What is the vertical displacement below any line?

A 0.29 kg particle moves in an xy plane according to x(t) = - 19 + 1 t - 3 t3 and y(t) = 20 + 7 t - 9 t2, with x and y in meters

Artist 52 [7]

Answer:

Part a)

$F = 7.76 N$

Part b)

$\theta = -137.7 degree$

Part c)

$\theta = -127.7 degree$

Explanation:

As we know that acceleration is rate of change in velocity of the object

So here we know that

$x = -19 + t - 3t^3$

$y = 20 + 7t - 9t^2$

Part a)

differentiate x and y two times with respect to time to find the acceleration

$a_x = \frac{d^2}{dt^2}(-19 + t - 3t^3)$

$a_x = \frac{d}{dt}(0 +1 - 9t^2)$

$a_x = -18t$

$a_y = \frac{d^2}{dt^2}(20 + 7t - 9t^2)$

$a_y = \frac{d}{dt}(0 +7 - 18t)$

$a_y = -18$

Now the acceleration of the object is given as

$\vec a = (-18t)\hat i + (-18)\hat j$

at t= 1.1 s we have

$\vec a = -19.8 \hat i - 18 \hat j$

now the net force of the object is given as

$\vec F = m\vec a$

$\vec F = (0.29 kg)(-19.8 \hat i - 18 \hat j)$

$\vec F = -5.74 \hat i - 5.22 \hat j$

now magnitude of the force will be

$F = \sqrt{5.74^2 + 5.22^2} = 7.76 N$

Part b)

Direction of the force is given as

$tan\theta = \frac{F_y}{F_x}$

$tan\theta = \frac{-5.22}{-5.74}$

$\theta = -137.7 degree$

Part c)

For velocity of the particle we have

$v_x = \frac{dx}[dt}$

$v_x = (0 +1 - 9t^2)$

$v_y = \frac{dy}{dt}$

$v_y = (0 +7 - 18t)$

now at t = 1.1 s

$\vec v = -9.89\hat i - 12.8 \hat j$

now the direction of the velocity is given as

$\theta = tan^{-1}(\frac{v_y}{v_x})$

$\theta = tan^{-1}(\frac{-12.8}{-9.89})$

$\theta = -127.7 degree$

7 0

3 years ago

What is the acceleration of a softball if it has a mass of 0.5 kg and hits the cathers glove with a force of 25n

ioda

Acceleration is found if we have the force and mass.

With the following equation: F = ma, we can find the missing values.

F = 25n
M = 0.5 kg
a = ?

a = f/m
a = 25/0.5
a = 50

a = 50 m/s

So, the acceleration is 50 m/s^2

3 0

3 years ago

Distance meters Time (seconds) Trial Tria 2 Trials Average 3 053 059 0.52 055 6 106 719 104 Uning the values trom the data table

kumpel [21]

Jsisisusuagahannananana this is worth 22 points g that’s crazy 25

5 0

2 years ago

The force of attraction between a ball is F=.........×10^-¹¹

DIA [1.3K]

Answer:

4.45×10¯¹¹ N

Explanation:

From the question given above, the following data were obtained:

Mass of ball (M₁) = 4 Kg

Mass of bowling pin (M₂) = 1.5 Kg

Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²

Distance apart (r) = 3 m

Force of attraction (F) =?

The force of attraction between the ball and the bowling pin can be obtained as follow:

F = GM₁M₂ / r²

F = 6.67×10¯¹¹ × 4 × 1.5 / 3²

F = 4.002×10¯¹⁰ / 9

F = 4.45×10¯¹¹ N

Therefore, the force of attraction between the ball and the bowling pin is 4.45×10¯¹¹ N

8 0

3 years ago

MathPhys i need your help please helpppo

Lena [83]

Answer:

6.77 m/s

Explanation:

First, in the x direction:

Given:

Δx = 3.17 m

v₀ = v cos 30.8° = 0.859 v

a = 0 m/s²

Δx = v₀ t + ½ at²

(3.17 m) = (0.859 v) t + ½ (0 m/s²) t²

3.17 = 0.859 v t

3.69 = v t

Next, in the y direction:

Given:

Δy = 0.432 m

v₀ = v sin 30.8° = 0.512 v

a = -9.81 m/s²

Δy = v₀ t + ½ at²

(0.432 m) = (0.512 v) t + ½ (-9.81 m/s²) t²

0.432 = 0.512 v t − 4.905 t²

Two equations, two variables. Solve for t in the first equation and substitute into the second equation:

t = 3.69 / v

0.432 = 0.512 v (3.69 / v) − 4.905 (3.69 / v)²

0.432 = 1.89 − 66.8 / v²

66.8 / v² = 1.458

v² = 45.8

v = 6.77

7 0

3 years ago