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Eva8 [605]
3 years ago
10

How is the acceleration of an object in newtons second law related to its mass and the net force on it?

Physics
2 answers:
ki77a [65]3 years ago
8 0

Answer:

C) Acceleration is directly proportional to net force divided by mass.

Explanation:

As per Newton's 2nd law we know that net force on the object is product of mass and acceleration of the object

so we will have

F = ma

so here we can say that the force and mass of the object is directly depends on each other and the equation is just product of mass and acceleration.

So in order to find the acceleration we have

a = \frac{F}{m}

so we will have

C) Acceleration is directly proportional to net force divided by mass.

9966 [12]3 years ago
4 0
B) Acceleration is directly proportional to the mass of the object
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Newtons second law states that Force=Mass × Acceleration.The acceleration of gravity in 9.8 m/s.How much force is gravity exerti
Naddika [18.5K]

.98 Newton’s because you convert 100 g to kg which is .1 kg them you multiply.1 kg by 9.8 and get .98 and the units of the force are in Newton’s

5 0
3 years ago
Q. At what point in a waterfall do the drops of water contain the most kinetic energy ?
antoniya [11.8K]

Answer:

2.When they reach the bottom of the fall

Explanation:

The potential energy of the waterfall is maximum at the maximum height and decreases with decrease in height. Based on the law of conservation of mechanical energy, as the potential energy of the water fall is decreasing with  decrease in height of the fall, its kinetic energy will be increasing and the kinetic energy will be maximum at zero height (bottom of the fall).

Thus, the correct option is "2" When they reach the bottom of the fall

7 0
2 years ago
A driver notices an upcoming speed limit change from 45 mi/h (20 m/s) to 25 mi/h (11 m/s). If she estimates
zloy xaker [14]

Answer:

-2.79 m/s²

Explanation:

Given:

v₀ = 20 m/s

v = 11 m/s

Δx = 50 m

Find: a

v² = v₀² + 2aΔx

(11 m/s)² = (20 m/s)² + 2a (50 m)

a = -2.79 m/s²

Round as needed.

8 0
3 years ago
A merry-go-round of radius 2 m is rotating at one revolution every 5 s. A
galben [10]

Answer:

a) The angular speed of the child is approximately 1.257 rad/s

b) The angular speed of the teenager is approximately 1.257 rad/s

c) The tangential speed of the child is approximately 1.257 m/s

d) For the child, r = 2 m

The tangential speed of the teenager is approximately 2.513 m/s

Explanation:

The revolutions per minute, r.p.m. of the merry-go-round = 1 revolution/(5 s)

The radius of the merry-go-round = 2 m

The location of the child = 1 m from the axis

The location of the teenager = 2 m from the axis

1 revolution = 2·π radians

Therefore, we have;

The angular speed, ω = (Angle turned)/(Time elapsed) = (2·π radians)/(5 s)

∴ The angular speed of the merry-go-round, ω = 2·π/5 radians/second

a) The angular speed of the child = The angular speed of the merry-go-round = 2·π/5 radians/second ≈ 1.257 rad/s

b) The angular speed of the teenager = The angular speed of the merry-go-round = 2·π/5 radians/second ≈ 1.257 rad/s

c) The tangential speed, v = r × The angular speed, ω

Where;

r = The radius of rotation of the object

For the child, r = 1 m

The tangential speed of the child = 1 m × 2·π/5 radians/second = 2·π/5 m/s ≈ 1.257 m/s

d) For the child, r = 2 m

The tangential speed of the teenager = 2 m × 2·π/5 radians/second = 4·π/5 m/s ≈ 2.513 m/s

8 0
2 years ago
A large spool in an electrician's workshop has 65 m of insulation-coated wire coiled around it. When the electrician connects a
Art [367]

Answer:

40.34\ \text{m}

Explanation:

L_1 = Length of wire = 65 m

I_1 = Initial current = 1.8 A

I_2 = Final current = 2.9 A

We know

R\propto \dfrac{1}{I}

and

R\propto L

\dfrac{V}{I}\propto L\\\Rightarrow L\propto \dfrac{1}{I}

so

\dfrac{L_2}{L_1}=\dfrac{I_1}{I_2}\\\Rightarrow L_2=\dfrac{I_1}{I_2}L_1\\\Rightarrow L_2=\dfrac{1.8}{2.9}\times 65\\\Rightarrow L_2=40.34\ \text{m}

The length of the wire remaining on the spool is 40.34\ \text{m}.

8 0
3 years ago
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