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Lera25 [3.4K]
1 year ago
15

What is time taken by the stone ?

Physics
1 answer:
SSSSS [86.1K]1 year ago
8 0

The time taken by the stone to hit the ground would be 5.12 seconds.

<h3>What are the three equations of motion?</h3>

There are three equations of motion given by  Newton

The first equation is given as follows

v = u + at

the second equation is given as follows

S = ut + 1/2×a×t²

the third equation is given as follows

v² - u² = 2×a×s

Keep in mind that these calculations only apply to uniform acceleration.

As given in the problem, a stone is dropped from the helicopter which is ascending at the speed of 19.6 m/s

height(S) = 156.8 meters

initial velocity(u) = -19.6 m/s

acceleration(a) = 9.81 m/s²

By using the second equation of motion given by newton

S = ut + 1/2at²

S = 156.8m ,u= -19.6 m/s , a= 9.81 m/s² and t =? seconds

156.8= -19.6t + 9.81t²

t = 5.12 seconds

Thus, the time taken by the stone to hit the ground would be 5.12 seconds.

Learn more about equations of motion from here,

brainly.com/question/5955789

#SPJ1

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two charges having the same charge magnitude experiencing an attracting force of 3.60N when the charges are 30cm apart.what is t
Tomtit [17]

The charges have opposite sign and magnitude 6 \mu C

Explanation:

The magnitude of the electrostatic force between two electric charges is given by Coulomb's law:

F=k\frac{q_1 q_2}{r^2}

where:

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

q_1, q_2 are the two charges

r is the separation between the two charges

In this problem, we have:

F = 3.60 N is the force between the two charges

r = 30 cm = 0.30 m is their separation

The two charges have same magnitude, so

q_1 = q_2 = q

So we can rewrite the equation as

F=\frac{kq^2}{r^2}

And solving for q:

q=\sqrt{\frac{Fr^2}{k}}=\sqrt{\frac{(3.60)(0.30)^2}{8.99\cdot 10^9}}=6\cdot 10^{-6} C = 6\mu C

Moreover, the force between the charges is attractive: we know that charges of same sign repel each other while charges of opposite sign attract each other, therefore the charges in this problem have opposite sign, so

q_1 = 6 \mu C\\q_2 = -6 \mu C

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3 0
3 years ago
How fast is the sixth cosmic velocity?
denpristay [2]
<span>25,000 miles per hour

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3 0
3 years ago
A youngster having a mass of 50.0 kg steps off a 1.00 m high platform. If she keeps her legs fairly rigid and comes to rest in 1
attashe74 [19]

The average act on her during the deceleration is 4.47 meters per second.                                                                

<u>Explanation</u>:

<u>Given</u>:

youngster mass m = 50.0 kg

She steps off a 1.00 m high platform that is s = 1 meter

She comes to rest in the 10-meter second

<u>To Find</u>:

The average force and momentum

<u>Formulas</u>:

p = m * v

F * Δ t = Δ p

vf^2= vi^2+2as

<u>Solution</u>:

a = 9.8 m/s

vi = 0

vf^2= 0+2(9.8)(1)

vf^2 = 19.6

vf = 4.47 m/s .

Therefore the average force is 4.47 m/s.                                          

                                 

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3 years ago
A descending elevator of mass 1,000 kg is uniformly decelerated to rest over a distance of 8 m by a cable in which the tension i
Stolb23 [73]

The speed  V_{i} of the elevator at the beginning of the 8 m descent is nearly 4 m/s. Hence, option A is the correct answer.

We are given that-

the mass of the elevator (m) = 1000 kg ;

the distance the elevator decelerated to be y = 8m ;

the tension is T = 11000 N;

let us determine the acceleration 'a' by using Newton's second law of motion.

∑Fy = ma

W - T = ma

(1000kg x 9.8 m/s² ) - 11000N = 1000 kg x a

9800 - 11000 = 1000

a = - 1.2 m/s²

Using the equation of kinematics to determine the initial velocity.

V_{f} ² = V_{i}² + 2ay

V_{i} = √ ( 2 x 1.2m/s² x 8 m )

V_{i} = √19.2 m²/s²

V_{i} = 4.38 m/s   ≈ 4 m/s

Hence, the initial velocity of the elevator is 4m/s.

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8 0
1 year ago
In which state of matter are water molecules measured as having a comparatively high temperature?
AnnZ [28]
Liquid water because if it said very high then it would be water vapor but it didn’t say that so the answer is B liquid water
8 0
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