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3 years ago

When you stretch a spring 20 cm past its natural length, it exerts a force of 8

Physics

2 answers:

Nastasia [14]3 years ago

6 0

Answer:

40 N/m

Explanation:

F = -kx (This is the Hooke's Law equation)

F is the force the spring exerts = 8 N

-k = spring constant

x = displacement (The distance stretched past it's natural length) = 20cm

x needs to be in meters, and 20 cm is = to 0.2 meters

Finally:

8N = -k (0.2m)

-k = 8N / 0.2 m

k = -40 N/m

Lina20 [59]3 years ago

5 0

Answer:

0.4 N/cm

Explanation:

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A motorist traveling at 12 m/s encounters a deer in the road 39 m ahead. If the maximum acceleration the vehicle’s brakes are ca

Vera_Pavlovna [14]

Answer:

Explanation:

Given

Motorcyclist speed=12 m/s

maximum acceleration $=-6 m/s^2$

distance=39 m

Let x be the distance traveled by motorist in his reaction time

therefore remaining 39-x will be traveled with $-6m/s^2$ acceleration

$v^2-u^2=2as$

s=39-x

v=0

u=12 m/s

$0-12^2=2\left ( -6\right )\left ( 39-x\right )$

x=27 m

Therefore he traveled 27 m in his reaction time

$27=12\times t$

t=2.25 s

(b)If his reaction time is 2.56 sec

then distance traveled in his reaction time

$x_0=12\times 2.56=30.72 m$

Remaining distance 39-30.72=8.28 m

therefore its velocity when it reaches the deer

$v^2-u^2=2as$

$v^2=12^2+2\times \left ( -6\right )\times 8.28=44.64$

v=6.681 m/s

8 0

3 years ago

Two satellites, A and B are in different circular orbits

jek_recluse [69]

Answer:

The ratio of the orbital time periods of A and B is $\frac{1}{2}$

Solution:

As per the question:

The orbit of the two satellites is circular

Also,

Orbital speed of A is 2 times the orbital speed of B

$v_{oA} = 2v_{oB}$ (1)

Now, we know that the orbital speed of a satellite for circular orbits is given by:

$v_{o} = \farc{2\piR}{T}$

where

R = Radius of the orbit

Now,

For satellite A:

$v_{oA} = \farc{2\piR}{T_{a}}$

Using eqn (1):

$2v_{oB} = \farc{2\piR}{T_{a}}$ (2)

For satellite B:

$v_{oB} = \farc{2\piR}{T_{b}}$ (3)

Now, comparing eqn (2) and eqn (3):

$\frac{T_{a}}{T_{b}} = \farc{1}{2}$

6 0

2 years ago

A scientist prepares a colloidal solution and pours it into a glass tank. She then flashes a beam of white light into one end of

balu736 [363]

The answer is C) <span>The higher frequencies of visible light were scattered by the colloid particles.</span>

5 0

2 years ago

Read 2 more answers

List small/average stars<br><br>

mario62 [17]

Answer:

Lol, you should do Nate, Bobby, Cindy, Joe, and Beth

Jk, if you want to be series and probably not fail go for these:

If it wants types of small/average stars, then go with

Small star names:

OGLE-TR-122B

Gliese 229 B

TRAPPIST-1

Teegarden's Star

Luyten 726-8 (A and B)

Proxima Centauri

Wolf 359 111400

Ross 248

Barnard's Star

CM Draconis B

Ross 154 167000

CM Draconis A

Kapteyn's Star

7 0

2 years ago

A piece of steel is 11.5cm long at 22C. It is heated to 1221C, close to its melting point. How long is it, in cm, at the high te

Nataly [62]

Answer:

The length at the final temperature is 11.7 cm.

Explanation:

We need to use the thermal expansion equation:

$\Delta L=\alpha L_{0}\Delta T$

Where:

L(0) is the initial length
ΔT is the differential temperature, final temperature minus initial temperature (T(f)-T(0))
ΔL is the final length minus the initial length (L(f)-L(0))
α is the coefficient of linear expantion of steel (12.5*10⁻⁶ 1/°C)

So, we have:

$L_{f}-L_{0}=\alpha L_{0}(T_{f}-T_{0})$

$L_{f}=L_{0}+\alpha L_{0}(T_{f}-T_{0})$

$L_{f}=0.115+(12.5*10^{-6})(0.115)(1221-22)$

$L_{f}=0.117\: m$

Therefore, the length at the final temperature is 11.7 cm.

I hope it helps you!

7 0

2 years ago