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3 years ago

9

A paper weight is dropped from the roof of a block of multi story flats. Each storey being 3 metre high. It passes the ceiling o

f the 20th storeys at 30m/sec. If g = 10.0m/sec^2, How many storeys does the flat have.

1 answer:

docker41 [41]3 years ago

6 0

Use equations of motion to get the correct answer!

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Three point charges are placed at distances of d , 2 d , and 3 d from a point P. The particle that is 2 d away from P has a char

mel-nik [20]

Answer:

at d the charge will be 3q and at 3d it will be 9q

Explanation:

for V=Vp-V2d

V=KQ/d=K*6q/2d=3kq/d for potential to 2d at 6q be zero the Vp will equal 3kq/d; hence at d, Q=3q and at 3d, Q=9q

7 0

3 years ago

A charge +q is located at the origin, while an identical charge is located on the x axis at x = +0.57 m. A third charge of +2q i

Jlenok [28]

Answer:

The third charge placed is 0.80 m.

Explanation:

Given that,

Distance = 0.57 m

First charge = q

Third charge = 2q

We need to calculate the electrostatic force on charge q₁ due to q₂

Using formula of electrostatic force

$F_{21}=\dfrac{kq_{1}q_{2}}{r_{1}^2}$

When placed another charge q₃ at certain distance from origin, then the net force on charge q₁ due to both charges is

$F_{net}=F_{21}+F_{31}$

The net electrostatic force on the charge at the origin doubles.

$2F_{21}=F_{21}+F_{31}$

$F_{31}=F_{21}$

$\dfrac{kq_{3}q_{1}}{r_{2}^2}=\dfrac{kq_{2}q_{1}}{r_{1}^2}$

$\dfrac{q_{3}}{r_{2}^2}=\dfrac{q_{2}}{r_{1}^2}$

$r_{2}^2=\dfrac{q_{3}}{q_{2}\timesr_{1}^2}$

$r_{2}=\sqrt{\dfrac{q_{3}}{q_{2}}}r_{1}$

Put the value into the formula

$r_{2}=\sqrt{\dfrac{2q}{q}}\times0.57$

$r_{2}=\sqrt{2}\times0.57$

$r_{2}=0.80\ m$

Hence, The third charge placed is 0.80 m from origin in x-axis.

4 0

3 years ago

Modifiable strength improvement factors include all of the following except...??

Alla [95]

Answer:

Muscle size and training specifiicty

Explanation:

It depends on all other factors such as anatomy and psychology, muscle fiber recruitment, motor pattern development and muscle memory but to increase modifiable strength, size plays no big role.

5 0

4 years ago

An ammeter is connected in series with a resistor of unknown resistance R and a DC power supply of known emf e. A student uses t

rosijanka [135]

Answer:

B:The actual power dissipated by the resistor is less than P because the ammeter had some resistance.

Explanation:

Here,power has been calculated using current I and total EMF \ε . So,P=EMF*current= ε I will represent total power dissipated in resistor and ammeter.

Now, this total power P has been dissipated in both resistor and ammeter. So, power dissipated in resistor must be less than P as some power is also dissipated in ammeter because it has non-zero resistance.

So, the answer is B:The actual power dissipated by the resistor is less than P because the ammeter had some resistance.

Note that option A,C and E are ruled out as they state power dissipated by resistor is greater than or equal to P which is false.

Also,option D is ruled out as ammeter is connected in series.

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4 years ago

How are temperature, pressure, and volume related dealing with the behavior of a gas

fredd [130]

Pressure and volume of a gas are inversely related. As one goes up, the other goes down, and vice-versa.

7 0

4 years ago