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2 years ago

5

if an object starts at rest and moves 60 meters north along a straight line in 2 seconds, what is the average velocity?

2 answers:

VashaNatasha [74]2 years ago

8 0

30 m/s because you would divide 60 meters by 2 seconds. 60 divided by two equals 30

Thepotemich [5.8K]2 years ago

5 0

Velocity =
(distance between start-point and end-point) / (time to run the course)
in the direction of
(direction from the start-point to the end-point).

Distance from start-point to end-point = 60 meters
Time to run the course = 2 seconds
Direction = north

Velocity = 30 m/s north

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Otrada [13]

If no other forces act on the object, according to Newton’s first law, the spacecraft will continue moving at a constant velocity, assuming that a planet or something with large mass doesn’t cross its path. Forces are not required to continue the motion of an object on a frictionless plane at a constant rate.

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3 years ago

Read 2 more answers

A wagon wheel consists of 8 spokes of uniform diamter, each of mass m, and length L. The outer ring has a mass m rin. What is th

Genrish500 [490]

Answer:

$L^2(\dfrac{8m}{3}+m_r)$

Explanation:

m = Mass of each rod

L = Length of rod = Radius of ring

$m_r$ = Mass of ring

Moment of inertia of a spoke

$\dfrac{mL^2}{3}$

For 8 spokes

$8\dfrac{mL^2}{3}$

Moment of inertia of ring

$m_rL^2$

Total moment of inertia

$8\dfrac{mL^2}{3}+m_rL^2\\\Rightarrow L^2(\dfrac{8m}{3}+m_r)$

The moment of inertia of the wheel through an axis through the center and perpendicular to the plane of the ring is $L^2(\dfrac{8m}{3}+m_r)$ .

3 0

3 years ago

In your daily life you come across a range of motion in which acceleration is in the direction of motion

irakobra [83]

That makes no sense to me somehow

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3 years ago

If a system has 225 kcal of work done to it, and releases 5.00 × 102 kj of heat into its surroundings, what is the change in int

vovikov84 [41]

We can solve the problem by using the first law of thermodynamics:

$\Delta U = Q-W$

where

$\Delta U$ is the change in internal energy of the system

$Q$ is the heat absorbed by the system

$W$ is the work done by the system on the surrounding

In this problem, the work done by the system is

$W=-225 kcal=-941.4 kJ$

with a negative sign because the work is done by the surrounding on the system, while the heat absorbed is

$Q=-5 \cdot 10^2 kJ=-500 kJ$

with a negative sign as well because it is released by the system.

Therefore, by using the initial equation, we find

$\Delta U=Q-W=-500 kJ+941.4 kJ=441.4 kJ$

8 0

3 years ago

A sinusoidally oscillating current I ( t ) with an amplitude of 9.55 A and a frequency of 359 cycles per second is carried by a

UNO [17]

Answer:

$P_{avg} = 6.283*10^{-9} \ W$

Explanation:

Given that;

I₀ = 9.55 A

f = 359 cycles/s

b = 72.2 cm

c = 32.5 cm

a = 80.2 cm

Using the formula;

$\phi = \frac{\mu_o Ic }{2 \pi} In (\frac{b+a}{b})$

where;

$E= \frac{d \phi}{dt}$

$E = \frac{\mu_o}{2 \pi}c In (\frac{b+a}{a}) I_o \omega cos \omega t$

$E_{rms} = \frac { {\frac{\mu_o \ c}{2 \pi} In (\frac{b+a}{a}) I_o (2 \pi f)}}{\sqrt{2}}$

Replacing our values into above equation; we have:

$E_{rms} = \frac { {\frac{4 \pi*10^{-7}*0.325}{2 \pi} In (\frac{72.2+80.2}{80.2}) *9.55 (2 \pi *359)}}{\sqrt{2}}$

$E_{rms} = \frac {8.98909588*10^{-4} }{\sqrt{2}}$

$E_{rms} = 6.356*10^{-4} \ V$

Then the $P_{avg$ is calculated as:

$P_{avg} = \frac{E^2}{R}$

$P_{avg} = \frac{(6.356*10^{-4})^2}{64.3}$

$P_{avg} = 6.283*10^{-9} \ W$

6 0

3 years ago