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jolli1 [7]
3 years ago
13

A mass of 1 kg is moving in a circle of radius 3.3 meters. What is the liner velocity v m/s that would give a Centripetal Force

of 23 N?
Physics
1 answer:
Ghella [55]3 years ago
4 0

Answer:

The linear velocity of the object is 8.71 m/s.

Explanation:

Given;

mass of the object, m = 1 kg

radius of the circle, r = 3.3 meters

centripetal force, F = 23 N

Centripetal force is given by;

F_c = \frac{mv^2}{r}\\\\

where;

v is the linear velocity of the object

F_c = \frac{mv^2}{r}\\\\mv^2 = F_cr\\\\v^2 = \frac{F_cr}{m}\\\\v= \sqrt{\frac{F_cr}{m}} \\\\v= \sqrt{\frac{23*3.3}{1}}\\\\v = 8.71 \ m/s

Therefore, the linear velocity of the object is 8.71 m/s.

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A 60 kg acrobat is in the middle of a 10 m long tightrope. The center of the rope dropped 30 cm in relation to the ends that are
Zigmanuir [339]

Answer:

The tension in each half of the rope, is approximately 4,908.8 N

Explanation:

The mass of the acrobat, m = 60 kg

The length of the rope, l = 10 m

The extent by which the center dropped = 30 cm = 0.3 m

Let, 'T' represent the tension in each half of the rope

Weight, W = Mass, m × The acceleration due to gravity, g

∴ W = m × g

The acceleration due to gravity, g ≈ 9.8 m/s²

∴ The weight of the acrobat, W = 60 kg × 9.8 m/s² ≈ 588 N

The angle the dropped rope makes with the horizontal, θ is given as follows;

θ = arctan((0.3 m)/(5 m)) = arctan(0.06) ≈ 3.434°

At equilibrium, the sum of vertical forces, \Sigma F_y = 0

The vertical component of the tension, T_y, in each half of the rope is given as follows;

T_y = T × sin(θ)

∴ \Sigma F_y = W + T × sin(θ) + T × sin(θ) = W + 2 × T × sin(θ)

Plugging in the values, with θ = arctan(0.06) for accuracy, we get;

588 N + 2 × T × sin(arctan(0.06) = 0

∴ 2 × -T × sin(arctan(0.06) = 588 N

-T= 588 N/(2 × sin(arctan(0.06)) = 4,908.81208 N ≈ 4,908.8 N

The tension in each half of the rope, T ≈ 4,908.8 N.

4 0
2 years ago
Graphs help you see ______.
Anastaziya [24]
D- if i was right please mark me as a brainliest answer and a thank you
4 0
3 years ago
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An electron moves with a speed of 8.0×106m/s along the -z-axis. It enters a region where there is a uniform magnetic field B = (
Crazy boy [7]

Answer:

Acceleration, a=9.36\times 10^{18}\ m/s^2

Explanation:

It is given that,

Speed of electron, v=8\times 10^6\ m/s

Charge on an electron, q=1.6\times 10^{-19}\ C

Mass of electron, m=9.1\times 10^{-31}\ kg

Magnetic field, B=5.5i-3.7j

Magnitude, |B|=\sqrt{5.5^2+(-3.77)^2}=6.66\ T

Magnetic force is given by :

F=qvB

Also, F = ma

a=\dfrac{qvB}{m}

a=\dfrac{1.6\times 10^{-19}\times 8\times 10^6\times 6.66}{9.1\times 10^{-31}}

a=9.36\times 10^{18}\ m/s^2

So, the acceleration of the electron is 9.36\times 10^{18}\ m/s^2. Hence, this is the required solution.

5 0
3 years ago
How do we find an atom's mass number?
Scorpion4ik [409]

Answer:

To calculate the atomic mass of a single atom of an element, add up the mass of protons and neutrons. Example: Find the atomic mass of an isotope of carbon that has 7 neutrons. You can see from the periodic table that carbon has an atomic number of 6, which is its number of protons.

Explanation:

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3 years ago
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hope this helps
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2 years ago
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