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3 years ago

List three reasons that rain predictions made by meteorologists are sometimes incorrect. Explain your response.

1 answer:

5 0

1. <span>the low pressure is moving slower than expected.
This make the meteorologist receive premature data which make them fail to interpret the data correctly and make the wronf prediction.
2. Sudden change in wind direction, which transfer the natural occurence into other region than where it initially predicted
3. We still haven't developed the methodology to 100% predict natural occurence</span>

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In a physics lab, a 0.500-kg cart (Cart A) moving with a speed of 129 cm/s encounters a magnetic collision with a 1.50-kg cart (

omeli [17]

Answer:

58 cm/s

Explanation:

0.5×129=0.5×(-45)+1.5×V

V=58

7 0

2 years ago

How much would you have to eat for your appendix to explode ? I ate so much on thanksgiving and i need to know i’m very scared

dimulka [17.4K]

Answer:

I don't think your appendix can explode because you ate too much honestly. It's not even possible to eat so much that your appendix explodes, and if you're feeling any pain it definitely isn't because your appendix is about to explode, believe me. Also you could just type it into the internet, that'd be a much faster solution.

6 0

2 years ago

In February 1955, a paratrooper fell 370 m from an airplane without being able to open his chute but happened to land in snow, s

nevsk [136]

a) 0.94 m

The work done by the snow to decelerate the paratrooper is equal to the change in kinetic energy of the man:

$W=\Delta K\\-F d = \frac{1}{2}mv^2 - \frac{1}{2}mu^2$

where:

$F=1.1 \cdot 10^5 N$ is the force applied by the snow

d is the displacement of the man in the snow, so it is the depth of the snow that stopped him

m = 68 kg is the man's mass

v = 0 is the final speed of the man

u = 55 m/s is the initial speed of the man (when it touches the ground)

and where the negative sign in the work is due to the fact that the force exerted by the snow on the man (upward) is opposite to the displacement of the man (downward)

Solving the equation for d, we find:

$d=\frac{1}{2F}mu^2 = \frac{(68 kg)(55 m/s)^2}{2(1.1\cdot 10^5 N)}=0.94 m$

b) -3740 kg m/s

The magnitude of the impulse exerted by the snow on the man is equal to the variation of momentum of the man:

$I=\Delta p = m \Delta v$

where

m = 68 kg is the mass of the man

$\Delta v = 0-55 m/s = -55 m/s$ is the change in velocity of the man

Substituting,

$I=(68 kg)(-55 m/s)=-3740 kg m/s$

7 0

2 years ago

In a lab experiment, a student is trying to apply the conservation of momentum. Two identical balls, each with a mass of 1.0 kg,

Studentka2010 [4]

Answer:

Second Trial satisfy principle of conservation of momentum

Explanation:

Given mass of ball A and ball B $=\ 1.0\ Kg.$

Let mass of ball $A$ and $B\ is\ m$

Final velocity of ball $A\ is\ v_1$

Final velocity of ball $B\ is\ v_2$

initial velocity of ball $A\ is\ u_1$

Initial velocity of ball $B\ is\ u_2$

Momentum after collision $=mv_1+mv_2$

Momentum before collision $= mu_1+mu_2$

Conservation of momentum in a closed system states that, moment before collision should be equal to moment after collision.

Now, $mu_1+mu_2=mv_1+mv_2$

Plugging each trial in this equation we get,

First Trial

$mu_1+mu_2=mv_1+mv_2\\1(1)+1(-2)=1(-2)+1(-1)\\1-2=-2-1\\-1=-3$

momentum before collision $\neq$ moment after collision

Second Trial

$mu_1+mu_2=mv_1+mv_2\\1(.5)+1(-1.5)=1(-.5)+1(-.5)\\.5-1.5=-.5-.5\\-1=-1$

moment before collision $=$ moment after collision

Third Trial

$mu_1+mu_2=mv_1+mv_2\\1(2)+1(1)=1(1)+1(-2)\\2+1=1-2\\3=-1$

momentum before collision $\neq$ moment after collision

Fourth Trial

$mu_1+mu_2=mv_1+mv_2\\1(.5)+1(-1)=1(1.5)+1(-1.5)\\.5-1=1.5-1.5\\-.5=0$

momentum before collision $\neq$ moment after collision

We can see only Trial- 2 shows the conservation of momentum in a closed system.

4 0

3 years ago

Read 2 more answers

Find the net electric force that the two charges would exert on an electron placed at point on the xx-axis at xx = 0.200 mm. Exp

UkoKoshka [18]

Answer:

The question has some details missing, here is the complete question ; A -3.0 nC point charge is at the origin, and a second -5.0nC point charge is on the x-axis at x = 0.800 m. Find the net electric force that the two charges would exert on an electron placed at point on the x-axis at x = 0.200 m.

Explanation:

The application of coulonb's law is used to approach the question as shown in the attached file.

6 0

2 years ago