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NeX [460]
2 years ago
12

A shopping cart given an initial velocity of 2.0 m/s north undergoes a constant acceleration of 3.0 m/s2 north. what is the dist

ance the cart moved after the first 4.0 s of its motion?
Physics
1 answer:
morpeh [17]2 years ago
7 0

Following the initial 4.0 seconds of travel, the cart moved 32m.

<h3>What is an equation of motion?</h3>

Physicists use equations of motion to describe how a physical system behaves in terms of how its motion changes over time.

The behavior of a physical system is described by the equations of motion in more detail as a collection of mathematical functions expressed in terms of dynamic variables. These variables typically comprise time and spatial coordinates, but they could also have momentum components. The most flexible option is generalized coordinates, which can be any useful variable that is a component of the physical system. In classical mechanics, the functions are defined in a Euclidean space, while curved spaces are used in relativity instead. The equations are the answers to the differential equations describing the motion of the dynamics of the dynamics of a system are known. The amount of motion changes according to the strength of the force and does so in the direction of the force's applied straight line.

To know more about equations of motion, click here:

brainly.com/question/14355103

#SPJ4

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If the velocity of a long jumper during take-offat an angle of 30 degrees is 42 f/s, how fast is he moving forward (Vx) and how
Arturiano [62]

Answer:

Vx=  11.0865(m/s)

Vy=  6.4008(m/s)

Explanation:

Taking into account that 1m is equal to 0.3048 ft, the takeoff speed in m / s will be:

V= 42(ft/s) × 0.3048(m/ft) = 12.8016(m/s)

The take-off angle is equal to 30 °, taking into account the Pythagorean theorem the velocity on the X axis will be:

Vx= 12.8016 (m/s) × cos(30°)= 11.0865(m/s)

And for the same theorem the speed on the Y axis will be:

Vy= 12.8016 (m/s) × sen(30°)= 6.4008(m/s)

5 0
3 years ago
What’s the difference between gravitational force &amp; gravitational field strength?
Zepler [3.9K]

Answer:

Gravitational field strength is the force experienced by a unit mass. Gravitational force is the amount of force acting on a body. It is the product of field strength times the mass under consideration. Gravitational pull is just a more colloquial name for gravitational force.

Explanation:

hope it helps u

5 0
3 years ago
Read 2 more answers
A television camera lens has a 17-cm focal length and a lens diameter of 6.0 cm. what is its number?
IRINA_888 [86]

Answer:

= 2.83

Explanation:

F number (N) is given by the formula;

  F- number = f/D

where f = focal length of lens and D = diameter of the aperture  

Therefore;

F number = 17 cm/6 cm

                <u> = 2.83</u>

3 0
3 years ago
Use the table below to calculate the acceleration of the object. Time (s) 0.0 3.0 6.0 9.0 Velocity (m/s) 0.0 1.2 2.4 3.6
antiseptic1488 [7]

explanation

a=average velocity/average time

average velocity=0.0+1.2+2.4+3.6/4

average velocity=7.2/4

average velocity=1.8 m/s

average time=0.0+3.0+6.0+9.0/4

average time=18/4

average time=4.5 s

a= average velocity/average time

a=1.8/4.5

a=0.4 m/s²

8 0
3 years ago
A produce distributor uses 800 packing crates a month, which it purchases at a cost of $10 each. The manager has assigned an ann
strojnjashka [21]

Answer:

$364.29

Explanation:

given,

Packing of crates per month (u)= 800

annual carrying cost of 35 percent of the purchase price per crate.

Ordering cost(S) = $ 28

D = 800 x 12 = 9600 crates/year

H = 0.35 P

H = 0.35 x $10

H = $3.50/crate per yr.

Present Total cost

= \dfrac{800}{2}\times 3.50 + \dfrac{9600}{800}\times 28

= 1400 + 336

= $ 1,736

Q_0 = \sqrt{\dfrac{2DS}{H}}

Q_0 = \sqrt{\dfrac{2\times 9600 \times 28}{3.50}}

Q_0 =\$ 391.92

Total cost at EOQ

= \dfrac{391.92}{2}\times 3.50 + \dfrac{9600}{391.92}\times 28

= 685.86 + 685.85

= $ 1,371.71

the firm save annually in ordering and carrying costs by using the EOQ

    = $ 1,736 - $ 1,371.71

    = $364.29

4 0
3 years ago
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