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USPshnik [31]
3 years ago
8

When a huge block of rock is pushed up at a normal fault, a ___ mountain is usually formed. A.fault -block B.dome C.folded D.bas

in
Physics
2 answers:
katrin2010 [14]3 years ago
8 0
The answer is A. Block mountain
DerKrebs [107]3 years ago
3 0

Answer: Option 'A' is correct.

Explanation:

When a huge block of rock is pushed up at a normal fault, a fault block mountain is usually formed.

They are formed by the movement blocks large blocks of crustal when forces in the crust of Earth pull it apart.

Some parts are pushed upward whereas other collapse down.

This led to formation of fault block mountain.

Hence, Option 'A' is correct.

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A cart is pulled by a force of 250 N at an angle of 35° above the horizontal. The cart accelerates at 1.4 m/s2. The free-body di
Pachacha [2.7K]

Answer:

m=146.277kg which is rounded to 146kg

Explanation:

Remember that F=ma

But F represents not 250N, but 250cos(35)N since the force is being pulled above the horizontal.

So 250cos(35)=204.7880111 approximately, and since a=1.4m/s^2, we have 204.7880111=m(1.4m/s^2). Then we divide both sides by the acceleration to get the mass. So m=146.2771508kg which the nearest number is 146kg

Mass is always in kg, unless stated otherwise.

4 0
2 years ago
Read 2 more answers
Three charges are enclosed inside a spherical closed surface. The net flux through the surface is −216 N · m2/C. If two of the c
sladkih [1.3K]

Answer:

q₃ = -4.81 nC

Explanation:

We can use the Gauss Law here:

∅ = q/∈₀

where,

∅ = Net Flux = - 216 N.m²/C

q = total charge enclosed inside sphere = ?

∈₀ = permittivity of free space = 8.85 x 10⁻¹² C/N.m²

Therefore,

- 216 N.m²/C = q / 8.85 x 10⁻¹² C²/N.m²

q = (-216 N.m²/C)(8.85 x 10⁻¹² C²/N.m²)

q = - 1.91 nC

So, the total charge will be sum of all three charges:

q = q₁ + q₂ + q₃

- 1.91 nC = 1.74 nC + 1.16 nC + q₃

q₃ = - 1.91 nC - 1.74 nC - 1.16 nC

<u>q₃ = -4.81 nC</u>

5 0
2 years ago
The equation Bold r (t )equals(8 t plus 9 )Bold i plus (2 t squared minus 8 )Bold j plus (6 t )Bold k is the position of a parti
KATRIN_1 [288]

Explanation:

It is given that, the position of a particle as as function of time t is given by :

r(t)=(8t+9)i+(2t^2-8)j+6tk

Let v is the velocity of the particle. Velocity of an object is given by :

v=\dfrac{dr(t)}{dt}

v=\dfrac{d[(8t+9)i+(2t^2-8)j+6tk]}{dt}

v=(8i+4tj+6k)\ m/s

So, the above equation is the velocity vector.

Let a is the acceleration of the particle. Acceleration of an object is given by :

a=\dfrac{dv(t)}{dt}

a=\dfrac{d[8i+4tj+6k]}{dt}

a=(4j)\ m/s^2

At t = 0, v=(8i+0+6k)\ m/s

v(t)=\sqrt{8^2+6^2} =10\ m/s

Hence, this is the required solution.

7 0
3 years ago
A liquid is poured into a vessel to a depth of 16cm when viewed from the top, the bottom appears to be raised 4cm. What is the r
zalisa [80]

Answer:

Solution

Verified by Toppr

Correct option is

C

3 cm

RI=apparent depthreal depth

Substituting, 34=apparentdepth12

Therefore, apparent depth=412×3=9

The height by which it appears to be raised is 12−9=3cm

Was this answer helpful?

71

0

SIMILAR QUESTIONS

A coin is placed at the bottom of a glass tumbler and then water is added. It appeared that the depth of the coin has been reduced because

Medium

View solution

>

A tank is filled with water to a height of 12.5 cm. The apparent depth of a needle lying at the bottom of the tank is measured by a microscope to be 9.4 cm. What is the refractive index of water? If water is replaced by a liquid of refractive index 1.63 up to the same height, by what distance would the microscope have to be moved to focus on the needle again?

3 0
2 years ago
A 500 kg table with 4 legs rests on solid flat ground. I place 3 books on the center of the table with masses of 10 kg, 20 kg an
BARSIC [14]

Answer:

1,373.4 N

Explanation:

The mass of the table acts at the centre in addition to the books since that is the centre of gravity of the table.

Mass of books will be 10kg+20kg+30kg=60 kg

Total mass of table and books will be 500kg+60kg=560 kg

This mass is evenly distributed into the four legs hence 560kg/4 legs=140 kg per leg

Force is product of mass and acceleration due to gravity hence F=gm

Taking g as 9.81 m/s2 then

F=140*9.81=1,373.4 N

Therefore, rhe normal force is equivalent to 1,373.4 N

6 0
2 years ago
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