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3 years ago

11

When can a high speed velocity cause damage?'

1 answer:

sweet-ann [11.9K]3 years ago

7 0

Answer:

50 Mph.

Explanation:

According to the National Severe Storms Laboratory, winds can really begin to cause damage when they reach <em><u>50 mph</u></em>. But here’s what happens before and after they reach that threshold, according to the Beaufort Wind Scale (showing estimated wind speeds): - at 19 to 24 mph, smaller trees begin to sway.

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Y=mx+b if y=8.18 m=1.31 b=17.2 then find x

Dafna1 [17]

Plug in the corresponding values into y = mx + b

8.18 in for y

1.31 in for m

17.2 in for b

8.18 = 1.31x + 17.2

Now bring 17.2 to the left side by subtracting 17.2 to both sides (what you do on one side you must do to the other). Since 17.2 is being added on the right side, subtraction (the opposite of addition) will cancel it out (make it zero) from the right side and bring it over to the left side.

8.18 - 17.2 = 1.31x

-9.02 = 1.31x

Then divide 1.31 to both sides to isolate x. Since 1.31 is being multiplied by x, division (the opposite of multiplication) will cancel 1.31 out (in this case it will make 1.31 one) from the right side and bring it over to the left side.

-9.02/1.31 = 1.31x/1.31

x ≈ -6.8855

x is roughly -6.89

Hope this helped!

~Just a girl in love with Shawn Mendes

8 0

3 years ago

Read 2 more answers

How do I solve this

lesantik [10]

multiply grav pull by mass of astro maybe with a calculator

7 0

4 years ago

A particle of mass m collides with a second particle of mass m. Before the collision, the first particle is moving in the x-dire

oee [108]

Answer:

a) v, v

b) 2mv^2

c) Elastic collion

Explanation:

(a) The velocity of the second particle after the collision is (v2x,v2y)=(v,−v). From momentum conservation in x-direction

Here x, y represent direction.They are not variable. 1 and 2 represent before and after.

2vm=v1xm+v2xm, we find v1x=v.

From momentum conservation in y-direction

0 =v1ym+v2ym, we findv1y=v.

(b) By energy conservation principle

Before: K=1/2m(2v)^2=2mv^2.

After: K=1/2m(v^2(1x)+v^2(1y))+12m(v22x+v22y)=2mv^2

(c) The collision is elastic

6 0

3 years ago

Which type of mixture could this illustration represent?

Oksanka [162]

Answer:

A homogeneous mixture

Explanation:

4 0

3 years ago

Read 2 more answers

If the electron just misses the upper plate as it emerges from the field, find the magnitude of the electric field.

Damm [24]

Answer:

The magnitude of the electric field be 171.76 N/C so that the electron misses the plate.

Explanation:

As data is incomplete here, so by seeing the complete question from the search the data is

vx_0=1.1 x 10^6

ax=0 As acceleration is zero in the horizontal axis so

Equation of motion in horizontal direction is given as

$s_x=v_x_0 t$

$t=\frac{s_x}{v_x}\\t=\frac{2 \times 10^{-2}}{1.1 \times 6}\\t=1.82 \times 10^{-8} s$

Now for the vertical distance

vy_o=0

than the equation of motion becomes

$s_y=v_y_0 t+\frac{1}{2} at^2\\s_y=\frac{1}{2} at^2\\0.5 \times 10^{-2}=\frac{1}{2} a(1.82 \times 10^{-8})^2\\a=3.02 \times 10^{13} m/s^2$

Now using this acceleration the value of electric field is calculated as

$E=\frac{F}{q}\\E=\frac{ma}{q}\\E=\frac{m_ea}{q_e}\\$

Here a is calculated above, m is the mass of electron while q is the charge of electron, substituting values in the equation

$E=\frac{9.1\times 10^{-31} \times 3.02 \times 10^{13} }{1.6 \times 10^{-19}}\\E=171.76 N/C$

So the magnitude of the electric field be 171.76 N/C so that the electron misses the plate.

5 0

3 years ago