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3 years ago

7

Atomic radiusmoving from left to right across a period,

1 answer:

Ilya [14]3 years ago

6 0

Answer:

Atomic radius decreases moving from left to right across a period.

Explanation:

When we move left to right across a period, the size of atoms generally decreases. It is because within the period the outer electrons are in same valence shell and the number of electrons and proton increases moving from left to right across the the period. It increases the effective nuclear charge resulting in the increased attraction of electron to the nucleus that causes the decreased radius of the atoms.

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When a sound wave encounters a barrier, what happens?

Tema [17]

Sound waves will either be reflected (echo effect) or absorbed (dissipated) depending upon the material make-up of the barrier

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4 years ago

Dwight uses a spring (k = 70 N/m) on a horizontal surface to make a launcher for model cars. The spring is attached to a holder

kramer

Part A)

As we know that spring force is given by

F = kx

here x = stretch in the spring from natural length

So here when spring reaches to its natural length

Force due to spring = 0

so acceleration = 0

Part b)

When spring is compressed from its natural length it will have elastic potential energy in it

so it is given by

$U = \frac{1}{2}kx^2$

now we know that there is no friction in it so maximum kinetic energy of the launcher must be equal to the elastic potential energy of the spring

$KE = \frac{1}{2}kx^2$

here we have

k = 70 N/m

x = 0.4 m

$KE = \frac{1}{2}(70)(0.4)^2$

$KE = 5.6 J$

Part c)

Now to find the speed we know that

$KE = \frac{1}{2} mv^2$

$5.6 = \frac{1}{2}0.3v^2$

$v = 6.11 m/s$

so its speed is 6.11 m/s

8 0

3 years ago

Read 2 more answers

Who water rocket starts from rest and roses straight up with an acceleration of 5 m/s until it runs out of water 2.5 seconds lat

Kitty [74]

Answer:

23. 4375 m

Explanation:

There are two parts of the rocket's motion

1 ) accelerating (assume it goes upto h1 height )

using motion equations upwards

$s = ut+\frac{1}{2}*a*t^{2} \\h_1=0+\frac{1}{2}*5*2.5^{2} \\=15.625 m$

Lets find the velocity after 2.5 seconds (V1)

V = U +at

V1 = 0 +5*2.5 = 12.5 m/s

2) motion under gravity (assume it goes upto h2 height )

now there no acceleration from the rocket. it is now subjected to the gravity

using motion equations upwards (assuming g= 10m/s² downwards)

V²= U² +2as

0 = 12.5²+2*(-10)*h2

h2 = 7.8125 m

maximum height = h1 + h2

= 15.625 + 7.8125

= 23. 4375 m

3 0

3 years ago

A 2.98-kg object oscillates on a spring with an amplitude of 8.05 cm. Its maximum acceleration is 3.55 m/s2. Calculate the total

Aloiza [94]

Answer:

a = ω^2 A formula for max acceleration (ignoring sign)

V = ω A formula for max velocity

V^2 = ω^2 A^2 = a A from first equation

E = 1/2 M V^2 = 1/2 * 2.98 * 3.55 * .0805 = .426 J

(kg * m/sec^2 * m = kg m^2 / sec^2 = Joule

6 0

2 years ago

A 1000 kg racecar, which is capable of a top speed of 125 m/s, is sitting in a

k0ka [10]

Sitting = no movement
KE=0

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3 years ago