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My name is Ann [436]

3 years ago

14

Does the polarity of the rocks in the Aragorn Sea suggest that it is currently expanding? Explain. Thanks This is the last quest

ion

1 answer:

d1i1m1o1n [39]3 years ago

7 0

The answer is no. Hope this helps

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If an object force of 50 N is used to move an object a distance of 20 m, what distance must the object be moved if the input for

steposvetlana [31]

Answer:

$\ d_{out} = 100 \ m.$

Explanation:

Given data:

$F_{in} = 50 \ \rm N$

$F_{out} = 10 \ \rm N$

$d_{in} = 20 \ m$

Let the distance traveled by the object in the second case be $d_{out}.$

In the given problem, work done by the forces are same in both the cases.

Thus,

$W_{in} = W_{out}$

$F_{in}.d_{in} = F_{out}.d_{out}$

$\Rightarrow \ d_{out} = \frac{F_{in}.d_{in}}{F_{out}}$

$\ d_{out} = \frac{50 \times 20}{10}$

$\ d_{out} = 100 \ m.$

5 0

3 years ago

Initially sliding with a speed of 1.9 m/s, a 1.8 kg block collides with a spring and compresses it 0.35 m before coming to rest.

Alika [10]

Let k = the force constant of the spring (N/m).

The strain energy (SE) stored in the spring when it is compressed by a distance x=0.35 m is
SE = (1/2)*k*x²
= 0.5*(k N/m)*(0.35 m)²
= 0.06125k J

The KE (kinetic energy) of the sliding block is
KE = (1/2)*mass*velocity²
= 0.5*(1.8 kg)*(1.9 m/s)²
= 3.249 J

Assume that negligible energy is lost when KE is converted into SE.
Therefore
0.06125k = 3.249
k = 53.04 N/m

Answer: 53 N/m (nearest integer)

3 0

3 years ago

two metal plates are near each other. explain why they will not become charged until connected to a souce of potential differenc

scZoUnD [109]

Answer:

idk

Explanation:

7 0

3 years ago

Think of an animal and list two resources that it might compete for in its community . Then describe what adaptions the animals

Viefleur [7K]

An animal is a lion, it may compete for food and it’s territory. The animal may have to fight other animals to get these things

6 0

4 years ago

A 750 g air-track glider attached to a spring with spring constant 14.0 N/m is sitting at rest on a frictionless air track. A 20

alexandr402 [8]

Answer:

the amplitude of the subsequent oscillations is 0.11 m

the period of the subsequent oscillations is 1.94 s

Explanation:

given Information:

the mass of air-track glider, $m_{1}$ = 750 g = 0.75 kg

spring constant, k = 13.0 N/m

the mass of glider, $m_{2}$ = 200 g = 0.2 kg

the speed of glider, $v_{2}$ = 170 cm/s = 1.7 m/s

the amplitude of the subsequent oscillations is A = 0.11 m

according to mechanical enery equation, we have

$A = \sqrt{\frac{m_{1} +m_{2} }{k} }v_{f}$

where

A is the amplitude and $v_{f}$ is the final speed.

to find $v_{f}$ , we can use momentum conservation lwa, where the initial momentum is equal to the final momentum.

$P_{f} = P_{i}$

$(m_{1} +m_{2} )v_{f} = m_{1} v_{1} +m_{2}v_{2}$

$v_{1}$ = 0, thus

(0.75+0.2) $v_{f}$ = (0.75)(0)+(0.2)(1.7)

0.95 $v_{f}$ = 0.34

$v_{f}$ = 0.36 m/s

Now we can calculate the amplitude

$A = \sqrt{\frac{0.75 +0.2 }{10} }0.36$

A = 0.11 m

the period of the subsequent oscillations is T = 1.94 s

the equation for period is

T = 2π $\sqrt{\frac{m_{1}+m_{2} }{k} }$

T = 2π $\sqrt{\frac{0.75+0.2 }{10} }$

T = 1.94 s

7 0

3 years ago