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Bogdan [553]
3 years ago
6

True or false?

Physics
2 answers:
Ghella [55]3 years ago
8 0

Answer:

true

Explanation:

sukhopar [10]3 years ago
3 0
I believe the answer is true.
You might be interested in
The most common isotope of hydrogen contains a proton and an electron 'separated by about -11-27 5.0 x 10 m. The mass of proton
Brrunno [24]

Answer:

A)   F_g = 4.05 10⁻⁴⁷ N, B)   F_e = 9.2 10⁻⁸N, C)    \frac{F_e}{F_g} = 2.3 10³⁹

Explanation:

A) It is asked to find the force of attraction due to the masses of the particles

Let's use the law of universal attraction

            F = G \frac{m_1m_2}{r^2}

let's calculate

            F = 6.67 \ 10^{-11} \ \frac{9.1 \ 10^{-31} \ 1.67 \ 10 ^{-27} }{(5 \ 10^{-11})^2 }

            F_g = 4.05 10⁻⁴⁷ N

B) in this part it is asked to calculate the electric force

Let's use Coulomb's law

            F = k \  \frac{q_1q_2}{r^2}

let's calculate

            F = 9 \ 10^9 \  \frac{(1.6 \ 10^{-19} )^2}{(5 \ 10^{-11})^2}

             F_e = 9.2 10⁻⁸N

C) It is asked to find the relationship between these forces

        \frac{F_e}{F_g} = \frac{9.2 \ 10^{-8} }{4.05 \ 10^{-47} }

        = 2.3 10³⁹

therefore the electric force is much greater than the gravitational force

4 0
3 years ago
A 7600 kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.35 m/s2 and feels no appreci
ollegr [7]

Answer:

a) The rocket reaches a maximum height of 737.577 meters.

b) The rocket will come crashing down approximately 17.655 seconds after engine failure.

Explanation:

a) Let suppose that rocket accelerates uniformly in the two stages. First, rocket is accelerates due to engine and second, it is decelerated by gravity.

1st Stage - Engine

Given that initial velocity, acceleration and travelled distance are known, we determine final velocity (v), measured in meters per second, by using this kinematic equation:

v = \sqrt{v_{o}^{2} +2\cdot a\cdot \Delta s} (1)

Where:

a - Acceleration, measured in meters per square second.

\Delta s - Travelled distance, measured in meters.

v_{o} - Initial velocity, measured in meters per second.

If we know that v_{o} = 0\,\frac{m}{s}, a = 2.35\,\frac{m}{s^{2}} and \Delta s = 595\,m, the final velocity of the rocket is:

v = \sqrt{\left(0\,\frac{m}{s} \right)^{2}+2\cdot \left(2.35\,\frac{m}{s^{2}} \right)\cdot (595\,m)}

v\approx 52.882\,\frac{m}{s}

The time associated with this launch (t), measured in seconds, is:

t = \frac{v-v_{o}}{a}

t = \frac{52.882\,\frac{m}{s}-0\,\frac{m}{s}}{2.35\,\frac{m}{s} }

t = 22.503\,s

2nd Stage - Gravity

The rocket reaches its maximum height when final velocity is zero:

v^{2} = v_{o}^{2} + 2\cdot a\cdot (s-s_{o}) (2)

Where:

v_{o} - Initial speed, measured in meters per second.

v - Final speed, measured in meters per second.

a - Gravitational acceleration, measured in meters per square second.

s_{o} - Initial height, measured in meters.

s - Final height, measured in meters.

If we know that v_{o} = 52.882\,\frac{m}{s}, v = 0\,\frac{m}{s}, a = -9.807\,\frac{m}{s^{2}} and s_{o} = 595\,m, then the maximum height reached by the rocket is:

v^{2} -v_{o}^{2} = 2\cdot a\cdot (s-s_{o})

s-s_{o} = \frac{v^{2}-v_{o}^{2}}{2\cdot a}

s = s_{o} + \frac{v^{2}-v_{o}^{2}}{2\cdot a}

s = 595\,m + \frac{\left(0\,\frac{m}{s} \right)^{2}-\left(52.882\,\frac{m}{s} \right)^{2}}{2\cdot \left(-9.807\,\frac{m}{s^{2}} \right)}

s = 737.577\,m

The rocket reaches a maximum height of 737.577 meters.

b) The time needed for the rocket to crash down to the launch pad is determined by the following kinematic equation:

s = s_{o} + v_{o}\cdot t +\frac{1}{2}\cdot a \cdot t^{2} (2)

Where:

s_{o} - Initial height, measured in meters.

s - Final height, measured in meters.

v_{o} - Initial speed, measured in meters per second.

a - Gravitational acceleration, measured in meters per square second.

t - Time, measured in seconds.

If we know that s_{o} = 595\,m, v_{o} = 52.882\,\frac{m}{s}, s = 0\,m and a = -9.807\,\frac{m}{s^{2}}, then the time needed by the rocket is:

0\,m = 595\,m + \left(52.882\,\frac{m}{s} \right)\cdot t + \frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right)\cdot t^{2}

-4.904\cdot t^{2}+52.882\cdot t +595 = 0

Then, we solve this polynomial by Quadratic Formula:

t_{1}\approx 17.655\,s, t_{2} \approx -6.872\,s

Only the first root is solution that is physically reasonable. Hence, the rocket will come crashing down approximately 17.655 seconds after engine failure.

7 0
3 years ago
Someone please help
saul85 [17]

Based on the attached image:

  • The name of the longitude line that passes through point A is the International Date Line
  • The longitude 180° is experiencing solar noon because the rays of the sun are parallel to it.
  • The longitude for 6 pm is 90° W, 12 midnight is 0°, and 6 am is 90° E
  • Longitude 120° is B
  • Solar time at Point B is 4 pm
  • The location will correspond to any point on the same latitude as A

<h3>What are lines of longitude?</h3>

Lines of longitude are imaginary lines which run along the earth from the North pole. to the South pole.

Longitude lines divide the earth into semi-circles.

Longitude lines are known as meridians and each meridian measures one arc degree of longitude.

Considering the attached image:

  • The name of the longitude line that passes through point A is the International Date Line
  • The longitude 180° is experiencing solar noon because the rays of the sun are parallel to it.
  • The longitude for 6 pm is 90° W, 12 midnight is 0°, and 6 am is 90° E
  • Longitude 120° is B
  • Solar time at Point B is 4 pm
  • the location will correspond to any point on the same latitude as A

In conclusion, longitude lines are imaginary lines and run from North to South on the earth.

Learn more about lines of longitude at: brainly.com/question/1939015

#SPJ1

8 0
2 years ago
If a roller coaster cart, with a mass of 100 kg, traveled this coaster, how much kinetic energy would it have at point 'E'?
zzz [600]

Answer:

Explanation:

Assuming no friction between the roller coaster car and the hill, and neglecting air resistance, the kinetic energy the roller coaster car would have at the bottom of the hill would be equal to its gravitational potential energy at the top of the hill, by conservation of energy.

8 0
3 years ago
In order to attain orbit around earth, the ATLAS-V rocket must accelerate up to a
IrinaK [193]
A) average acceleration = final velocity - initial velocity / time

= 7700 - 0 / 11

= 700ms^-2


B) force = mass x acceleration

= (3.05 x 105) x 700

= 320.25 x 700

= 224,175N
7 0
3 years ago
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