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3 years ago

What are the top three osha cited ladder violations

Physics

1 answer:

bonufazy [111]3 years ago

6 0

Answer:

Explanation:

1. FALL PROTECTION-GENERAL REQUIREMENTS (29 CFR 1926.501) 6,010 VIOLATIONS

2. HAZARD COMMUNICATION (29 CFR 1910.1200). 3,671

3. SCAFFOLDING (29 CFR 1926.451). 2,813

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A football player kicks a football in a field goal attempt. When the football reaches its maximum height, what is the relationsh

Stels [109]

At position of maximum height we know that the vertical component of its velocity will become zero

so the object will have only horizontal component of velocity

so at that instant the motion of object is along x direction

while if we check the acceleration of object then it is due to gravity

so the acceleration of object is vertically downwards

so it is along y axis

so here these two physical quantities are perpendicular to each other

so correct answer would be

<em>C)At the maximum height, the velocity and acceleration vectors are perpendicular to each other. </em>

3 0

3 years ago

Read 2 more answers

In the aerials competition in skiing, the competitors speed down a ramp that slopes sharply upward at the end. The sharp upward

11111nata11111 [884]

Answer:

u = 11.6 m/s

Explanation:

The end of a launch ramp is directed 63° above the horizontal. A skier attains a height of 10.9 m above the end of the ramp.

Maximum height, H = 10.9

Let v is the launch speed of the skier. The maximum height attained by the projectile is given by :

$H=\dfrac{u^2\ sin^2\theta}{g}$

$10.9=\dfrac{u^2\ sin^2(63)}{9.8}$

u = 11.6 m/s

So, the launch speed of the skier is 11.6 m/s. Hence, this is the required solution.

4 0

3 years ago

Electrons and protons travel from the Sun to the Earth at a typical velocity of 3.83 ✕ 105 m/s in the positive x-direction. Thou

Leona [35]

Answer:

$F=2.84*10^{-26}N$ & -y direction

$F=2.84*10^{-26}N$ & +y direction

Explanation:

From the question we are told that:

Speed of electron $V_e=3.83 * 10^5 m/s$ +x direction

Earths magnetic field $B_e=3.04 * 10^-^8$ +z direction

Generally the equation for magnetic force $F_m$ is mathematically given by

$F=q(V_e*B_e)$

where

$q=1.6*10^{-19}c\\\=i*\=z=-\=j$

$F=1.6*10^{-19}(3.83 * 10^5 m/s*3.04 * 10^-^8)$

$F=-2.84*10^{-26}N \=j$

Magnitude & Direction

$F=2.84*10^{-26}N$ -y direction

Generally the equation for magnitude and direction of the magnetic force on an electron. is mathematically given by

$\=F'=-1.6*10^{-19}(3.83 * 10^5 m/s*3.04 * 10^-^8)$

$\=F'=-2.84*10^{-26}N \=j$

Magnitude & Direction

$F=2.84*10^{-26}N$ & +y direction

5 0

3 years ago

The football player running toward the goal line has

soldi70 [24.7K]

Kinetic energy.

Kinetic energy is the type of energy observed in moving objects. In this case the football player is running, ie moving, so he/she must have kinetic energy.

5 0

3 years ago

When the distance between two charges is halved, the electrical force between them?

Llana [10]

If the distance between two charges is halved, the electrical force between them increases by a factor 4.

In fact, the magnitude of the electric force between two charges is given by:
$F= k \frac{q_1 q_2}{r^2}$
where
k is the Coulomb's constant
q1 and q2 are the two charges
r is the separation between the two charges

We see that the magnitude of the force F is inversely proportional to the square of the distance r. Therefore, if the radius is halved:
$r'= \frac{r}{2}$
the magnitude of the force changes as follows:
$F'=k \frac{q_1 q_2}{r'^2}=k \frac{q_1 q_2}{( \frac{r}{2})^2 }=k \frac{q_1 q_2}{ \frac{r^2}{4} } =4k \frac{q_1 q_2}{r^2}=4 F$
so, the force increases by a factor 4.

3 0

3 years ago