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Simora [160]
3 years ago
11

While surveying a cave, a spelunker follows a passage 190m straight west, then 210m in a direction 45.0east of south, and then 2

70m at 30.0 east of north. After a fourth unmeasured displacement, she finds herself back where she started. I need to use vector components to find the magnitude and direction of the fourth displacement.
Physics
1 answer:
hodyreva [135]3 years ago
3 0
<span>The first thing you need to consider all the displacements in y direction.
S1 has no displacement in y direction.
S2 has a displacement of (-210 sin 45)
S3 has a displacement of (270 sin30)
but
S1 + S2+ S3+ S4 = 0 0 + (-210 sin 45) + (270 sin30) + Sy=0 Sy= +105 *sqrt2 - 135 = 148.05 -135 = 13.05
 
You need to consider also the displacement in x direction
S1 has a displacement of -190 m
S2 has a displacement of 210 cos45
S3 has a displacement of 270 cos 30 S1 + S2 + S3 + S4 = 0 -190 + 105 * sqrt2 + 135* sqrt3 +S4 = 0 S4 =190 - 148.05 - 233.55= - 191.6
Final displacement would be the result of 191.6i - 13.05j
I hope this can help</span>
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8 0
3 years ago
SOMEONE PLEASE HELP ME ASAP PLEASE!!!!!​
Zina [86]

Answer:

x₁ = 58.09 m

Explanation:

Displacement is calculated by finding the final distance away from a point then subtracting the initial distance.

Given that initial position of object is x=25.89 m

Displacement Δx=32.2 m

Final position x₁=?

Δx = x₁-x

32.2= x₁-25.89

32.2+25.89 = x₁

58.09 m =x₁

5 0
3 years ago
You see the boy next door trying to push a crate down the sidewalk. He can barely keep it moving, and his feet occasionally slip
Murrr4er [49]

Answer:

Explanation:

Given that the weight of the crate is

M=48kg

Then, weight of boy is

W=mg=48×9.8=470.4N

ΣF = ma ,Along y axis

Since the body is not moving in y direction then, a=0 along y axis

N-W=0

N=W

The normal equals weight of boy

N=470.4N

Then, applying frictional force opposing the boy motion.

Fr=μN

Fr=0.5×470.4

Fr=235.2N

Then, this is the forward force that the boy try using to pull the crate.

Since he can barely move his foot he has not yet overcome the coefficient of static friction.

ΣF = ma. , along x-axis

a along x-axis is 0 since he cannot move his foot, I.e he was not moving

F-Fr= 0

F=Fr=235.2N

The forward force the boy apply is 235.2N on the crate

Also, analysing the crate.

The forward force is now F=235.2N

Then the frictional force =Fr

Then,

ΣF = ma. , along x-axis

a along x-axis is 0 since the crate did not move,

F-Fr=0

Fr=F=235.2N

This is the frictional force on the crate.

Then using frictional law

Fr=μN

N=Fr/μ

N=235.2/0.9

N=261.33N

Now this normal is equal to the weight of the crate

ΣF = ma ,Along y axis

Since the body is not moving in y direction then, a=0 along y axis

N-W=0

W=N=261.33N

Then, since weight is given as

W=mg

m=W/g

m=261.33/9.81

m=26.64kg

The mass of the crate is 26.64kg

7 0
3 years ago
Read 2 more answers
A weather balloon is partially inflated with helium gas to a volume of 2.0 m³. The pressure was measured at 101 kPa and the temp
Yanka [14]

Answer:

4.7 m³

Explanation:

We'll use the gas law P1 • V1 / T1 = P2 • V2 / T2

* Givens :

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( We must always convert the temperature unit to Kelvin "K")

* What we want to find :

V2 = ?

* Solution :

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V2 × 40 / 283.15 ≈ 0.67

V2 = 0.67 × 283.15 / 40

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7 0
2 years ago
What is the magnitude and direction of the electric field atradiaConsider a coaxial conducting cable consisting of a conductingr
Alexxx [7]

Answer:

 E = 9.4 10⁶ N / C ,     The field goes from the inner cylinder to the outside

Explanation:

The best way to work this problem is with Gauss's law

             Ф = E. dA = qint / ε₀

 

We must define a Gaussian surface, which takes advantage of the symmetry of the problem. We select a cylinder with the faces perpendicular to the coaxial.

The flow on the faces is zero, since the field goes in the radial direction of the cylinders.

The area of ​​the cylinder is the length of the circle along the length of the cable

         dA = 2π dr L

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They indicate that the distance at which we must calculate the field is

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The radius of the outer shell is

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When comparing these two values ​​we see that the field must be calculated between the two housings.

Gauss's law states that the charge is on the outside of the Gaussian surface does not contribute to the field, the charged on the inside of the surface is

         λ = q / L

         Qint = λ L

Let's replace

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       E = 1 / 2piε₀  λ / r

Let's calculate

         E = 1 / 2pi 8.85 10⁻¹²  3.4 10-12 / 6.5 10-3

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The field goes from the inner cylinder to the outside

5 0
3 years ago
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