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Westkost [7]
3 years ago
7

A box of mass 115 kg sits on an inclined surface with an angle of 59º. What

Physics
1 answer:
LenKa [72]3 years ago
8 0

Answer:

966 N

Explanation:

The component of the weight along the direction perpendicular to the plane is:

mg cos \theta

where

m is the mass of the box, g the acceleration of gravity, \theta the angle of the inclined plane.

Similarly, the component of the weight of the box along the surface is

mg sin \theta

acting down along the incline.

Here we have:

m = 115 kg

g = 9.8 m/s^2

\theta=59^{\circ}

Substitting in the last equation, we find the component of the weight of the box along the surface:

mg sin \theta = (115)(9.8)(sin 59)=966 N

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A 30kg mass is brought from the earth to the moon what is the weight on the earth
ozzi

Answer:

Solution

verified

Verified by Toppr

Given:

Mass of body = 30 kg

gravitational acceleration on the moon = 1.62 m/s

2

Weight of the body on the moon = Mass of the body×gravitational acceleration on the moon=30×1.62=48 N

8 0
1 year ago
8.) If a car moving at 50km/h skids 15m with locked brakes, how far does the same car moving at 100km/h
pantera1 [17]

(8) A car starting with a speed <em>v</em> skids to a stop over a distance <em>d</em>, which means the brakes apply an acceleration <em>a</em> such that

0² - <em>v</em>² = 2 <em>a</em> <em>d</em> → <em>a</em> = - <em>v</em>² / (2<em>d</em>)

Then the car comes to rest over a distance of

<em>d</em> = - <em>v</em>² / (2<em>a</em>)

Doubling the starting speed gives

- (2<em>v</em>)² / (2<em>a</em>) = - 4<em>v</em>² / (2<em>a</em>) = 4<em>d</em>

so the distance traveled is quadrupled, and it would move a distance of 4 • 15 m = 60 m.

Alternatively, you can explicitly solve for the acceleration, then for the distance:

A car starting at 50 km/h ≈ 13.9 m/s skids to a stop in 15 m, so locked brakes apply an acceleration <em>a</em> such that

0² - (13.9 m/s)² = 2 <em>a</em> (15 m) → <em>a</em> ≈ -6.43 m/s²

So the same car starting at 100 km/h ≈ 27.8 m/s skids to stop over a distance <em>d</em> such that

0² - (27.8 m/s)² = 2 (-6.43 m/s²) <em>d</em> → <em>d</em> ≈ 60 m

(9) Pushing the lever down 1.2 m with a force of 50 N amounts to doing (1.2 m) (50 N) = 60 J of work. So the load on the other end receives 60 J of potential energy. If the acceleration due to gravity is taken to be approximately 10 m/s², then the load has a mass <em>m</em> such that

60 J = <em>m g h</em>

where <em>g</em> = 10 m/s² and <em>h</em> is the height it is lifted, 1.2 m. Solving for <em>m</em> gives

<em>m</em> = (60 J) / ((10 m/s²) (1.2 m)) = 5 kg

(10) Is this also multiple choice? I'm not completely sure, but something about the weight of the tractor seems excessive. It would help to see what the options might be.

4 0
3 years ago
75 kg + 1352 g = ______________g?
JulsSmile [24]
Answer is 76,352 just look it up
8 0
2 years ago
Read 2 more answers
An object of mass m is traveling in a circle with centripetal force Fc. If the velocity of the object is v, what is the radius o
borishaifa [10]

Hi there!

Recall the equation for centripetal force:
F_c = \frac{mv^2}{r}

We can rearrange the equation to solve for 'r'.

Multiply both sides by r:
r * F_c = mv^2

Divide both sides by Fc:
\boxed{ r= \frac{mv^2}{F_c}}

7 0
2 years ago
How much elastic potential energy is stored in a bungee cord with a spring constant of 10.0 N/m when the cord is stretched 2.00
Aleonysh [2.5K]

Answer:

<em> The elastic potential energy stored in the bungee cord = 20 J</em>

Explanation:

potential energy: This is the energy possessed by a body due to its position. The S.I unit of energy is Joules. The mathematical expression for elastic potential energy is given below

E = 1/2ke²................ Equation 1

Where E = elastic potential energy of the spring, k = force constant of the spring, e = extension

<em>Given: K = 10 N/m, e = 2.00 m</em>

<em>Substituting these values into Equation 1</em>

<em>E = 1/2(10)(2)²</em>

<em>E = 5×4</em>

<em>E = 20 Joules.</em>

<em>Therefore the elastic potential energy stored in the bungee cord = 20 J</em>

<em></em>

6 0
3 years ago
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