A car starts from rest and drives at 45 m/s in 20 s. how far did it travel

If the mass of an object increases, how is its acceleration affected, assuming the net force acting on the object remains the sa

vovikov84 [41]

Based on Newton's second law of motion, the net force applied to an object is equal to the product of the mass of the object and the acceleration it experiences. That is,

F = ma

If we are to assume that the net force is constant and that the mass is increased, the acceleration should therefore decrease in order to make constant the value at the right-hand side of the equation.

7 0

3 years ago

Two cars are traveling along a straight line in the same direction, the lead car at 25 m/s and the other car at 35 m/s. At the m

Phoenix [80]

Answer:

a. $t_1=12.5\ s$

b. $a_2=-13.61\ m.s^{-2}$ must be the minimum magnitude of deceleration to avoid hitting the leading car before stopping

c. $t_2=2.5714\ s$ is the time taken to stop after braking

Explanation:

Given:

speed of leading car, $u_1=25\ m.s^{-1}$

speed of lagging car, $u_{2}=35\ m.s^{-1}$

distance between the cars, $\Delta s=45\ m$

deceleration of the leading car after braking, $a_1=-2\ m.s^{-2}$

a.

Time taken by the car to stop:

$v_1=u_1+a_1.t_1$

where:

$v_1=0$ , final velocity after braking

$t_1=$ time taken

$0=25-2\times t_1$

$t_1=12.5\ s$

b.

using the eq. of motion for the given condition:

$v_2^2=u_2^2+2.a_2.\Delta s$

where:

$v_2=$ final velocity of the chasing car after braking = 0

$a_2=$ acceleration of the chasing car after braking

$0^2=35^2+2\times a_2\times 45$

$a_2=-13.61\ m.s^{-2}$ must be the minimum magnitude of deceleration to avoid hitting the leading car before stopping

c.

time taken by the chasing car to stop:

$v_2=u_2+a_2.t_2$

$0=35-13.61\times t_2$

$t_2=2.5714\ s$ is the time taken to stop after braking

7 0

3 years ago

Which skateboarder has greater momentum?

Verizon [17]

Answer:

skateboard b

Explanation:

p=mv

skateboard a

p=(60kg)(1.5m/s)=90kg*m/s

skateboard b

p=(50kg)(2m/s)=100kg*m/s

5 0

3 years ago

List small/average stars<br><br>

mario62 [17]

Answer:

Lol, you should do Nate, Bobby, Cindy, Joe, and Beth

Jk, if you want to be series and probably not fail go for these:

If it wants types of small/average stars, then go with

Small star names:

OGLE-TR-122B

Gliese 229 B

TRAPPIST-1

Teegarden's Star

Luyten 726-8 (A and B)

Proxima Centauri

Wolf 359 111400

Ross 248

Barnard's Star

CM Draconis B

Ross 154 167000

CM Draconis A

Kapteyn's Star

7 0

3 years ago

n a downhill ski race, surprisingly, little advantage is gained by getting a running start. (This is because the initial kinetic

Annette [7]

Answer:

Explanation:

a ) starting from rest , so u = o and initial kinetic energy = 0 .

Let mass of the skier = m

Kinetic energy gained = potential energy lost

= mgh = mg l sinθ

= m x 9.8 x 70 x sin 30

= 343 m

Total kinetic energy at the base = 343 m + 0 = 343 m .

b )

In this case initial kinetic energy = 1/2 m v²

= .5 x m x 2.5²

= 3.125 m

Total kinetic energy at the base

= 3.125 m + 343 m

= 346.125 m

c ) It is not surprising as energy gained due to gravitational force by the earth is enormous . So component of energy gained due to gravitational force far exceeds the initial kinetic energy . Still in a competitive event , the fractional initial kinetic energy may be the deciding factor .

7 0

4 years ago