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azamat
3 years ago
15

What should we do to be a computer engineer ‍ ‍?​

Physics
2 answers:
ycow [4]3 years ago
6 0

Answer:

to be an eginere u would have to go to college and study hard

Explanation:

____ [38]3 years ago
5 0
Study and go to university/college
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A ball rolls onto the path of your car as you drive down a quiet neighborhood street. To avoid hitting the child that runs to re
ehidna [41]

Answer:

F = 7.143 kN

Explanation:

given,

time taken to apply break = 1.05 s

car slows down from 15 m/s to 9 m/s

mass of the car = 1250 Kg

force is equal to the change in momentum with respect to time.

F = \dfrac{\Delta P}{t}

F = \dfrac{m(v_f - v_i)}{t}

F = \dfrac{1250\times (9 - 15)}{1.05}

F = \dfrac{-7500}{1.05}

F = -7142.85 N

F = - 7.143 kN

Force is acting opposite direction of velocity of car i.e. the sign is negative.

7 0
3 years ago
True or false
madam [21]
That's true, you can't get any more precise than what's measured to the least precision.
8 0
3 years ago
VERY EASY QUESTION FOR HIGH SCHOOL STUDENTS:
Elina [12.6K]

Answer:

Explanation: 6000z

5 0
3 years ago
Find the electric field at a point midway between two charges of +40.0 x 10^-9 c and.+60.0 x 10^-9 c
PIT_PIT [208]
Missing part in the text: "...the charges are <span>separated by a distance of 30.0 cm."
</span>
Solution:
The point midway between the two charges is located 15.0 cm from one charge and 15.0 from the other charge. The electric field generated by each of the charges is
E=k_e \frac{q}{r^2}
where
ke is the Coulomb's constant
Q is the value of the charge
r is the distance of the point at which we calculate the field from the charge (so, in this problem, r=15.0 cm=0.15 m).

Let's calculate the electric field generated by the first charge:
E_1 = (8.99 \cdot 10^9 Nm^2 C^{-2} ) \frac{+40.0 \cdot 10^{-9} C}{(0.15 m)^2}=1.6 \cdot 10^4 N/C

While the electric field generated by the second charge is
E_2 = (8.99 \cdot 10^9 N m^2 C^{-2} ) \frac{+60.0 \cdot 10^{-9} C}{(0.15 m)^2}=2.4 \cdot 10^4 N/C

Both charges are positive, this means that both electric fields are directed toward the charge. Therefore, at the point midway between the two charges the two electric fields have opposite direction, so the total electric field at that point is given by the difference between the two fields:
E=E_2 - E_1 = 2.4 \cdot 10^4 N/C - 1.6 \cdot 10^4 N/C = 8000 N/C
4 0
3 years ago
A 4.00 kg stone is dropped from a height of 145 m. What is the stone’s potential and kinetic energy respectively when it is 50.0
Lorico [155]

Answer:

C - 3,720 J, 1,950 J

Explanation:

Hello

Epotential = M*G*H

Ep = (4)(9.8)(145-50)

Ep = 3729 J

Then

Ec = 1950 J

Best regards

6 0
3 years ago
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