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san4es73 [151]
3 years ago
11

_____ is the frictional force needed to slow an object in motion

Physics
2 answers:
s2008m [1.1K]3 years ago
8 0

Answer:

<u>Drag force</u> is the frictional force needed to slow an object in motion

Explanation:

const2013 [10]3 years ago
8 0

Answer:

Kinetic friction

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2)It is known that the connecting rodS exerts on the crankBCa 2.5-kN force directed down andto the left along the centerline ofA
11111nata11111 [884]

Answer:

M_c = 100.8 Nm

Explanation:

Given:

F_a = 2.5 KN

Find:

Determine the moment of this force about C for the two cases shown.

Solution:

- Draw horizontal and vertical vectors at point A.

- Take moments about point C as follows:

                        M_c = F_a*( 42 / 150 ) *144

                        M_c = 2.5*( 42 / 150 ) *144

                        M_c = 100.8 Nm

- We see that the vertical component of force at point A passes through C.

Hence, its moment about C is zero.

5 0
3 years ago
Carpet can keep a room quiet by:
taurus [48]
Hard surfaces reflect sound back into the room, while carpets help to absorb the sound so it reflects less
6 0
3 years ago
Acceleration is generally defined as the time rate of change of velocity. When can it be defined as the time rate of change of s
Lena [83]

Answer:

When the velocity doesn't change its direction

Explanation:

Since velocity vector has 2 components: direction and magnitude, and speed is the velocity's magnitude. So if the velocity doesn't change its direction, we essentially use its magnitude, aka speed, to calculate the rate of change for acceleration.

8 0
3 years ago
Read 2 more answers
A golf ball strikes a hard, smooth floor at an angle of 39.8 ° and, as the drawing shows, rebounds at the same angle. The mass o
tangare [24]

Answer:

The magnitude of the impulse is 1.33 kg m/s

Explanation:

please look at the solution in the attached Word file

Download docx
7 0
3 years ago
Read 2 more answers
With detailed explaniation
belka [17]
  • Ø=37°
  • Initial velocity=u=20m/s
  • g=10m/s²

#A

\\ \rm\Rrightarrow H_{max}=\dfrac{u^2sin^2\theta}{2g}

\\ \rm\Rrightarrow H_{max}=\dfrac{20^2(sin37)^2}{2(10)}

\\ \rm\Rrightarrow H_{max}=\dfrac{400sin^237}{20}

\\ \rm\Rrightarrow H_{max}=20sin^237

\\ \rm\Rrightarrow H_{max}=7.2m

#B

\\ \rm\Rrightarrow R=\dfrac{u^2sin2\theta}{g}

\\ \rm\Rrightarrow R=\dfrac{20^2sin74}{10}

\\ \rm\Rrightarrow R=40sin74

\\ \rm\Rrightarrow R=38.5m

#C

\\ \rm\Rrightarrow T=\dfrac{2usin\theta}{g}

\\ \rm\Rrightarrow T=\dfrac{2(20)sin37}{10}

\\ \rm\Rrightarrow T=4sin37

\\ \rm\Rrightarrow T=2.4s

Now

\\ \rm\Rrightarrow v=u-gt

\\ \rm\Rrightarrow v=20-10(2.4)

\\ \rm\Rrightarrow v=20-24

\\ \rm\Rrightarrow v=-4m/s

4 0
2 years ago
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