A push pull or other action that has the ability to change motion

A toddler pushes his 7.0 kg toy box at a relatively constant velocity across the tiled floor of the family room applying a horiz

xxMikexx [17]

Answer:The coefficient of friction between the box and the floor, = 1.456 × 10⁻²

Explanation:

8 0

2 years ago

I will mark as the brainliest answer

erastova [34]

Answer:

8) 1500 feet

9) 20 miles

10) 4 Days

11) 2250 miles

12) 1 hour and 5 minutes

13) 27.27miles per hour

Explanation

8) There are 60 seconds in one minute so 60x25=1500 feet

9) 30 minutes is half of an hour so 40 miles ÷ 2 = 20 miles

10) 12x4=48

11) 500 miles x 4.5 hours is 2250 miles

12) Train leaves at 3pm after 60 miles it will be 4 pm and after 5 more miles 4:05 pm so 1 hour and 5 minutes

13)

Elmo = 40 minutes and 5 miles

Bert and Ernie = 45 minutes and 15 miles

Cookie Monster = 20 minutes and 10 miles

Home = 5 minutes abd 20 miles

Average Speed including stops is 27.27 miles per hour

5 0

3 years ago

What is the resistance of a voltage of 65 V and a current of 2.2 A? Include units.

kozerog [31]

Answer:

29.5 ohms

Explanation:

R= V/I

= 65 / 2.2

= 29.5 ohms

4 0

2 years ago

The first car has twice the mass of a second car, but only half as much kinetic energy. When both cars increase their speed by 9

kirill [66]

Answer:

Speed of the car 1 = $V_1=8.98m/s$

Speed of the car 2 = $V_2=17.96m/s$

Explanation:

Given:

Mass of the car 1 , M₁ = Twice the mass of car 2(M₂)

mathematically,

M₁ = 2M₂

Kinetic Energy of the car 1 = Half the kinetic energy of the car 2

KE₁ = 0.5 KE₂

Now, the kinetic energy for a body is given as

$KE =\frac{1}{2}mv^2$

where,

m = mass of the body

v = velocity of the body

thus,

$\frac{1}{2}M_1V_1^2=0.5\times \frac{1}{2}M_2V_2^2$

or

$\frac{1}{2}2M_2V_1^2=0.5\times \frac{1}{2}M_2V_2^2$

or

$2M_2V_1^2=0.5\times M_2V_2^2$

or

$2V_1^2=0.5\times V_2^2$

or

$4V_1^2= V_2^2$

or

$2V_1= V_2$ .................(1)

also,

$\frac{1}{2}M_1(V_1+9.0)^2=\frac{1}{2}M_2(V_2+9.0)^2$

or

$\frac{1}{2}2M_2(V_1+9.0)^2=\frac{1}{2}M_2(2V_1+9.0)^2$

or

$2(V_1+9.0)^2=(2V_1+9.0)^2$

or

$\sqrt{2}(V_1+9.0)=(2V_1+9.0)$

or

$(\sqrt{2}V_1+ \sqrt{2}\times 9.0)=(2V_1+9.0)$

or

$(\sqrt{2}V_1+ 12.72)=(2V_1+9.0)$

or

$(2V_1-\sqrt{2}V_1)=(12.72-9.0)$

or

$(0.404V_1)=(3.72)$

or

$V_1=8.98m/s$

and, from equation (1)

$V_2=2\times 8.98m/s = 17.96m/s$

Hence,

Speed of car 1 = $V_1=8.98m/s$

Speed of car 2 = $V_2=17.96m/s$

8 0

2 years ago

Read 2 more answers

If the distance between a neutral atom and a point charge is quadrupled, by what factor does the force on the atom by the point

IgorLugansk [536]

New force or Old Force

8 0

3 years ago