How much force does it take to accelerate a 2000 kg car at 1m/s^2

You are driving a 2400.0-kg car at a constant speed of 14.0 m/s along a wet, but straight, level road. As you approach an inters

olya-2409 [2.1K]

Answer:0.43

Explanation:

Given

mass of car $m=2400 kg$

Speed of car $u=14 m/s$

Distance traveled before coming to halt $s=23.2 m$

Let $\mu$ the coefficient of friction

Maximum deceleration road can provide during motion is

$a=\mu g$

using $v^2-u^2=2 as$

$0-14^2=2\cdot (-\mu g)\cdot 23.2$

$\mu =\frac{196}{454.72}$

$\mu =0.431$

7 0

2 years ago

#1 a person weighs 540n on earth what is the persons mass.

bixtya [17]

Answer:

G on earth =10N/Kg

W= mg

540= m x 10

m = 540/10

m= 54 Kg

W= mg

W= 66 x 10

W= 660 N

G on moon =1.6 N/Kg

#1

W= mg

W= 54 x 1.6

W=86.4 N

#2

W=mg

W= 66 x 1.6

W= 105.6 N

8 0

2 years ago

Suppose a clay model of a koala bear has a mass of 0.205 kg and slides on ice at a speed of 0.720 m/s. It runs into another clay

Westkost [7]

Answer:

Final velocity, v = 0.28 m/s

Explanation:

Given that,

Mass of the model, $m_1=0.205\ kg$

Speed of the model, $u_1=0.72\ m/s$

Mass of another model, $m_2=0.32\ kg$

Initial speed of another model, $u_2=0$

To find,

Final velocity

Solution,

Let V is the final velocity. As both being soft clay, they naturally stick together. It is a case of inelastic collision. Using the conservation of linear momentum to find it as :

$m_1u_1+m_2u_2=(m_1+m_2)V$

$V=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}$

$V=\dfrac{0.205\times 0.72+0.32\times 0}{(0.205+0.32)}$

V = 0.28 m/s

So, their final velocity is 0.28 m/s. Hence, this is the required solution.

6 0

3 years ago

Convection currents in this layer drive the tectonic plates above them.

Sphinxa [80]

Hello, hello? Hey! Hey, wow, day 4. I knew you could do it.

Uh, hey, listen, I may not be around to send you a message tomorrow. *banging* It's-It's been a bad night here for me. Um, I-I'm kinda glad that I recorded my messages for you *clears throat* uh, when I did.

Uh, hey, do me a favor. *banging* Maybe sometime, uh, you could check inside those suits in the back room? *banging* I'm gonna to try to hold out until someone checks. Maybe it won’t be so bad. *bang bang* Uh, I-I-I-I always wondered what was in all those empty heads back there. *chime plays*.

You know...*deep moan* oh, no - *noises followed by a loud screech and static*

Explanation:

7 0

2 years ago

When a golfer tees off, the head of her golf club, which has a mass of 160 g, is traveling 50 m/s just before it strikes a 46 g

olya-2409 [2.1K]

Answer:

v1=21.81m/s

Explanation:

<em>When a golfer tees off, the head of her golf club, which has a mass of 160 g, is traveling 50 m/s just before it strikes a 46 g golf ball at rest on a tee. Immediately after the collision, the club head continues to travel in the same direction but at a reduced speed of 44 m/s. Neglect the mass of the club handle and determine the speed (in m/s) of the golf ball just after impact.</em>

According to the law of conservation of momentum, if the net external force on a system is zero, then the linear momentum of the system is conserved.

During collision of two particles, the external force on the system of two colliding particles is zero (only internal force acts between the colliding particles), therefore, the momentum is conserved during the collision.

Answer and Explanation:

Given :

head of the golf club=160g

velocity of the golf club=50 m/s

golf ball mass=46g

velocity=om/s

m1u1+m2u2=m1v1+m2v2.........................................1

160*50 +46*0=160*44+46*v1

8000=7040+46v1

960=46v1

v1=960/46

v1=21.81m/s

3 0

3 years ago