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Hunter-Best [27]
2 years ago
6

In the design of a supermarket, there are to be several ramps connecting different parts of the store. Customers will have to pu

sh grocery carts up the ramps and it is obviously desirable that this not be too difficult. The engineer has done a survey and found that almost no one complains if the force directed up the ramp is no more than 20N .
Ignoring friction, at what maximum angle θ should the ramps be built, assuming a full 30kg grocery cart?

Express your answer using two significant figures.
Physics
1 answer:
tatuchka [14]2 years ago
6 0

Answer:

3.90 degrees

Explanation:

Let g= 9.81 m/s2. The gravity of the 30kg grocery cart is

W = mg = 30*9.81 = 294.3 N

This gravity is split into 2 components on the ramp, 1 parallel and the other perpendicular to the ramp.

We can calculate the parallel one since it's the one that affects the force required to push up

F = WsinΘ

Since customer would not complain if the force is no more than 20N

F = 20

294.3sin\theta = 20

sin\theta = 20/294.3 = 0.068

\theta = sin^{-1}0.068 = 0.068 rad = 0.068*180/\pi \approx 3.90^0

So the ramp cannot be larger than 3.9 degrees

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LUCKY_DIMON [66]
The answer to this question is "Greater than". As per the OSHA or Organizational Health and Health Organization, a noise level of 95 DB or decibels is greater than the lowest level at which hearing protection is required in 85 decibels and the person's exposure should be limited only to six hours or less than of it. 
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2 years ago
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The different in size of each of the rope's pullers, correspond to a difference in the magnitude of the applied force, such that
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Answer:

F = - 50 N

Hence, the magnitude of resultant force is 50 N and its direction is leftwards.

Explanation:

The magnitude of the resultant force is always equal to the sum of all forces. While, the direction of resultant force will be equal to the direction of the force with greater magnitude:

Resultant\ Force = F = F_{1} - F_{2}

considering right direction to be positive:

F₁ = Force applied on right rope = 150 N

F₂ = Force applied on left rope = 200 N

Therefore, the resultant force can be found by using these values in equation:

F = 150\ N - 200\ N

<u>F = - 50 N</u>

<u>Hence, the magnitude of resultant force is 50 N and its direction is leftwards.</u>

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2 years ago
A ball weighing 1 lb is attached to a string 2 feet long and is whirled in a vertical circle at a constant speed of 10 ft/sec.
fredd [130]

Explanation:

It is given that,

Mass of the ball, m = 1 lb

Length of the string, l = r = 2 ft

Speed of motion, v = 10 ft/s

(a) The net tension in the string when the ball is at the top of the circle is given by :

F=\dfrac{mv^2}{r}-mg

F=m(\dfrac{v^2}{r}-g)

F=1\ lb\times (\dfrac{(10\ ft/s)^2}{2}-1\ lb\times 32\ ft/s^2)

F = 18 N

(b) The net tension in the string when the ball is at the bottom of the circle is given by :

F=\dfrac{mv^2}{r}+mg

F=m(\dfrac{v^2}{r}+g)

F=1\ lb\times (\dfrac{(10\ ft/s)^2}{2}+1\ lb\times 32\ ft/s^2)

F = 82 N

(c) Let h is the height where the ball at certain time from the top. So,

T=mg(\dfrac{r-h}{r})+\dfrac{mv^2}{r}

T=\dfrac{m}{r}(g(r-h)+v^2)

Since, v^2=u^2-2gh

T=\dfrac{m}{r}(u^2-3gh+gr)

Hence, this is the required solution.

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2 years ago
What units are given to the right of the equals sign
zhuklara [117]
The answer

2y + 14 = 17

The 17 is to the right of the = sign
It is also the answer
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gaseous h2 and br2 are added to an evacuated 1.15L container kept at 298K. The intial partial pressurre of H2(g) is 0.782 atm an
Nastasia [14]

The partial pressures of HBr when the system reaches equilibrium is 2.4 X 10⁻¹¹ atm

<u>Explanation:</u>

H₂ + Br₂ ⇒ 2HBr

PH₂ = 0.782atm

PBr₂ = 0.493atm

Kp = (PHBr)²/ (PH₂) (PBr₂) = 1.4 X 10⁻²¹

At equilibrium:

Let 2x = pressure of HBr

PH₂ = 0.782 -x

PBr₂ = 0.493 - x

Kp = (2x)^2 / (0.782-x)(0.493-x)

Now, because Kp is very small, x will be very small compared to 0.782 and 0.493.

Then,

Kp = 1.4X10⁻²¹ = (4x²) / (0.782)(0.493)

x = 1.2X10⁻¹¹

PHBr = 2x = 2.4 X 10⁻¹¹ atm

Therefore, the partial pressures of HBr when the system reaches equilibrium is 2.4 X 10⁻¹¹ atm

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