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2 years ago

What are 5 examples of expansion and contraction? What are their benefits?

Physics

1 answer:

liubo4ka [24]2 years ago

6 0

Answer:

If you hold a hot glass tumbler under cold water, it will crack

Explanation:

This is because the outer surface of the glass comes in direct contact with cold water and contracts more compared to the inner surface.

This is one example I know.

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QUESTION 2 DOK 3 A Thompson's gazelle has a maximum acceleration of 4.5 m/s2 At this acceleration, how much time is required for

Dominik [7]

Answer:

2.47 s

Explanation:

Convert the final velocity to m/s.

40 km/h → 11.1111 m/s

We have the acceleration of the gazelle, 4.5 m/s².

We can assume the gazelle starts at an initial velocity of 0 m/s in order to determine how much time it requires to reach a final velocity of 11.1111 m/s.

We want to find the time t.

Find the constant acceleration equation that contains all four of these variables.

v = v₀ + at

Substitute the known values into the equation.

11.1111 = 0 + (4.5)t
11.1111 = 4.5t
t = 2.469133333

The Thompson's gazelle requires a time of 2.47 s to reach a speed of 40 km/h (11.1111 m/s).

5 0

3 years ago

You are taking a turn at 30.0 m/s on a ramp of radius 25.0 m. What is your acceleration?

Nat2105 [25]

Acceleration = v^2 / r

Acceleration = 30^2/25

Acceleration = 900/25

Acceleration = 36 m/s^2

6 0

3 years ago

Two Velocities in a Traveling Wave? Wave motion is characterized by two velocities: the velocity with which the wave moves in th

Lyrx [107]

For the answer to the question above,
<span>There is nothing in the equations to suggest that the string moves in the x direction so D) v_x(x,t)=0.
</span>
y(x,t) = A sin(kx-omega t)
d{y(x,t)}/d{x} = A k cos(kx - omega t)

3 0

3 years ago

A 21.0 kg shopping cart is moving with a velocity of 6.0 m/s. It strikes a 11.0 kg box that is initially at rest. They stick tog

DedPeter [7]

Answer:

a) 126 kgm/s

b) 0 kgm/s

c) 3.9 m/s

Explanation:

To solve this question, we will use the law of conservation of momentum.

Momentum before collision = momentum after collision

m1v1 + m2v2 = (m1 + m2)v, where

m1 = mass of the shopping cart, 21 kg

m2 = mass of the box, 11 kg

v1 = initial velocity of the shopping cart, 6 m/s

v2 = initial velocity of the box, 0 m/s

v = final velocity of the box+cart

Momentum of the shopping cart before collision = P

P = mv

P = 21 * 6

P = 126 kgm/s = c

Momentum of the box before collision

Like in question a above, the momentum of the box is P

P = mv

P = 11 * 0

P = 0 kgm/s = b

Velocity of the combined shopping cart wreckage after collision is

m1v1 + m2v2 = (m1 + m2)v

(21 * 6) + (11 * 0) = (21 + 11)v

126 + 0 = 32v

32v = 126

v = 126/32

v = 3.9375 m/s, on approximating to 1 decimal place, we have 3.9 m/s and option b as the answer.

Please vote brainliest

8 0

3 years ago

Dos cargas Q1=2pc y Q2=4pc estan separadas por una distancia de 6cm ¿con que fuerza se atraen?

noname [10]

Here we can use coulomb's law to find the force between two charges

As per coulombs law

]tex]F = \frac{kq_1q_2}{r^2}[/tex]

here we have

$k = 9 * 10^9$

$q_1 = 2pC$

$q_2 = 4pC$

$r = 6cm = 0.06 m$

now by using the above equation we have

$F = \frac{9*10^9 * 2*10^{-12} * 4*10^{-12}}{0.06^2}$

$F = 2 * 10^{-11} N$

so here the force between two charges is of above magnitude and this will be repulsive force between them as both charges are of same sign.

3 0

3 years ago