Which part of the following graph shows the object going at the fastest speed?

A model rocket blasts off and moves upward with an acceleration of 12 m/s2 until it reaches a height of 26 m. at that height, it

Diano4ka-milaya [45]

initial acceleration of rocket is given as

a = 12 m/s^2

h = 26 m

now we can use kinematics to find its speed

$v_f^2 - v_i^2 = 2 a h$

$v_f^2 - 0 = 2 * 12*26$

$v_f = 24.98 m/s$

now after this it will be under free fall

so now again using kinematics

$v_f = 0$

at maximum height

$v_f^2 - v_i^2 = 2 a s$

$0 - 24.98^2 = 2 * (-9.8)* h$

$h = 31.8 m$

total height from the ground = 31.8 + 26 = 57.8 m

Part b)

now after reaching highest height it will fall to ground

So in order to find the speed we can use kinematics again

$v_f^2 - v_i^2 = 2 a d$

$v_f^2 - 0 = 2*9.8*57.8$

$v_f = 33.67 m/s$

Part c)

first rocket accelerate to reach height 26 meter and speed becomes 24.98 m/s

now we have

$v_f - v_i = a t$

$24.98 - 0 = 12*t_1$

$t_1 = 2.1 s$

after this it will reach to highest point and final speed becomes zero

$v_f - v_i = at$

$0 - 24.98 = -9.8 * t$

$t_2 = 2.55 s$

now from this it will fall back to ground and reach to final speed 33.67 m/s

now we have

$v_f - v_i = at$

$33.67 - 0 = 9.8 * t$

$t_3 = 3.44 s$

so total time is given as

t = 3.44 + 2.55 + 2.1 = 8.1 s

5 0

3 years ago

A 6 kg mass collides with a body at rest .After collision they travel together with a velocity equal

Mars2501 [29]

Answer:

12 kg

Explanation:

Momentum before collision = momentum after collision

m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂

After the collision, they have the same velocity, so v₁ = v₂ = v:

m₁ u₁ + m₂ u₂ = m₁ v + m₂ v

m₁ u₁ + m₂ u₂ = (m₁ + m₂) v

We know that m₁ = 6 kg, u₂ = 0 m/s, and v = u₁ / 3.

(6 kg) u₁ + m₂ (0 m/s) = (6 kg + m₂) (u₁ / 3)

(6 kg) u₁ = (6 kg + m₂) (u₁ / 3)

6 kg = (6 kg + m₂) (1/3)

18 kg = 6 kg + m₂

m₂ = 12 kg

3 0

4 years ago

The inclined plane in the figure above has two sections of equal length and different roughness. The dashed line shows where sec

saveliy_v [14]

The static friction exerted on the block by the incline is $\mu _ s _1 Mgcos \ \theta$ .

The given parameters;

mass of the block, = M
coefficient of static friction in section 1, = $\mu_s_1$
angle of inclination of the plane, = θ

The normal force on the block is calculated as follows;

Fₙ = Mgcosθ

The static friction exerted on the block by the incline is calculated as follows;

$F_s = \mu_s F_n\\\\F_s = \mu _s_1(Mg cos\ \theta)\\\\F_s = \mu _s_1 Mgcos\ \theta$

Thus, the static friction exerted on the block by the incline is $\mu _ s _1 Mgcos \ \theta$

Learn more here:brainly.com/question/17237604

3 0

3 years ago

What is the moment of inertia of a cube with mass M=0.500kg and side lengths s=0.030m about an axis which is both normal (perpen

prisoha [69]

Answer:

The moment of inertia I is

I = 2.205x10^-4 kg/m^2

Explanation:

Given mass m = 0.5 kg

And side lenght = 0.03 m

Moment of inertia I = mass x radius of rotation squared

I = mr^2

In this case, the radius of rotation is about an axis which is both normal (perpendicular) to and through the center of a face of the cube.

Calculating from the dimensions of the the box as shown in the image below, the radius of rotation r = 0.021 m

Therefore,

I = 0.5 x 0.021^2 = 2.205x10^-4 kg/m^2

8 0

4 years ago

One day on her walk home from school, Leann noticed some rust on an old car Leann knows that when iron and oxygen chemically rea

Katena32 [7]

Answer:

reactants

Explanation:

The reaction model is given as shown below:

4Fe + 3O₂ → 2Fe₂O₃

The name that describes the iron and oxygen found on the left hand side of Leann's model is reactants.

Reactants in a chemical expression are the species that are combining together to give a product.

Some reactions only have one reactant and such a chemical process is often termed decomposition reaction.

The two moles of Fe₂O₃ is the product of the reaction.

4 0

3 years ago