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3 years ago

Given: Q1 = 3.0 × 10-5 C

Physics

2 answers:

Igoryamba3 years ago

8 0

F=KQ1Q2/R2
F=9 x 10^9 x 3 x 10^-5 x 4 x 10^-5 / 9
F=12 x 10^-1
F=1.2 N

Mariulka [41]3 years ago

4 0

$F=k\dfrac{Q_1Q_2}{r^2}= \\ \\ =9\cdot 10^9\cdot \dfrac{3\cdot 10^{-5}\cdot 4\cdot 10^{-5}}{9}=...$

You just do the calculations.

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A 25 kg child runs at a speed of 5.0 m/s and jumps onto a stationary shopping cart and holds on for dear life. The cart has mass

makkiz [27]

Answer:

3.38 m/s

Explanation:

Mass of child = m₁ = 25

Initial speed of child = u₁ = 5 m/s

Initial speed of cart = u₂ = 0 m/s

Mass of cart = m₂ = 12 kg

Velocity of cart with child on top = v

This is a case of perfectly inelastic collision

$m_1u_1+m_2u_2=(m_1+m_2)v\\\Rightarrow v=\frac{m_1u_1+m_2u_2}{m_1+m_2}\\\Rightarrow v=\frac{25\times 5+12\times 0}{25+12}\\\Rightarrow v=\frac{125}{37}\\\Rightarrow v=3.38\ m/s$

Velocity of cart with child on top is 3.38 m/s

7 0

3 years ago

A 4000 kg satellite is placed 2.60 x 10^6 m above the surface of the Earth.

mash [69]

a) The acceleration of gravity is $4.96 m/s^2$

b) The critical velocity is 6668 m/s (24,006 km/h)

c) The period of the orbit is 8452 s

d) The satellite completes 10.2 orbits per day

e) The escape velocity of the satellite is 9430 m/s

f) The escape velocity of the rocket is 11,191 m/s

Explanation:

The acceleration of gravity for an object near a planet is given by

$g=\frac{GM}{(R+h)^2}$

where

G is the gravitational constant

M is the mass of the planet

R is the radius of the planet

h is the height above the surface

In this problem,

$M=5.98\cdot 10^{24} kg$ (mass of the Earth)

$R=6.37\cdot 10^6 m$ (Earth's radius)

$h=2.60\cdot 10^6 m$ (altitude of the satellite)

Substituting,

$g=\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6 + 2.60\cdot 10^6)^2}=4.96 m/s^2$

The critical velocity for a satellite orbiting around a planet is given by

$v=\sqrt{\frac{GM}{R+h}}$

where we have again:

$M=5.98\cdot 10^{24} kg$ (mass of the Earth)

$R=6.37\cdot 10^6 m$ (Earth's radius)

$h=2.60\cdot 10^6 m$ (altitude of the satellite)

Substituting,

$v=\sqrt{\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6 + 2.60\cdot 10^6)}}=6668 m/s$

Converting into km/h,

$v=6668 m/s \cdot \frac{3600 s/h}{1000 m/km}=24,006 km/h$

The period of the orbit is given by the circumference of the orbit divided by the velocity:

$T=\frac{2\pi (R+h)}{v}$

where

$R=6.37\cdot 10^6 m$

$h=2.60\cdot 10^6 m$

v = 6668 m/s

Substituting,

$T=\frac{2\pi (6.37\cdot 10^6 + 2.60\cdot 10^6)}{6668}=8452 s$

One day consists of:

$t = 24 \frac{hours}{day} \cdot 60 \frac{min}{hours} \cdot 60 \frac{s}{min}=86400 s$

While the period of the orbit is

T = 8452 s

So, the number of orbits completed by the satellite in one day is

$n=\frac{t}{T}=\frac{86400}{8452}=10.2$

The escape velocity for an object in the gravitational field of a planet is given by

$v=\sqrt{\frac{2GM}{R+h}}$

where here we have:

$M=5.98\cdot 10^{24} kg$

$R=6.37\cdot 10^6 m$

$h=2.60\cdot 10^6 m$

Substituting, we find

$v=\sqrt{\frac{2(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6 + 2.60\cdot 10^6)}}=9430 m/s$

We can apply again the formula to find the escape velocity for the rocket:

$v=\sqrt{\frac{2GM}{R+h}}$

Where this time we have:

$M=5.98\cdot 10^{24} kg$

$R=6.37\cdot 10^6 m$

$h=0$ , because the rocket is located at the Earth's surface, so its altitude is zero.

And substituting,

$v=\sqrt{\frac{2(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6)}}=11,191 m/s$

Learn more about gravitational force:

brainly.com/question/1724648

brainly.com/question/12785992

#LearnwithBrainly

6 0

3 years ago

Calculate the acceleration of a 270000-kg jumbo jet just before takeoff when the thrust on the aircraft is 160000 N .

Radda [10]

Answer:

<h3>The answer is 0.59 m/s²</h3>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

$a = \frac{f}{m} \\$

f is the force

m is the mass

From the question we have

$a = \frac{160000}{270000} = \frac{16}{27} \\ = 0.592592...$

We have the final answer as

Hope this helps you

7 0

3 years ago

A(n) ________ is a specialized bursa that wraps around a tendon to reduce friction.

Kisachek [45]

Tendon Sheath - is a specialized bursa that wraps around a tendon to reduce friction.

<h3>What is Tendon Sheath ?</h3>

Tendon Sheath is a thin layer of tissue, surrounds each tendon in our body. The tendon sheath can also be called synovial lining or fibrous sheath. Tendon sheaths help to protect tendons from abrasive damage as they move.

Connection between Bursa and Tendon Sheath : Bursae are small fluid-filled sacs that can lie under a tendon, cushioning the tendon and protecting it from the injury. Bursae also provides an extra cushioning to adjacent structures that otherwise might rub against each other, which will cause wear and tear ( example, between a bone and a ligament ) .

So, lastly we can say that Tendon Sheath is the specialized bursa that wraps around a tendon to reduce friction.

To know more about Tendon Sheath please click here : brainly.com/question/17087116

#SPJ4

8 0

10 months ago

23 grams of sodium reacted with 35.5 grams of chlorine. Calculate the mass of the sodium chloride compound formed

xxTIMURxx [149]

According to law of conservation of mass within a reaction,
The mass of the compound formed is (23+35.5) grams means 58.5 grams of sodium chloride[NaCl] will be formed.

3 0

3 years ago

Read 2 more answers