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Nat2105 [25]
3 years ago
9

What pair of atoms forms ionic bonds

Physics
1 answer:
WARRIOR [948]3 years ago
8 0

K,F potassium , fluorine form ionic bond,


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A remote controlled toy car starts from rest and begins to accelerate in a straight line. The figure below represents "snapshots
Nikolay [14]

(a). The car's average velocity between t = 1.0s to t = 1.5s will be - 1\;m/s

(b). The car's acceleration at t = 1.5s will be - 0.4\;m/s^{2}

(c). Car's speed is increasing with time.

We have a a remote controlled toy car that starts from rest and begins to accelerate in a straight line.

We have to determine -

  • The car's average velocity (in m/s) in the interval between -

        t = 1.0 s  to  t = 1.5 s.

  • The car's acceleration at t = 1.5 s.
  • Determining whether car's speed increasing or decreasing with time.

<h3>What is Acceleration?</h3>

The rate of change of velocity with respect to time is called Acceleration. Mathematically -

$a=\frac{dv}{dt}

According to the question, we have the following data for the Car -

t = 0s → x = 0m

t = 0.5s → x = 0.1m

t = 1.0s → x = 0.4m

t = 1.5s → x = 0.9m

t = 2.0s → x = 1.6m

PART - A

The car's average velocity between t = 1.0s to t = 1.5s will be -

$v_{avg} = \frac{0.9-0.4}{1.5-1}= 1 m/s

PART - B

Velocity at t = 1.5 s will be -

$v(1.5)=\frac{0.9}{1.5}= 0.6\;m/s

The car's acceleration at t = 1.5s will be -

$a(1.5) = \frac{v}{t} = \frac{0.6}{1.5} = 0.4\;m/s^{2}

PART - C

Since, the acceleration of the car is positive, this means that the car is accelerating in the forward direction. Hence, its speed is increasing with time.

[ The following data was missing in your answer. The complete question would include this data also -

t = 0s → x = 0m

t = 0.5s → x = 0.1m

t = 1.0s → x = 0.4m

t = 1.5s → x = 0.9m

t = 2.0s → x = 1.6m ]

To solve more questions on Kinematics, visit the link below-

brainly.com/question/17272824

#SPJ2

7 0
1 year ago
Read 2 more answers
Which of the device has the highest resistance?
-BARSIC- [3]
It would be c) Rheostat
5 0
3 years ago
Read 2 more answers
A car is traveling at a speed of 38.0 m/s on an interstate highway where the speed limit is 75.0 mi/h. Is the driver exceeding t
lidiya [134]

Answer:

Yes, the car driver is exceeding the given limit.

Explanation:

<u>Given:</u>

  • Speed of the car, v = 38.0 m/s.
  • Speed limit of the highway, \rm v_o=75.0\ mi/h.

<h2><u>Converting the speed limit from mi/h to m/s:</u></h2>

We know,

1 mi = 1.60934 km.

1 km = 1000 m.

Therefore, 1 mi = 1.60934 × 1000 m = 1609.34 m.

1 hour = 60 minutes.

1 minute = 60 seconds.

Therefore, 1 hour = 60 × 60 seconds = 3600 seconds.

Using these values,

\rm 1\ \dfrac{mi}{h}=\dfrac{1609.34\ m}{3600\ s}=0.447\ m/s.

Therefore,

\rm v_o = 75.0\ mi/h=75.0\times 0.447=33.52\ m/s.

Clearly,

\rm v_o

which means, the car driver is exceeding the given speed limit.

6 0
3 years ago
Suppose you observe two stars and you know they have the same luminosity. If one star is twice as far away as the other, the mor
rosijanka [135]

Answer:

The farther star will appear 4 times fainter than the star that is near to the observer.

Explanation:

Since it is given that the luminosity of the 2 stars is same thus they radiate the same energy per unit time

Consider a spherical wave front of energy 'E' that leaves both the stars (Both radiate 'E' as they have same luminosity)

This Energy is spread over the whole surface area of sphere Thus when the wave front is at a distance 'r' the energy per unit surface area is given by

e_{1}=\frac{E}{4\pi r^{2}}

For the star that is twice away from the earth the distance is '2r' thus we will receive an energy given by

e_{2}=\frac{E}{4\pi (2r)^{2}}=\frac{E}{8\pi r^{2}}=\frac{e_{1}}{4}

Hence we sense it as 4 times fainter than the nearer star.

5 0
2 years ago
Mathphys Help help help
Keith_Richards [23]

Answer:

1030 mph

Explanation:

The new velocity equals the initial velocity plus the wind velocity.

First, in the x (east) direction:

vₓ = 335 mph + 711 cos 19° mph

vₓ = 1007 mph

And in the y (north) direction:

vᵧ = 0 mph + 711 sin 19° mph

vᵧ = 231 mph

The net speed can be found with Pythagorean theorem:

v² = vₓ² + vᵧ²

v² = (1007 mph)² + (231 mph)²

v ≈ 1030 mph

6 0
2 years ago
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