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gayaneshka [121]
3 years ago
11

Plz help i dont understand

Physics
1 answer:
maks197457 [2]3 years ago
7 0

Answer:

B.

Explanation:

A position-time graph is a graph of the position of an object against (versus) time. The slope of the line of a position-time graph is typically used to determine or calculate the velocity of an object.  

Velocity can be defined as the rate of change in displacement (distance) with time. Velocity is a vector quantity and as such it has both magnitude and direction.

Mathematically, velocity is given by the equation;

Velocity = \frac{distance}{time}

Uniform velocity can be defined as a situation in which a physical object or body travels equal distances in equal intervals of time along a fixed direction, regardless of how small the interval may be.

During uniform velocity, both the magnitude and direction of the physical object in motion do not change with respect to time. Also, an object in uniform velocity has zero tangential acceleration and its graph is a straight line parallel to the time axis.

Hence, a vehicle with a uniform velocity of 10 meters per seconds (m/s) is represented by Graph B.

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3 years ago
What exerts a centripetal force on a person running around a curve?
Eduardwww [97]

Answer:

The inertial force of the body

Explanation:

Everybody that is moving in a curved path has an inertial force called centrifugal force.

The counterforce of the centrifugal force is called the centripetal force. It also acts on every rotating body.

This force is always directed towards the center of the origin of the curve.

The velocity of the object changes its direction and magnitude at any instant of time. But the speed and angular velocity of the object remains the same for uniform circular motion.

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blagie [28]
Assuming Adam is on earth g= 9.8 m/s and m= weight/ gravity = 667/9.8 = 68 kg
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Describe the shape of a convex lens and explain what it does to light.
alexira [117]
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3 years ago
Find the shear stress and the thickness of the boundary layer (a) at the center and (b) at the trailing edge of a smooth flat pl
melomori [17]

Answer:

a) The shear stress is 0.012

b) The shear stress is 0.0082

c) The total friction drag is 0.329 lbf

Explanation:

Given by the problem:

Length y plate = 2 ft

Width y plate = 10 ft

p = density = 1.938 slug/ft³

v = kinematic viscosity = 1.217x10⁻⁵ft²/s

Absolute viscosity = 2.359x10⁻⁵lbfs/ft²

a) The Reynold number is equal to:

Re=\frac{1*3}{1.217x10^{-5} } =246507, laminar

The boundary layer thickness is equal to:

\delta=\frac{4.91*1}{Re^{0.5} }  =\frac{4.91*1}{246507^{0.5} } =0.0098 ft

The shear stress is equal to:

\tau=0.332(\frac{2.359x10^{-5}*3 }{1}  )(246507)^{0.5} =0.012

b) If the railing edge is 2 ft, the Reynold number is:

Re=\frac{2*3}{1.215x10^{-5} } =493015.6,laminar

The boundary layer is equal to:

\delta=\frac{4.91*2}{493015.6^{0.5} } =0.000019ft

The sear stress is equal to:

\tau=0.332(\frac{2.359x10^{-5}*3 }{2}  )(493015.6^{0.5} )=0.0082

c) The drag coefficient is equal to:

C=\frac{1.328}{\sqrt{Re} } =\frac{1.328}{\sqrt{493015.6} } ==0.0019

The friction drag is equal to:

F=Cp\frac{v^{2} }{2} wL=0.0019*1.938*(\frac{3^{2} }{2} )(10*2)=0.329lbf

7 0
3 years ago
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