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VLD [36.1K]
3 years ago
15

A dog walks 14 meters to the west and then 20 meters back

Physics
1 answer:
Phoenix [80]3 years ago
3 0

Answer:

34 meters

Explanation:

20 meters +14 meters =34 meters

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I NEED HELP ON 2 QUESTIONS PLEASEEE
tino4ka555 [31]

Answer:

2) c) give-way vessel

3) a) With one short blast

Explanation:

2) A vessel that is required to take early substantial action to ensure avoiding  collision called Give way vessel

In overtaking, the vessel intending to overtake is the Give-Way Vessel the vessel that is going to be overtaken is the Stand-On Vessel

Therefore, the correct option is c) give-way vessel

3) When vessels use sound signals in a meeting head on situation both vessel are Give-Way vessels and both vessel pass the each other by turning to the starboard side therefore they intend to pass each other on their port side requiring one short blast

Therefore, the correct option is a) With one short blast.

4 0
4 years ago
A person in a kayak starts paddling, and it accelerates from 0 to 0.65 m/s in a distance of 0.40 m. If the combined mass of the
SCORPION-xisa [38]

Answer:

Magnitude of the net force acting on the kayak = 39.61 N

Explanation:

Considering motion of kayak:-

Initial velocity, u =  0 m/s

Distance , s = 0.40 m

Final velocity, v = 0.65 m/s

We have equation of motion v² = u² + 2as

Substituting

   v² = u² + 2as

    0.65² = 0² + 2 x a x 0.4

    a = 0.53 m/s²

We have force, F  = ma

Mass, m = 75 kg

    F  = ma = 75 x 0.53 = 39.61 N

Magnitude of the net force acting on the kayak = 39.61 N

6 0
4 years ago
Silva set up a dog-walking business. He earned $60 in June. He is projected to earn 110% more in July. How much is Silva project
Ivanshal [37]

Answer:

D = $660 because it says %110 more.

hope it helps


8 0
3 years ago
Read 2 more answers
A 964-kg car starts from rest at the bottom of a drive way and has a speed of 3.00 m/s at a point where the drive way has risen
Aleks [24]

Answer:

Work done by the engine is <u>12876.32 joules</u>.

Explanation:

We know that,

\mathrm{W}=\mathrm{KE}+\mathrm{PE}+\mathrm{W}_{\mathrm{f}}

Where,

W = work done by engine

\mathrm{KE}=\mathrm{kinetic} \text { energy at top of driveway }=\frac{1}{2}\left(\mathrm{mV}^{2}\right)

m = mass of the car = 964kg.

V = speed of car at top of driveway = 3 m/sec

PE = potential energy at top of driveway = m × g × h

\mathrm{g}=\text { acceleration due to gravity }=9.8 \mathrm{m} / \mathrm{s}^{2}

h = vertical height of driveway = 0.6 m

\mathrm{W}_{\mathrm{t}}=\text { work done against friction }=2870

Substituting values,

W=\frac{1}{2} \times(964) \times\left(3^{2}\right)+964 \times(9.8) \times(0.6)+2870

W=0.5 \times 964 \times 9+964 \times 9.8 \times 0.6+2870

\mathrm{W}=4338+5668.32+2870

W = 12876.32 joules  

Work done by the engine is <u>12876.32 joules</u>.

8 0
4 years ago
Anyone know the answer to this?
KonstantinChe [14]
Pull, aim, squeeze, spread
3 0
3 years ago
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