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liberstina [14]
3 years ago
10

Two children stretch a jump rope between them and send wave pulses back and forth on it. The rope is 2.6 m long, its mass is 0.5

kg, and the force exerted by the children is 44 N. What is the speed of the waves on the rope
Physics
1 answer:
Irina-Kira [14]3 years ago
6 0

Answer:

15.13 m/s

Explanation:

The wave speed of the stretched rope can be calculated using the following formula

v = \sqrt{\frac{F_T}{\mu}}

where F_T = 44N is the tension on the rope and \mu = m/L = 0.5 / 2.6 = 0.1923 kg/m is the density of the rope per unit length

v = \sqrt{\frac{44}{0.1923}} = \sqrt{228.8} = 15.13 m/s

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Ya'll please I need help only on #5 Its confusing<br><br> (Grade 10 Physics, Potential Energy)
skad [1K]

Answer:

Potential energy = work done on the spring

= Force x displacement

= 15 N x 0.025 m

= 0.375 J

3 0
3 years ago
⦁ A 68 kg crate is dragged across a floor by pulling on a rope attached to the crate and inclined 15° above the horizontal. (a)
GREYUIT [131]

Answer:

303.29N and 1.44m/s^2

Explanation:

Make sure to label each vector with none, mg, fk, a, FN or T

Given

Mass m = 68.0 kg

Angle θ = 15.0°

g = 9.8m/s^2

Coefficient of static friction μs = 0.50

Coefficient of kinetic friction μk =0.35

Solution

Vertically

N = mg - Fsinθ

Horizontally

Fs = F cos θ

μsN = Fcos θ

μs( mg- Fsinθ) = Fcos θ

μsmg - μsFsinθ = Fcos θ

μsmg = Fcos θ + μsFsinθ

F = μsmg/ cos θ + μs sinθ

F = 0.5×68×9.8/cos 15×0.5×sin15

F = 332.2/0.9659+0.5×0.2588

F =332.2/1.0953

F = 303.29N

Fnet = F - Fk

ma = F - μkN

a = F - μk( mg - Fsinθ)

a = 303.29 - 0.35(68.0 * 9.8- 303.29*sin15)/68.0

303.29-0.35( 666.4 - 303.29*0.2588)/68.0

303.29-0.35(666.4-78.491)/68.0

303.29-0.35(587.90)/68.0

(303.29-205.45)/68.0

97.83/68.0

a = 1.438m/s^2

a = 1.44m/s^2

7 0
3 years ago
A 140 g baseball is moving horizontally to the right at 35 mis when it is hit by the bat. the ball fl ies off to the le ft at 55
Anit [1.1K]

Answer:

J = 12.32kg*m/s

Explanation:

Assumptions: I'm assuming mis is m/s

Given: The baseball's mass is 140g, so convert to kg, 0.140kg. (Good rule of thumb, in physics convert grams to kilograms). It is initially traveling 35 m/s to the right.

When it hits the bat, it flies left with a velocity of 55 m/s at an angle of 25°. To reiterate:

mass = 0.140kg, Initial: 35m/s, Final: 55m/s at an angle of 25°

For this problem, we have to use the impulse equation, but before that lets break the velocity into components (It will be apparent towards the end):

The initial velocity is moving only in the horizontal direction, so:

v_{0x} = 35 m/s

The final velocity has an x and a y component:

v_{fx} = 55cos(25) = 49.84692829m/s\\v_{fy} = 55sin(25) = 23.2440044m/s

Now the equation for impulse is (dp is Δp, which is difference in momentum; dv is Δv, the difference in velocity; J is impulse):

J = dp= m*dv

To get Δv, we have to find the difference of velocity, that is why we broke it into components. I'm going to define right as positive and left as negative. After that, we find the velocity vector:

dv_{x} = (35-(-49.84692829)) = 84.84692829 m/s\\dv_{y} = (0-(23.2440044)) = -23.2440044 m/s\\dv = \sqrt{(84.84692829)^2+(-23.2440044)^2} = 87.97320604m/s

Finally, substitute into the equation

J = m*dv\\J = 0.140kg * 87.97320604m/s\\J = 12.31624885kg*m/s\\J = 12.32 kg*m/s

J = 12.32kg*m/s

8 0
3 years ago
An object is placed at 0 on a number line. It moves 3 units to the right, then 4 units to the left, and then 6 units to the righ
vaieri [72.5K]
The answer is +5 units.
8 0
3 years ago
A ball of mass 0.50 kg is fired with velocity 120 m/s into the barrel of a spring gun of mass 1.6 kg initially at rest on a fric
Gre4nikov [31]

Answer:

The fraction of the ball's kinetic energy stored in the spring is 3.6KJoules

Explanation:

Given

Mass of ball =0.5kg

Velocity of ball =120m/s

The kinetic energy stored by the ball is expressed as

K.E=1/2(m*v²)

Substituting our data into the expression we have

K. E=(0.5*120²)/2

K. E=7200/2

K.E=3,600 JOULE

K.E=3.6KiloJoules

4 0
3 years ago
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