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bulgar [2K]
4 years ago
7

According to newtons first law which characteristics of a moving object would remain constant if there were no other forces acti

ng on it
Physics
2 answers:
mars1129 [50]4 years ago
6 0

Answer:

Speed ** is the answer

Explanation:

jolli1 [7]4 years ago
4 0

Answer:

velocity

Explanation:

According to Newton's first law of motion, a moving object's <u>velocity </u> remains constant if there were no forces acting on it. If no external, unbalanced force acts on it:

F_{ext} = m a = 0\\ \Rightarrow a = 0 \\ \Rightarrow  \frac{dv}{dt} = 0 \\ \Rightarrow v = constant

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If a car is moving 7.0m/s and has 62J of energy, how much mass does it have ?
Ede4ka [16]

Answi am sorry but i do not know the answer

Explanation:

7 0
2 years ago
Sophie applies a 50 n force to push a box 2 meter across the floor calculate the smount of work done in the box
kicyunya [14]
Work done = force x distance

Work done = 50 x 2

Work done = 100J
4 0
3 years ago
If you ride your bike 20 miles and it takes you 120 minutes,what is your average speed
Alexus [3.1K]
120 minutes=2 hours
20/2= 10mph
5 0
3 years ago
As a pole of a 2nd-order discrete-time system moves away from the origin in the z-plane, while its phase remains constant, the d
Rudik [331]

Answer:

False

Explanation:

When the location of the poles changes in the z-plane, the natural or resonant frequency (ω₀) changes which in turn changes the damped frequency (ωd) of the system.

As the poles of a 2nd-order discrete-time system moves away from the origin then natural frequency (ω₀) increases, which in turn increases damped oscillation frequency (ωd) of the system.

ωd = ω₀√(1 - ζ)

Where ζ is called damping ratio.

For small value of ζ

ωd ≈ ω₀

4 0
3 years ago
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
4 years ago
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