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LuckyWell [14K]
3 years ago
8

7. An airplane travels 4988.966 Km in 57 minutes. Calculate the average speed in km/sec.

Physics
1 answer:
zubka84 [21]3 years ago
6 0

Answer:

wow easy

Explanation:

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This particle is found in the nucleus and adds mass to the nucleus, but changing its number does not change what the element is.
Sliva [168]
Adding neutron to the nucleus creates new isotopes of that element. Mass of the nucleus will increase if you add neutron to it but element will remain the same. If you add proton to the nucleus, mass of the nucleus will change but also element will also change. That is, atom of the element will change. 

5 0
3 years ago
A 1.20 kg solid ball of radius 40 cm rolls down a 5.20 m long incline of 25 degrees. Ignoring any loss due to friction, how fast
sladkih [1.3K]

Answer:

1.6s

Explanation:

Given that A 1.20 kg solid ball of radius 40 cm rolls down a 5.20 m long incline of 25 degrees. Ignoring any loss due to friction,

To know how fast the ball will roll when it reaches the bottom of the incline, we need to calculate the acceleration at which it is rolling.

Since the frictional force is negligible, at the top of the incline plane, the potential energy = mgh

Where h = 5.2sin25

h = 2.2 m

P.E = 1.2 × 9.8 × 2.2

P.E = 25.84 j

At the bottom, K.E = P.E

1/2mv^2 = 25.84

Substitutes mass into the formula

1.2 × V^2 = 51.69

V^2 = 51.69/1.2

V^2 = 43.07

V = 6.56 m/s

Using the third equation of motion

V^2 = U^2 + 2as

Since the object started from rest,

U = 0

6.56^2 = 2 × a × 5.2

43.07 = 10.4a

a = 43.07/10.4

a = 4.14 m/s^2

Using the first equation of motion,

V = U + at

Where U = 0

6.56 = 4.14t

t = 6.56/4.14

t = 1.58s

Therefore, the time the ball rolls when it reaches the bottom of the incline is approximately 1.6s

6 0
3 years ago
What type of galaxy is this?
stiks02 [169]

Answer:

Barred Spiral.

8 0
2 years ago
A positive charge of 18nC is evenly distributed along a straight rod of length 4.0 m that is bent into a circular arc with a rad
sergey [27]

Explanation:

Formula for angle subtended at the center of the circular arc is as follows.

           \theta = \frac{S}{r}

where,   S = length of the rod

              r = radius

Putting the given values into the above formula as follows.      

                \theta = \frac{S}{r}

                             = \frac{4}{2}

                             = 2 radians (\frac{180^{o}}{\pi})

                             = 114.64^{o}

Now, we will calculate the charge density as follows.

                 \lambda = \frac{Q}{L}

                            = \frac{18 \times 10^{-9} C}{4 m}

                            = 4.5 \times 10^{-9} C/m

Now, at the center of arc we will calculate the electric field as follows.

                 E = \frac{2k \lambda Sin (\frac{\theta}{2})}{r}

                     = \frac{2(9 \times 10^{9} Nm^{2}/C^{2})(4.5 \times 10^{-9}) Sin (\frac{114.64^{o}}{2})}{2 m}

                      = 34.08 N/C

Thus, we can conclude that the magnitude of the electric eld at the center of curvature of the arc is 34.08 N/C.

5 0
3 years ago
During a single-replacement reaction, a more-active metal will replace a less-active metal in a compound. Which single-replaceme
son4ous [18]

Answer:

Mg will replace Ag in a compound

Explanation:

A single replacement reaction is driven by the position of ions on the activity series.

As a rule of thumb, the position of metal ions on the activity series determines their reactivity.

Metal ions that are above another are more reactive and they will displace those that are lower.

Generally, activity increases as we go up the group.

Mg ions are higher than Ag ions on the series so, Mg will displace Ag from a solution.

8 0
3 years ago
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