3 years ago

A linebacker runs at 3 m/s and collides with a running back running towards him at 3.2

Physics

1 answer:

Anit [1.1K]3 years ago

3 0

Answer:b

Explanation:

im in fifth grade so i guessd. gl tho heh.

You might be interested in

Just the the moon seems to shine and the two below it.

katrin [286]

The first one is revolves

5 0

3 years ago

You go to another forest and measure the shadow of a tree as being 6 meters long. The shadow of your meter

krok68 [10]

Well, 6/n=3/1, so n=2.

6 0

3 years ago

Pleaaaseeeee helppppp meeeee<br>pleaaaseeeee helppppp meeeee<br>pleaaaseeeee helppppp meeeee<br>

Aleksandr-060686 [28]

Answer:

The Republican party was formed because they wanted??????

Explanation:

7 0

3 years ago

If the force on a spring is 2 N and it stretched 0.5 m, what is the spring constant?

andriy [413]

<h2>It solved by the Hooke's law states F=kx</h2>

answer is

6 0

3 years ago

A force F = (2.75 N)i + (7.50 N)j + (6.75 N)k acts on a 2.00 kg mobile object that moves from an initial position of d = (2.75 m

Natasha2012 [34]

Answer:

The work done on the object by the force in the 5.60 s interval is 40.93 J.

Explanation:

Given that,

Force $F=(2.75\ N)i+(7.50\ N)j+(6.75\ N)k$

Mass of object = 2.00 kg

Initial position $d=(2.75)i-(2.00)j+(5.00)k$

Final position $d=-(5.00)i+(4.50)j+(7.00)k$

Time = 4.00 sec

We need to calculate the work done on the object by the force in the 5.60 s interval.

Using formula of work done

$W=F\cdot d$

$W=F\cdot(d_{f}-d_{i})$

Put the value into the formula

$W=(2.75)i+(7.50)j+(6.75)k\cdot (-(5.00)i+(4.50)j+(7.00)k-(2.75)i+(2.00)j-(5.00)k)$

$W=(2.75)i+(7.50)j+(6.75)k\cdot((-7.75)i+6.50j+2.00k)$

$W=2.75\times-7.75+7.50\times6.50+6.75\times2.00$

$W=40.93\ J$

Hence, The work done on the object by the force in the 5.60 s interval is 40.93 J.

4 0

3 years ago