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Svetlanka [38]
3 years ago
8

Sinkholes can form when

Physics
2 answers:
Sergio [31]3 years ago
8 0

Rain water containing carbon dioxide dissolves underground rock

kherson [118]3 years ago
7 0
Land surfaces change, erosion happens, the ground collapses, etc.
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Find the minor measurement of the vernier scale by taking 49, 1mm divisions of the main scale and dividing it into 50 vernier di
olga nikolaevna [1]
<h2>♨ANSWER♥</h2>

length of V-50 = 49mm

length of V-1 = 49/50mm

= 0.98mm

so,

minor measurement = (M-1) - (V-1)

= 1mm -0.98mm

= 0.02mm

☆ Therefore,

The minor measurement of the vernier scale is 0.02mm.

<u>☆</u><u>.</u><u>.</u><u>.</u><u>hope this helps</u><u>.</u><u>.</u><u>.</u><u>☆</u>

_♡_<em>mashi</em>_♡_

6 0
2 years ago
If the velocity of an object is zero, then that object cannot be accelerating. | True or False
valkas [14]

Answer: False

Explanation:

3 0
3 years ago
It takes a minimum distance of 57.46 m to stop a car moving at 13.0 m/s by applying the brakes (without locking the wheels). Ass
vivado [14]

Answer:

The minimum stopping distance when the car is moving at

29.0 m/sec = 285.94 m

Explanation:

We know by equation of motion that,

v^{2}=u^{2}+2\cdot a \cdot s

Where, v= final velocity m/sec

u=initial velocity m/sec

a=Acceleration m/Sec^{2}

s= Distance traveled before stop m

Case 1

u=  13 m/sec, v=0, s= 57.46 m, a=?

0^{2} = 13^{2}  + 2 \cdot a \cdot57.46

a = -1.47 m/Sec^{2} (a is negative since final velocity is less then initial velocity)

Case 2

u=29 m/sec, v=0, s= ?, a=-1.47 m/Sec^{2} (since same friction force is applied)

v^{2} = 29^{2}  - 2 \cdot 1.47 \cdot S

s = 285.94 m

Hence the minimum stopping distance when the car is moving at

29.0 m/sec = 285.94 m

4 0
2 years ago
A point charge is placed 100 m from a 6 uC charge generating an electric field. What is the
kvv77 [185]

The strength of the electric field is 5 N/C

Explanation:

The magnitude of the electric field produced by a single-point charge is given by:

E=\frac{kQ}{r^2}

where

k=8.99\cdot 10^9 Nm^{-2}C^{-2} is the Coulomb's constant

Q is the magnitude of the charge

r is the distance from the charge

In this problem, we have:

Q=6 \mu C = 6\cdot 10^{-6}C is the charge producing the field

r = 100 m is the distance from the charge at  which we want to calculate the field

Substituting into the equation, we find the s trength of the electric field:

E=\frac{(8.99\cdot 10^9)(6\cdot 10^{-6})}{(100)^2}=5.4 N/C \sim 5 N/C

Learn more about electric field:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

3 0
3 years ago
Kali left school and traveled toward her friend's house at an average speed of 40 km/h. Matt left one hour later and traveled in
zlopas [31]

Answer:

t = 5 hr

Explanation:

Let kali moves toward east with velocity= V₁= 40 km/ h

Mat moves toward west with velocity = V₂= 50 km/hr

As Klai left one hour earlier = t₁= 1 hr

distance traveled in 1st hour = s₁ = v * t = 40 * 1 = 40 km

Remaining distance = 400 - 40 = 360 km

As they move in the opposite directions:

Relative speed= 40 + 50 = 90 km/ h

s = v * t

⇒ t = s / v

⇒ t₂ = 360 / 90

⇒ t₂ = 4 hr

Total time = t = t₁ + t₂

t = 1 hr + 4 hr

t = 5 hr

5 0
3 years ago
Read 2 more answers
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