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NeTakaya
3 years ago
10

A car is driven 215 km west and then 85 km southwest. What is the displacement of the car from the point of origin.

Physics
1 answer:
9966 [12]3 years ago
4 0

displacement=  281 , 59 k m  (magnitude)

θ  =  77 , 68  degrees  from  S to SW    (angle of direction)

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A mover loads a crate onto a truck bed 1.6m from the street using a ramp that is 4.6m long. What is a mechanical advantage?
Ne4ueva [31]

Answer:

Mechanical advantage = 2.875

Explanation:

Given:

A diagram is shown below for the above scenario.

Length of ramp (Effort arm) = 4.6 m

Height of truck bed ( Resistance length) = 1.6 m

Mechanical advantage (MA) is the ratio of effort arm and resistance length.

So, mechanical advantage is given as,

MA=\frac{\textrm{Effort arm}}{\textrm{Resistance length}}= \frac{4.6}{1.6}=2.875

6 0
3 years ago
If a car is moving to the left with constantvelocity, one can conclude thatthere mustbe no forces applied to the car.the netforc
Allisa [31]

Answer:

the net force applied to the car is zero.

Explanation:

According to Newton's second law, the acceleration of an object (a) is directly proportional to the net force applied (F):

a=\frac{F}{m}

where m is the object's mass.

In this problem, the car is moving with constant velocity: this means that the acceleration is zero, a = 0. Therefore, according to the previous equation, the net force must also be zero: F = 0. So, the correct answer is

the net force applied to the car is zero.

3 0
3 years ago
What is the relationship between an object’s mass and its gravitational potential energy?.
antiseptic1488 [7]

Answer:

If we’re talking about objects on the Earth, the gravitational potential energy is given by:

Explanation:

PEg=mgh

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5 0
2 years ago
What can be contracted through contact with saliva?
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One of the most important properties of materials in many applications is strength. Two of the qualitative measures of the stren
katrin2010 [14]

To solve this exercise it is necessary to take into account the concepts related to Tensile Strength and Shear Strenght.

In Materials Mechanics, generally the bodies under certain loads are subject to both Tensile and shear strenghts.

By definition we know that the tensile strength is defined as

\sigma = \frac{F}{A}

Where,

\sigma =Tensile strength

F = Tensile Force

A = Cross-sectional Area

In the other hand we have that the shear strength is defined as

\sigma_y = \frac{F_y}{A}

where,

\sigma_y =Shear strength

F_y = Shear Force

A_0 =Parallel Area

PART A) Replacing with our values in the equation of tensile strenght, then

311*10^6 = \frac{F}{(15*10^{-6})(30*10^{-2})}

Resolving for F,

F= 1399.5N

PART B) We need here to apply the shear strength equation, then

\sigma_y = \frac{F_y}{A}

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In such a way that the material is more resistant to tensile strength than shear force.

6 0
3 years ago
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