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3 years ago

14

Which of the following would be an example of basic research

2 answers:

abruzzese [7]3 years ago

5 0

Basic research, also called pure research or fundamental research, is scientific research aimed to improve scientific theories for improved understanding or prediction of natural or other phenomena.

(source: wikipedia)

olga55 [171]3 years ago

4 0

Answer: Newton's discoveries of the laws of motion.

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The current and the potential difference in an inductor are in phase. B. The current lags the potential difference by π/2 in an

slava [35]

Answer:

The current lags the potential difference by π/2 in an inductor

Explanation:

The potential difference leads to the current by $\frac{\pi}{2}$ . Alternate signals such as current and voltage -in this case- are periodic, this means that this signals are repeated at fixed spaces of time. Thus, In an inductor the current lags the potential difference by $\frac{\pi}{2}$ .

6 0

3 years ago

Help mee! Physics i think!

xxTIMURxx [149]

Answer:

Use the form of equation:

Q=mL

You have the specific latent heat of vaporization L = 2.260*10^{6}

And Q, the heat energy supplied, which equals 1695 KJ = 1695*10^{3} J

So you can get the mass by substitution in the formula below.

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3 years ago

A moving curling stone, A, collides head on with stationary stone, B. Stone B has a larger mass than stone A. If friction is neg

Kitty [74]

Answer:

The correct answer is option 'c': Smaller stone rebounds while as larger stone remains stationary.

Explanation:

Let the velocity and the mass of the smaller stone be 'm' and 'v' respectively

and the mass of big rock be 'M'

Initial momentum of the system equals

$p_i=mv+0=mv$

Now let after the collision the small stone move with a velocity v' and the big roch move with a velocity V'

Thus the final momentum of the system is

$p_f=mv'+MV'$

Equating initial and the final momenta we get

$mv=mv'+MV'\\\\m(v-v')=MV'.....i$

Now since the surface is frictionless thus the energy is also conserved thus

$E_i=\frac{1}{2}mv^2$

Similarly the final energy becomes

$E_f=\frac{1}{2}mv'^2+\frac{1}{2}MV'^2$ \

Equating initial and final energies we get

$\frac{1}{2}mv^2=\frac{1}{2}mv'^2+\frac{1}{2}MV'^2\\\\mv^2=mv'^2+MV'^2\\\\m(v^2-v'^2)=MV'^2\\\\m(v-v')(v+v')=MV'^2......(ii)$

Solving i and ii we get

$v+v'=V'$

Using this in equation i we get

$v'=\frac{v(m-M)}{(M-m)}=-v$

Thus putting v = -v' in equation i we get V' = 0

This implies Smaller stone rebounds while as larger stone remains stationary.

4 0

3 years ago

1. A 12 kg barrel is pulled up by a rope. The barrel accelerates at 1.2 m/s2. Find the force exerted by the rope. Show all your

allochka39001 [22]

F-mg=ma
F-(12kg)(9.8m/s^2)=(12kg)(1.2m/s^2)
F-117.6N=14.4N
F=132 Newtons

5 0

4 years ago

A seesaw pivots as shown in below. What is the net torque about the pivot point?

rodikova [14]

Answer:

2.3 Nm clockwise

Explanation:

Take counterclockwise to be positive and clockwise to be negative.

∑τ = (3 N) (2.5 m) − (7 N) (1.4 m)

∑τ = 7.5 Nm − 9.8 Nm

∑τ = -2.3 Nm

The net torque is 2.3 Nm clockwise.

7 0

4 years ago