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SpyIntel [72]
3 years ago
12

A bowling ball with a mass of 4.5 kg travels at a velocity of 37 m/s for 2.5 s until it is stopped at the end of the lane by the

ball return.
What additional information is required to determine the weight of the bowling ball?

A.
the speed at which the ball return captured the ball

B.
the amount of friction from the bowling lane acting on the bowling ball

C.
the acceleration due to gravity acting on the bowling ball

D.
the force the bowler applied to the bowling ball
Physics
2 answers:
Goryan [66]3 years ago
8 0
Weight = (mass) x (acceleration of gravity) We know the mass. All we need now is the acceleration of gravity. All that other stuff about velocity and time is chaff, launched to confuse our radar.
sineoko [7]3 years ago
3 0

The answer is C.  the acceleration due to gravity acting on the bowling ball

I took the quiz and got 90%

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A truck starts from rest with an acceleration of 0.3 m/ S^2 find its speed in km/h when it has moves through 150 m​
Nadya [2.5K]

Answer:

9.5 m/s

Explanation:

Distance, S = 150m

Acceleration, a = 0.3 m/s^2

Initial velocity, u = 0 m/s

Final velocity, v

Use kinematics equation

v^2 - u^2 = 2aS

v^2 - 0 = 2*0.3*150 = 90

v = sqrt(90) = 9.49 m/s

3 0
3 years ago
Read 2 more answers
According to the general theory of relativity, what are consequences of the curvature of space-time? Check all that apply. The d
Svetlanka [38]

Answer:

The dilation of time.

The falling of objects.

The changing of paths of light.

Explanation:

I have explained in the image attached below.

From the explanation, the correct ones are;

The dilation of time.

The falling of objects.

The changing of paths of light.

4 0
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
What is the acceleration due to gravity at an altitude of 116?
Schach [20]
Gravitational acceleration (Ga) is  inversely proportional to  k / Distance^2

so Ga * Distance^2 = K

On the surface of Earth acceleration due to gravity is about 9.8m/s^2 with an average distance to the earths core of about 6371 km (Wolfram alpha).

So k = 9.8 * 6371^2
I'm presuming that your distance of 116 is km

As 
Ga = k / distance^2 

Ga = ((9.8 * 6371^2) / (6371 + 116)^2 ) = 397778481.8 / 42081169

= 9.45 m/s^2 to 2sf
7 0
4 years ago
The Chartered Institute of Management Accountants defines activity-based cost management as follows: "An approach to the costing
egoroff_w [7]

Answer:

ABC is the activity based accounting. It is the costing done for each separate activity which maybe unit level, batch level, product sustaining or facility sustaining.

Explanation:

The basic difference between ABC and traditional costing systems can be explained with the help of the following diagram.

Traditional Costing System

Overhead Cost Accounts (For each individual expense e.g. tax)

First Stage Allocation

  1. Cost Centers ( normally Departments)
  2. Cost Centers ( normally Departments)

N.  Cost Centers ( normally Departments)

Second Stage Allocation ( Direct Labor Or Machine Hours)

Cost Objects ( products, services and customers)

ACTIVITY BASED COSTING SYSTEMS

Overhead Cost Accounts (For each individual expense e.g. tax)

First Stage Allocation ( resource cost drivers)

  1. Activity Cost Centers
  2. Activity Cost Centers

N. Activity  Cost Centers

Second Stage Allocation ( activity cost drivers)

Cost Objects ( products, services and customers) (Direct Costs)

Four steps are involved in the design of ABC systems.

  1. identifying the major activities that take place in an organization
  2. assigning cost to cost pools / cost centers for each activity
  3. determining the cost for each major activity
  4. assigning the cost of activities to products according to the product's demand for activities.

The first two steps relate to the first stage and final two steps to the second stage of the two - stage allocation process shown above.

5 0
3 years ago
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