If two opposing forces are equal, then the net force is 0 N.<br><br> true or false?

Which statement correctly describes magnetic field lines?

skelet666 [1.2K]

The answer is A. the fields lines never cross, if you bring another magnet near it, the lines work just compress

4 0

2 years ago

A rope is tied to a tree 4.5 feet from the ground and then run through a pulley hooked to a vehicle 33 feet from the tree. If a

Angelina_Jolie [31]

Answer:

The vehicle displacement is 9.90 feet.

Explanation:

Given that,

Height of tree = 4.5 feet

Distance = 33 feet

According to figure,

We need to calculate the value of l

Using Pythagorean theorem

$l=\sqrt{(33)^2+(4.5)^2}$

We need to calculate the vehicle displacement

Using horizontal component

Vehicle displacement =horizontal component of pulled rope

$Vehicle\ displacement= d\cos\theta$

Where, $\theta$ is angle between rope and ground

d = pulled length of rope

$Vehicle\ displacement=10\times\dfrac{33}{\sqrt{(33)^2+(4.5)^2}}$

$Vehicle\ displacement=9.90\ feet$

Hence, The vehicle displacement is 9.90 feet.

3 0

2 years ago

Two go-carts, A and B, race each other around a 1.0 track. Go-cart A travels at a constant speed of 20.0 /. Go-cart B accelerate

maria [59]

Complete Question

Q. Two go-carts, A and B, race each other around a 1.0km track. Go-cart A travels at a constant speed of 20m/s. Go-cart B accelerates uniformly from rest at a rate of 0.333m/s^2. Which go-cart wins the race and by how much time?

Answer:

Go-cart A is faster

Explanation:

From the question we are told that

The length of the track is $l = 1.0 \ km = 1000 \ m$

The speed of A is $v__{A}} = 20 \ m/s$

The uniform acceleration of B is $a__{B}} = 0.333 \ m/s^2$

Generally the time taken by go-cart A is mathematically represented as

$t__{A}} = \frac{l}{v__{A}}}$

=> $t__{A}} = \frac{1000}{20}$

=> $t__{A}} = 50 \ s$

Generally from kinematic equation we can evaluate the time taken by go-cart B as

$l = ut__{B}} + \frac{1}{2} a__{B}} * t__{B}}^2$

given that go-cart B starts from rest u = 0 m/s

So

$1000 = 0 *t__{B}} + \frac{1}{2} * 0.333 * t__{B}}^2$

=> $1000 = 0 *t__{B}} + \frac{1}{2} 0.333 * t__{B}}^2$

=> $t__{B}} = 77.5 \ seconds$

Comparing $t__{A}} \ and \ t__{B}}$ we see that $t__{A}}$ is smaller so go-cart A is faster

3 0

2 years ago

The planet should move around the elliptical orbit, and two segments of the orbit should become shaded in green. What aspect(s)

strojnjashka [21]

Answer: not sure

Explanation:

4 0

2 years ago

Carbonic acid _____.

artcher [175]

Dissolves limestone and other rocks.

3 0

2 years ago

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If two opposing forces are equal, then the net force is 0 N. true or false?