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Damm [24]
3 years ago
12

- With the device starting from a still position (0 m/s), you drop it, allowing gravity to accelerate it towards the ground at 9

.8 m/s2. Assuming it takes 1.5 seconds for the device to hit the ground, what is it’s velocity? [V = a*t]
Physics
1 answer:
IrinaVladis [17]3 years ago
3 0
Starting velocity (u)=0
acceleration (a) = 9.8
time (t) = 1.5
final velocity (v) =?

v=at+u
v=9.8(1.5)+0
v=14.7m/s2
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Alekssandra [29.7K]
According to another source this is what I got
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5 0
3 years ago
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Three objects have velocities that vary with time
Solnce55 [7]

#21

  • initial velocity=u=5m/s
  • Time=t=2s
  • Acceleration=a=1.3m/s²

According to first equation of kinematics

  • v=u+at
  • v=5+1.3(2)
  • v=5+2.6
  • v=7.6m/s

#22

#a

  • v1=2+3(3)=2+9=11m/s
  • v2=-8-4(3)=-20m/s
  • v3=1-5(3)=-14m/s

The order is

  • v2<V3<v1

#b

for speed find absolute velocity

  • S1=|11|=11m/s
  • S2=|-20|=20m/s
  • S3=|-14|=14m/s

So order is

S1<S3<S2

7 0
2 years ago
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Taking the resistivity of platinoid as 3.3 x 10-7 m, find the resistance of 7.0 m of platinoid wire of average diameter 0.14 cm.
mr_godi [17]

Answer:

1.5 \Omega

Explanation:

The resistance of a wire is given by the equation:

R=\rho \frac{L}{A}

where

\rho is the resistivity of the material

L is the length of the wire

A is the cross-sectional area of the wire

In this problem, we have a wire of platinoid, whose resistivity is

\rho = 3.3\cdot 10^{-7} \Omega m

The length of the wire is

L = 7.0 m

And its radius is

r=\frac{0.14 cm}{2}=0.07 cm = 7\cdot 10^{-4} m, so the cross-sectional area is

A=\pi r^2=\pi(7\cdot 10^{-4})^2=1.54\cdot 10^{-6}m^2

Solving for R, we find the resistance of the wire:

R=(3.3\cdot 10^{-7})\frac{7.0}{1.54\cdot 10^{-6}}=1.5 \Omega

3 0
3 years ago
Consider the circuit shown in the figure to find the power delivered to 6 Ohm resistance (in W). Given that Vs= 30
vekshin1

66.7 Watts

Explanation:

Let R_{1}=1.0 ohms, R_{2}=3.0ohms and R_{3}=6.0\:ohms. Since R_{2} and R_{3} are in parallel, their combined resistance R_{23} is given by

\dfrac{1}{R_{23}}=\dfrac{1}{R_{2}}+\dfrac{1}{R_{3}}

or

R_{23}=\dfrac{R_{2}R_{3}}{R_{2}+ R_{3}}=2.0\:ohms

The total current flowing through the circuit <em>I</em> is given

I=\dfrac{V_{s}}{R_{Total}}

where

R_{Total}=R_{1}+R_{23}= 3.0\:ohms

Therefore, the total current through the circuit is

I=\dfrac{30\:V}{3.0\:ohms}=10\:A

In order to find the voltage drop across the 6-ohm resistor, we first need to find the voltage drop across the 1-ohm resistor V_{1}:

V_{1}=(10\:A)(1.0\:ohms)=10\:V

This means that voltage drop across the 6-ohm resistor V_{3} is 20 V. The power dissipated <em>P</em> by the 6-ohm resitor is given by

P=I^{2}R_{3}= \dfrac{V^{2}}{R_{3}}=\dfrac{(20\:V)^{2}}{6\:ohms}= 66.7 W

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