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miss Akunina [59]
2 years ago
15

The mass of 04kg is rotated by a spring at a constant speed V . If the length of the string is 1m and the minimum tension is 3N,

what is value of V​
Physics
1 answer:
wlad13 [49]2 years ago
3 0

I think you mean "rotated by a string", which is to say a 0.4-kg mass is suspended from a string and swung around in a circle with constant speed.

The centripetal acceleration a of the object is

a = V² / R

where R is the length of the string, 1 m, so that

a = V² / (1 m) = V² m⁻¹

As the mass goes through its uniform circular motion, the only relevant force acting on it is tension with magnitude T, pointing in the direction in which the mass is accelerating (the center of its circular path). By Newton's second law,

T = ma

where m is the object's mass, 0.4 kg. Then

3 N = (0.4 kg) (V² m⁻¹)

⇒   V² = (3 N) / (0.4 kg) • (1 m) = 7.5 m²/s²

⇒   V ≈ 2.74 m/s

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4 0
2 years ago
The speed of a projectile when it reaches its maximum height is 0.58 times its speed when it is at half its maximum height. What
Ne4ueva [31]

Speed of the projectile at its maximum height is only along horizontal direction

so at highest point

v_1 = v_x

now when he is at half of the maximum height the speed will be in x and y direction both

v_2 = \sqrt{v_y^2 + v_x^2}

here it is given that

v_1 = 0.58 v_2

v_x = 0.58\sqrt{v_x^2 + v_y^2}

2.97 v_x^2 = v_x^2 + v_y^2

1.97 v_x^2 = v_y^2

also we know that

v_y^2 = v_{iy}^2 - 2 g \frac{H}{2}

here we know that maximum height is given as

H = \frac{v_{iy}^2}{2g}

v_y^2 = v_{iy}^2 - 2 g\frac{v_{iy}^2}{4g}

v_y^2 = \frac{v_{iy}^2}{2}

now from above

1.97 v_x^2 = \frac{v_{iy}^2}{2}

1.98 v_x = v_{iy}

also we know that angle of projection is

tan\theta = \frac{v_{iy}}{v_x}

tan\theta = \frac{1.98v_x}{v_x}

so angle is

\theta = tan^{-1} 1.98

\theta = 63.3 degree

6 0
3 years ago
Please help! Will give a lot of points
krek1111 [17]
Answer:

Chemical and Biological weathering.

Explanation:

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3 0
2 years ago
PL-1) A spring that hangs vertically is 25 cm long when no weight is attached to its lower end. Steve adds 250 g of mass to the
Anuta_ua [19.1K]

Answer:

20.42 N/m

Explanation:

From hook's law,

F = ke ......................... Equation 1

Where F = Force applied to the spring., k = spring constant, e = extension.

Make k the subject of the equation,

k = F/e ................. Equation 2

Note: The force on the spring is equal to the weight of the mass hung on it.

F = W = mg.

k = mg/e................ Equation 3

Given: m = 250 g = 0.25 kg, e = 37-25 = 12 cm = 0.12 m.

Constant: g = 9.8 m/s²

Substitute into equation 3

k = (0.25×9.8)/0.12

k = 20.42 N/m.

Hence the spring constant = 20.42 N/m

7 0
3 years ago
How does area affect the pressure?​
dybincka [34]

The smaller the area the greater the pressure, while the bigger the smaller the pressure. So they are inversely proportional

5 0
3 years ago
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