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Step2247 [10]
3 years ago
13

Distance = 6 km south 60 minutes What was the average velocity

Physics
2 answers:
shutvik [7]3 years ago
8 0

For this case we have to by definition:

v = \frac {x} {t}

Where:

v: Velocity

x: Distance traveled

t: Time

According to the data we have:

x = 6 \ km = 6000 \ m\\t = 60 \ min = 3600 \ s

So:

v = \frac {6} {60} = 0.1  \frac {km} {min}

Equivalent to:

v = \frac {6000} {3600} = 1.67 \frac {m} {s}

Answer:

0.1 \frac {km} {min}

blsea [12.9K]3 years ago
3 0

v = x/t

v = average velocity, x = displacement, t = elapsed time

Given values:

x = 6km south, t = 60min

Plug in and solve for v:

v = 6/60

v = 0.1km/min south

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lesya692 [45]

\\ \bull\sf\dashrightarrow F=ma

\\ \bull\sf\dashrightarrow a=\dfrac{F}{m}

Now

\\ \bull\sf\dashrightarrow a\propto\dfrac{1}{m}

\\ \bull\sf\dashrightarrow a\propto F

  • Lower mass=Higher acceleration
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Option B has lowest mass and highest force hence its correct

8 0
2 years ago
A ball is falling after rolling off a tall roof. The ball has what type of energy.
Firlakuza [10]

Answer:

Correct answer: Kinetic energy of rotation, kinetic energy of translation and potential energy

Explanation:

Before it began to rotate it possessed potential energy. After being released from the rest it starts to roll on the roof and get kinetic energy of rotation, kinetic energy of translation along with potential energy.

God is with you!!!

7 0
3 years ago
A 1000 N object is pulled along a level surface with a horizontal force of 200 N. The object moves with a constant velocity of 2
krek1111 [17]

0.2 is the value of coefficient of friction (k)

F=kN

F=horizontal force

n=Normal Force

k=coefficient of friction

k=F/N

k=200/1000

k=0.2

The ratio of the normal force pushing two surfaces together to the frictional force preventing motion between them is known as the friction coefficient. Usually, the Greek letter mu is used to indicate it .N is the normal force, and F is the frictional force, hence F = N/N.

Due to the fact that both F and N are measured in units of force, the coefficient of friction has no dimensions (such as newtons or pounds). The coefficient of friction can have a variety of values for both static and dynamic friction. Static friction occurs when an object encounters friction that resists any applied force, keeping the object at rest until the static frictional force is released. In kinetic friction, the frictional force resists the motion of the object.

To know more about  coefficient of friction visit brainly.com/question/136431

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7 0
10 months ago
A watermelon is thrown down from a skyscraper with a speed of 7.0\,\dfrac{\text m}{\text s}7.0 s m ​ 7, point, 0, space, start f
Degger [83]

Answer:

y = 17,89 m

Explanation:

Let us fixate the reference point in top of the building, from where the watermelon is thrown down. We will assume also that the positive axis of our system points up. We describe the watermelon’s motion with the equation:

 v_y^2 =v_0^2 + 2ay

Clearing the equation so we isolate y we have that:

 y = (v_y^2 - v_0^2 )/2a

Making a substitution with the values from the statement we have:

y = ((20 m/s)^2 - (7 m/s)^2)/(2*9,81 m/s^2) = 17,89 m]

So, this skyscraper is about 17,89 m tall; which is not very tall for a skyscraper but who am I to judge.  17,89 m is also the displacement of the watermelon from the point it was thrown down.

I hope everything was clear with my explanation. If I can help with anything else, just let me know. Have an awesome day :D

7 0
3 years ago
Read 2 more answers
A 1.0 kg block is placed against an ideal spring with spring constant 800 N/m and initially compressed 0.20 m. The spring and bl
densk [106]

Answer:

the release will be at 3.266 m distance

Explanation:

mass = 1 Kg

spring constant (k) = 800 N/m

initial compression = 0.20 m

θ = 30⁰

U= \dfrac{1}{2}kx^2\\U= \dfrac{1}{2}800\times 0.2^2\\U= 16 J

U=mgh\\h=\dfrac{U}{mg}\\h=\dfrac{16}{1 \times 9.8}\\h = 1.633m

d=\dfrac{h}{sin \theta}\\d=\dfrac{1.633}{sin 30}\\\\d= 3.266m

hence the release will be at 3.266 m distance.

5 0
2 years ago
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