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Anarel [89]
4 years ago
11

What's the mass show the work?

Physics
1 answer:
Alja [10]4 years ago
6 0
You do this one just like the other one that I just solved for you.

For this one ...

The density of the object is 2.5 gm/cm³.
We know that every cm³ of it we have contains 2.5 gm of mass.
We have to find out how many cm³ we have.

The question tells us:  We have  2.0 cm³.

Each cm³ of space that the object occupies contains 2.5 gm of mass.

So the 2.0 cm³ that we have contains (2 x 2.5 gm) = 5 gms.
That's the mass of our object.
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A 16.0kg canoe moving to the left at 12.5m/s makes an elastic head-on collision with a 14.0kg raft moving to the right at 16.0m/
kherson [118]

The canoe is moving at 14.1 m/s to the right after the collision.

Explanation:

According to the law of conservation of momentum, in absence of external forces the total momentum of the system must be 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 = 16.0 kg is the mass of the canoe

u_1 = -12.5 m/s is the initial velocity of canoe (we take right as positive direction, and since the canoe is moving to the left, its velocity is negative)

v_1 is the final velocity of the canoe

m_2 = 14.0 kg is the mass of the raft

u_2 = +16.0 m/s is the initial velocity of the raft

v_2 = -14.4 m/s is the final velocity of the raft

Re-arranging the equation and substituting the values, we find: the final velocity of the canoe:

v_1 = \frac{m_1 u_1 + m_2 u_2-m_2 v_2}{m_1}=\frac{(16.0)(-12.5)+(14.0)(16.0)-(14.0)(-14.4)}{16.0}=+14.1 m/s

So, the canoe is moving at 14.1 m/s to the right after the collision.

Learn more about momentum:

brainly.com/question/7973509

brainly.com/question/6573742

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#LearnwithBrainly

5 0
3 years ago
A 320 g air track cart traveling at 1.25 m/s collides with a stationary 270 g cart. What is the speed of the 270 g cart after th
Nutka1998 [239]

Answer:

The speed of the 270g cart after the collision is 0.68m/s

Explanation:

Mass of air track cart (m1) = 320g

Initial velocity (u1) = 1.25m/s

Mass of stationary cart (m2) = 270g

Velocity after collision (V) = m1u1/(m1+m2) = 320×1.25/(320+270) = 400/590 = 0.68m/s

7 0
3 years ago
usted / el partido de fútbol Usted va al partido de fútbol. Question 1 with 1 blankyo / la biblioteca Question 2 with 1 blanknos
34kurt

Answer:

1- Yo VOY A la biblioteca

2- Nosotros VAMOS A la piscina

3- Mis primas VAN AL gimnasio

4- Tú VAS de excursión

5- Ustedes VAN AL parque municipal

6- Alejandro VA AL museo de ciencias

Explanation:

Las preguntas refieren en su totalidad a la conjugación temporal del verbo "ir". Así, refiere a conjugar dicho verbo en tiempo presente simple, el cual se conjuga de la siguiente manera:

Yo voy, el va, tú vas, nosotros vamos, ellos van, ustedes van

Así, debe interpretarse la persona o personas que realizan la acción, para determinar así como se conjuga dicho verbo.

4 0
3 years ago
How much potassium nitrate, KNO3, would completely dissolve in 100g of water at 40℃?
const2013 [10]
Ok I know this from other stuff potassium nitrate would completely dissolve in a 100 g on was at 30 c would be 60 but this is 40 so I’m not really sure and I don’t what to ok give you a bad grade but if I had to guess I would go with 65 grams
8 0
3 years ago
A 1500-kg sedan goes through a wide intersection traveling from north to south when it is hit by a 2200-kg SUV traveling from ea
astra-53 [7]

Answer:

Velocity of Sedan = 21 m/s

Velocity of SUV = 12 m/s

Explanation:

As we know that deceleration due to friction force is given as

a = - \mu g

so we have

a = -(0.75)(9.81)

a = -7.36 m/s^2

now the two cars comes to rest at a point which is at position of 5.39 m West and 6.43 m South

so net displacement of the car is given as

d = \sqrt{5.39^2 + 6.43^2}

d = 8.39 m

now the velocity of the two cars just after the impact is given as

v^2 - v_i^2 = 2a d

0 - v_i^2 = 2(-7.36)(8.39)

v_i = 11.11 m/s

direction of the motion is given as

tan\theta = \frac{6.43}{5.39}

\theta = 50 degree South of West

now we can use momentum conservation as there is no external force on it

Momentum conservation in North to south direction

m_1 v_1 = (m_1 + m_2) vsin\theta

1500 v_1 = (1500 + 2200) (11.11) sin50

v_1 = 21 m/s

Similarly momentum conservation towards West direction

m_2 v_2 = (m_1 + m_2) vsin\theta

2200 v_2 = (1500 + 2200) (11.11) cos50

v_2 = 12 m/s

5 0
3 years ago
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