A motorboat is a lot heavier than a pebble. Why does the boat float?

Answer:

lithium

Explanation:

I took physical science 2 years ago and passed with an A

Analysing the question:

Since the stone was dropped, there was no initial velocity applied on it and hence it's initial velocity of the stone is 0 m/s

We are given:

height of the tower (h) = 66 m

mass of the stone (m) = 0.5 kg

initial velocity of the stone (u) = 0 m/s

time taken by the stone to reach the ground (t) = t seconds

acceleration due to gravity = 10 m/s²

** Neglecting air resistance**

Finding the time taken by the stone to reach the ground:

from the second equation of motion

h = ut + 1/2at²

replacing the variables

66 = (0)(t) + 1/2 (10)(t)²

66 = 5t²

t² = 13.2

t = 3.6 seconds

I initially wanted to subtract the height of the student from the height of the tower since the time i calculated is the time taken by the stone to reach the ground and that means that the stone has already hit the student before 3.6 seconds

but since we were NOT given the height of a student, the person who posed this question wants the time taken by the stone to reach the ground and that is what we solved

Answer:

hello your question is incomplete attached below is the complete question

21) The current at points B and C would be the same ( identical bulbs) while the current at Point A will be greater than the currents at point B and C. i.e. twice the current at either point B or point C

22) The potential difference across the bulbs will be the same and this is because the bulbs are connected in parallel to the the power source ( battery)

hence the voltage in the battery will be equal to the voltage across each bulb

Explanation:

The current at points B and C would be the same ( identical bulbs) while the current at Point A will be greater than the currents at point B and C. i.e. twice the current at either point B or point C

The potential difference across the bulbs will be the same and this is because the bulbs are connected in parallel to the the power source ( battery)

hence the voltage in the battery will be equal to the voltage across each bulb

Answer:

F = 58.35 [N]

Explanation:

To solve this problem we must use Newton's second law, which tells us that force is equal to the product of mass by acceleration. But first we must use the following equation of kinematics.

We have to convert speeds from kilometers per hour to meters per second

[tex]\frac{26km}{hr}*\frac{1000m}{1km}*\frac{1hr}{3600s}=\frac{7.22m}{s} \\\frac{54km}{hr}*\frac{1000m}{1km}*\frac{1hr}{3600s}=15\frac{m}{s}[/tex]

[tex]v_{f}=v_{o}+(a*t) \\[/tex]

where:

Vf = final velocity = 15 [m/s]

Vi = initial velocity = 7.22 [m/s]

a = acceleration [m/s^2]

t = time = 4 [s]

Note: the positive sign of the above equation is because the car increases its speed

15 = 7.22 + (a*4)

a = 1.945 [m/s^2]

Now we can use the Newton's second law:

F = m*a

F = 30*1.945

F = 58.35 [N]

Answer:

8.93*10^13 N.

Explanation:

       [tex]F_{g}= \frac{G*m_{1}*m_{s}}{r_{s}^{2} }[/tex]

       [tex]F_{g}= \frac{6.67e-11Nm^2/kg^2*1.99e30 kg*67.3 kg}{10e4m^2} = 8.93e13 N[/tex]

Answer:

The charge pass through your hair dryer is 3000 C.

Explanation:

Given that,

Power = 1200 W

Voltage = 120 V

Flow time = 5 min

We need to calculate the current

Using formula of power

[tex]P=VI[/tex]

[tex]I=\dfrac{P}{V}[/tex]

Put the value into the formula

[tex]I=\dfrac{1200}{120}[/tex]

[tex]I=10\ A[/tex]

We need to calculate the charge pass through your hair dryer

Using formula of current

[tex]I=\dfrac{Q}{t}[/tex]

[tex]Q=It[/tex]

Put the value into the formula

[tex]Q=10\times5\times60[/tex]

[tex]Q=3000\ C[/tex]

Hence, The charge pass through your hair dryer is 3000 C.

Answer:

The voltage is   [tex]V =   0.993V_b[/tex]

Explanation:

From the question we are told that

   The time that has passed is  [tex]t = \frac{\tau}{2}[/tex]

 Here [tex]\tau[/tex] is know as the time constant

    The voltage of the  power source is   [tex]V_b[/tex]

Generally the voltage equation for charging a capacitor is mathematically represented as

       [tex]V =  V_b  [1 - e^{- \frac{t}{\tau} }][/tex]

=>   [tex]V =  V_b  [1 - e^{- \frac{\frac{\tau}{2}}{\tau} }][/tex]

=>   [tex]V =  V_b  [1 - e^{- \frac{\tau}{2\tau} }][/tex]

=>   [tex]V =  V_b  [1 - e^{- \frac{1}{2} }][/tex]

=>   [tex]V =   0.993V_b[/tex]    

[tex]a = \frac{vf - vi}{t} [/tex]

here initial velocity vi=0 as ball release from rest

the final velocity is vf=4.0

time is t=6

so putting all these values in above equation

[tex]a = \frac{ 4.0- 0}{6} [/tex]

[tex]a = 0.6667m \s {}^{2} [/tex]