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Veseljchak [2.6K]

2 years ago

7

PLS HELP!!! Response: Develop a unique example (not posted on the discussion board by anyone else) and calculate the resultant d

isplacement between two points when there are two legs or distinct parts to the trip. Include the displacement of each leg of the trip as well as the resultant displacement of the entire trip. Don't forget: include both the direction and the magnitude as part of the displacement. Research the actual distance between these two points on the globe. Compare the resultant displacements and account for the error. Don't forget: include both the direction and the magnitude as part of the displacement.

1 answer:

Jlenok [28]2 years ago

8 0

Answer:

he is the first person can be 3in and a 6man is the best not a man of the hartford

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PLZ ANSWER ASAP IF YOU KNOW THIS...At the county fair, Carrie and Sam climbed up on the carousel horses. Around and around they

Maslowich

The answer is (a) because movement is acceleration

8 0

2 years ago

It is easier to see clothes with pointed needle than a blunt one

topjm [15]

Answer:

Explanation:

It is easier to see clothes with pointed needle than a blunt one because pressure exerted is more in a pointed needle as it occupies less space compared to blunt needle, A blunt has more surface area so the pressure exerted will less as compared to a pointed needle.

IF YOU FOUND MY ANSWER USEFUL THEN PLEASE MARK ME BRAINLIEST.

7 0

3 years ago

The vertical force f acts downward at A on the two membered frames. Determine the magnitude of the two components of F directed

galben [10]

Answer:

The magnitude will be "353.5 N". A further solution is given below.

Explanation:

The given values is:

F = 500 N

According to the question,

In ΔABC,

⇒ $\angle BCA = (90-30)$

⇒ $=60^{\circ}$

then,

⇒ $\angle BAC=(180-45-60)$

⇒ $=75^{\circ}$

Now,

The corresponding angle will be:

⇒ $\angle FAC=60^{\circ}$

⇒ $\angle FAB=70+60$

⇒ $=135^{\circ}$

Aspect of F across the AC arm will be:

= $F\times cos(60)$

On putting the values of F, we get

= $500\times (.5)$

= $200 \ Newton$

Component F along the AC (in magnitude) will be:

= $F\times cos(135)$

= $500\times (-.707)$

= $-353.5 \ N \$

4 0

2 years ago

A 10 kg box hangs from a rope. What is the tension in the rope (in Newtons) if the box is stationary

Archy [21]

Answer:

T = 98 N

Explanation:

The gravity of the earth is known to be 9.8 m/s²

Data:

m = 10 kg
g = 9.8 m/s²
T = ?

Use formula:

$\boxed{\bold{T=m*g}}$

Replace and solve:

$\boxed{\bold{T=10\ kg*9.8\frac{m}{s^{2}}}}$

$\boxed{\boxed{\bold{T=98\ N}}}$

The tension in the rope is <u>98 Newtons.</u>

Greetings.

5 0

2 years ago

A force is applied to a block to move it up a 30° incline. The incline is frictionless, the block is initially at rest, F= 65.0

muminat

Answer:

0.96 m, upward

Explanation:

$\theta=30^{\circ}$

$F=65 N$

$M=5 kg$

Initial velocity, u=0

$t=0.550 s$

$Fcos\theta-Mgsin\theta=Ma$

Where $g=9.8 m/s^2$

Substitute the values

$65cos30-5\times 9.8sin30=5a$

$31.8=5a$

$a=\frac{31.8}{5}=6.36 m/s^2$

$S=ut+\frac{1}{2}at^2$

$S=0+\frac{1}{2}(6.36)(0.55)^2=0.96 m$

Hence,the block moves upward because displacement is positive.

7 0

3 years ago