m/s)}{(2.9m/s)^2}=1.52J$

hence, the change in Er is about 1.52J times the initial rotational energy

5 0

2 years ago

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HELP PLEASE ASAP!!! I CAN'T FAIL PHYSICS xoxo thank you!