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2 years ago

A boat is stationary at 12 meters away from a dock. The boat then begins to move toward the dock with

Physics

1 answer:

just olya [345]2 years ago

8 0

Answer: 2.19 seconds

Explanation:

<u>Given:</u>

Initial speed, u = 0 m/s

Acceleration of the boat, a = 5 m/s^2

Distance between the boat and dock, d = 12 m

Using the third kinematics equation to solve for time:

d = u*t + (1/2)*a*t^2

12 = 0*t + (1/2)*5*t^2

t = sqrt (12*2/5)

t = 2.19 seconds

Therefore, it will take the boat approximately 2.19 seconds to reach the dock

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Will give brainliest! how does an engineer use physical science?

pentagon [3]

Answer: gravity, circuits

Explanation:

3 0

3 years ago

If it takes 100 N to get a 10 kg object to accelerate at 10m/s/s, how much force will it take to get a 20 kg object to accelerat

yan [13]

Answer: C) 200 N

Explanation:

The force $F$ is defined as:

$F=m.a$

Where:

$m=20 kg$ is the mass of the object

$a=10 m/s^{2}$ is the acceleration

Then:

$F=(20 kg)(10 m/s^{2})$

Finally:

$F=200 N$

Hence, the correct option is C.

7 0

3 years ago

During an episode of turbulence in an airplane you feel 210 n heavier than usual.if your mass is 72 kg, what are the magnitude a

lana66690 [7]

According to Newton's Second Law of Motion, the net force experienced by the system is equal to the mass of the system in question times the acceleration in motion. In this case, the net force is the difference of gravitational force and the force experience by the motion of the airplane. This difference is already given to be 210 N.

Net force = ma
210 N = (73 kg)(a)
a = +2.92 m/s²

Thus, the acceleration of the airplane's motion is 2.92 m/s² to the positive direction which is upwards.

8 0

4 years ago

the net external force on the 24-kg mower is stated to be 51 N. If the force of friction opposing the motion is 24 N, what force

-Dominant- [34]

Answer:

PART A)

External force will be 75 N

PART B)

distance moved will be 1.125 m

Explanation:

PART A)

Given that net force on the mower is

$F_{net} = 51 N$

now we also know that friction force due to ground is given as

$F_f = 24 N$

now we have

$F_{net} = F_{ext} - F_f$

$51 = F_{ext} - 24$

$F_{ext} = 75 N$

so external force will be 75 N

PART B)

deceleration due to friction when external force is removed from it

$a = \frac{F_f}{m}$

$a = \frac{24}{24} = 1 m/s^2$

now we can find the distance by kinematics

$v_f^2 - v_i^2 = 2 a d$

$0 - 1.5^2 = 2(-1)d$

$d = 1.125 m$

so the distance moved will be 1.125 m

6 0

3 years ago

Which situation is the best analogy for the doppler effect?

Rudiy27

The best scenario to describe the doppler effect would be listening to the siren of a passing ambulance or fire truck

then it is coming towards you, the pitch is higher, it gets higher as it approaches and peaks as it gets right in front of you. then it drop at once when it passes you and continues to drop till it fades away. this is a classic descrption of the doppler effect

8 0

3 years ago