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Fynjy0 [20]
3 years ago
6

Offenders who are under house arrest may not leave their residences for any reason

Physics
1 answer:
WITCHER [35]3 years ago
6 0
This is false. I hope this helps.
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The weight of an object on the surface of mass is 1850 Newton and its mass is 500kg.Find the acceleration due to gravity on mars
Fudgin [204]

F = m.a

a = F/m

a = 1850/500

a = 3.7 m/s²

3 0
2 years ago
The original meaning of the word "essay' is which of these? treatise dialogue attempt thought
Elina [12.6K]
The answer to this is, attempt.
8 0
3 years ago
Read 2 more answers
A person of mass 55 kg swings on a rope length 4 m from rest (when the rope makes an angle of 30 degrees with the vertical) and
vovangra [49]

Answer:

θ = 19.66°

Explanation:

To determine the angle that the rope makes with the vertical for the two people, you first take into account the potential energy of the first person before he swings on the rope:

U=mgh

h: distance to the ground

g: gravitational acceleration = 9.8m/s^2

m: mass of the first person = 55 kg

In the image attache below you can notice that the height h is:

h=4-4cos(30\°)=0.53m

Then, the potential energy is:

U=(55kg)(9.8m/s^2)(0.53m)=285.67J

When the first person picks up the second person (when the rope is exactly vertical), all the potential energy becomes kinetic energy. Next, when both people reaches the maximum height h' the energy must be equal to the initial potential energy of the first person:

U'=(m_1+m_2)gh'=285.67\ J

From the previous equation you can get h':

h'=\frac{285.67J}{(55kg+70kg)(9.8m/s^2)}=0.2332m

Finally, you obtain the angle between the rope at the height h,' and the vertical, by calculating the following:

h'=4-4cos(\theta)\\\\\theta=cos^{-1}(\frac{4-h'}{4})=cos^{-1}(\frac{4-0.2332}{4})=19.66\°

hence, the angle between the rope and the vertical, when the two people are in the rope is 19.66°

8 0
3 years ago
A solid block, with a mass of 0.15kg, on a frictionless surface is pushed directly onto a horizontal spring, with a spring const
iren [92.7K]

Answer:

16.1 m/s

Explanation:

We can solve the problem by using the law of conservation of energy.

At the beginning, the spring is compressed by x = 35 cm = 0.35 m, and it stores an elastic potential energy given by

U=\frac{1}{2}kx^2

where k = 316 N/m is the spring constant. Once the block is released, the spring returns to its natural length and all its elastic potential energy is converted into kinetic energy of the block (which starts moving). This kinetic energy is equal to

K=\frac{1}{2}mv^2

where m = 0.15 kg is the mass of the block and v is its speed.

Since the energy must be conserved, we can equate the initial elastic energy of the spring to the final kinetic energy of the block, and from the equation we obtain we can find the speed of the block:

\frac{1}{2}kx^2=\frac{1}{2}mv^2\\v=\sqrt{\frac{kx^2}{m}}=\sqrt{\frac{(316 N/m)(0.35 m)^2}{0.15 kg}}=16.1 m/s

4 0
3 years ago
A sports car accelerates at 2.80 m/s2 along a straight road. At t1 = 5.00s and t2 = 5.50 s it passes two marks that are 35.0 m a
harina [27]

Explanation:

Using kinematics,

s = ut + 0.5at².

We have s = 35.0m, t = 0.50s, a = 2.80m/s².

Substitute in the variables, we have u = 69.3m/s.

This u is only the initial velocity at t1. To find the velocity at t0, we apply a different kinematics equation with different values:

We have t = 5.00s, a = 2.80m/s², v = 69.3m/s.

Using kinematics,

v = u + at.

Substitute in the variables, we have u = 55.3m/s.

Hence the car's initial velocity is 55.3m/s.

7 0
3 years ago
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