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adelina 88 [10]

3 years ago

5

What are anticlines and synclines?

1 answer:

vesna_86 [32]3 years ago

3 0

<span>d. They are types of folds created by compressional stresses.</span>.

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CAN SOMEONE PLZ TELL ME HOW TO WRITE THIS DATA IN THE THE TABLE INTO A GRAPH?

vitfil [10]

Here is a helpful video https://www.khanacademy.org/_render

6 0

2 years ago

The record distance in the sport of throwing cowpats is 81.1 m. This record toss was set by Steve Urner of the United States in

N76 [4]

Answer:

28.2 m/s

Explanation:

The range of a projectile launched from the ground is given by:

$d=\frac{v^2}{g}sin 2\theta$

where

v is the initial speed

g = 9.8 m/s^2 is the acceleration of gravity

$\theta$ is the angle at which the projectile is thrown

In this problem we have

d = 81.1 m is the range

$\theta=45^{\circ}$ is the angle

Solving for v, we find the speed of the projectile:

$v=\sqrt{\frac{dg}{sin 2 \theta}}=\sqrt{\frac{(81.1 m)(9.8 m/s^2)}{sin (2\cdot 45^{\circ})}}=28.2 m/s$

7 0

2 years ago

In 2013, the world record depth for free diving without assistance was 101 m. What is the pressure in the water (assumed freshwa

serg [7]

Answer:

The pressure in the water is $1.1\times10^{6}\ Pa$

Explanation:

Given that,

Depth = 101 m

Let P be the pressure at the bottom of water at a depth

We need to calculate the pressure in water

Using formula of pressure

$P=P_{w}+P_{atm}$

Where, $P_{atm}$ = atmospheric pressure

$P_{w}$ = pressure in water

Put the value of $P_{w}$ in to the formula

$P=\rho hg+P_{atm}$

Put the value into the formula

$P=1000\times101\times9.8+101325$

$P=1.1\times10^{6}\ Pa$

Hence, The pressure in the water is $1.1\times10^{6}\ Pa$

6 0

2 years ago

Read 2 more answers

An incident ray strikes a mirror with an angle of 30 degrees to the surface of the mirror. what is the angle of the reflected ra

kkurt [141]

With 30degree because incident angle = reflected angle

7 0

2 years ago

A star with the mass (M=2.0×1030kg) and size (R=3.5×108m) of our sun rotates once every 30.0 days. After undergoing gravitationa

Gre4nikov [31]

Answer:

$r=97.22x10^{3} m$

Explanation:

Using the angular formulas can determine the radius using both values neutron star and the the knowing star so

$L=I*w$

$L_{1}=I_{1}*w_{1}=L_{2}=I_{2}*w_{2}$

$I_{1}*w_{1}=I_{2}*w_{2}$

I=Inertia of the star

w=angular velocity

$I=\frac{2*m*r^{2}}{5}$

$w=\frac{2\pi}{t}$

Notice the angular velocity determinate by the time and the Inertia have the radius value so

$\frac{2}{5}*m*r_{sn}^{2}*\frac{2\pi }{t_{1}}=\frac{2}{5}*m*r_{s}^{2}*\frac{2\pi }{t_{2}}$

$r_{sn}^{2}*\frac{1}{t_{1}}=r_{s}^{2}*\frac{1}{t_{2}}$

$r_{sn}^{2}=r_{s}^{2}*\frac{t_{1}}{t_{2}}$

$t_{1}=0.2s\\t_{2}=30day*\frac{24hr}{1day}*\frac{60minute}{1hr}*\frac{60seg}{1minute}=2.592x10^{6}s$

$r_{sn}=3.5x10^{8}m*\sqrt{\frac{0.2s}{2.592x^{6}s}}$

$r_{sn}=97.22x10^{3} m$

8 0

2 years ago