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2 years ago

11

A uniform thin circular rubber band of mass M and spring constant k has an original radius R?

1 answer:

Gemiola [76]2 years ago

6 0

Answer: Yes

Explanation:

A ruber band can be described by his mas, M, the spring constant K, and his rest radius, wich is the radius of the circular rubber band when there is no force applied to it, so yes, a uniform thin circular rubber band of mass M and spring constant k has an original radius R

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: The truck is to be towed using two ropes. Determine the magnitudes of forces FA and FB acting on each rope in order to develop

Sholpan [36]

Answer:

Fa=774 N

Fb=346 N

Explanation:

We will solve this problem by equating forces on each axis.

On x-axis let forces in positive x-direction be positive and forces in negative x-direction be negative
On y-axis let forces in positive y-direction be positive and forces in negative y-direction be negative

While towing we know that car is mot moving in y-direction so net force in y-axis must be zero

⇒∑Fy=0

⇒ $Fa*sin(50)-Fb*sin(20)=0$

⇒ $Fa*sin(50)=Fb*sin(20)$

⇒ $Fa=2.24Fb$

Given that resultant force on car is 950N in positive x-direction

⇒∑Fx=950

⇒ $Fa*cos(20)+Fb*cos(50)=950$

⇒ $2.24*Fb*cos(20)+Fb(50)=950$

⇒ $Fb*(2.24*cos(20)+cos(50))=950$

⇒ $Fb=\frac{950}{2.24*cos(20)+cos(50)}$

⇒ $Fb=\frac{950}{2.24*0.94+0.64}$

⇒ $Fb=\frac{950}{2.75}=345.5$

⇒ $Fa=2.24*Fb$

$=2.24*345.5$

$=773.93$

Therefore approximately, Fa=774 N and Fb=346 N

5 0

2 years ago

A marble is dropped straight down from a distance h above the floor.

Mrac [35]

Fm=Fe and am>ae
Hopefully this helps

7 0

2 years ago

When were Earth’s landmasses first recognizable as the continents we know today? 10 million years ago 135 million years ago 180

Bess [88]

Answer:

b

Explanation:

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2 years ago

Read 2 more answers

A wire carrying a current of 10 A and 2 m in length is placed in a field of flux density 0.15 T. What’s the force on the wire if

saul85 [17]

Explanation:

I = 10A

l = 2m

B = 0.15T

F = ?

a) ¶ = 90

F = BILsin¶

F = 0.15×10×2×sin90

F = 3N

b) ¶ = 45 degree

F = BILsin¶

F = 0.15×10×2×sin45

F = 2.12N

c) ¶ = 0 degree

F = BILsin¶

F = 0.15×10×2×sin0

F = 0

Goodluck

7 0

2 years ago

How much energy is needed to change the speed of a 1600 kg sport utility vehicle from 15.0 m/s to 40.0 m/s?

podryga [215]

Answer:

Energy needed = 1100 kJ

Explanation:

Energy needed = Change in kinetic energy

Initial velocity = 15 m/s

Mass, m = 1600 kg

$\texttt{Initial kinetic energy = }\frac{1}{2}mv^2=0.5\times 1600\times 15^2=180000J$

Final velocity = 40 m/s

$\texttt{Final kinetic energy = }\frac{1}{2}mv^2=0.5\times 1600\times 40^2=1280000J$

Energy needed = Change in kinetic energy = 1280000-180000 = 1100000J

Energy needed = 1100 kJ

4 0

2 years ago