Ask question

Ask question

All categories

3 years ago

10

The total momentum of the two cars before a collision is 100 kg m/s. what is the total momentum of the two cars after they colli

de?

2 answers:

OlgaM077 [116]3 years ago

4 0

The answer is c. 100 kg m/s

Dahasolnce [82]3 years ago

3 0

total momentum of the two cars after they collide= 100 kg m/s

Explanation:

We use law of conservation of momentum which says that , for an isolated system , total momentum before collision is equal to the total momentum after the collision.

Total momentum of two cars before collision= 100 kg m/s

so the total momentum of the two cars after the collision= 100 kg m/s

You might be interested in

Darryl throws a basketball at the gym floor. The ball bounces once on the floor and comes to rest in his coach’s hands. At which

mina [271]

Answer:

when the ball is at rest in his coach's hands.

Explanation:

The forces on the basketball are balanced when the basketball is not experiencing any acceleration. This happens when the ball is in his coach's hand: in fact, at that moment the ball is at rest, so it means that its acceleration is zero. According to Newton's second law, this also mean that the net force on the basketball is zero, so the forces on the ball are balanced:

$F=ma$

where F is the net force, m is the mass of the ball and a is the acceleration.

8 0

3 years ago

if quasars often resemble little blue stars, what was it about them that so surprised astronomers when they were discovered?

Anna11 [10]

Quasar is famous for being an intergalactic object which is billions of years away from the earth yet can still be seen, unlike the other star body, unlike giant galaxies.

Hence, the fact that quasars can be detected from distances where even the biggest and most luminous galaxies cannot be seen means that "they must be intrinsically far more luminous than the brightest galaxies."

This condition, including other related evidence gotten in recent years concerning our galaxy, has shown that quasars are probably the central nuclei of very distant, very active galaxies.

The surprising thing was that quasars and active galaxies have a lot of mass in the center of the very small volume of the space.

Therefore, the surprising thing about quasars was that due to this mass and energy they are 100 times more luminous than Milky Way which means they have high recession velocity and a very large amount of red-shifting.

To learn more about quasars, refer: brainly.com/question/9965257

#SPJ4

8 0

9 months ago

Many Amtrak trains can travel at a top speed of 42.0 m/s. Assuming a train maintains that speed for several hours, how many kilo

777dan777 [17]

Answer:

605 km

Explanation:

Hello

the same units of measure should be used, then

Step 1

convert 42 m/s ⇒ km/h

1 km =1000 m

1 h = 36000 sec

$42 \frac{m}{s}*\frac{1\ km}{1000\ m}=0.042\ \frac{km}{s}\\ 0.042\ \frac{km}{s}\\$

$0.042\ \frac{km}{s}*\frac{3600\ s}{1\ h} =151.2 \frac{km}{h}\\ \\Velocity =151.2\ \frac{km}{h}$

Step 2

find kilometers traveled after 4 hours

$V=\frac{s}{t}\\ \\$

V,velocity

s, distance traveled

t. time

now, isolating s

$V=\frac{s}{t} \\s=V * t\\$

and replacing

$s=V * t\\s=151.2\frac{km}{h}*4 hours\\ s=604.8 km\\$

S=604.8 Km

Have a great day

4 0

3 years ago

one group of atoms on the periodic tables are known as "noble gases". why have they been given this name?

maria [59]

This group is called “noble gases” because they do not react with other elements. This is because they have a full valence shell.

8 0

3 years ago

A merry-go-round with a rotational inertia of 600 kg m2 and a radius of 3. 0 m is initially at rest. A 20 kg boy approaches the

Margaret [11]

Hi there!

$\boxed{\omega = 0.38 rad/sec}$

We can use the conservation of angular momentum to solve.

$\large\boxed{L_i = L_f}$

Recall the equation for angular momentum:

$L = I\omega$

We can begin by writing out the scenario as a conservation of angular momentum:

$I_m\omega_m + I_b\omega_b = \omega_f(I_m + I_b)$

$I_m$ = moment of inertia of the merry-go-round (kgm²)

$\omega_m$ = angular velocity of merry go round (rad/sec)

$\omega_f$ = final angular velocity of COMBINED objects (rad/sec)

$I_b$ = moment of inertia of boy (kgm²)

$\omega_b$ = angular velocity of the boy (rad/sec)

The only value not explicitly given is the moment of inertia of the boy.

Since he stands along the edge of the merry go round:

$I = MR^2$

We are given that he jumps on the merry-go-round at a speed of 5 m/s. Use the following relation:

$\omega = \frac{v}{r}$

$L_b = MR^2(\frac{v}{R}) = MRv$

Plug in the given values:

$L_b = (20)(3)(5) = 300 kgm^2/s$

Now, we must solve for the boy's moment of inertia:

$I = MR^2\\I = 20(3^2) = 180 kgm^2$

Use the above equation for conservation of momentum:

$600(0) + 300 = \omega_f(180 + 600)\\\\300 = 780\omega_f\\\\\omega = \boxed{0.38 rad/sec}$

8 0

2 years ago