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3 years ago

Which surface feature of the moon is characterized by mountainous areas?

Physics

2 answers:

solmaris [256]3 years ago

6 0

Crates is the correct answer because they create dents that make the bumps of mountains.

jeyben [28]3 years ago

3 0

Answer:

Terrae

Explanation:

The terrae of the moon is composed of many mountains. It is also called the highlands of the moon.

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An atom is chemically stable when it’s outer_____ is completely filled with _____.

Svetlanka [38]

an atom is chemically stable when its outer layer is completely filled with energy. im pretty sure this is right

8 0

3 years ago

A person of mass m will bungee-jump from a bridge over a river where the height from the bridge to the river is h. The bungee co

Likurg_2 [28]

Answer: $k_{min}=\frac{18mg}{h}$

Explanation:

Given

mass of person is m

Distance between bridge and river is h

chord has an un-stretched length of $\frac{2h}{3}$

Let spring constant be k

Person will just stop before hitting the river

Conserve energy i.e. Potential Energy of Person is converted in to elastic energy of chord

$mgh=\frac{kx^2}{2}$

$x=h-\frac{2h}{3}=\frac{h}{3}$

$mgh=\frac{kh^2}{18}$

$k=\frac{18mg}{h}$

Thus $k_{min}=\frac{18mg}{h}$

5 0

4 years ago

A 10-cm-long spring is attached to theceiling. When a 2.0 kg mass is hung from it,the spring stretches to a length of 15 cm.a.Wh

alekssr [168]

(a) 392 N/m

Hook's law states that:

$F=k\Delta x$ (1)

where

F is the force exerted on the spring

k is the spring constant

$\Delta x$ is the stretching/compression of the spring

In this problem:

- The force exerted on the spring is equal to the weight of the block attached to the spring:

$F=mg=(2.0 kg)(9.8 m/s^2)=19.6 N$

- The stretching of the spring is

$\Delta x=15 cm-10 cm=5 cm=0.05 m$

Solving eq.(1) for k, we find the spring constant:

$k=\frac{F}{\Delta x}=\frac{19.6 N}{0.05 m}=392 N/m$

(b) 17.5 cm

If a block of m = 3.0 kg is attached to the spring, the new force applied is

$F=mg=(3.0 kg)(9.8 m/s^2)=29.4 N$

And so, the stretch of the spring is

$\Delta x=\frac{F}{k}=\frac{29.4 N}{392 N/m}=0.075 m=7.5 cm$

And since the initial lenght of the spring is

$x_0 = 10 cm$

The final length will be

$x_f = x_0 +\Delta x=10 cm+7.5 cm=17.5 cm$

8 0

3 years ago

Read 2 more answers

What mRNA sequence would result from the following DNA sequence?

Nezavi [6.7K]

Answer:

UAC CUG AGG AUC

Explanation:

The mRNA sequence from ATG GAC TCC TAG DNA sequence would be UAC CUG AGG AUC.

According to Chargaff's base pairing rule, the purine bases always pair with pyrimidine bases. Specifically, Adenine base must pair with Thymine base while Guanine base must pair with Cytosine base. In RNA, Thymine base is replaced with Uracil base.

Hence:

ATG GAC TCC TAG will pair with

UAC CUG AGG AUC

5 0

3 years ago

A 7300 kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.20 m/s2 and feels no appreci

Simora [160]

Answer:

Explanation:

We shall first calculate the velocity at height h = 575 m .

acceleration a = 2.2 m /s²

v² = u² + 2 a s

u is initial velocity , v is final velocity , s is height achieved

v² = 0 + 2 x 2.2 x 575

v = 50.3 m /s

After 575 m , rocket moves under free fall so g will act on it downwards

If it travels further by height H

from the relation

v² = u² - 2 g H

v = 0 , u = 50.3 m /s

H = ?

0 = 50.3² - 2 x 9.8 H

H = 129.08 m

Total height attained by rocket

= 575 + 129.08

= 704.08 m .

4 0

3 years ago