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3 years ago

8

If an object is dropped from a tall building and hits the ground 3.0 s later, how tall is the building?

1 answer:

allsm [11]3 years ago

6 0

D = 1/2 g t^2. It works out to 44.1 meters.

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A car with an initial velocity of 5.0 is acelerated at 3.0for 4.0seconds what os the velocity of tje car ofter the 4.0seconds

sasho [114]

Vf =Vi+at
Vf= 5 + 3*4
Vf=5+12
Vf=17m/sec

6 0

3 years ago

Find the magnitude: || 5-3i || ...?

____ [38]

The magnitude would be :

$\sqrt{5^2 + 3^2}$

= √34

Hope this helps

5 0

2 years ago

What is the kinetic energy of an object with a mass of 50kg and is and it is traveling at a rate of 60m/s.

liq [111]

Answer:

90,000 J

Explanation:

Kinetic energy can be found using the following formula.

$KE=\frac{1}{2}mv^2$

where <em>m </em>is the mass in kilograms and <em>v</em> is the velocity in m/s.

We know the object has a mass of 50 kilograms. We also know it is a traveling at a rate of 60 m/s. Velocity is the speed of something, so the velocity of the object is 60 m/s.

<em>m</em>=50

<em>v</em>=60

Substitute these values into the formula.

$KE=\frac{1}{2}*50*60^2$

First, evaluate the exponent: 60^2. 60^2 is the same as multiplying 60, 2 times.

60^2=60*60=3,600

$KE=\frac{1}{2}*50*3,600$

Multiply 50 and 3,600

$KE=\frac{1}{2}*180,000$

Multiply 1/2 and 3,600, or divide 3,600 by 2.

$KE=90,000$

Add appropriate units. Kinetic energy uses Joules, or J.

$KE=90,000 Joules$

The kinetic energy of the object is 90,000 Joules

6 0

3 years ago

Six artificial satellites circle a space station at constant speed. The mass m of each satellite, distance L from the space stat

nikklg [1K]

Answers:

a) $T_{2}>T_{5}>T_{1}>T_{3}=T_{6}>T_{4}$

b) $a_{4}>a_{6}>a_{1}>a_{3}>a_{5}>a_{2}$

Explanation:

a) Since we are told the satellites circle the space station at constant speed, we can assume they follow a uniform circular motion and their tangential speeds $V$ are given by:

$V=\omega L=\frac{2\pi}{T} L$ (1)

Where:

$\omega$ is the angular frequency

$L$ is the radius of the orbit of each satellite

$T$ is the period of the orbit of each satellite

Isolating $T$ :

$T=\frac{2 \pi L}{V}$ (2)

Applying this equation to each satellite:

$T_{1}=\frac{2 \pi L}{V_{1}}=261.79 s$ (3)

$T_{2}=\frac{2 \pi L}{V_{2}}=1570.79 s$ (4)

$T_{3}=\frac{2 \pi L}{V_{3}}=196.349 s$ (5)

$T_{4}=\frac{2 \pi L}{V_{4}}=98.174 s$ (6)

$T_{5}=\frac{2 \pi L}{V_{5}}=785.398 s$ (7)

$T_{6}=\frac{2 \pi L}{V_{6}}=196.349 s$ (8)

Ordering this periods from largest to smallest:

$T_{2}>T_{5}>T_{1}>T_{3}=T_{6}>T_{4}$

b) Acceleration $a$ is defined as the variation of velocity in time:

$a=\frac{V}{T}$ (9)

Applying this equation to each satellite:

$a_{1}=\frac{V_{1}}{T_{1}}=0.458 m/s^{2}$ (10)

$a_{2}=\frac{V_{2}}{T_{2}}=0.0254 m/s^{2}$ (11)

$a_{3}=\frac{V_{3}}{T_{3}}=0.4074 m/s^{2}$ (12)

$a_{4}=\frac{V_{4}}{T_{4}}=1.629 m/s^{2}$ (13)

$a_{5}=\frac{V_{5}}{T_{5}}=0.101 m/s^{2}$ (14)

$a_{6}=\frac{V_{6}}{T_{6}}=0.814 m/s^{2}$ (15)

Ordering this acceerations from largest to smallest:

$a_{4}>a_{6}>a_{1}>a_{3}>a_{5}>a_{2}$

4 0

3 years ago

What is the speed of a wave with 3Hz frequency and a wavelength of 9 m

SIZIF [17.4K]

Answer:

Given

Frequency (f) = 3Hz

Wavelength = 9 m

Speed = ?

Explanation:

we know

Speed = wavelength * frequency

= 9*3

= 27 m/ s

7 0

3 years ago