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3 years ago

What can stay constant if there were no forces

1 answer:

6 0

If an object experiences no net force, its velocity will remain constant. The object is either at rest and the velocity is zero, or it moves in a straight line with a constant speed.

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Examine the following equation.

alex41 [277]

Answer:

E)brain decay

Explanation:

Looking at the question causes it.

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2 years ago

8892 ml to grams then to mg

andrew11 [14]

Answer:

8892 ml = 8892 gm = 8892000 mg

1 ml = 1 gram

8892 ml = 8892 gram

1 gram or ml = 1000 milligram

8892 ml = 8892 × 1000 = 8892000 milligram

hope this helps

have a good day :)

Explanation:

remember: 1 kilogram = 1000 gram = 1000000 milligram.

Milliliter is expressed same as gram and liter is expressed same as kilogram.

1 meter = 100 cm, 1 kilometer = 1000 meter,

1 cm = 10 millimeter.

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2 years ago

Read 2 more answers

The measure of how much salt will dissolve into 100g of water is _______________ .

Bingel [31]

The measure of how much salt will dissolve into 100g of water is _solution_ .

8 0

3 years ago

Read 2 more answers

20 POINTS!

CaHeK987 [17]

Potential energy is the answer

6 0

3 years ago

In a game of angry birds you launch a bird with an angle of 53 degrees to horizontal. Unfortunatly, its not a good shot and the

Alisiya [41]

Answer:

The maximum height covered is 3.25 m.

The horizontal distance covered is 9.81 m.

The total time in the air is 1.63 seconds.

Explanation:

The launch speed, $u_0= 10 m/s$ .

Angle of launch with the horizontal, $\theta = 53 ^{\circ}$

So, the vertical component of the initial velocity,

$u_0\sin\theta=10 \sin 53 ^{\circ}\cdots(i)$ .

The horizontal component of the initial velocity,

$u_0\cos\theta=10 \cos 53 ^{\circ}$

Let, t be the time of flight, to the horizontal distance covered

$D=10 \cos (53 ^{\circ})t\cdots(ii)$ .

Not, applying the equation of motion in the vertical direction.

$s= ut +\frac 1 2 at^2$

Where s is the displacement in time t, u is the initial velocity and a is the acceleration.

In this case, $u =10 \sin 53 ^{\circ}$ (from equation (i), s=0 (as the final height is same as the launch height) and $a = -9.81 m/s^2$ (negative sign is due to the downward direction).