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wel
3 years ago
6

Lisa traveled a total of 440 miles. Her trip took 8 hours. What was her average speed

Physics
2 answers:
zaharov [31]3 years ago
8 0

Her average speed was 55MPH. Hope this helps!


VashaNatasha [74]3 years ago
7 0

not 100% sure but i think her average speed is 55


sorry if i'm wrong

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A car is traveling north. can its acceleration vector ever Point South? explain​
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Yes.

The acceleration vector WILL point south when the car is slowing down while traveling north

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Apply a force of 50N to the left describe the motion of the box
GenaCL600 [577]
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(force =mass×acceleration)

So the box has a constant acceleration and a changing velocity.
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A black stone was discovered to have magnetic properties.
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3 years ago
a 2.0 kg ball is dropped from a height of 20 m onto a soft surface and rebounds to a height of 5.0 m . what is the magnitude of
Yuki888 [10]

Based on the data provided, the impulse of the floor on the ball is 59.4 Ns.

<h3>What is the impulse of the floor on the ball?</h3>

Using the equation of motion to determine the velocity at the end of the fall

  • v^2 = u^2 + 2gh

Where v is velocity at the end of fall

u is initial velocity = 0

g is acceleration due to gravity = 9.81 m/s^2

h is height = 20

  • Taking downward velocity as negative and up as positive

v^2 = 0 + 2 (9.81)(20)

v^2 = 392.4

v = - 19.8 m/s

The velocity, v after bouncing is calculated also:

u = 0

g = 9.81 m/s^2

h = 5.0 m

v^2 = 0 + 2(9.81)(5)

v^2 = 98.1

v = 9.904 m/s

  • Impulse = change in momentum
  • Impulse = m(v- u)

Impulse = 2.0 × (9.9 -(-19.8)

Impulse = 59.4 Ns

Therefore, the impulse of the floor on the ball is 59.4 Ns.

Learn more about impulse at: brainly.com/question/904448

5 0
2 years ago
A 2 kg mass connected to a spring with spring constant k = 10 N/m oscillates in simple harmonic motion with an amplitude of A =
insens350 [35]

Answer:

Kinetic energy at 0.05 m is 0.037 J

Explanation:

Given:

Mass, m = 2 kg

Spring constant, k = 10 N/m

Amplitude, A = 0.1 m

Angular frequency, ω = √k/m

Substitute the suitable values in the above equation.

\omega = \sqrt\frac{10}{2}

ω = 2.24 s⁻¹

Simple harmonic equation is represent by the equation:

x = A cos ωt

Substitute 0.05 m for x, 0.1 m for A and 2.24 s⁻¹ for ω in the above equation.

0.05 = 0.1\cos(2.24t)

t = 0.47 s

Kinetic energy at x = 0.05 is determine by the relation:

E=\frac{1}{2} kA^{2}\sin^{2}(\omega t)

Substitute the suitable values in the above equation.

E=\frac{1}{2} \times 10 \times 0.1^{2} \sin^{2}(2.24\times0.47)

E = 0.037 J

5 0
2 years ago
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