PHYSICS HIGH SCHOOL

you exert a force of 30 n on the head of a thumbtack. the head of the thumbtack has a radius of 5 mm. what is the pressure on your thumb?

Answers

Answer 1
Answer:

The pressure exerted on your thumb by the thumbtack is approximately 3.818 x 10^5 Pascal (Pa).

To calculate the pressure on your thumb, we can use the formula:

Pressure = Force / Area

Force (F) = 30 N

Radius of the thumbtack head (r) = 5 mm = 0.005 m

First, we need to calculate the area of the thumbtack head. Since the head is circular, the area can be found using the formula for the area of a circle:

Area = \pi * r^2

Area = \pi * (0.005 m)^2

Area ≈ 7.854 x 10^((-5)) {m^2}

Now we can calculate the pressure:

Pressure = Force / Area

Pressure = 30 N / 7.854 x 10^(-5) m^2

Pressure ≈ 3.818 x 10^5 Pa

To know more about pressure refer here

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8. A 740 kg car traveling 19 m/s comes to a complete stop in 2.0 s. What is the force exerted on the car during this stop?

Answers

Answer:

Force exerted on the car is 7030 N.

Explanation:

It is given that,

Mass of the car, m = 740 kg

Initial speed of the car, u = 19 m/s

Final speed of the car, v = 0

Time taken, t = 2 s

Let F is the force exerted on the car during this stop. We know that it is equal to the product of force and acceleration. Mathematically, it is given as :

F=m* (v-u)/(t)

F=740* (0-19)/(2)

F = -7030 N

So, the force exerted on the car during this stop is 7030 N. Hence, this is the required solution.
force = mass * acceleration 

acceleration = change_in_velocity / time

so:

force = 740 kg * (19 m/s - 0 m/s) / 2.0 s
= 740 * 19 / 2 kg m per second^{2}
= 7030 kg m per second^{2}
= 7030 newtons of force

An object is 6.0cm in front of a converging lens with a focal length of 10cm .Part A
Use ray tracing to determine the location of the image.
Express your answer using two significant figures.
q =
Part B
Is the image upright or inverted?
Part C
Is the image real or virtual?

Answers

Answer:  A) see attach file .

B) the iamgen is uprigth

C) the imagen is virtual

Explanation: see attach file.

A proton moves along the x axis according to the equation x = 38 t + 14 t2, where x is in meters and t is in seconds. Calculate (a) the average velocity of the proton during the first 3.0 s of its motion, (b) the instantaneous velocity of the proton at t = 3.0 s, and (c) the instantaneous acceleration of the proton at t = 3.0 s.

Answers

Answer:

a) 240 m

b) 122 m/s

c) 28 m/s²

Explanation:

Given:

Equation for motion

x = 38t + 14t²

a) average velocity during first 3 seconds

average velocity = \frac{\textup{change in displacement}}{\textup{cahnge in time}}

now,

distance, at t = 0 s

x = 38 × 0 + 14 × 0² = 0 m

distance, at t = 3 s

x = 38 × 3 + 14 × 3² = 240 m

therefore,

average velocity = (240-0)/(3-0)  = 80 m/s

b) instantaneous velocity of the proton at t = 3.0 s

Instantaneous velocity, v = (dx)/(dt)=38+28* t

or

Instantaneous velocity, v = (dx)/(dt)=38+28* 3

= 122 m/s

c) instantaneous acceleration of the proton at t = 3.0 s

Now,

Acceleration = (dv)/(dt)  = 0 + 28 = 28 m/s²

The height of a wave from the equilibrium is 1 meter. Which parameter of the wave does this distance represent

Answers

Answer:

The amplitude of the wave

Explanation:

A wave can be defined by using several characteristics:

- The amplitude of a wave is the maximum displacement of the wave from its equilibrium position

- The wavelength of a wave corresponds to the distance between two adjacent points of the wave with same shape (e.g. the distance between two consecutive crests)

- The period of a wave corresponds to the time taken for one complete oscillation of the wave to occur

- The frequency of a wave corresponds to the number of complete oscillations of the wave in one second

In this problem, we are told that the height of the wave from equilibrium is 1 meter: based on the definitions above, we can say that this parameter corresponds to the amplitude of the wave.
The maximum height that a wave reaches above or below
its equilibrium level is the "amplitude" of the wave.

Two cylindrical resistors are made from same material and have the same length. When connected across the same battery, one dissipates twice as much power as the other. How do their diameters compare ?

Answers

Answer:

(d_2)/(d_1) = \sqrt2 = 1.41

Explanation:

As resistor is connected to the battery of constant EMF then the power across the resistor is given as

P = (E^2)/(R)

now if two resistors are made up of same material and of same length then due to different cross sectional area they both have different resistance

Due to different resistance they both will have different power

Since power is inversely depends on the resistance

So if the power is twice that of the other then the resistance must be half

so we have

R_1 = \rho (L)/(A_1)

R_2 = \rho(L)/(A_2)

since one resistance is half that of other resistance

So the area of one must be twice that of other

so we have

(A_2)/(A_1) = 2

(\pi d_2^2)/(\pi d_1^2) = 2

d_2 = 1.41 d_1

A cheetah runs 900 meters in 30 seconds. how fast is it going?

Answers

900 meters/30 seconds= 30 meters/second
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