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irina [24]
3 years ago
14

An example of Repression would be?

Physics
1 answer:
mart [117]3 years ago
3 0

child who is abused by a parent later has no recollection of the events, but has trouble forming relationships

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Calculate the speed of an 8.0*10^4kg airliner with a kinetic energy of 1.1*10^9J.
Burka [1]

Answer:

The velocity will be v = 165.83[m/s]

Explanation:

This is a problem where the definition of kinetic energy can be applied, which can be determined with the following equation.

E_{k}=\frac{1}{2}*m*v^{2}\\   where:\\m = mass = 80000[kg]\\v = velocity [m/s]\\E_{k}= kinetic energy [J]=1100000000[J]\\Replacing:\\v=\sqrt{\frac{2*E_{k} }{m} } \\v=\sqrt{\frac{2*1100000000 }{80000} }\\v=165.83[m/s]

8 0
3 years ago
What is the final velocity of a drag racer that has constant acceleration and finishes a
Svetradugi [14.3K]

Answer:

120 mph

Explanation:

Given:

Δx = 0.25 mi

v₀ = 0 mi/s

t = 15 s

Find: v

Δx = ½ (v + v₀) t

0.25 mi = ½ (v + 0 mi/s) (15 s)

v = 0.0333 mi/s

v = 120 mi/h

8 0
3 years ago
A box is being pulled across a horizontal surface by a 20 N force to the right.
Oksana_A [137]

Answer:

Im pretty sure it's C

Explanation:

The answer A talks about constant speed, which does not correspond to Net Force. The answer B, talks about constant velocity, but talks about how much force apposes the box. The answer C talks about the value of the net force acting on the box, so im pretty sure the answer is C.

6 0
3 years ago
Read 2 more answers
PLEASE HELP MEEEE <br>marking brainliest ​
aniked [119]

Answer:

Option C

Explanation:

Centripetal acceleration formula regarding velocity and radius is,

a_c=\frac{v^2}{r}

now we know the centripetal acceleration is 9 and the radius is 16 so we plug these values into our formula,

a_c=\frac{v^2}{r}\\\\9=\frac{v^2}{16} \\\\144=v^2\\\\\sqrt{144}=v \\\\v=12\ m/s

so velocity is 12 m/s

Now for the angular velocity, the formula of centripetal acceleration regarding angular velocity and radius is,

a_c=rw^2

we know the centripetal acceleration is 9 and the radius is 16 so plug these values into the formula,

a_c=rw^2\\\\9=16w^2\\\\0.5625=w^2\\\\\sqrt{0.5625}=w \\\\0.75\ rad/s=w\\

so angular velocity is 0.75 rad/s

8 0
3 years ago
A 4.4-µF capacitor is initially connected to a 5.1-V battery. Once the capacitor is fully charged the battery is removed and a 2
Grace [21]

Question is incomplete. Missing part:

Find the charge on the capacitor at the following times:

1) t = 0 mu S  

2) t = 1 mu S

3) t = 50 mu S

1) 22.4 \mu C

We start by calculating the initial charge on the capacitor. For this, we can use the following relationship:

C=\frac{Q_0}{V_0}

where

C is the capacitance

Q0 is the initial charge stored

V0 is the initial potential difference across the capacitor

When the capacitor is connected to the battery, we have:

C=4.4\mu F = 4.4\cdot 10^{-6}F

V_0 = 5.1 V

Solving for Q_0,

Q_0 = CV_0 = (4.4\cdot 10^{-6})(5.1)=2.24 \cdot 10^{-5} C = 22.4 \mu C

So, when the battery is disconnected, this is the charge on the capacitor at time t = 0.

2) 20.0\mu C

To find the charge on the capacitor at any other time t, we use the equation:

Q(t) = Q_0 e^{-\frac{t}{RC}}

where

Q_0 = 22.4 \mu C

t is the time

R=2.0 \Omega is the resistance

C=4.4\mu F is the capacitance

Therefore, at time t=1 \mu s, we have:

Q(t) = (22.4) e^{-\frac{1}{(2.0)(4.4)}}=20.0 \mu C

3) 0.08 \mu C

As before, we use again the equation:

Q(t) = Q_0 e^{-\frac{t}{RC}}

However, here the time to consider is

t=50 \mu C

Substituting into the formula,

Q(t) = (22.4) e^{-\frac{50.0}{(2.0)(4.4)}}=0.08 \mu C

4 0
3 years ago
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