Please I need help on this

Help me with this! I do not understand it! I will give brainliest!<br> Is it A B C or D

Firdavs [7]

The answer is C. Click C.

3 0

3 years ago

A pyrotechnical releases a 3 kg firecracker from rest. at t=0.4 s, the firecracker is moving downward with a speed 4 m/s. At the

olga2289 [7]

Answer:

a) F = 30 N, b) I = 12 N s , c) I = -12 N s , d) ΔI = 0 N s

Explanation:

This exercise is a case at the moment, let's define the system formed by the firecracker and its two parts, in this case the forces during the explosion are internal and the moment is conserved

Initial, before the explosion

p₀ = m v

The speed can be found by kinematics

v = v₀ - g t

v = 0 - 10 0.4

v = -4.0 m / s

Final after division

pf = m₁ v₁f + m₂ v₂f

p₀ = pf

M v = m₁ v₁f + m₂ v₂f

Where M is the initial mass (M = 3 kg), m₁ is the mass mtop (m₁ = 1 kg) and m₂ in the mass m botton (m₂ = 2kg) and the piece that moves up (v₁f = 6m/s )

a) before the explosion the only force acting on the body is gravity

F = mg

F = 3 10 = 30 N

b) The expression for momentum is

I = Ft

Before the explosion the only force that acts is the weight

I = mg t

I = 3 10 0.4

I = 12 N s

c) To calculate this part we use the conservation of the moment and calculate the speed of the body that descends body 2

M v = m₁ v₁f + m₂ v₂f

v₂f = (M v - m₁ v₁f) / m₂

v₂f = (3 (-4) - 1 6) / 2

v₂f = - 9 m / 2

The negative sign indicates that body 2 (botton) is descending

Now we can use the momentum and momentum relationship for the body during the explosion

I = F t = Dp

F t = pf –po)

F t= [m₁ v₁f + m₂ v₂f]

I = [1 6 + 2 (-9) -0]

I = -12 N s

This is the impulse during the explosion the negative sign indicates that it is headed down

d) impulse change

I₀ = Mv

I₀ = 3 *4

I₀ =-12 N s

ΔI =If – I₀

ΔI = - 12 – (-12)

ΔI = -0 N s

3 0

4 years ago

17.

Zolol [24]

Answer:

I don't know this answer at all

Explanation:

I don't know about these problems

6 0

3 years ago

What is the magnitude of the total acceleration of point A after 2 seconds? The bar starts from rest and has a constant angular

monitta

Answer:

a_total = 2 √ (α² + w⁴) , a_total = 2,236 m

Explanation:

The total acceleration of a body, if we use the Pythagorean theorem is

a_total² = a_T²2 + $a_{c}$ ²

where

the centripetal acceleration is

a_{c} = v² / r = w r²

tangential acceleration

a_T = dv / dt

angular and linear acceleration are related

a_T = α r

we substitute in the first equation

a_total = √ [(α r)² + (w r² )²]

a_total = 2 √ (α² + w⁴)

Let's find the angular velocity for t = 2 s if we start from rest wo = 0

w = w₀ + α t

w = 0 + 1.0 2

w = 2.0rad / s

we substitute

a_total = r √(1² + 2²) = r √5

a_total = r 2,236

In order to finish the calculation we need the radius to point A, suppose that this point is at a distance of r = 1 m

a_total = 2,236 m

7 0

3 years ago

A truck on a straight road starts from rest, accelerating at 2.00m/s^2 until it reaches a speed of 20.0m/s. Then the truck trave

Dennis_Churaev [7]

Answer:

Explanation:

a )

Time to reach the speed of 20 m/s with an acceleration of 2 m/s² can be calculated as follows .

v = u + a t

20 = 0 + 2 t

t = 20 /2 = 10 s .

Total time = 10 s + 20 s + 5 s = 35 s .

b) Average velocity = Total distance travelled / total time

Distance travelled in first 10 s

S₁ = ut + 1/2 a t²

= 0 + .5 x 2 x 10²

= 100 m

Distance travelled in next 20 s

S₂= 20s x 20 m/s = 400 m

Distance travelled in last 5 s .

deceleration in last 5 s

v = u + at

0 = 20 m/s + a x 5

a = - 4 m/s²

v² = u² - 2 a s

0 = (20 m/s)² - 2 x 4 m/s² x s

s = 50 m

S₃ = 50 m

Total distance = S₁ + S₂ + S₃

= 100 m + 400 m + 50 m

= 550 m .

Average velocity = 550 m / 35 s

= 15.71 m /s .

3 0

3 years ago