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2 years ago

You're driving your new sports car at 85 mph over the top of a hill that has a radius of curvature of 525 m.

Physics

1 answer:

Bumek [7]2 years ago

5 0

Explanation:

It is given that,

Speed of the sports car, v = 85 mph = 37.99 m/s

The radius of curvature, r = 525 m

Let $W_N$ is the normal weight and $W_A$ is the apparent weight of the person. Its apparent weight is given by :

$W_A=mg-\dfrac{mv^2}{r}$

So, $\dfrac{W_A}{W_N}=\dfrac{mg-\dfrac{mv^2}{r}}{mg}$

$\dfrac{W_A}{W_N}=\dfrac{g-\dfrac{v^2}{r}}{g}$

$\dfrac{W_A}{W_N}=\dfrac{9.8-\dfrac{(37.99)^2}{525}}{9.8}$

$\dfrac{W_A}{W_N}=0.719$

$\dfrac{W_A}{W_N}=71.9\%$

Hence, this is the required solution.

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Why do most objects tend to contain nearly equal numbers of positive and negative charges?

mart [117]

Most objects tend to contain the same numbers of positive and negative charge because this is the most stable situation. In fact, if an object has an excess of positive charge, it tends to attract an equal number of negative charges to balance this effect and restore neutrality: the attracted negative charges combine with the excess of positive charges, leaving the object electrically neutral.

3 0

2 years ago

Read 2 more answers

A point charge Q is held at a distance r from the center of a dipole that consists of two charges ±qseparated by a distance s. T

atroni [7]

Answer:

The magnitude of the force on the dipole due to the charge Q = $\rm \dfrac{1}{\epsilon_o}\times \dfrac{1}{4\pi }\dfrac{2qQs}{r^3}.$

The magnitude of the torque on the dipole = $\rm \dfrac{1}{\epsilon_o}\times \dfrac{1}{4\pi}\dfrac{2qQs^2}{r^3}.$

Explanation:

Given that a point charge Q is held at a distance r from the center of a dipole that consists of two charges ±q, separated by a distance s and the charge Q is located in the plane that bisects the dipole.

The magnitude of the electric field that the dipole exerts at the position where the charge Q is held is given by

$\rm E = \dfrac{k2qs}{(r^2+s^2)^{3/2}}.$

where,

k is the Coulomb's constant, having value = $\dfrac{1}{4\pi \epsilon_o}$

$\epsilon_o$ is the electrical permittivity of free space.

Also, r>>s, therefore, $\rm r^2+s^2\approx r^2.$

$\rm E = \dfrac{k2qs}{(r^2)^{3/2}}=\dfrac{k2qs}{r^3}.$

The magnitude of the electric force F on a charge q placed at a point and the magnitude of the electric field E at that point are related as

$\rm F=qE$

Therefore, the electric force on the charge Q due to the dipole is given by

$\rm F=Q\dfrac{k2qs}{r^3}=\dfrac{1}{4\pi \epsilon_o}\dfrac{2qQs}{r^3}.$

According to Newton's third law of motion, the magnitude of the force exerted by the dipole on the charge Q is same as the magnitude of the force exerted by the charge on the dipole.

Thus, the magnitude of the force on the dipole due to the charge Q = $\dfrac{1}{\epsilon_o}\times \dfrac{1}{4\pi }\dfrac{2qQs}{r^3}.$

The magnitude of the torque on the dipole is given by

$\rm \tau = Fs\ \sin\theta$

Since the charge Q is placed in the plane that bisects the dipole, therefore, $\theta = 90^\circ$ .

$\rm \tau = \dfrac{1}{4\pi \epsilon_o}\dfrac{2qQs}{r^3}\cdot s\cdot 1=\dfrac{1}{4\pi \epsilon_o}\dfrac{2qQs^2}{r^3}.$

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2 years ago

Four wrestlers step on a scale for a tournament. Wrestler A weighs in at 170 pounds; wrestler B weighs in at 168 pounds; wrestle

OLga [1]

The force due to gravity is equal to the product of the mass and acceleration due to gravity. Since the acceleration due to gravity is constant, we just have to rely on the comparison of their masses to determine the mass effect. Thus, the answer to this item is Wrestler C weighing 171 pounds.

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2 years ago

Read 2 more answers

Over a period of more than 30 years, albert klein of california drove 2.5 × 106 km in one automobile. consider two charges, q1 =

Semenov [28]

For q3 to be in equilibrium the total force acting on it has to be zero.
Let's say that total distance traveled by car is L (this is just for the convenience).
We can set up a system of equations to find an answer. Let's say that from q1 to q3 the distance is r_1 and from q3 to q2 the distance is r_2, we know that this distance has to be equal to:
$r_1+r_2=L km$
The second equation is going to the total force acting on the charge q3:
$F_{q3}=F_{q3q1}+F_{q3q2}=0\\ 0=k_c\frac{q_1q_3}{r_1^2}+k_c\frac{q_3q_2}{r^2}$
k_c is the Coulomb's constant. Since left-hand side is zero we just divide whole equation with k_c to get rid of it:
$0=\frac{q_1q_3}{r_1^2}+\frac{q_3q_2}{r^2}$
Let's solve this for r_1^2:
$0=\frac{8}{r_1^2}+\frac{24}{r^2}\\ \frac{1}{r_1^2}=-\frac{3}{r^2}\\ r_1^2=-\frac{r^2}{3};r_2=L-r_1\\ r_1^2=\frac{(L-r_1)^2}{3}\\ r_1^2=\frac{L^2-2Lr_1+r_1^2}{3}\\ 3r_1^2=L^2-2Lr_1+r_1^2\\ 2r_1^2+2Lr_1-L^2=0$
Now we have a quadratic equation with following parameter:
$a=2\\ b=2L\\ c=-L^2$
We know that two solutions are:
$r_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ r_{1,\:2}=\frac{-2L\pm \sqrt{4L^2+8L^2}}{4}\\ r_{1,\:2}=\frac{-2L\pm \sqrt{12L^2}}{4}\\$
We need a positive solution. When we plug in all the numbers we get:
$r_1=0.915\cdot 10^6$km$

6 0

2 years ago

A submarine dives from rest a 100-m distance beneath the surface of an ocean. Initially, the submarine moves at a constant rate

tiny-mole [99]

Answer:

a. Time = 16.11 s

b. Gauge Pressure = 1009400 Pa = 1 MPa

c. Absolute Pressure = 1110725 Pa + 1.11 MPa

d. Force = 2.22 MN

Explanation:

For the accelerated part of motion of submarine we can use equations of motion.

Using 1st equation of motion:

Vf = Vi + at₁

t₁ = (Vf - Vi)/a

where,

t₁ = time taken during accelerated motion = ?

Vf = final velocity = 4 m/s

Vi = Initial Velocity = 0 m/s (Since, it starts from rest)

a = acceleration = 0.3 m/s²

Therefore,

t₁ = (4 m/s - 0 m/s)/(0.3 m/s²)

t₁ = 13.33 s

Now, using 2nd equation of motion:

d₁ = (Vi)(t₁) + (0.5)(a)(t₁)²

where,

d₁ = the depth covered during accelerated motion

Therefore,

d₁ = (0 m/s)(13.33 s) + (0.5)(0.3 m/s²)(13.33 s)²

d₁ = 88.89 m

Hence,

d₂ = d - d₁

where,

d₂ = depth covered during constant speed motion

d = total depth = 100 m

Therefoe,

d₂ = 100 m - 88.89 m

d₂ = 11.11 m

So, for uniform motion:

s₂ = vt₂

where,

v = constant speed = 4 m/s

t₂ = time taken during constant speed motion

11.11 m = (4 m/s)t₂

t₂ = 2.78 s

Therefore, total time taken by submarine to move down 100 m is:

t = t₁ + t₂

t = 13.33 s + 2.78 s

t = 16.11 s

The gauge pressure on submarine can be calculated by the formula:

Pg = ρgh

where,

Pg = Gauge Pressure = ?

ρ = density of salt water = 1030 kg/m³

g = 9.8 m/s²

h = depth = 100 m

Therefore,

Pg = (1030 kg/m³)(9.8 m/s²)(100 m)

Pg = 1009400 Pa = 1 MPa

The absolute pressure is given as:

P = Pg + Atmospheric Pressure

where,

P = Absolute Pressure = ?

Atmospheric Pressure = 101325 Pa

Therefore,

P = 1009400 Pa + 101325 Pa

P = 1110725 Pa + 1.11 MPa

Since, the force to open the door must be equal to the force applied to the door by pressure externally.

Therefore, the force required to open the door can be found out by the formula of pressure:

P = F/A

F = PA

where,

P = Absolute Pressure on Door = 1110725 Pa

A = Area of door = 2 m²

F = Force Required to Open the Door = ?

Therefore,

F = (1.11 MPa)(2 m²)

F = 2.22 MN

8 0

2 years ago