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3 years ago

At 12 seconds what is the direction of the velocity of the rocket what is the direction of the acceleration rocket

Physics

1 answer:

algol [13]3 years ago

8 0

Answer:

rocket at the moment of clearance, the velocity and acceleration have the same direction, vertical upwards.

Explanation:

In a rocket at the moment of clearance, the velocity and acceleration have the same direction, vertical upwards.

In the case of landing we have a downward acceleration due to the force of gravity and upward velocity.

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Describe asexual reproduction

Brrunno [24]

“Asexual reproduction is a type of reproduction by which offspring arise from a single organism, and inherit the genes of that parent only; it does not involve the fusion of gametes, and almost never changes the number of chromosomes.” -Wikipedia

4 0

3 years ago

An object initially at rest experiences an acceleration

lidiya [134]

Answer:

1.4m/s

Explanation:

Average velocity is the total distance covered divided by the total time taken.

Average velocity = $\frac{total distance }{time }$

Total time taken = 5s + 6s = 11s

The first distance covered = velocity x time = 1.4 x 5 = 7m

second distance covered = velocity x time = 1.4 x 6 = 8.4m

So;

Average velocity = $\frac{7 + 8.4}{11}$ = 1.4m/s

5 0

3 years ago

Charge g is distributed in a spherically symmetric ball of radius a. (a) Evaluate the average volume charge density p. (b) Now a

nasty-shy [4]

Answer:

Explanation:

The volume of a sphere is:

V = 4/3 * π * a^3

The volume charge density would then be:

p = Q/V

p = 3*Q/(4 * π * a^3)

If the charge density depends on the radius:

p = f(r) = k * r

I integrate the charge density in spherical coordinates. The charge density integrated in the whole volume is equal to total charge.

$Q = \int\limits^{2*\pi}_0\int\limits^\pi_0 \int\limits^r_0 {k * r} \, dr * r*d\theta* r*d\phi$

$Q = k *\int\limits^{2*\pi}_0\int\limits^\pi_0 \int\limits^r_0 {r^3} \, dr * d\theta* d\phi$

$Q = k *\int\limits^{2*\pi}_0\int\limits^\pi_0 {\frac{r^4}{4}} \, d\theta* d\phi$

$Q = k *\int\limits^{2*\pi}_0 {\frac{\pi r^4}{4}} \, d\phi$

$Q = \frac{\pi^2 r^4}{2}}$

Since p = k*r

Q = p*π^2*r^3 / 2

Then:

p(r) = 2*Q / (π^2*r^3)

3 0

3 years ago

Two balls of clay, with masses M1 = 0.49 kg and M2 = 0.47 kg, are thrown at each other and stick when they collide. Mass 1 has a

malfutka [58]

Answer:

a) $p_i=1.568\hat{i}+0.752 \hat{j}$

b) $v_{fx}=1.668\ m.s^{-1}$

c) $v_{fy}=0.7999\ m.s^{-1}$

Explanation:

Given masses:

$m_1=0.49\ kg$

$m_2=0.47\ kg$

Velocity of mass 1, $v_1=3.2 \hat{i}\ m.s^{-1}$

Velocity of mass 2, $v_2=1.6 \hat{j}\ m.s^{-1}$

Initial momentum:

$p_i=m_1.v_1+m_2.v_2$

$p_i=0.49\times 3.2 \hat{i}+0.47\times 1.6 \hat{j}$

$p_i=1.568\hat{i}+0.752 \hat{j}$

magnitude of initial momentum:

$p_i=\sqrt{1.568^2+0.752 ^2}$

$p_i=1.739\ kg.m.s^{-1}$

From the conservation of momentum:

$p_f=p_i$

$m_f.v_f=1.739$

$v_f=\frac{1.739}{0.49+0.47}$

$v_f=1.85\ m.s^{-1}$ is the magnitude of final velocity.

Direction of final velocity will be in the direction of momentum:

$tan\theta=\frac{0.752 }{1.568}$

$\theta=25.62^{\circ}$

$\therefore v_{fx}=1.85\ cos25.62^{\circ}$

$v_{fx}=1.668\ m.s^{-1}$

Vertical component of final velocity:

$v_{fy}=1.85\ sin 25.62^{\circ}$

$v_{fy}=0.7999\ m.s^{-1}$

6 0

3 years ago

9. A Veggie meatball with v = 5.0 m/s rolls off a 1.0 m high table. How long does it take to hit the floor if no one sneezes? Wh

Paha777 [63]

By Considering the vertical distance and both vertical and horizontal final velocity, the time t = 0.45 s and Velocity V = 6.7 m/s

Given that a Veggie meatball with v = 5.0 m/s rolls off a 1.0 m high table.

Height h = 1.0 m

As the ball rolls off the table, it will be fallen under gravity. Where

g = 9.8 m/ $s^{2}$

Initial vertical velocity $u_{y}$ = 0

Initial horizontal velocity $u_{x}$ = 5 m/s

Considering the vertical distance, the formula to use to calculate the time will be;

h = ut + 1/2g $t^{2}$

1 = 0 + 1/2 x 9.8 $t^{2}$

1 = 4.9 $t^{2}$

$t^{2}$ = 1/4.9

t = $\sqrt{0.204}$

t = 0.45 seconds

It takes 0.45 seconds to hit the floor if no one sneezes.

To calculate its velocity when it hits the floor, we will need to calculate for both vertical and horizontal final velocity and find the resultant velocity of the two.

Vertical component

$V_{y}$ = $U_{y}$ + gt