A ball rotates on a string as shown below.



If the string in the image above breaks at point 1, which of the following images best shows the direction the ball will travel?

Question 8 options:

Diagram A

Diagram B

Diagram C

Diagram D

Diagram E

The ball is hit with an initial vertical velocity of

(147 m/s) sin(38º) ≈ 90.5 m/s

Recall that

v ² - u ² = 2 a ∆y

where u and v are initial and final velocities, respectively; a is acceleration; and ∆y is displacement.

Vertically, the ball is in freefall, so it is only subject to acceleration due to gravity, with magnitude g = 9.80 m/s² in the downward direction. At its maximum height, the ball has zero vertical velocity (v = 0) and the displacement is equal to the maximum height, so

0² - (90.5 m/s)² = 2 (-g) ∆y

∆y = (90.5 m/s)² / (2g)

∆y ≈ 418 m

Answer:

Vx = 35.31 [km/h]

Vy = 18.77 [km/h]

Explanation:

In order to solve this problem, we must decompose the velocity component by means of the angle of 28° using the cosine function of the angle.

[tex]v_{x} = 40*cos(28)\\V_{x} = 35.31 [km/h][/tex]

In order to find the vertical component, we must use the sine function of the angle.

[tex]V_{y}=40*sin(28)\\V_{y} = 18.77 [km/h][/tex]

Answer:

3.33 m/s

Explanation:

Using the formula: s = d / t (where s is speed, d is distance, and t is time)

[Convert minutes to seconds then solve]

Half a minute is 30 seconds. Therefore:

s = 100 / 30 = 3.33 m/s

Answer:

Yes

Explanation:

But I feel also 2 is correct but your answer is right

Answer:

-71 V

Explanation:

Given that,

An electron starts from rest at point A and has a speed of 5.0 × 10⁶ m/s at point B.

We need to find the electric potential difference [tex]V_A-V_B[/tex]. It can be calculated by using the conservation of charges as follows :

[tex]qV=\dfrac{1}{2}mv^2[/tex]

m and q are mass and charge on electrons

[tex]V=\dfrac{mv^2}{2(-q)}\\\\V=-\dfrac{9.1\times 10^{-31}\times (5\times 10^6)^2}{2\times 1.6\times 10^{-19}}\\\\V=-71.09\ V[/tex]

So, the electric potential difference is (-71 V). So, the correct option is (a).

Answer:separate

Explanation:

Answer:

Explanation:

The constant acceleration of any object, neglecting air resistance and friction, is always -9.8. Proof:

[tex]s(t)=-4.9t^2+13[/tex] which is the position of this feather before it is dropped. The first derivative of position is velocity, so the velocity function for the feather is:

[tex]v(t)=-9.8t[/tex]. We could find the velocity that the feather is experiencing at 3 seconds, but that is not what we are being asked. What we are being asked is the acceleration of the feather at 3 seconds. So we find the acceleration function of the feather:

a(t) = -9.8

It turns out that the time doesn't matter to acceleration due to gravity because feathers and elephants all fall under the same pull.

Explanation:

so sorry

don't know but please mark me as brainliest please

Answer:

Approximately [tex]3.5\; \rm kN[/tex] in each of the two ropes.

Explanation:

Let [tex]F_1[/tex] and [tex]F_2[/tex] denote the tension in each of the two ropes.

Refer to the diagram attached. The tension force in each rope may be decomposed in two directions that are normal to one another.  

The first direction is parallel to resultant force on the barge.

These two components would be in the same direction. The resultant force in that direction would be the sum of these two components: [tex]\displaystyle \frac{F_1 + F_2}{\sqrt{2}}[/tex]. That force should be equal to [tex]5\; \rm kN[/tex].

The second direction is perpendicular to the resultant force on the barge.

These two components would be in opposite directions. The resultant force in that direction would be the difference of these two components: [tex]\displaystyle \frac{F_1 - F_2}{\sqrt{2}}[/tex]. However, the net force on the barge is normal to this direction. Therefore, the resultant force should be equal to zero.

That gives a system of two equations and two unknowns:

The second equation suggests that [tex]F_1 = F_2[/tex]. Hence, replace the [tex]F_2[/tex] in the first equation with [tex]F_1[/tex] and solve for [tex]F_1\![/tex]:

[tex]F_1 = \displaystyle \frac{5\; \rm kN}{2\, \sqrt{2}} \approx 3.5\; \rm kN[/tex].

Because [tex]F_1 = F_2[/tex] (as seen in the second equation,) [tex]F_2 = F_1 \approx 3.5\; \rm kN[/tex].

In other words, the tension in each of the two ropes is approximately [tex]3.5\; \rm kN[/tex].

The tension in each of the rope is 3,535.5 N

The given expression:

the resultant force, R = 5 kN = 5000 N

the angle in between the forces, α = 45

To find:

The tension in  each of the ropes  is calculated as follows;

The tension in vertical direction

[tex]F_y = F \times sin(\alpha)\\\\F_y = 5000 \times sin(45)\\\\F_y = 5000 \times 0.7071\\\\F_y = 3,535.5 \ N[/tex]

The tension in horizontal direction;

[tex]F_x = F \times cos(\alpha)\\\\F_x = 5000 \times cos(45)\\\\F_x = 5000 \times 0.7071\\\\F_x = 3,535.5 \ N[/tex]

Thus, the tension in each of the rope is 3,535.5 N

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Answer:

Apparent depth = 45 cm

Explanation:

The refractive index of water in a pool, n = 4/3

Real depth, d = 60 cm

We need to find its apparent  depth when viewed vertically through  air.​ The ratio of real depth to the apparent depth is equal to the refractive index of the material. Let the apparent depth is d'. So,

[tex]n=\dfrac{d}{d'}\\\\d'=\dfrac{d}{n}\\\\d'=\dfrac{60}{\dfrac{4}{3}}\\\\d'=45\ cm[/tex]

So, the apparent depth is 45 cm.