The motion of an object looks different to observers in different

The diagram shows two vectors that point west and north. What is the magnitude of the resultant vector? 13 miles 17 miles 60 mil

Kipish [7]

Using the formula A squared plus B squared equals C squared, we can find the solution by substituting 5 for A and 12 for B.

By squaring 5, we get 25, and by squaring 12, we get 144. Adding these, we get 169. The square root of this is 13.

6 0

2 years ago

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A 120-kg refrigerator that is 2.0 m tall and 85 cm wide has its center of mass at its geometrical center. You are attempting to

Mademuasel [1]

To solve this problem it is necessary to apply the concepts related to Force of Friction and Torque given by the kinematic equations of motion.

The frictional force by definition is given by

$F= \mu mg$

Our values are here,

$\mu=0.3$

$m=120kg$

$g=9.8m/s^2$

Replacing,

$F=0.30*120*9.8 = 352.8N$

Consider the center of mass of the body half its distance from the floor, that is d = 0.85 / 2 = 0.425m. The torque about the lower farther corner of the refrigerator should be zero to get the maximum distance, then

$F*x = mg*d$

Re-arrange for x,

$x= \frac{mg*d}{F}$

$x= \frac{mg*d}{\mu mg}$

$x= \frac{d}{\mu}$

$x= \frac{0.425}{0.3}$

$x = 1.42m$

Then we can conclude that 1.42m is the distance traveled before turning.

6 0

2 years ago

A passenger bus is travelling 28.0 m/s to the right when the driver applies the brakes. The bus stops in 5.00 s. What is the acc

MAVERICK [17]

Change in velocity = d(v)
d(v) = v2 - v1 where v1 = initial speed, v2 = final speed
v1 = 28.0 m/s to the right
v2 = 0.00 m/s
d(v) = (0 - 28)m/s = -28 m/s to the right

Change in time = d(t)
d(t) = t2 - t1 where t1 = initial elapsed time, t2 = final elapsed time
t1 = 0.00 s
t2 = 5.00 s
d(t) = (5.00 - 0.00)s = 5.00s

Average acceleration = d(v) / d(t)
(-28.0 m/s) / (5.00 s)
(-28.0 m)/s * 1 / (5.00 s) = -5.60 m/s² to the right

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2 years ago

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A convex mirror with a focal length of 0.25 m forms a 0.080 m tall image of an automobile at a distance of 0.24 m behind the mir

Semmy [17]

Answer:

The distance and height of the object is 6 m and 2 m.

The image is virtual and upright.

Explanation:

Given that,

Focal length = 0.25 m

Length of image = 0.080 m