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4 years ago

What are safety goggles used for

Physics

2 answers:

77julia77 [94]4 years ago

7 0

Answer:

It is used to protect your eyes from chemical during science or used during construction since dust might go into eye. So both for eye protection.

Greeley [361]4 years ago

6 0

They are used to protect your eyes from any substance

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Suppose a plane accelerates from rest for 30 s, achieving a takeoff speed of 80 m/s after traveling a distance of 1200 m down th

Margaret [11]

Answer:

300 m

Explanation:

The train accelerate from the rest so u = 0 m/sec

Final speed that is v = 80 m/sec

Time t = 30 sec

The distance traveled by first plane = 1200 m

We know the equation of motion $S=ut+\frac{1}{2}at^2$ where s is distance a is acceleration and u is initial velocity

Using this equation for first plane $1200=0\times 30+\frac{1}{2}a30^2$

$a=2.67\frac{m}{sec^2}$

As the acceleration is same for both the plane so a for second plane will be 2.67 $\frac{m}{sec^2}$

The another equation of motion is $v^2=u^2+2as$ using this equation for second plane $40^2=0+2\times 2.67\times s$

s = 300 m

5 0

3 years ago

A motor keep a Ferris wheel (with moment of inertia 6.97 × 107 kg · m2 ) rotating at 8.5 rev/hr. When the motor is turned off, t

Talja [164]

Answer:

$P = 133.13 Watt$

Explanation:

Initial angular speed of the ferris wheel is given as

$\omega_i = 2\pi f$

$\omega_i = 2\pi(8.5/3600)$

$\omega_i = 0.015 rad/s$

final angular speed after friction is given as

$\omega_f = 2\pi f$

$\omega_f = 2\pi(7.5/3600)$

$\omega_f = 0.013 rad/s$

now angular acceleration is given as

$\alpha = \frac{\omega_f - \omega_i}{\Delta t}$

$\alpha = \frac{0.015 - 0.013}{15}$

$\alpha = 1.27 \times 10^{-4} rad/s^2$

now torque due to friction on the wheel is given as

$\tau = I \alpha$

$\tau = (6.97 \times 10^7)(1.27 \times 10^{-4})$

$\tau = 8875.3 N m$

Now the power required to rotate it with initial given speed is

$P = \tau \omega$

$P = 8875.3 \times 0.015$

$P = 133.13 Watt$

8 0

3 years ago

Match the indices of refraction with the corresponding effects on the waves.

Yanka [14]

Answer:

B) waves speed up

C) waves bend away from the normal

Explanation:

The index of refraction of a material is the ratio between the speed of light in a vacuum and the speed of light in that medium:

$n=\frac{c}{v}$

where

c is the speed of light in a vacuum

v is the speed of light in the medium

We can re-arrange this equation as:

$v=\frac{c}{n}$

So from this we already see that if the index of refraction is lower, the speed of light in the medium will be higher, so one correct option is

B) waves speed up

Moreover, when light enters a medium bends according to Snell's Law:

$n_1 sin \theta_1 = n_2 sin \theta_2$

where

$n_1 ,n_2$ are the index of refraction of the 1st and 2nd medium

$\theta_1,\theta_2$ are the angles made by the incident ray and refracted ray with the normal to the interface

We can rewrite the equation as

$sin \theta_2 = \frac{n_1}{n_2}sin \theta_1$

So we see that if the index of refraction of the second medium is lower ( $n_2$ ), then the ratio $\frac{n_1}{n_2}$ is larger than 1, so the angle of refraction is larger than the angle of incidence:

$\theta_2>\theta_1$

This means that the wave will bend away from the normal. So the other correct option is

C) waves bend away from the normal

3 0

4 years ago

When a piece of iron or mass 78 gm is put in a graduated cylinder containing 100 cm cube of water the reading of the cylinder be

Wittaler [7]

Answer:

Ro = 7.8 [g/cm³]

Explanation:

According to the principle of Archimedes, the volume of a body immersed in a liquid is equal to the volume displaced by water. That is, in this problem The displacement volume is equal to the new volume minus the original volume.

$V_{n}=110[cm^{3} ]\\V_{o}=100[cm^{3} ]\\V_{d}=110-100 = 10 [cm^{3} ]$

We now know that density is defined as the relationship between mass and volume.

$Ro = m/V_{d}$

where:

Ro = density [g/cm³]

m = mass = 78 [g]

Vd = displacement volume [cm³]

$Ro = 78/10\\Ro = 7.8 [g/cm^{3} ]$

7 0

3 years ago

Suppose you exert a 400-N force on a wall but the wall does not move. The work you are doing on the wall is : unknown, because t

Ilya [14]

Answer:

W = 0 :The work done on the wall is zero,because the wall is not moving

Explanation:

Work theory

Work is the product of a force applied to a body and the displacement of the body in the direction of this force.

W= F*d Formula (1)

W: Work (Joules) (J)

F: force applied (N)

d=displacement of the body (m)

The work is positive (W+) if the force goes in the same direction of movement.

The work is negative (W-)if the force goes in the opposite direction to the movement

Data

F= 400-N

d= 0

Problem development

We apply formula (1) to calculate the work done on the wall:

W= 400*0

W=0

6 0

3 years ago