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3 years ago

13

Consider to cars traveling directly towards each other. Car A has twice the mass of Car B. In order for the cars to completely s

top when they collide, which of the following is true? (Remember, Momentum = mass * velocity)
Car A should have twice the velocity of Car B

Car B should have twice the velocity of Car A

Car A should have 4x the velocity of Car B

Car B should have 4x the velocity of Car A

1 answer:

Greeley [361]3 years ago

7 0

Please mark brainliest you answer is B).

Have a good night!

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Two parallel wires are separated by 6.10 cm, each carrying 2.85 A of current in the same direction. (a) What is the magnitude of

Westkost [7]

Answer:

The force per unit length is $2.66 \times 10^{-5} \ N/m$

Explanation:

The current carrying by each wires = 2.85 A

The current in both wires flows in same direction.

The gap between the wires = 6.10 cm

Now we will use the below expression for the force per unit length. Moreover, before using the below formula we have to change the unit centimetre into meter. So, we just divide the centimetre with 100.

$F/l = \frac{\mu _0i_1 i_2}{2\pi d} \\i_1 = 2.85 \\i_ 2 = 2.85 \\\mu _0 = 4\pi \times 10^{-7} \\d = 0.061 \\F/l = \frac{4\pi \times 10^{-7} \times 2.85 \times 2.85}{2 \pi \times 0.061} \\= 2.66 \times 10^{-5} \ N/m$

4 0

3 years ago

A solid block is attached to a spring scale. When the block is suspended in air, the scale reads 20.1 N; when it is completely i

Paha777 [63]

Answer:

A) $V = 4.92 \cdot 10^{-4} m^{3} = 492 cm^{3}$

B) $d = 4181.49 kg/m^{3} = 4.18 g/cm^{3}$

Explanation:

A) Using the Archimedes' force we can find the weight of water displaced:

$W_{d} = W_{a} - W_{w}$

Where:

$W_{a}$ : is the weight of the block in the air = 20.1 N

$W_{w}$ : is the weight of the block in the water = 15.3 N

$W_{d} = W_{a} - W_{w} = 20.1 N - 15.3 N = 4.8 N$

Now, the mass of the water displaced is:

$m = \frac{W_{d}}{g} = \frac{4.8 N}{9.81 m/s^{2}} = 0.49 kg$

The volume of the block can be found using the mass of water displaced and the density of the water:

$V = \frac{m}{d} = \frac{0.49 kg}{997 kg/m^{3}} = 4.92 \cdot 10^{-4} m^{3} = 492 cm^{3}$

B) The density of the block can be found as follows:

$d = \frac{W_{a}}{g*V} = \frac{20.1 N}{9.81 m/s^{2}*4.92 \cdot 10^{-4} m^{3}} = 4181.49 kg/m^{3} = 4.18 g/cm^{3}$

I hope it helps you!

6 0

3 years ago

Mercury, a planet in our Solar System, and Callisto, a moon of the planet Jupiter, are about the same size. However, Mercury’s m

TiliK225 [7]

You may jump higher because the more the mass of the planet, the more gravitational force. There is less mass(and gravity) on Callisto so you wouldn’t be weighed down as much and can jump higher. Whereas on Jupiter there is more weight holding you down.

3 0

3 years ago

Read 2 more answers

IMPORTANT 3 QUESTIONS!

Brums [2.3K]

Answer:

7. 20,000,000 mL.

8. 8,000 m.

9. 120,000 secs.

10. 4

Explanation:

7. Determination of the volume in millilitres (mL)

Volume in litre (L) = 20,000 L

Volume in millilitres (mL) =..?

The volume in mL can be obtained as follow:

1 L = 1,000 mL

Therefore,

20,000 L = 20,000 x 1,000 = 20,000,000 mL.

Therefore, 20,000 L is equivalent to 20,000,000 mL.

8. Determination of the distance in metre (m)

Distance in mile = 5 mile

Distance in metre =?

First, we shall convert from mile to kilometre.

This can be done as follow:

1 mile = 1.6 km

Therefore,

5 mile = 5 x 1.6 = 8 km

Finally, we shall convert 8 km to metre (m).

This is illustrated below:

1 km = 1,000 m

Therefore,

8 km = 8 x 1,000 = 8,000 m

Therefore, 5 miles is equivalent to 8,000 m.

9. Determination of the time in seconds.

Time = 400 minutes for 5 days.

First, we shall convert 400 mins to hour.

This is illustrated below:

60 minutes = 1 hour

Therefore,

400 mins = 400/60 = 20/3 hours

The time (hours) is 20/3 hours in 1 day.

Therefore, the time (hours) in 5 days will be = 20/3 x 5 = 100/3 hours.

Next, we shall convert 100/3 hours to minutes.

This is illustrated below:

1 hour = 60 minutes

Therefore,

100/3 hours = 100/3 x 60 = 2000 mins

Finally, we shall convert 2000 mins to seconds.

This is illustrated below:

1 mins = 60 secs

2000 mins = 2000 x 60 = 120,000 secs.

Therefore, the time is 120,000 secs.

10. Determination of the number of significant figures.

To obtain the significant figures of a number, we simply count all the numbers available.

Therefore, the number of significant figures for 9876 is 4.

6 0

3 years ago

A tree falls in a forest. How many years must pass before the 14C activity in 1.03 g of the tree's carbon drops to 1.02 decay pe

Illusion [34]

Answer:

t = 5.59x10⁴ y

Explanation:

To calculate the time for the ¹⁴C drops to 1.02 decays/h, we need to use the next equation:

$A_{t} = A_{0}\cdot e^{- \lambda t}$ (1)

<em>where $A_{t}$ : is the number of decays with time, A₀: is the initial activity, λ: is the decay constant and t: is the time.</em>

To find A₀ we can use the following equation:

$A_{0} = N_{0} \lambda$ (2)

<em>where N₀: is the initial number of particles of ¹⁴C in the 1.03g of the trees carbon </em>

From equation (2), the N₀ of the ¹⁴C in the trees carbon can be calculated as follows:

$N_{0} = \frac{m_{T} \cdot N_{A} \cdot abundance}{m_{^{12}C}}$

<em>where $m_{T}$ : is the tree's carbon mass, $N_{A}$ : is the Avogadro's number and $m_{^{12}C}$ : is the ¹²C mass. </em>

$N_{0} = \frac{1.03g \cdot 6.022\cdot 10^{23} \cdot 1.3\cdot 10^{-12}}{12} = 6.72 \cdot 10^{10} atoms ^{14}C$

Similarly, from equation (2) λ is:

$\lambda = \frac{Ln(2)}{t_{1/2}}$

<em>where t 1/2: is the half-life of ¹⁴C= 5700 years </em>

$\lambda = \frac{Ln(2)}{5700y} = 1.22 \cdot 10^{-4} y^{-1}$

So, the initial activity A₀ is:

$A_{0} = 6.72 \cdot 10^{10} \cdot 1.22 \cdot 10^{-4} = 8.20 \cdot 10^{6} decays/y$

Finally, we can calculate the time from equation (1):

$t = - \frac{Ln(A_{t}/A_{0})}{\lambda} = - \frac {Ln(\frac{1.02decays \cdot 24h \cdot 365d}{1h\cdot 1d \cdot 1y \cdot 8.20 \cdot 10^{6} decays/y})}{1.22 \cdot 10^{-4} y^{-1}} = 5.59 \cdot 10^{4} y$

I hope it helps you!

4 0

3 years ago