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Hunter-Best [27]
3 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]3 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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tatuchka [14]

Answer:

I would love to be a gardener :)

Explanation:

Plants are kool!!!

4 0
3 years ago
A 107 gram apple falls from a branch that is 2 meters above the ground. (a) How much time elapses before the apple hits the grou
andreev551 [17]

Answer:

The time of motion is 0.64 s.

Explanation:

Given;

mass of the apple, m = 107 g

height of fall, h = 2 m

The velocity of the apple when it hits the ground is calculated from the law of conservation of energy;

P.E = K.E\\\\mgh = \frac{1}{2} mv^2\\\\gh = \frac{1}{2} v^2\\\\v^2 = 2gh\\\\v = \sqrt{2gh} \\\\v = \sqrt{2\times 9.8\times 2} \\\\v = 6.261 \ m / s

The time of motion is calculated;

v = u + gt

6.261 = 0 + 9.8t

6.261 = 9.8t

t = 6.261 / 9.8

t = 0.64 s

Therefore, the time of motion is 0.64 s

7 0
3 years ago
How long with it take an object to accelerate from 6 m/s to 13 m/s at a rate of 1.4<br> m/s2. *
Blizzard [7]

Explanation:

hello,

a = ( v - u ) / t

where u is the initial velocity.

and v is the final velocity.

t represents time,

and a represents acceleration.

in this case,

a = 1.4 m/s²

u = 6 m/s

v = 13 m/s

hence,

1.4 = (13 - 6)/t

1.4t = 7

t = 7/1.4

t = 5 s

thank you!

7 0
2 years ago
Ice skaters often end their performances with spin turns, where they spin very fast about their center of mass with their arms f
nekit [7.7K]

Answer:

Her angular speed (in rev/s) when her arms and one leg open outward is 1.4 rev/s

Explanation:

given information:

moment inertia of arm and leg when in, I₁ = 0.9 kgm²

moment inertia of arm and leg when extended, I₂ = 2.9 kgm²

angular speed when in, ω₁ = 4.5 rev/s

so, her angular speed (in rev/s) when her arms and one leg open outward is

L₁ = L₂

I₁ω₁ = I₂ω₂

ω₂ = I₁ω₁/I₂

     = 0.9 x 4.5/2,9

     = 1.4 rev/s

4 0
3 years ago
Calculate the acceleration due to gravity inside earth as a function of the radial distance r from the planet's center. (hint: i
denis-greek [22]
An estimated value for gravity at a distance r from the middle of the Earth can be gotten by supposing that the Earth's density is spherically symmetric. The gravity hinge on only on the mass inside the sphere of radius r. All the assistances from outside cancel out as a fall out of the inverse-square law of gravitation. Another result is that the gravity is the same as if all the mass were concentrated at the midpoint. Therefore, the gravitational acceleration at this radius is 
g(r) = GM(r) / r² 
M(r) = mass enclosed by radius r. 
If the Earth had a continual density ρ, the mass would be M(r) = (4/3)πρr³ and the dependence of gravity on distance would be 
g(r) = (4/3)πGρr 
G = 6.674e-11 m³/kgs²
4 0
3 years ago
Read 2 more answers
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