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dimulka [17.4K]

3 years ago

8

How is the mass of air in a hot air balloon affected when it is floating?

1 answer:

Bad White [126]3 years ago

3 0

<h2>sueunxicnrjcnexjsexmjemvjsmzj</h2>

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The velocity of a car increases by 40 m/s in 80 seconds. What is the acceleration of the car

yan [13]

Given that the velocity of a car increases by 40 m/s in 80 seconds, the acceleration of the car will be given by:
a=(final velocity-initial velocity)/(time)
thus;
final velocity=40 m/s
initial velocity=0
time=80 seconds
hence;
a=(40-0)/80
=0.5m/s^2

7 0

3 years ago

a 282 kg bumper car moving right at 3.50 m/s collides with a 155 kg bumper car moving 1.88 m/s left. afterwards, the 282 kg car

Vaselesa [24]

The momentum of the 155 kg car afterwards is 469.7 kg m/s to the right

Explanation:

We can solve the problem by using the law of conservation of momentum: the total momentum of the system is conserved before and after the collision, so we can write:

$p_i = p_f\\m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$

where:

$m_1 = 282 kg$ is the mass of the bumper car

$u_1 = 3.50 m/s$ is the initial velocity of the bumper car (we take the right as positive direction)

$v_1 = 0.800 m/s$ is the final velocity of the bumper car

$m_2 = 155 kg$ is the mass of the second bumper car

$u_2 = -1.88 m/s$ is the initial velocity of the second car (moving to the left)

$v_2$ is the final velocity of the second car

Solving for $v_2$ ,

$v_2 = \frac{m_1 u_1+m_2 u_2 - m_1 v_1}{m_2}=\frac{(282)(3.50)+(155)(-1.88)-(282)(0.800)}{155}=3.03 m/s$

where the positive sign means the direction is to the right.

And now we can find the momentum of the 155 kg afterwards, which is

$p_2 = m_2 v_2 = (155)(3.03)=469.7 kg m/s$ (to the right)

Learn more about momentum:

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#LearnwithBrainly

7 0

3 years ago

in a billiards game, the white cue ball hits the black ball, moving the black ball to the right while the cue ball moves to the

Gemiola [76]

Answer:

The black ball moving to the right after being struck by the white ball

Explanation:

4 0

3 years ago

An electron enters a region of uniform perpendicular en and bn fields. it is observed that thevelocity nv of the electron is una

konstantin123 [22]

<span>v is perpendicular to both E and B and has a magnitude E/B</span>

5 0

3 years ago

An SUV has a mass of 6000 kg and a compact car has a mass of 2000 kg. They both go around the same unbanked circular curve on th

Evgen [1.6K]

Answer:

Therefore, the centripetal acceleration of the SUV is <u>3 times</u> ao of the compact car.

Explanation:

FOR SUV:

$ac_{suv} = \frac{mv^2}{r}\\$

where,

ac_suv = centripetal acceleration of the SUV = ?

m = mass of SUV = 6000 kg

v = speed of SUV = vo

r = radius of path

Therefore,

$ac_{suv} = \frac{6000\ vo^2}{r}\\$ ---------------- equation (1)

FOR CAR:

$ao = \frac{mv^2}{r}\\$

where,

ao = centripetal acceleration of the car = ?

m = mass of car = 2000 kg

v = speed of car = vo

r = radius of path

Therefore,

$ao = \frac{2000\ vo^2}{r}\\$ --------------------- equation (2)

Dividing equation (1) by eq(2):

$\frac{ac_{suv}}{ao} = \frac{6000}{2000}\\\\ac_{surv} = 3ao$

Therefore, the centripetal acceleration of the SUV is <u>3 times</u> ao of the compact car.

8 0

3 years ago