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aleksandr82 [10.1K]
3 years ago
13

What is a locus of points

Physics
2 answers:
svetlana [45]3 years ago
7 0

A locus is the set of all points (usually forming a curve or surface) satisfying some condition. For example, the locus of points in the plane equidistant from a given point is a circle, and the set of points in three-space equidistant from a given point is a sphere.

Gnesinka [82]3 years ago
5 0

"A locus is the set of all points (usually forming a curve or surface) satisfying some condition. For example, the locus of points in the plane equidistant from a given point is a circle, and the set of points in three-space equidistant from a given point is a sphere." -mathworld.wolfram.com/Locus.html

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A soccer player carries the ball for a distance of 40.0 m in the direction 42.0° west of south. find the westward component of t
Maksim231197 [3]

Actually what the problem meant about the westward component of the ball’s displacement is the horizontal component of the displacement. To help us better understand the problem, I attached a figure of the situation.

We can see from the figure that to solve for the value of the horizontal component, we have to make use of the sin function. That is:

sin θ = side opposite to the angle / hypotenuse of the triangle

sin 42 = x / 40 m

x = (40 m) sin 42

x = 26.77 m

 

Therefore the ball has a westward displacement of about 26.77 m

3 0
3 years ago
A skateboarder rolls down the sidewalk with an initial velocity of 0 m/s. If her acceleration is
VikaD [51]

Answer:

80m/s

Explanation:

to find it you have to work it out by using the formula distance divided by speed to find time.

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2 years ago
A 5 kg box drops a distance of 10 m to the ground. If 70% of the initial potential energy goes into increasing the internal ener
Elina [12.6K]

Answer:

Explanation:

From the given information:

The initial PE (PE)_i = m×g×h

= 5 kg × 9.81 m/s² × 10 m

= 490.5 J

The change in Potential energy P.E of the box is:

ΔP.E = P.E_f -P.E_i

ΔP.E = 0 - (PE)_i

ΔP.E = -P.E_i

If we take a look at conservation of total energy for determining the change in the internal energy of the box;

\Delta P.E + \Delta K.E + \Delta U = 0

\Delta U = -\Delta P.E - \Delta K.E

this can be re-written as:

\Delta U =- (-\Delta P.E_i) - \Delta K.E

Here, K.E = 0

Also, 70% goes into raising the internal energy for the box;

Thus,

\Delta U =(70\%) \Delta P.E_i-0

\Delta U =(0.70) (490.5)

ΔU = 343.35  J

Thus, the magnitude of the increase is = 343.35 J

7 0
3 years ago
Interactive Solution 9.37 presents a method for modeling this problem. Multiple-Concept Example 10 offers useful background for
viva [34]

Answer:

21.67 rad/s²

208.36538 N

Explanation:

\omega_f = Final angular velocity = \dfrac{1}{6}78=13\ rad/s

\omega_i = Initial angular velocity = 78 rad/s

\alpha = Angular acceleration

\theta = Angle of rotation

t = Time taken

r = Radius = 0.13

I = Moment of inertia = 1.25 kgm²

From equation of rotational motion

\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\dfrac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\dfrac{13-78}{3}\\\Rightarrow \alpha=-21.67\ rad/s^2

The magnitude of the angular deceleration of the cylinder is 21.67 rad/s²

Torque is given by

\tau=I\alpha\\\Rightarrow \tau=1.25\times -21.67\\\Rightarrow \tau=-27.0875

Frictional force is given by

F=\dfrac{\tau}{r}\\\Rightarrow F=\dfrac{-27.0875}{0.13}\\\Rightarrow F=-208.36538\ N

The magnitude of the force of friction applied by the brake shoe is 208.36538 N

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3 years ago
Patrick Mahomes is throwing a football with an applied force of 230 N but is experiencing air resistance of 30 N. The ball has a
Ganezh [65]

Answer:

www.chiefs.com

Web results

Patrick Mahomes

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3 years ago
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