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Gala2k [10]
4 years ago
6

A test rocket is launched vertically from ground level (y = 0 m), at time t = 0.0 s. The rocket engine provides constant upward

acceleration during the burn phase. At the instant of engine burnout, the rocket has risen to 49 m and acquired a velocity of 30m/s. How long did the burn phase last?
Physics
1 answer:
SVETLANKA909090 [29]4 years ago
3 0
The important thing to note here is the direction of motion of the test rocket. Since it mentions that the rocket travels vertically upwards, then this motion can be applied to rectilinear equations that are derived from Newton's Laws of Motions.These useful equations are:

y = v₁t + 1/2 at²
a = (v₂-v₁)/t

where
y is the vertical distance travelled
v₁ is the initial velocity
v₂ is the final velocity
t is the time 
a is the acceleration

When a test rocket is launched, there is an initial velocity in order to launch it to the sky. However, it would gradually reach terminal velocity in the solar system. At this point, the final velocity is equal to 0. So, v₂ = 0. Let's solve the second equation first.

a = (v₂-v₁)/t
a = (0-30)/t
a = -30/t

Let's substitute a to the first equation:
y = v₁t + 1/2 at²
49 = 30t + 1/2 (-30/t)t²
49 = 30t -15t
49 = 15 t
t = 49/15
t = 3.27 seconds

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Homework help plz it would help a lot ​
blagie [28]

#82

here we know that

acceleration = 2 m/s/s

time = 5 s

initial speed = 4 m/s

now we can use kinematics to find the final speed

v_f = v_i + at

v_f = 4 + 2(5)

v_f = 14 m/s

So correct answer will be option D)

#83

here we know that

acceleration = 3 m/s/s

time = 4 s

initial speed = 5 m/s

now we can use kinematics to find the final speed

v_f = v_i + at

v_f = 5 + 3(4)

v_f = 17 m/s

So correct answer will be option C)

#84

here we know that

acceleration = 7 m/s/s

time = 3 s

initial speed = 8 m/s

now we can use kinematics to find the final speed

v_f = v_i + at

v_f = 8 + 7(3)

v_f = 29 m/s

So correct answer will be option C)

6 0
3 years ago
A 2. 0 μf and a 4. 0 μf capacitor are connected in series across an 8. 0-v dc source. what is the charge on the 2. 0 μf capacito
Nezavi [6.7K]

voltage across 2.0μf capacitor is 5.32v

Given:

C1=2.0μf

C2=4.0μf

since two capacitors are in series there equivalent capacitance will be

[tex] \frac{1}{c} = \frac{1}{c1} + \frac{1}{c2} [/tex]

c =  \frac{c1 \times c2}{c1 + c2}

=  \frac{2 \times 4}{2 + 4}

=1.33μf

As the capacitance of a capacitor is equal to the ratio of the stored charge to the potential difference across its plates, giving: C = Q/V, thus V = Q/C as Q is constant across all series connected capacitors, therefore the individual voltage drops across each capacitor is determined by its its capacitance value.

Q=CV

given,V=8v

= 1.33 \times 10 {}^{ - 6}  \times 8

= 10.64 \times 10 {}^{ - 6}

charge on 2.0μf capacitor is

\frac{Qeq}{2 \times 10 {}^{ - 6} }

=  \frac{10.64 \times 10 {}^{ - 6} }{2 \times 10 {}^{ - 6} }

=5.32v

learn more about series capacitance from here: brainly.com/question/28166078

#SPJ4

3 0
2 years ago
What is the force on a person’s hand, which is using a rope to accelerate a 5 kg block upward with an acceleration of 2.2 m/s 2
slavikrds [6]

Answer:

F = 11 N

Explanation:

Given,

Mass of a block, m = 5 kg

Acceleration of the block, a = 2.2 m/s²

We need to find the force on the person's hand. Let it is F. We know that force is given by the product of mass and acceleration as follows :

F = ma

F = 5 kg × 2.2 m/s²

F = 11 N

So, the force on a person's hand is 11 N.

5 0
3 years ago
You cheated on a test and you know it was the wrong thing to do according to the psychology theory what helped you know this
Alexandra [31]

Answer:

You cheated on a test, and you know it was the wrong thing to do. According to the psychoanalytic theory, what helped you know this?

The id controlled your biological response and made you sweaty because you were scared of getting caught.

The superego acted as your moral conscience; you just knew that it wasn't right to cheat.

The ego made you feel guilty as a defense mechanism. both B and C

Explanation:

the  answer is both B,C

6 0
3 years ago
A hockey stick of mass ms and length L is at rest on the ice (which is assumed to be frictionless). A puck with mass mp hits the
krek1111 [17]

Answer:

L = mp*v₀*(ms*D) / (ms + mp)

Explanation:

Given info

ms = mass of the hockey stick

uis = 0 (initial speed of the hockey stick before the collision)

xis = D (initial position of center of mass of the hockey stick before the collision)

mp = mass of the puck

uip = v₀ (initial speed of the puck before the collision)

xip = 0 (initial position of center of mass of the puck before the collision)

If we apply

Ycm = (ms*xis + mp*xip) / (ms + mp)

⇒  Ycm = (ms*D + mp*0) / (ms + mp)

⇒  Ycm = (ms*D) / (ms + mp)

Now, we can apply the equation

L = m*v*R

where m = mp

v = v₀

R = Ycm

then we have

L = mp*v₀*(ms*D) / (ms + mp)

5 0
3 years ago
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