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krek1111 [17]
3 years ago
8

An object travels a distance d with acceleration a over a period of time t according to the equation: d = at² After 2.3 seconds

with an acceleration of 9.8 m/s2, how far does an object travel (rounding the answer to two significant figures)?​

Physics
1 answer:
matrenka [14]3 years ago
5 0

Answer:

A 26m

Explanation:

firstly the answer will not be part c as distance unit will be meters (m)

next, substitute the values of a and t as shown

\frac{1}{2}  (9.8)(2.3)^2 = 25.921

the answer rounded off to 2 significant figures will be hence 26m

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Question 24 1 pts Find the voltage in an extension cord having a 0.0600 22 resistance and through which a 5.00A current is flowi
Neporo4naja [7]

The voltage in the extension cord is 30 V.

The problem above can be solved using ohm's law

⇒ Formula:

V = IR.................. Equation 1

⇒ Where:

  • V = Voltage in the extension cord
  • I = Current flowing through the extension cord
  • R = Resistance of the extension cord.

From the question, I think there was a slight error in the value of the current given it suppose to be 500 A, and not 5.00 A

⇒ Given:

  • I = 500 A
  • R = 0.06 ohms

⇒ Substitute these values into equation 1

  • V = 500(0.06)
  • V = 30 V

Hence the voltage in the extension cord is 30 V

Learn more about voltage here: brainly.com/question/4429782

6 0
2 years ago
The masses are m1 = m, with initial velocity 2v0, and m2 = 7.4m, with initial velocity v0. Due to the collision, they stick toge
lesya [120]

Answer:

Loss, \Delta E=-10.63\ J

Explanation:

Given that,

Mass of particle 1, m_1=m =0.66\ kg

Mass of particle 2, m_2=7.4m =4.884\ kg

Speed of particle 1, v_1=2v_o=2\times 6=12\ m/s

Speed of particle 2, v_2=v_o=6\ m/s

To find,

The magnitude of the loss in kinetic energy after the collision.

Solve,

Two particles stick together in case of inelastic collision. Due to this, some of the kinetic energy gets lost.

Applying the conservation of momentum to find the speed of two particles after the collision.

m_1v_1+m_2v_2=(m_1+m_2)V

V=\dfrac{m_1v_1+m_2v_2}{(m_1+m_2)}

V=\dfrac{0.66\times 12+4.884\times 6}{(0.66+4.884)}

V = 6.71 m/s

Initial kinetic energy before the collision,

K_i=\dfrac{1}{2}(m_1v_1^2+m_2v_2^2)

K_i=\dfrac{1}{2}(0.66\times 12^2+4.884\times 6^2)

K_i=135.43\ J

Final kinetic energy after the collision,

K_f=\dfrac{1}{2}(m_1+m_2)V^2

K_f=\dfrac{1}{2}(0.66+4.884)\times 6.71^2

K_f=124.80\ J

Lost in kinetic energy,

\Delta K=K_f-K_i

\Delta K=124.80-135.43

\Delta E=-10.63\ J

Therefore, the magnitude of the loss in kinetic energy after the collision is 10.63 Joules.

7 0
3 years ago
Light always travels in a straight line
Licemer1 [7]

 Light travels in straight lines. Once a light has been produced, it will keep moving in a straight line until it hits something else. Shadows are evidence of light traveling in straight lines. An object blocks light so that it can’t reach the surface where we see the shadow.

8 0
3 years ago
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Why do some athletes get injuries before and after the game?<br>​
natita [175]

Answer:

they don't strech so they tear a muscle when they perform

Explanation:

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3 years ago
Question 18 (2 points)
fomenos

Answer:

A.Moving electric charges (electrons in a circuit) creates a magnetic field and

a magnetic field can cause an electric charge to move (electricity).

Explanation:

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3 years ago
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