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3 years ago

A vector is 9.55m long and points in a -48.0 degree direction. find the x component

Physics

1 answer:

stich3 [128]3 years ago

8 0

Answer:

x=6.39 m

Explanation:

Rectangular Components of a Vector

Vectors can be expressed in several forms, including the magnitude-angle form, and the rectangular-coordinates form.

If the magnitude r and angle β are given, we can find the rectangular coordinates (x,y) as follows:

$x=r\cos\beta$

$y=r\sin\beta$

The vector of the question has a magnitude of r=9.55 m and an angle β=-48°. The x-component is:

$x=9.55\cdot\cos(-48^\circ)$

Using a digital calculator:

x=6.39 m

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Nuclear Reactions: Please answer numbers 4,5,6.

Vikentia [17]

Answer:

4. ¹₁H + ¹³₆C —> ¹⁴₇N + γ

5. ¹₀n + ¹⁰₅B —> ⁷₃Li + ⁴₂He

6. ⁴₂He + ¹⁴₇N —> ¹⁷₈O + ¹₁H

Explanation:

4. ¹₁H + ¹³₆C —> ¹⁴₇N + __

Let ᵇₐX be the unknown.

Thus, the equation becomes:

¹₁H + ¹³₆C —> ¹⁴₇N + ᵇₐX

Next, we shall determine b, a and X. This is illustrated below:

For b:

1 + 13 = 14 + b

14 = 14 + b

Collect like terms

b = 14 – 14

b = 0

For a:

1 + 6 = 7 + a

7 = 7 + a

Collect like terms

a = 7 – 7

a = 0

Therefore,

ᵇₐX => ⁰₀X => γ

Thus, the balanced equation is

¹₁H + ¹³₆C —> ¹⁴₇N + γ

5. ¹₀n + ¹⁰₅B —> __ + ⁴₂He

Let ˣᵧA be the unknown.

Thus, the equation becomes:

¹₀n + ¹⁰₅B —> ˣᵧA + ⁴₂He

Next, we shall determine x, y and A. This can be obtained as follow:

For x:

1 + 10 = x + 4

11 = x + 4

Collect like terms

x = 11 – 4

x = 7

For y:

0 + 5 = y + 2

5 = y + 2

Collect like terms

y = 5 – 2

y = 3

Therefore,

ˣᵧA =>⁷₃A => ⁷₃Li

Thus, the balanced equation is:

¹₀n + ¹⁰₅B —> ⁷₃Li + ⁴₂He

6. ⁴₂He + ¹⁴₇N —> __ + ¹₁H

Let ᶜₑG be the unknown.

Thus, the equation becomes:

⁴₂He + ¹⁴₇N —> ᶜₑG + ¹₁H

Next, we shall determine c, e and G. This can be obtained as follow:

For c:

4 + 14 = c + 1

18 = c + 1

Collect like terms

c = 18 – 1

c = 17

For e:

2 + 7 = e + 1

9 = e + 1

Collect like terms

e = 9 – 1

e = 8

Therefore,

ᶜₑG => ¹⁷₈G => ¹⁷₈O

Thus, the balanced equation is:

⁴₂He + ¹⁴₇N —> ¹⁷₈O + ¹₁H

8 0

3 years ago

Calculate the tension (in N) in a vertical strand of spiderweb if a spider of mass 5.00 ✕ 10-5 kg hangs motionless on it.

Mandarinka [93]

Answer:

4.9 x 10⁻⁴N

Explanation:

Given parameters:

Mass of the vertical strand of spiderweb = 5 x 10⁻⁵kg

Unknown:

Tension in the vertical strand = ?

Solution:

The tension is the vertical strand will be the weight of the strand.

Weight of strand = mg

m is the mass

g is the acceleration due to gravity = 9.8m/s²

Tension in the vertical strand = 5 x 10⁻⁵kg x 9.8m/s²

Tension in the vertical strand = 4.9 x 10⁻⁴N

3 0

3 years ago

While walking between gates at an airport, you notice a child running along a moving walkway. Estimating that the child runs at

Mashcka [7]

Answer:

The speed of the moving walkway is 1.50 m/s

Explanation:

The position of the child can be calculated using the following equation:

x = x0 + v · t

Where :

x = position of the child at time t.

v = velocity of the child.

t = time.

When the child runs in the same direction as the walkway, the velocity of the child will be its velocity relative to the walkway plus the velocity of the walkway. Then, if we place the origin of the frame of reference at the start of the walkway:

x = x0 + v · t

25 m = 0 m + (2.8 m/s + v) · t₁

Where v is the velocity of the walkway

On its way back, the velocity of the child relative to the walkway is in the opposite direction to the velocity of the walkway. Then:

x = x0 + v · t

0 m = 25 m + (-2.8 m/s + v) · t₂

We also know that t₁ + t₂ = 25 s

Then: t₁ = 25 - t₂

So, we can write the following system of equations:

25 m = (2.8 m/s + v) · (25 s - t₂)

-25 m = (-2.8 m/s + v) · t₂

Let´s take the second equation and solve it for t₂

-25 m / (-2.8 m/s + v) = t₂

Now, let´s replace t₂ in the first equation:

25 m = (2.8 m/s + v) · (25 s + 25 m / (-2.8 m/s + v))

Let´s sum the fraction: 25 s + 25 m / (-2.8 m/s + v)

25 m = (2.8 m/s + v) · (25 s ·(-2.8 m/s + v) + 25 m) / (-2.8 m/s + v)

multiply by (-2.8 m/s + v) both sides of the equation:

25 m(-2.8 m/s + v) = (2.8 m/s + v) · (-70 m + 25 s · v + 25 m)

Apply distributive property:

-70 m²/s +25 m·v = -196 m²/s +70 m·v +70 m²/s -70 m·v +25 s ·v² + 25 m v

56 m²/s = 25 s · v²

56 m²/s / 25 s = v²

v = 1.50 m/s

The speed of the moving walkway is 1.50 m/s

7 0

3 years ago

(V) What are the objectives of the space mission?

Svetach [21]

Explanation:

I hope this helps, I got this from another person

3 0

3 years ago

A seagull flying horizontally over the ocean at a constant speed of 2.60 m/s carries a small fish in its mouth. It accidentally

Ivenika [448]

(a) +2.60 m/s

The motion of the fish dropped by the seagul is a projectile motion, which consists of two independent motions:

- a horizontal uniform motion, at constant speed

- a vertical motion, at constant acceleration (acceleration of gravity, $g=-9.8 m/s^2$ , downward)

In this part we are only interested in the horizontal motion. As we said the horizontal component of the fish's velocity does not change, therefore its value when the fish reaches the ocean is equal to its initial value, which is the speed at which the seagull was flying (because it was flying horizontally):

$v_x = +2.60 m/s$

(b) -17.2 m/s

The vertical component of the fish's velocity instead follows the equation:

$v_y = u_y +gt$

where

$u_y = 0$ is the initial vertical velocity, which is zero

$g=-9.8 m/s^2$ is the acceleration of gravity

t is the time

Since the fish reaches the ocean at t = 1.75 s, we can substitute this time into the formula to find the final vertical velocity:

$v_y = 0+(-9.8)(1.75)=-17.2 m/s$

where the negative sign indicates the direction (downward).

(c)

The horizontal component of the fish's velocity would increase

The vertical component of the fish's velocity would stay the same.

As we said from part (a) and (b):

- The horizontal component of the fish's velocity is constant during the motion and it is equal to the initial velocity of the seagull -> so if the seagull's initial speed increases, the horizontal velocity of the fish will increase too

- The vertical component of the fish's velocity does not depend on the original speed of the seagull, therefore it is not affected.

4 0

3 years ago

A vector is 9.55m long and points in a -48.0 degree direction. find the x component​

A vector is 9.55m long and points in a -48.0 degree direction. find the x component