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zysi [14]
3 years ago
11

Someone pls help! My physics quiz is today:(

Physics
2 answers:
lisabon 2012 [21]3 years ago
8 0

Answer:

it should be -2m/s

Explanation:

Because it is going down from minutes 6-8. And down in most cases means negative. And since it is going down 2 it should be -2m/s.

Hope this helped!!

Vladimir [108]3 years ago
3 0

Answer:

0m/s^2

Explanation:

This is a position vs time graph, so the slop of the line represents the velocity (if you do not understand why that is then feel free to ask).  Since the slope is constant (a straight line), we know that the velocity is not changing, in other words, the object is not accelerating.

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An object whose weight is 10kg is placed on smooth plane inclined at 30° to the horizontal. find the acceleration of the object
velikii [3]

Answer:

4.9 m/s²

Explanation:

Draw a free body diagram.  There are two forces on the object:

Weight force mg pulling straight down,

and normal force N pushing perpendicular to the plane.

Sum the forces in the parallel direction.

∑F = ma

mg sin θ = ma

a = g sin θ

a = (9.8 m/s²) (sin 30°)

a = 4.9 m/s²

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3 years ago
Define torque qualitatively and quantitatively.
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When you use a wrench to tighten or loosen a nut on a bolt, you are applying torque. It is measured in units of force times distance.  A force of F newtons pulling on a handle of L meters in length would supply a torque of F L newton-meters. More technically, torque is the vector cross product of force times perpendicular distance from the object, F x r = F r sin @
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To overcome the problems that blur images and don't provide the best resolution from Earth, astronomers have started using flexi
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simple

3 0
3 years ago
A migrating robin flies due north with a speed of 12 m/s relative to the air. The air moves due east with a speed of 6.7 m/s rel
mafiozo [28]

Here it is given that speed of migrating Robin is 12 m/s relative to air

so we can say that

\vec v_{ra} = 12 m/s North

so it will be

Let North direction is along Y axis and East direction is along X axis

\vec v_{ra} = 12\hat j

also it is given that speed of air is 6.7 m/s relative to ground

\vec v_a = 6.7 \hat i

now as we know by the concept of relative motion

\vec v_{ab} = \vec v_a - \vec v_b

\vec v_{ra} = \vec v_r  - \vec v_a

now by rearranging the terms

\vec v_r = \vec v_{ra} + \vec v_a

\vec v_r = 12 \hat j + 6.7 \hat i

now we need to find the speed of Robin which means we need to find the magnitude of its velocity which we found above

So here we will say

v_r = \sqrt{12^2 + 6.7^2}

v_r = 13.7 m/s

so the net speed of Robin with respect to ground will be 13.7 m/s

7 0
3 years ago
A satellite orbiting Earth at an orbital radius r has a velocity v. Which represents the velocity if the satellite is moved to a
satela [25.4K]

Answer:

The correct answer is V√5

Explanation:

Let V be the velocity of the satellite orbiting at radius r.

Let V(5r) be the velocity of the satellite orbiting at radius 5r.

Recall:

Escape velocity is given by:

V = √(2gr)

Where V is the escape velocity

g is the acceleration due to gravity

r is the radius of the earth.

With the above equation, we can obtain the answer to the question as follow:

V = √(2gr)

V(5r) = √(2g5r)

Next, we'll obtained the ratio of V(5r) to V as shown below

V(5r) : V => V(5r)/V

V(5r)/V = √(2g5r) / √(2gr)

V(5r)/V = √5

Cross multiply

V(5r) = V√5

From the above illustration, we can see that when the satellite is moved to 3r, then the expression for the velocity will be V√5

3 0
3 years ago
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