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2 years ago

What will be net force be if net forces are unbalanced?

1 answer:

6 0

Answer: If the forces on an object are balanced, the net force is zero. If the forces are unbalanced forces, the effects don't cancel each other. Any time the forces acting on an object are unbalanced, the net force is not zero, and the motion of the object changes.

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Select the correct answer.

skelet666 [1.2K]

Answer:

B is the best answer for the question

6 0

3 years ago

Read 2 more answers

A solid cube of side 5cm has a mass of 250g. What is its density in kgm ??

Mariulka [41]

Answer:

5kgm

Explanation:

convert cm to m and g to kg

250/1000=0.25kg

5/1000=0.05m

then find the density

density=mass/volume

=0.25kg/0.05m

=5kgm

7 0

3 years ago

A 1.5 kg cart is attached to a spring with spring constant of 5 N/m. The cart & spring is pulled to stretch the spring by 3

vaieri [72.5K]

22.5 J

Explanation:

Given:

x = 3 m

$k = 5\:\text{N/m}$

The spring potential energy $PE_s$ is

$PE_s = \frac{1}{2}kx^2 = \frac{1}{2}(5\:\text{N/m})(3\:\text{m})^2$

$\:\:\:\:\:\:\:=22.5\:\text{J}$

3 0

3 years ago

4: Question 2 - 10.0 pts possible

Orlov [11]

Answer:

5. All of the statements are true; non is false

Explanation:

8 0

3 years ago

A tube 1.20 m long is closed at one end. A stretched wire is placed near the open end. The wire is 0.350 m long and has a mass o

Ksju [112]

Answer:

71.4583 Hz

67.9064 N

Explanation:

L = Length of tube = 1.2 m

l = Length of wire = 0.35 m

m = Mass of wire = 9.5 g

v = Speed of sound in air = 343 m/s

The fundamental frequency of the tube (closed at one end) is given by

$f=\frac{v}{4L}\\\Rightarrow f=\frac{343}{4\times 1.2}\\\Rightarrow f=71.4583\ Hz$

The fundamental frequency of the wire and tube is equal so he fundamental frequency of the wire is 71.4583 Hz

The linear density of the wire is

$\mu=\frac{m}{l}\\\Rightarrow \mu=\frac{9.5\times 10^{-3}}{0.35}\\\Rightarrow \mu=0.02714\ kg/m$

The fundamental frequency of the wire is given by

$f=\frac{1}{2l}\sqrt{\frac{T}{\mu}}\\\Rightarrow f^2=\frac{1}{4l^2}\frac{T}{\mu}\\\Rightarrow T=f^2\mu 4l^2\\\Rightarrow T=71.4583^2\times 0.02714\times 4\times 0.35^2\\\Rightarrow T=67.9064\ N$

The tension in the wire is 67.9064 N

7 0

3 years ago