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kaheart [24]
2 years ago
14

Jim stands beside a wide river and wonders how wide it is. he spots a large rock on the bank directly across from him. he then w

alks upstream 65 strides and judges that the angle between him and the rock, which he can still see, is now at an angle of 30o . jim measures his stride to be about 0.8 m long. estimate the width of the river.
Physics
1 answer:
LuckyWell [14K]2 years ago
7 0

To solve this problem, we must imagine that Jim’s initial position, the position of the rock, and Jim’s final position all connects to form a triangle. Now we can imagine that the triangle is a right triangle with the 90° angle on the initial position.

The angle of 30° is directly opposite to the length of his total stride while the width of the river is the side adjacent to the angle. Therefore can use the tan function to solve for the width of the river:

tan θ = opposite side / adjacent side

tan 30 = total stride distance / width of river

where total stride distance = 65 * 0.8 = 52 m

width of river = 52 m / tan 30

<span>width of river = 90.07 m</span>

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UNO [17]

Anything that's not supported and doesn't hit anything, and
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be moving up, then it loses 9.8 m/s of its upward speed every
second, on account of gravity.

                (64.2 m/s)  -  [ (9.8 m/s² ) x (1.5 sec) ] 

            =  (64.2 m/s)  -       [      14.7 m/s      ]

            =             49.5 m/s  .  (upward)

7 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
Mickey, a daredevil mouse of mass m , m, is attempting to become the world's first "mouse cannonball." He is loaded into a sprin
Sati [7]

Answer:

  h = v₀² / 2g ,      h = k/4g     x²

Explanation:

In this exercise we can use the law of conservation of energy at two points, the lowest, before the shot and the highest point that the mouse reaches

Starting point. Lower compressed spring

              Em₀ = K = ½ m v²

Final point. Highest on the path

             Em_{f} = U = mg h

             

As or no friction the energy is conserved  

              Em₀ =  Em_{f}

              ½ m v₀²² = m g h

             h = v₀² / 2g

We can also use as initial energy the energy stored in the spring that will later be transferred to the mouse

                  ½ k x² = 2 g h

                  h = k/4g     x²

8 0
2 years ago
Read 2 more answers
Two electrodes, separated by a distance d, in a vacuum are maintained at a constant potential difference. An electron, accelerat
Alja [10]

Answer:

Explanation:

Given that, the distance between the electrode is d.

The electron kinetic energy is Ek when the electrode are at distance "d" apart.

So, we want to find the K.E when that are at d/3 distance apart.

K.E = ½mv²

Note: the mass doesn't change, it is only the velocity that change.

Also,

K.E = Work done by the electron

K.E = F × d

K.E = W = ma × d

Let assume that if is constant acceleration

Then, m and a is constant,

Then,

K.E is directly proportional to d

So, as d increase K.E increase and as d decreases K.E decreases.

So,

K.E_1 / d_1 = K.E_2 / d_2

K.E_1 = E_k

d_1 = d

d_2 = d/3

K.E_2 = K.E_1 / d_1 × d_2

K.E_2 = E_k × ⅓d / d

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K.E_2 = ⅓E_k

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7 0
2 years ago
Charge q1 is distance r from a positive point charge Q. Charge q2=q1/3 is distance 2r from Q. What is the ratio U1/U2 of their p
worty [1.4K]

We have that The ratio U1/U2 of their potential energies due to their interactions with Q is

  • U1/U2=6
  • U1/U2=6

From the question we are told that

Question 1

Charge q1 is distance r from a positive point charge Q.

Question 2

Charge q2=q1/3 is distance 2r from Q.

Charge q1 is distance s from the negative plate of a parallel-plate capacitor.

Charge q2=q1/3 is distance 2s from the negative plate.

Generally the equation for the potential energy  is mathematically given as

U=\frac{-k*qQ}{r}

Therefore

The Equations of U1 and U2 is

For U1

U1=\frac{-k*q_1Q}{r}

For U2

U2=\frac{-k*q_1Q}{3*2r}

Since

U is a function of q and  q2=q1/3

Therefore

U1/U2=6

For Question 2

For U1

U1=\frac{-k*q_1Q}{s}\\\\For U2\\\\U2=\frac{-k*q_1Q}{3*2r}

Therefore

U1/U2=6

For more information on this visit

brainly.com/question/23379286?referrer=searchResults

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