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3 years ago

Your friend claims that weather predictions are just guesses. What would you say to change your friend's mind?

1 answer:

4 0

To change my friends mind, after they've already said weather predictions are just guesses, I would say they are correct, but not only to they use guesses to predict what will happen, but they also, before hand, use atmospheric tools. They then predict from the data collected. They do both.

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Whenever the forces applied to an object that is at rest are balanced, it will..l

lana [24]

Inertia is a force which keeps stationary objects at rest and moving objects in motion at ... False - Pounds is a unit of force commonly used in the British system of ... When a chemistry student places a beaker on a balance and determines it to be ... In this case, an object moving to the right could have a balance of forces if it is ...

5 0

3 years ago

wo lacrosse players collide in midair. Jeremy has a mass of 120 kg and is moving at a speed of 3 m/s. Hans has a mass of 140 kg

Julli [10]

2.71 m/s fast Hans is moving after the collision.

<u>Explanation</u>:

Given that,

Mass of Jeremy is 120 kg ( $M_J$ )

Speed of Jeremy is 3 m/s ( $V_J$ )

Speed of Jeremy after collision is ( $V_{JA}$ ) -2.5 m/s

Mass of Hans is 140 kg ( $M_H$ )

Speed of Hans is -2 m/s ( $V_H$ )

Speed of Hans after collision is ( $V_{HA}$ )

Linear momentum is defined as “mass time’s speed of the vehicle”. Linear momentum before the collision of Jeremy and Hans is

= $=\mathrm{M}_{1} \times \mathrm{V}_{\mathrm{J}}+\mathrm{M}_{\mathrm{H}} \times \mathrm{V}_{\mathrm{H}}$

Substitute the given values,

= 120 × 3 + 140 × (-2)

= 360 + (-280)

= 80 kg m/s

Linear momentum after the collision of Jeremy and Hans is

= $=\mathrm{M}_{\mathrm{J}} \times \mathrm{V}_{\mathrm{JA}}+\mathrm{M}_{\mathrm{H}} \times \mathrm{V}_{\mathrm{HA}}$

= 120 × (-2.5) + 140 × $V_{HA}$

= -300 + 140 × $V_{HA}$

We know that conservation of liner momentum,

Linear momentum before the collision = Linear momentum after the collision

80 = -300 + 140 × $V_{HA}$

80 + 300 = 140 × $V_{HA}$

380 = 140 × $V_{HA}$

380/140= $V_{HA}$

$V_{HA}$ = 2.71 m/s

2.71 m/s fast Hans is moving after the collision.

4 0

3 years ago

Earth’s polar ice caps contain about 2.3 × 1019 kg of ice. This mass contributes essentially nothing to the moment of inertia of

sp2606 [1]

Answer:

Explanation:

Initial moment of inertia of the earth I₁ = 2/5 MR² , M is mss of the earth and R is the radius . If ice melts , it forms an equivalent shell of mass 2.3 x 10¹⁹ Kg

Final moment of inertia I₂ = 2/5 M R² + 2/3 x 2.3 x 10¹⁹ x R²

For change in period of rotation we shall apply conservation of angular momentum law

I₁ ω₁ = I₂ ω₂ , ω₁ and ω₂ are angular velocities initially and finally .

I₁ / I₂ = ω₂ / ω₁

I₁ / I₂ = T₁ / T₂ , T₁ , T₂ are time period initially and finally .

T₂ / T₁ = I₂ / I₁

(2/5 M R² + 2/3 x 2.3 x 10¹⁹ x R²) / 2/5 MR²

1 + 5 / 3 x 2.3 x 10¹⁹ / M

= 1 + 5 / 3 x 2.3 x 10¹⁹ / 5.97 x 10²⁴

= 1 + .0000064

T₂ = 24 (1 + .0000064)

= 24 hours + .55 s

change in length of the day = .55 s .

3 0

3 years ago

a ball is projected horizontally with a velocity of 5 m per second from the top of a building 19.6 m high how long will the ball

zepelin [54]

Answer:

1.98s

Explanation:

The time taken to hit the ground is given by

h=ut+ 1/2 at^2

but u =0

so we have

h=1/2at^2

making t the subject

t=√2h/g

√2×19.6/10

1.98s

8 0

3 years ago

A wad of clay of mass m1 = 0.49 kg with an initial horizontal velocity v1 = 1.89 m/s hits and adheres to the massless rigid bar

notka56 [123]

Answer:

<h2>The angular velocity just after collision is given as</h2><h2> $\omega = 0.23 rad/s$ </h2><h2>At the time of collision the hinge point will exert net external force on it so linear momentum is not conserved</h2>

Explanation:

As per given figure we know that there is no external torque about hinge point on the system of given mass

So here we will have

$L_i = L_f$

now we can say

$m_1v_1\frac{L}{2} = (m_2L^2 + m_1(\frac{L}{2})^2)\omega$

so we will have

$0.49(1.89)(0.45) = (2.13(0.90)^2 + 0.49(0.45)^2)\omega$

$\omega = 0.23 rad/s$

Linear momentum of the system is not conserved because at the time of collision the hinge point will exert net external force on the system of mass

So we can use angular momentum conservation about the hinge point

6 0

3 years ago