PHYSICS COLLEGE

Bill and Janet are pulling on opposite sides of a table. Bill pulls with a force of450 N to the left, and Janet pulls with a force of 300 N to the right. What is
the net force on the table?
O
A. 450 N to the right
O
B. 450 N to the left
C. 150 N to the left
O
D. 150 N to the right

Answers

Answer 1

Answer:

Answer:

Explanation:

It is given that,

Force acting in left side, F = 450 N

Force acting in right side, F' = 300 N

Let left side is taken to be negative while right side is taken to be positive. So,

F = -450 N

F' = +300 N

The net force will act in the direction where the magnitude of force is maximum. Net force is given by :

$F_(net)=-450\ N+300\ N$

$F_(net)=-150\ N$

So, the net force on the table is 150 N and it is acting to the left side. Hence, the correct option is (c).

Answer 2

Answer:

Answer:

C. 150 N to the left

Explanation:

If we take right to be positive and left to be negative, then:

∑F = -450 N + 300 N

∑F = -150 N

The net force is 150 N to the left.

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Answers

Answer:

Explanation:

Each plate of a parallel‑plate capacitor is a square of side 0.0479 m, and the plates are separated by 0.479 × 10 − 3 m. The capacitor is charged and stores 8.11 × 10 − 9 J of energy. Find the electric field strength E inside the capacitor.

Answers

Explanation:

It is known that the relation between electric field and potential is as follows.

E = $(V)/(d)$

And, formula to calculate the capacitance is as follows.

C = $(\epsilon_(o) A)/(d)$

= $(8.854 * 10^(-12) * (0.479 m)^(2))/(0.479 * 10^(-3))$

= $4.24 * 10^(-9)$ F

Hence, energy stored in a capacitor is as follows.

W = $(1)/(2)CV^(2)$

V = $\sqrt{(2W)/(C)}$

E = $\sqrt{(2W)/(d^(2)C)}$

= $(2 * 8.11 * 10^(-9) J)/((0.479 * 10^(-3))^(2) * 4.24 * 10^(-9))$

= $16.687 * 10^(3) N/C$

Thus, we can conclude that electric field strength E inside the capacitor is $16.687 * 10^(3) N/C$ .

A container has a large cylindrical lower part with a long thin cylindrical neck open at the top. The lower part of the container holds 22.8 m^3 of water and the surface area of the bottom of the container is 9.10 m^2. The height of the lower part of the container is 2.50 m, and the neck contains a column of water 8.50 m high. The total volume of the column of water in the neck is 0.200 m^3.(a) What is the magnitude of the force exerted by the water on the bottom of the container? (b) Explain why it is not equal to the weight of the water.

Answers

Answer:

(a) 1045.5 KN

(b) 225.63 KN

Explanation:

Since Pressure, $P=\frac {F}{A}$ where F is force exerted and A is area of the bottom of container

Making F the subject then

F=PA

Height of container=8.5m+2.5m=11.0 m

Density of water $\rho_(water)=1000 Kg/m^(3)$

Surface area of the bottom of the container is $9.10 m^(2)$

Pressure at the bottom of container

$P=P_(atm)+h\rho_(water) g$ where $P_(atm)$ is atmospheric pressure taken as $101.3*10^(3) Kg/m.s^(2)$ , h is height which is 11 m, $\rho_(water)$ is density of water and g is acceleration due to gravity which is taken as $9.81 m/s^(2)$

$P=101.3*10^(3) Kg/m.s^(2) +11m*1000 Kg/m^(3) *9.81 m/s^(2)=209.1*10^(3) Kg/m.s^(2)$

Force exerted is then found by

$F=PA=209.1*10^(3) Kg/m.s^(2)*5 m^(2)=104.55*10^(4) N$

Therefore, force at the bottom is 1045.5 KN

(b)

Volume of container at lower part is given as 22.8 cubic meters hence mass of water =volume*density of water

Mass=22.8*1000=22800 Kg

Volume of water confined in the column is 0.2 cubic meters hence the mass of water confined in the column is 0.2*1000=200 Kg

Total mass=200+22800=23000 Kg

Weight of water, W=mg=23000*9.81=225630 N=225.63 KN

Therefore, the weight of water is less than force applied at the bottom of container since pressure exerted by atmosphere on the surface of water is considered during calculation of force exerted at the bottom of the container

The elasticity of demand for gasoline has been estimated to be 2.0, and the standard error is 1.0. The upper and lower bounds on the 95 percent confidence interval for the elasticity of demand for gasoline are:

Answers

Answer:

The confidence interval for the mean is given by the following formula:

$\bar X \pm z_(\alpha/2)(\sigma)/(√(n))$ (1)

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that $z_(\alpha/2)=1.96$

The standard error for this case is given:

$SE =(\sigma)/(√(n))=1$

Now we have everything in order to replace into formula (1):

$2-1.96*11=0.04$

$2+1.96*1=3.96$

Explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X =2$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size

Calculate the confidence interval

The confidence interval for the mean is given by the following formula:

$\bar X \pm z_(\alpha/2)(\sigma)/(√(n))$ (1)

The standard error for this case is given:

$SE =(\sigma)/(√(n))=1$

Now we have everything in order to replace into formula (1):

$2-1.96*11=0.04$

$2+1.96*1=3.96$

Two identical small charged spheres are a certain distance apart, and each one initially experiences an electrostatic force of magnitude F due to the other. With time, charge gradually leaks off of both spheres. Calculate the magnitude of the electrostatic force, when each of the spheres has lost half of its initial charge. (Your answer will be a function of F, since no values are giving)

Answers

Answer:

1/4F

Explanation:

We already know thatThe electrostatic force is directly proportional to the product of the charge, from Coulomb's law.

So F α Qq

But if it is now half the initial charges, then

F α (1/2)Q *(1/2)q

F α (1/4)Qq

Thus the resultant charges are each halved is (1/4) and the first initial force experienced at full charge.

Thus the answer will be 1/4F

A weather balloon is designed to expand to a maximum radius of 16.2 m when in flight at its working altitude, where the air pressure is 0.0282 atm and the temperature is â65âC. If the balloon is filled at 0.873 atm and 21âC, what is its radius at lift-off?

Answers

Answer:

5.78971 m

Explanation:

$P_1$ = Initial pressure = 0.873 atm

$P_2$ = Final pressure = 0.0282 atm

$V_1$ = Initial volume

$V_2$ = Final volume

$r_1$ = Initial radius = 16.2 m

$r_2$ = Final radius

Volume is given by

$(4)/(3)\pi r^3$

From the ideal gas law we have the relation

$(P_1V_1)/(T_1)=(P_2V_2)/(T_2)\n\Rightarrow (0.873* (4)/(3)\pi r_1^3)/(294.15)=(0.0282(4)/(3)\pi r_2^3)/(208.15)\n\Rightarrow (0.873r_1^3)/(294.15)=(0.0282* 16.2^3)/(208.15)\n\Rightarrow r_1=(0.0282* 16.2^3* 294.15)/(208.15* 0.873)\n\Rightarrow r_1=5.78971\ m$

The radius of balloon at lift off is 5.78971 m

Final answer:

To find the radius of the weather balloon at lift-off, the ideal gas law can be used. Using the equation P1V1 = P2V2, where P1 and V1 are the initial pressure and volume, and P2 and V2 are the pressure and volume at lift-off, the radius at lift-off can be calculated to be approximately 4.99 m.

Explanation:

To find the radius of the weather balloon at lift-off, we can use the ideal gas law, which states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.

In this case, we know that the number of moles is constant, as the balloon is filled with the same amount of helium at lift-off and in flight. Therefore, we can write the equation as P1V1 = P2V2, where P1 and V1 are the initial pressure and volume, and P2 and V2 are the pressure and volume at lift-off.

Plugging in the given values, we have (0.873 atm)(V1) = (0.0282 atm)(16.2 m)^3. Solving for V1, we find that the volume at lift-off is approximately 110.9 m^3. The radius can then be calculated using the formula for the volume of a sphere: V = (4/3) * π * r^3, where r is the radius.

Therefore, the radius at lift-off is approximately 4.99 m.