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levacccp [35]
3 years ago
14

What is the mass of a basketball that has Connecticut energy of 102 J and it's travelling at 5 m/s

Physics
1 answer:
Stella [2.4K]3 years ago
8 0

Answer:

8.16Kg

Explanation:

K.E= 1/2 mv²

102=1/2 m 5²

m=8.16 Kg

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In the mobile m1=0.42 kg and m2=0.47 kg. What must the unknown distance to the nearest tenth of a cm be if the masses are to be
LuckyWell [14K]

Complete Question

The complete question is shown on the first uploaded image

Answer:

Explanation:

From he question we are told that

    The first mass is   m_1 = 0.42kg

      The second mass is  m_2 = 0.47kg

From the question we can see that at equilibrium the moment about the point where the  string  holding the bar (where m_1 \ and \ m_2 are hanged ) is attached is zero  

   Therefore we can say that

               m_1 * 15cm  = m_2 * xcm

Making x the subject of the formula  

                x = \frac{m_1 * 15}{m_2}

                    = \frac{0.42 * 15}{0.47}

                     x = 13.4 cm

Looking at the diagram we can see that the tension T  on the string holding the bar where m_1  \  and   \ m_2 are hanged  is as a result of the masses (m_1 + m_2)

     Also at equilibrium the moment about the point where the string holding the bar (where (m_1 +m_2)  and  m_3 are hanged ) is attached is  zero

   So basically

          (m_1 + m_2 ) * 20  = m_3 * 30

          (0.42 + 0.47)  * 20 = 30 * m_3

 Making m_3 subject

          m_3 = \frac{(0.42 + 0.47) * 20 }{30 }

                m_3 = 0.59 kg

3 0
3 years ago
A 1.3-kg model airplane flies in a circular path on the end of a 23-m line. The plane makes
storchak [24]

(a) The plane makes 4.3 revolutions per minute, so it makes a single revolution in

(1 min) / (4.3 rev) ≈ 0.2326 min ≈ 13.95 s ≈ 14 s

(b) The plane completes 1 revolution in about 14 s, so that in this time it travels a distance equal to the circumference of the path:

(2<em>π</em> (23 m)) / (14 s) ≈ 10.3568 m/s ≈ 10 m/s

(c) The plane accelerates toward the center of the path with magnitude

<em>a</em> = (10 m/s)² / (23 m) ≈ 4.6636 m/s² ≈ 4.7 m/s²

(d) By Newton's second law, the tension in the line is

<em>F</em> = (1.3 kg) (4.7 m/s²) ≈ 6.0627 N ≈ 6.1 N

4 0
3 years ago
A roller coaster starts from rest at its highest point and then descends on its (frictionless) track. its speed is 40 m/s when i
Alenkinab [10]

Given that,

Initial velocity , Vi = 0

Final velocity , Vf = 40 m/s

Acceleration due to gravity , a = 9.81 m/s²

Distance can be calculated as,

2as = Vf² - Vi²

2 * 9.81 *s = 40² - 0²

s = 81.55 m

For half height, that is, s = 40.77m

Vf= ??

2as = Vf² - Vi²

2 * 9.81 * 40.77 = Vf² - 0²

Vf² = 800

Vf = 28.28 m/s

Therefore, speed of roller coaster when height is half of its starting point will be 28 m/s.  

5 0
3 years ago
A baseball hits a car, breaking its window and triggering its alarm which sounds at a frequency of 1210 Hz. What frequency (in H
IRINA_888 [86]

Answer:

The frequency of sound heard by the boy is 1181 Hz.

Explanation:

Given that,

Frequency of sound from alarm  f_{0} = 1210\ Hz

Speed = -8.25 m/s

Negative sign show the boy riding away from the car

Speed of sound = 343

We need to calculate the heard frequency

Using formula of frequency

f = f_{0}(\dfrac{v+v_{0}}{v-v_{s}})

Where, f_{0} = frequency of source

v_{0} = speed of observer

v_{s} = speed of source

v = speed of sound

Put the value into the formula

f=1210\times\dfrac{343+(-8.25)}{343-0}

here, source is at rest

f=1180.8\ Hz

f=1181\ Hz

Hence, The frequency of sound heard by the boy is 1181 Hz.

8 0
4 years ago
What is the density of a piece of quartz with a mass of 30g and a volume of 6cm^3
FromTheMoon [43]

Answer:

\boxed {\boxed {\sf d= 5 \ g/cm^3}}

Explanation:

Density can be found by dividing the mass by the volume.

d=\frac{m}{v}

The mass of the quartz is 30 grams and the volume is 6 cubic centimeters.

m=30 \ g \\v= 6 \ cm^3

Substitute the values into the formula.

d=\frac{30 \ g}{6 \ cm^3}

Divide.

d= 5 \ g/cm^3

The density of this piece of quartz is 5 grams per cubic centimeter.

7 0
3 years ago
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