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rosijanka [135]
2 years ago
7

Sebuah benda dijatuhkan bebas dari ketinggian 200 m jika grafitasi setempat 10 m/s maka hitunglah kecepatan dan ketinggian benda

saat ek=4ep
Physics
1 answer:
Pie2 years ago
7 0
Please post in English so i or someone else can help you.
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A horizontal spring with spring constant 750 N/m is attached to a wall. An athlete presses against the free end of the spring, c
ANTONII [103]

Answer:

37.5 N Hard

Explanation:

Hook's law: The force applied to an elastic material is directly proportional to the extension provided the elastic limit of the material is not exceeded.

Using the expression for hook's law,

F = ke.............. Equation 1

F = Force of the athlete, k = force constant of the spring, e = extension/compression of the spring.

Given: k = 750 N/m, e = 5.0 cm = 0.05 m

Substitute into equation 1

F = 750(0.05)

F = 37.5 N

Hence the athlete is pushing 37.5 N hard

4 0
2 years ago
Read 2 more answers
The drawing shows three particles far away from any other objects and located on a straight line. The masses of these particles
belka [17]

Answer:

F_a=5.67\times 10^{-5}\ N

<u />F_b=3.49\times 10^{-5}\ N

F_c=9.16\times 10^{-5}\ N

Explanation:

Given:

  • mass of particle A, m_a=363\ kg
  • mass of particle B, m_b=517\ kg
  • mass of particle C, m_c=154\ kg
  • All the three particles lie on a straight line.
  • Distance between particle A and B, x_{ab}=0.5\ m
  • Distance between particle B and C, x_{bc}=0.25\ m

Since the gravitational force is attractive in nature it will add up when enacted from the same direction.

<u>Force on particle A due to particles B & C:</u>

F_a=G. \frac{m_a.m_b}{x_{ab}^2} +G. \frac{m_a.m_c}{(x_{ab}+x_{bc})^2}

F_a=6.67\times 10^{-11}\times (\frac{363\times 517}{0.5^2}+\frac{363\times 154}{(0.5+0.25)^2})

F_a=5.67\times 10^{-5}\ N

<u>Force on particle C due to particles B & A:</u>

<u />F_c=G.\frac{m_c.m_b}{x_{bc}^2} +G.\frac{m_c.m_a}{(x_{ab}+x_{bc})^2}<u />

F_c=6.67\times 10^{-11}\times (\frac{154\times 517}{0.25^2}+\frac{154\times 363}{(0.25+0.5)^2} )

F_c=9.16\times 10^{-5}\ N

<u>Force on particle B due to particles C & A:</u>

<u />F_b=G.\frac{m_b.m_c}{x_{bc}^2} -G.\frac{m_b.m_a}{x_{ab}^2}<u />

<u />F_b=6.67\times 10^{-11}\times (\frac{517\times 154}{0.25^2}-\frac{517\times 363}{0.5^2}  )<u />

<u />F_b=3.49\times 10^{-5}\ N<u />

3 0
2 years ago
What is the mass of an object weighing 63 N on Earth?
avanturin [10]
Weight expressed in Newtons is expressed in the equation whereby Weight= the mass of an object * the force of gravity. The force of gravity on earth is a constant 9.8 meters per second squared. Therefore if weight (w) = 63 N and the force of gravity is 63 N then the mass must equal 6.43 kg. Because the equation for weight is w=mg so 63 N (w) = m * 9.8 m/s^2. 
3 0
2 years ago
Which statements describe vectors? Check all that apply. -Vectors have magnitude only. -Vectors have direction only. -Vectors ha
Natali [406]

Answer:

Vectors have both magnitude and direction

Explanation:

Vectors show how strong the force in because the bigger the arrow, the stronger the force.  Also, it obviously shows direction because its an arrow.

6 0
2 years ago
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For a long ideal solenoid having a circular cross-section, the magnetic field strength within the solenoid is given by the equat
andrezito [222]

Answer:

Radius of the solenoid is 0.93 meters.

Explanation:

It is given that,

The magnetic field strength within the solenoid is given by the equation,

B(t)=5t\ T, t is time in seconds

\dfrac{dB}{dt}=5\ T

The induced electric field outside the solenoid is 1.1 V/m at a distance of 2.0 m from the axis of the solenoid, x = 2 m

The electric field due to changing magnetic field is given by :

E(2\pi x)=\dfrac{d\phi}{dt}

x is the distance from the axis of the solenoid

E(2\pi x)=\pi r^2\dfrac{dB}{dt}, r is the radius of the solenoid

r^2=\dfrac{2xE}{(dE/dt)}

r^2=\dfrac{2\times 2\times 1.1}{(5)}

r = 0.93 meters

So, the radius of the solenoid is 0.93 meters. Hence, this is the required solution.

4 0
2 years ago
Read 2 more answers
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