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4vir4ik [10]
4 years ago
14

Josh starts at a position of x=0 meters at the left side of the front of the classroom. Over the course of the class he walks ba

ck and forth in front of the class 15 times, 6 meters to the right, and then 6 meters back to the left. What is the total distance (in meters) Josh walks? (Enter just the number, no units.)
Physics
1 answer:
Aneli [31]4 years ago
3 0
Answer:
180 meters

Explanation:
Josh walks 6 meters to the right and the 6 meters back. This means that in one time, Josh walks a total distance of 6+6 = 12 meters
Now, we are given that Josh walks this same distance 15 times.
Therefore, we will multiply this distance by 15 to get the total distance that he walks as follows:
Total distance = 12*15 = 180 meters

Based on the above calculations, Josh walks 180 meters over the course of the class

Hope this helps :)
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If the pendulum took longer to complete one oscillation, how would the graph change?
pickupchik [31]

We don't know what kind of graph it is.

For example, it might be a graph of the pendulum's distance from center,

angle from center, speed, acceleration, total distance swung since it was

started, mass, weight, temperature, etc.


If the graph shows the pendulum's distance from center, angle from center,

speed, or acceleration, then the graph will look like a wave, with the period

of the wave being the period of the pendulum's oscillation. If the pendulum

took longer to complete one oscillation, that means its PERIOD increased,

and the distance between the peaks of the graph would be longer.


If it was a graph of total distance the pendulum swung since it was started,

the graph wouldn't look like a wave, just a steadily rising wiggle line. If the

pendulum took longer to complete one oscillation, the wiggles in the line

would be farther apart, and the average slope of any large section of the

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7 0
3 years ago
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Which of the following is not an example of a polymer?
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Concrete is not a polymer which Nylon, and Kevlar are
8 0
4 years ago
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Two dogs fight over a bone. The larger of the two pulls on the bone to the right with a force of 42 N. The smaller one pulls to
tamaranim1 [39]

Answer:

Explanation:

A

35 N  Small Dog <=======BONE=========> Bigger Dog 42 N

B

Fnet = Large Dog - small dog The forces are subtracted because they are acting in opposite Directions.

Fnet = 42 - 35

Fnet = 7 N

C

m = 2.5 kg

F = 7 N

a = ?

F = m * a

7 = 2.5 a

a = 7 / 2.5

a = 2.8 m/s^2

4 0
2 years ago
How do you calculate the power in a circuit or device if given the current and the resistance?
Drupady [299]

Answer:

you use the Ohms law so to find the voltage you would need to multiply the current by the resistance which gives you the power.

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5 0
3 years ago
A 10-kg package drops from chute into a 25-kg cart with a velocity of 3 m/s. The cart is initially at rest and can roll freely w
amid [387]

Answer:

(a) the final velocity of the cart is 0.857 m/s

(b) the impulse experienced by the package is 21.43 kg.m/s

(c) the fraction of the initial energy lost is 0.71

Explanation:

Given;

mass of the package, m₁ = 10 kg

mass of the cart, m₂ = 25 kg

initial velocity of the package, u₁ = 3 m/s

initial velocity of the cart, u₂ = 0

let the final velocity of the cart = v

(a) Apply the principle of conservation of linear momentum to determine common final velocity for ineleastic collision;

m₁u₁  + m₂u₂ = v(m₁  +  m₂)

10 x 3   + 25 x 0   = v(10  +  25)

30  = 35v

v = 30 / 35

v = 0.857 m/s

(b) the impulse experienced by the package;

The impulse = change in momentum of the package

J = ΔP = m₁v - m₁u₁

J = m₁(v - u₁)

J = 10(0.857 - 3)

J = -21.43 kg.m/s

the magnitude of the impulse experienced by the package = 21.43 kg.m/s

(c)

the initial kinetic energy of the package is calculated as;

K.E_i = \frac{1}{2} mu_1^2\\\\K.E_i = \frac{1}{2} \times 10 \times (3)^2\\\\K.E_i = 45 \ J\\\\

the final kinetic energy of the package;

K.E_f = \frac{1}{2} (m_1 + m_2)v^2\\\\K.E_f = \frac{1}{2} \times (10 + 25) \times 0.857^2\\\\K.E_f = 12.85 \ J

the fraction of the initial energy lost;

= \frac{\Delta K.E}{K.E_i} = \frac{45 -12.85}{45} = 0.71

7 0
3 years ago
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