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3 years ago

8

Miles and Joselin are riding a two person long board down an extremely dangerous hill while Chloe watches in horror and Talia in

delight. At the end of the hill Alex and Jason are holing a crash pad to stop the dynamic duo from falling off the cliff. If the riders have a combined mass of 200 kg and an acceleration of 30 m/s^2, what Force will they crash the mattress with? Use the correct Force for full credit.

1 answer:

saw5 [17]3 years ago

8 0

Answer:

They will crash with the force of 6,000N

Explanation:

Force is equal to mass time acceleration. When we multiply 200 kg, the mass, and 30 m/s2, the acceleration, the answer is 6,000.

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Find the distance from a point charge q=100nC where the field intensity is equal to E=6kN/C. please include description of the r

madam [21]

Answer: 0.3872m

Explanation:

q= 100nC --> 100x10^-9 C

k= 9x10^9 Nm^2/C^2

E= 6kN/C --> 600 N/C

r=?

$E= K\frac{q}{r^{2} }$ --> $r=\sqrt{\frac{kq}{E} }$ Despejas "r"

Resuelves

<h3> $r=\sqrt{\frac{(9x10^9 Nm^2/C^2)(100x10x^{-9} C)}{6000N/C} }$ (la x es por, no es una variable)</h3><h3>r= 0.3872983346m</h3>

4 0

2 years ago

The cornea behaves as a thin lens of focal length approximately 1.80 {\rm cm}, although this varies a bit. The material of which

spayn [35]

Answer:

The height of the image will be "1.16 mm".

Explanation:

The given values are:

Object distance, u = 25 cm

Focal distance, f = 1.8 cm

On applying the lens formula, we get

⇒ $\frac{1}{v} -\frac{1}{u} =\frac{1}{f}$

On putting estimate values, we get

⇒ $\frac{1}{v} -\frac{1}{(-25)} =\frac{1}{1.8}$

⇒ $\frac{1}{v} =\frac{1}{1.8} -\frac{1}{25}$

⇒ $v=1.94 \ cm$

As a result, the image would be established mostly on right side and would be true even though v is positive.

By magnification,

$m=\frac{v}{u}$ and $m=\frac{h_{1}}{h_{0}}$

⇒ $\frac{v}{u} =\frac{h_{1}}{h_{0}}$

⇒ $\frac{1.94}{25}=\frac{{h_{1}}}{15}$

⇒ ${h_{1}}=1.16 \ mm$

8 0

3 years ago

At what distance of separation, r, must two 3.20 x 10-9 Coulomb charges be positioned in order for the repulsive force between t

STatiana [176]

I'm not 100% sure but I think the answer is 60.4, because you multiply 3.20 by 10, to get 32 - 9 = 23, then subtract 23 from 83.4. Hope this helps you, and good luck!!!

4 0

2 years ago

Read 2 more answers

A car driving at 20 m/s accelerates continuously 2m/s2 for 3 seconds. What is its final velocity

Lesechka [4]

Answer: V= u+ at

V= final velocity

u=initial velocity

a=acceleration

t=time taken

V= 20 + 2*3

V= 26m/s

Explanation:

5 0

2 years ago

A uniform thin wire is bent into a quarter-circle of radius a = 20.0 cm, and placed in the first quadrant. Determine the coordin

Mashcka [7]

Answer:

$r_{cm}=[12.73,12.73]cm$

Explanation:

The general equation to calculate the center of mass is:

$r_{cm}=1/M*\int\limits {r} \, dm$

Any differential of mass can be calculated as:

$dm = \lambda*a*d\theta$ Where "a" is the radius of the circle and λ is the linear density of the wire.

The linear density is given by:

$\lambda=M/L=M/(a*\pi/2)=\frac{2M}{a\pi}$

So, the differential of mass is:

$dm = \frac{2M}{a\pi}*a*d\theta$

$dm = \frac{2M}{\pi}*d\theta$

Now we proceed to calculate X and Y coordinates of the center of mass separately:

$X_{cm}=1/M*\int\limits^{\pi/2}_0 {a*cos\theta*2M/\pi} \, d\theta$

$Y_{cm}=1/M*\int\limits^{\pi/2}_0 {a*sin\theta*2M/\pi} \, d\theta$

Solving both integrals, we get:

$X_{cm}=2*a/\pi=12.73cm$

$Y_{cm}=2*a/\pi=12.73cm$

Therefore, the position of the center of mass is:

$r_{cm}=[12.73,12.73]cm$

5 0

2 years ago