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3 years ago

(25 Brainly Points.)

Physics

1 answer:

Advocard [28]3 years ago

3 0

Because an earthquake happens when two of the atomic plates rub together .

Hopes this helps. PLZ MARK BRAINIEST

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Two protons are 1 × 10−10 m apart (about one atomic radius). Which interaction between two protons is stronger, the gravitationa

anastassius [24]

Answer: The electric repulsion between the two protons is stronger than the gravitational attraction.

Explanation: Please see the attachments below

4 0

4 years ago

The element chlorine is found in nature as a molecule that can be represented as - A. Cl.

poizon [28]

A
Cl is the chemical symbol for chlorine numbers after it are isotopes

5 0

3 years ago

A small space probe of mass 170 kg is launched from a spacecraft near Mars. It travels toward the surface of Mars, where it will

alukav5142 [94]

Answer:

The change in momentum is $\Delta p = kg \cdot m/s$

Explanation:

From the question we are told that

The mass of the probe is $m = 170 kg$

The location of the prob at time t = 22.9 s is $A =$

The momentum at time t = 22.9 s is $p = < 51000, -7000, 0> kg m/s$

The net force on the probe is $F = N$

Generally the change in momentum is mathematically represented as

$\Delta p = F * \Delta t$

The initial time is 22.6 s

The final time is 22.9 s

Substituting values

$\Delta p = * (22.9 - 22.6)$

$\Delta p = * (0.3)$

$\Delta p = kg \cdot m/s$

6 0

3 years ago

To start an avalanche on a mountain slope, an artillery shell is fired with an initial velocity of 290 m/s at 57.0° above the ho

cupoosta [38]

Answer:

xf = 5.68 × 10³ m

yf = 8.57 × 10³ m

Explanation:

given data

vi = 290 m/s

θ = 57.0°

t = 36.0 s

solution

firsa we get here origin (0,0) to where the shell is launched

xi = 0 yi = 0

xf = ? yf = ?

vxi = vicosθ vyi = visinθ

ax = 0 ay = −9.8 m/s

now we solve x motion: that is

xf = xi + vxi × t + 0.5 × ax × t² ............1

simplfy it we get

xf = 0 + vicosθ × t + 0

put here value and we get

xf = 0 + (290 m/s) cos(57) (36.0 s)

xf = 5.68 × 10³ m

and

now we solve for y motion: that is

yf = yi + vyi × t + 0.5 × ay × t ² ............2

put here value and we get

yf = 0 + (290 m/s) × sin(57) × (36.0 s) + 0.5 × (−9.8 m/s2) × (36.0 s) ²

yf = 8.57 × 10³ m

5 0

3 years ago

HELP!!! 30 POINTS+BRAINLIEST!!!!!QUICK!!

tia_tia [17]

Answer:

A:7.2 $ms^{-1}$

B:14.25 $ms^{-1}$

C:1.45 $sec$

D:10.3 $m$

E:2.9 $sec$

F:20.88 $m$

Explanation:

Let $v$ be the velocity and $\alpha$ be the angle between the velocity and ground.

Question A:

Horizontal component of velocity is given by $vcos(\alpha )$ .

So,horizontal component is $16\times cos(63)=16\times 0.45=7.2ms^{-1}$

Question B:

Vertical component of velocity is given by $vsin(\alpha )$ .

So,vertical component is $16\times sin(63)=16\times 0.89=14.25ms^{-1}$

Question C:

Time required is given by $\frac{\text{vertical component of velocity}}{g}}=\frac{14.25}{9.8}=1.45 seconds$

Question D:

Maximum height is given by $\frac{\text{vertical component of velocity}^{2}}{2g}}=\frac{203.06}{19.6}=10.3m$

Question E:

Time of flight is twice the time required to reach maximum height= $2\times 1.45=2.9 seconds$ .

Question F:

The distance between the player and ball after landing is called range and is given by $\text{horizontal component of velocity}\times \text{time of flight}=$ $7.2\times 2.9=20.88m$

8 0

3 years ago