Describe, in terms of the motion of particles in an object

Physics

1 answer:

natita [175]3 years ago

8 0

Answer:

The motion of particles in any object moves randomly I think

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Calculate the potential energy of a 5 kg object sitting at the top of a 2 meter ramp.

trapecia [35]

The formula used to find potential energy is P.E. = M * G * H (P.E. is potential energy, M is mass, G is gravitational pull, and H is height). So the answer to your question is 5 * 9.8 * 2, which equals 98.

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A box mass of 24kg is being pulled horizontally on a rough surface by an applied force of 585N. The coefficient of kinetic frict

Elan Coil [88]

the normal force is the force applied opposite to the weight of the was box. So the normal force is equal to the weight of the box = 24 kg *(9.81 m/s2) = 235.44 N

the acceleration of the box be solve using newtons 2nd law of motion:

F = ma

a = F/ m = 585 N/ 24 kg = 24.38 m/s2

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3 years ago

Read 2 more answers

A diagram of a closed circuit with a power source on the left labeled 6 V. There are 3 resistors in parallel, separate paths, co

AfilCa [17]

Answer:

Current: 1.0 Amperes

The minimum current is flowing through path D

Explanation:

We first find the equivalent resistance to the three resistors in parallel ( which is the total resistance of the circuit) via the equation:

$\frac{1}{R_e} =\frac{1}{R_B}+\frac{1}{R_C}+\frac{1}{R_D}\\\frac{1}{R_e} =\frac{1}{10}+\frac{1}{20}+\frac{1}{50}=0.17\\R_e=(1/0.17)\Omega\\R_e=5.88 \Omega$

with this info, we can estimate the current going through branch A using Ohm's Law, and the information that the power source is 6 V:

$V=I*R\\6V=I*5.88\Omega\\I=\frac{6}{5.88} Amp\\I=1.02A$

where the current comes in units of Amperes since all other the quantities are given in the SI system, and we can round this answer to 1.0 Amp following the request to round it to the tenth.

The current will be the lowest through the branch with the largest resistor due to the fact that less current will flow through the path of more resistance.

Than means that the lowest current will be registered through branch D where the 50 $\Omega$ resistor is.