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2 years ago

12

A large crate is at rest on a ramp at a loading dock

1 answer:

VARVARA [1.3K]2 years ago

4 0

Answer:

yuhhh

Explanation:

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Un electrón en un tubo de rayos catódicos acelera desde el reposo con una aceleración constante de 5.33x10¹²m/s² durante 0.150μs

Andre45 [30]

Can you translate that in English ? I'll try to help you out with that..

3 0

2 years ago

A Styrofoam slab has thickness h and density ρs. When a swimmer of mass m is resting on it, the slab floats in fresh water with

sattari [20]

<span>A = area of styrofoam
M = mass of stryofoam = A*h*rho_s
m = mass of swimmer

Total mass = m + M = m + A*h*rho_s
Downward force = g*(total mass) = g*[m + A*h*rho_s]

The slab is completely submerged.
Buoyant force = g*(mass of water displaced) = g*[A*h*rho_w]

Equate these
g*[m + A*h*rho_s] = g*[A*h*rho_w]
m + A*h*rho_s = A*h*rho_w
A*h*[rho_w - rho_s] = m
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8 0

2 years ago

A large crate is suspended from the end of a vertical rope. Is the tension in the rope greater when the crate is at rest or when

choli [55]

Answer:

Part a)

the tension force is equal to the weight of the crate

Part b)

tension force is more than the weight of the crate while accelerating upwards

tension force is less than the weight of crate if it is accelerating downwards

Explanation:

Part a)

When large crate is suspended at rest or moving with uniform speed then it is given as

$F_t - mg = ma$

here since speed is constant or it is at rest

so we will have

$a = 0$

$F_t = mg$

so the tension force is equal to the weight of the crate

Part b)

Now let say the crate is accelerating upwards

now we can say

$F_t - mg = ma$

$F_t = mg + ma$

so tension force is more than the weight of the crate

Now if the crate is accelerating downwards

$F_t - mg = -ma$

$F_t = mg - ma$

so tension force is less than the weight of crate if it is accelerating downwards

4 0

2 years ago

In the sport of parasailing, a person is attached to a rope being pulled by a boat while hanging from a parachute-like sail. A r

scoray [572]

Answer:

W = 506.75 N

Explanation:

tension = 2300 N

Rider is towed at a constant speed means there no net force acting on the rider.

hence taking all the horizontal force and vertical force in consideration.

net horizontal force:

F cos 30° - T cos 19° = 0

F cos 30° = 2300 × cos 19°

F = 2511.12 N

net vertical force:

F sin 30° - T sin 19°- W = 0

W = F sin 30° - T sin 19°

W = 2511.12 sin 30° - 2300 sin 19°

W = 506.75 N

8 0

2 years ago

After an eye examination, you put some eyedrops on your sensitive eyes. The cornea (the front part of the eye) has an index of r

mina [271]

Answer:

t = 103.45 n m

Explanation:

given,

refractive index of cornea = 1.38

refractive index of eye drop = 1.45

wavelength of refractive index = 600 nm

refractive index of eye drop is greater than refractive index of cornea and the air.

Formula used in this case

for constructive interference

$2 n t = (m + \dfrac{1}{2})\lambda$

At m = 0 for the minimum thickness, so

$2\times 1.45 \times t = (0 + 0.5)\times 600$

$2.9 \times t =300$

$t =\dfrac{300}{2.9}$

t = 103.45 n m

the minimum thickness of the film of eyedrops t = 103.45 n m

6 0

2 years ago