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AlekseyPX
3 years ago
7

A car travels 60 miles per hour. how much time does it takes the car to travel 30 miles?

Physics
2 answers:
elena55 [62]3 years ago
7 0

1 hour= 60 mins

60 mins= 60 miles

30 miles= (60 miles÷60 minutes)×30 miles

             =30 mins


I am Lyosha [343]3 years ago
7 0
1 hr = 60 minutes 60 minutes / 60 miles = 1 minutes/miles 30 miles * 1 minute = 30 minutes 30 minutes
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A 125 g pendulum bob hung on a string of length 35 cm has the same period as when the bob is hung from a spring and caused to os
Bingel [31]

Answer:

k = 3.5 N/m

Explanation:

It is given that the time period the bob in pendulum is the same as its time period in spring mass system:

Time\ Period\ of\ Pendulum = Time\ Period\ of\ Spring-Mass\ System\\2\pi \sqrt{\frac{l}{g}} = 2\pi \sqrt{\frac{m}{k}

\frac{l}{g} = \frac{m}{k}\\\\ k = g\frac{m}{l}

where,

k = spring constant = ?

g = acceleration due to gravity = 9.81 m/s²

m = mass of bob = 125 g = 0.125 kg

l = length of pendulum = 35 cm = 0.35 m

Therefore,

k = (9.81\ m/s^2)(\frac{0.125\ kg}{0.35\ m})\\\\

<u>k = 3.5 N/m</u>

4 0
3 years ago
The escape speed from an object is v2 = 2GM/R, where M is the mass of the object, R is the object's starting radius, and G is th
Rom4ik [11]

Answer:

Approximate escape speed = 45.3 km/s

Explanation:

Escape speed

        v=\sqrt{\frac{2GM}{R}}

Here we have

   Gravitational constant = G = 6.67 × 10⁻¹¹ m³ kg⁻¹ s⁻²

   R = 1 AU = 1.496 × 10¹¹ m

   M = 2.3 × 10³⁰ kg

Substituting

    v=\sqrt{\frac{2\times 6.67\times 10^{-11}\times 2.3\times 10^{30}}{1.496\times 10^{11}}}=4.53\times 10^4m/s=45.3km/s

Approximate escape speed = 45.3 km/s

6 0
3 years ago
A bird sits on a high-voltage power line with its feet 3.87 cm apart. The wire is made from aluminum, is 2.11 cm in diameter, an
Svetlanka [38]

Answer:

ΔV=0.484mV

Explanation:

The potential difference across the end of conductor that obeys Ohms law:

ΔV=IR

Where I is current

R is resistance

The resistance of a cylindrical conductor is related to its resistivity p,Length L and cross section area A

R=(pL)/A

Given data

Length L=3.87 cm =0.0387m

Diameter d=2.11 cm =0.0211 m

Current I=165 A

Resistivity of aluminum p=2.65×10⁻⁸ ohms

So

ΔV=IR

=I(\frac{pL}{A})\\ =I(\frac{pL}{\pi r^{2} } )\\=I(\frac{pL}{\pi (d/2)^{2} } )\\=165A((\frac{(2.65*10^{-8})(0.0387m)}{\pi (0.0211m/2)^{2} } ))\\=4.84*10^{-4}V

ΔV=0.484mV  

3 0
3 years ago
An unstable atomic nucleus has a mass of 17.010-27kg, and starts out at rest. When it decays, it the original nucleus disintegra
slega [8]

Answer:

Part a)

v = -(8.33\hat j + 9.33\hat i)\times 10^6 m/s

Part b)

E = 4.4 \times 10^{-13} J

Explanation:

As per momentum conservation we know that there is no external force on this system so initial and final momentum must be same

So we will have

m_1v_1 + m_2v_2 + m_3v_3 = 0

(5 \times 10^{-27})(6 \times 10^6\hat j) + (8.4 \times 10^{-27})(4 \times 10^6\hat i) + (3.6 \times 10^{-27}) v = 0

(30\hat j + 33.6\hat i)\times 10^6 + 3.6 v = 0

v = -(8.33\hat j + 9.33\hat i)\times 10^6 m/s

Part b)

By equation of kinetic energy we have

E = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2 + \frac{1}{2}m_3v_3^2

E = \frac{1}{2}(5 \times 10^{-27})(6\times 10^6)^2 + \frac{1}{2}(8.4 \times 10^{-27})(4 \times 10^6)^2 + \frac{1}{2}(3.6 \times 10^{-27})(8.33^2 + 9.33^2) \times 10^{12}

E = 9\times 10^{-14} + 6.72 \times 10^{-14} + 2.82\times 10^{-13}

E = 4.4 \times 10^{-13} J

8 0
3 years ago
an aircraft landing on an air craft carrier is brought to a complete stop from an inital velocity of 215km/hr in 2.7 seconds. wh
worty [1.4K]

u= 215 km/hr = 215 * 1000/ 3600 = aprx 60m/s
v=0
t=2.7sec
v= u - at
u= at
60/2.7 = 22.23 m/s^2



Hope it helps
8 0
3 years ago
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