Which planet would float if placed in water

. (a) How long can you play tennis on the 800 kJ (about 200 kcal) of energy in a candy bar? (b) Does this seem like a long time?

galina1969 [7]

(a). The power of the candy bar is,

$P=440\text{ W}$

The time taken to play on 800 kJ energy of the candy bar is,

$t=\frac{E}{P}$

where E is the energy.

Substituting the known values,

$\begin{gathered} t=\frac{800\times10^3}{440} \\ t=1.818\times10^3\text{ s} \\ t=1818\text{ seconds} \end{gathered}$

As 1 minute is equal to 60 seconds,

Thus,

$\begin{gathered} t=1818\text{ seconds} \\ t=\frac{1818}{60} \\ t=30.3\text{ min} \end{gathered}$

Thus, the time taken to play tennis on the 800 kJ energy is 30.3 minutes.

(b). By doing the exercise, the process of digestion of food inside our body increases. Thus, the exercise does not helps us to burn the calories. But it helps us to diggest the heavy meal like candy bar easily.

The time taken to digest the canndy bar or to utilise its energy is large because it takes a lot of time to burn small amount of food and make it digest quickly.

8 0

1 year ago

An RC circuit consists of a resistor with resistance 1.0 kΩ, a 120-V battery, and two capacitors, C1 and C2, with capacitances o

Serhud [2]

Answer:

$Q_t= 8.3 * 10^3 C$

Explanation:

From the question we are told that:

Resistor $R=1000ohms$

Voltage $v=120_V$

Capacitance of c_1 $c_1=20 \mu F$

Capacitance of c_2 $c_2=60 \mu F$

Time $t=0$

Generally the equation for charges is mathematically given by

$For C_1\\Charge\ on\ C_1 = CV = 20*120 = 2400 μC = 2.4 x 10^-3 C\\Charge\ on\ C_1 = 2400 μC = 2.4 x 10^-3 C\\Charge\ on\ C_1 = 2.4 x 10^-3 C\\$

$ForC_2\\Charge on C_2 = 60*120 =7200 μC = 7.2 x 10^-3\\Charge on C_2 = 7.2 x 10^-3$

Generally the equation for voltage across capacitors is mathematically given by

$V_c(t)=V(1-e^{-t/RC})$

$C=C_1+C_2=80 \mu f\\t=2RC=>160000s$

$V_c(t)=120(1-e^{-(160000)/1000*(80)})$

$V_c(t)=103.7598$

Generally the equation for charges is mathematically given by

$Q1(t) = C1Vc(t)\\Q1(t) = 20*103.7598\\Q1(t) = 2075.196\\\\Q2(t) = 60*103.7598\\Q2(t) = 6225.6\\$

Generally the equation for total charges $Q_t$ is mathematically given by

$Q_t=Q1(t)+Q2(t)$

$Q_t= 8.3 * 10^3 C$

5 0

3 years ago

47. the beam is supported by two rods ab and cd that have cross-sectional areas of 12mm^2 and 8mm^2, respectively. determine the

Ugo [173]

Let the beam is of length L

Now the stress on both the end is same

now we can say that torque on the beam due to two forces must be zero

$N_1* x = N_2* (L - x)$

also we know that stress at both ends are same

$\frac{N_1}{12} = \frac{N_2}{8}$

$2*N_1 = 3*N_2$

Now from two equations we have

$\frac{3}{2}N_2*x = N_2* (L - x)$

solving above equation we have

$x = \frac{2}{5}L$

<em>so the load is placed at distance 0.4L from the end of 12 mm^2 area</em>

8 0

3 years ago

Candace is curious about which brand of spot remover will best remove ketchup stains from her carpet. What is the dependent vari

Artemon [7]

A dependent variable could be oxi clean because it needs water to work.

4 0

3 years ago

One long wire carries a current of 30 A along the entire x axis. A second long wire carries a current of 40 A perpendicular to t

WINSTONCH [101]

Complete question is;

One long wire carries a current of 30 A along the entire x axis. A second long wire carries a current of 40 A perpendicular to the xy plane and passes through the point (0, 4, 0) m. What is the magnitude of the resulting magnetic field at the point y = 2.0 m on the y axis?

Answer:

B_net = 50 × 10^(-7) T

Explanation:

We are told that the 30 A wire lies on the x-plane while the 40 A wire is perpendicular to the xy plane and passes through the point (0,4,0).

This means that the second wire is 4 m in length on the positive y-axis.

Now, we are told to find the magnitude of the resulting magnetic field at the point y = 2.0 m on the y axis.

This means that the position we want to find is half the length of the second wire.

Thus, at this point the net magnetic field is given by;

B_net = √[(B1)² + (B2)²]

Where B1 is the magnetic field due to the first wire and B2 is the magnetic field due to the second wire.

Now, formula for magnetic field due to very long wire is;

B = (μ_o•I)/(2πR)

Thus;

B1 = (μ_o•I_1)/(2πR_1)

Also, B2 = (μ_o•I_2)/(2πR_2)

Now, putting the equation of B1 and B2 into the B_net equation, we have;

B_net = √[((μ_o•I_1)/(2πR_1))² + ((μ_o•I_2)/(2πR_2))²]

Now, factorizing out some common terms, we have;

B_net = (μ_o/2π)√[((I_1)/R_1))² + ((I_2)/R_2))²]

Now,

μ_o is a constant and has a value of 4π × 10^(−7) H/m

I_1 = 30 A

I_2 = 40 A

Now, as earlier stated, the point we are looking for is 2 metres each from wire 2 end and wire 1.

Thus;

R_1 = 2 m

R_2 = 2 m

So, let's calculate B_net.

B_net = ((4π × 10^(−7))/2π)√[(30/2)² + (40/2)²]

B_net = 50 × 10^(-7) T

5 0

3 years ago