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Black_prince [1.1K]

3 years ago

5

An object starts at position 12 on a horizontal line with a reference point of 0. What is the position of the object if it moves

14 units to the left?

2 answers:

Wittaler [7]3 years ago

8 0

-2 is the right answer

kolbaska11 [484]3 years ago

8 0

Answer:

The position of the object is $-2$

Explanation:

I add a picture of the situation in order to explain the question.

We are going to state the ''zero'' on a horizontal line. Then, we are going to state as positive the sense from left to right.

Now, if the object moves 14 units to the left , we need to subtract to the original position 14 units (because it will be a negative move given that the object is moving to the left) ⇒

$12-14=-2$

It new position will be at $-2$ .

In general, the first step to solve this kind of exercises is to set a reference system. Then, it is very important to draw the situation.

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On a certain planet, which is perfectly spherically symmetric, the free-fall acceleration has magnitude g = go at the north pole

ohaa [14]

The reason why there is a difference between free-fall acceleration is a centrifugal force.
I attached a diagram that shows how this force aligns with the force of gravity.
From the diagram we can see that:
$F=F_g-F_{cf}=mg'-mw^2r'cos(\alpha)\\ ma=mg'-mw^2r'cos(\alpha)\\ a=g'-w^2rcos^2(\alpha)\\$
Where g' is the free-fall acceleration when there is no centrifugal force, r is the radius of the planet, and w is angular frequency of planet's rotation. $\alpha$ is the latitude.
We can calculate g' and wr^2 from the given conditions in the problem.
$g(90)=g_0;\ g_0= g'-w^2rcos^2(90)\\
g_0=g'\\
g(0)=ag_0;\ ag_0=g_0-w^2rcos^2(0)\\
ag_0=g_0-w^2r\\
w^2r=g_0(a-1)
$
Our final equation is:
$g=g_0-g_0(a-1)cos^2(\alpha)$
Colatitude is:
$\alpha_c=90^\circ-\alpha$
The answer is:
$g=g_0-g_0(a-1)cos^2(90-9)=g_0-g_0(a-1)sin^2(9)$

5 0

4 years ago

A: How far did she travel? <br><br><br><br> B: How long did she take?

allochka39001 [22]

A. 60 miles
B. 5 hours

Unless you are looking for slope, in which case the answer is different

7 0

3 years ago

you have been called to testify as a as an expert witness in a trial involving a head-on collision Car A weighs 1515 pounds and

GrogVix [38]

Answer:

70.6 mph

Explanation:

Car A mass= 1515 lb

Car B mass=1125 lb

Speed of car B is 46 miles/h

Distance before locking, d=19.5 ft

Coefficient of kinetic friction is 0.75

Initial momentum of car B=mv where m is mass and v is velocity in ft/s

46 mph*1.46667=67.4666668 ft/s

$Momentum_B=1125*67.4666668 ft/s$

Initial momentum of car A is given by

$Momentum_A=1515v_a$ where $v_a$ is velocity of A

Taking East as positive and west as negative then the sum of initial momentum is

$1515v_a-(1125*67.4666668 ft/s)$

The common velocity is represented as $v_c$ hence after collision, the final momentum is

$Momentum_final=(m_a+m_b)v_c=(1515+1125)v_c=2640v_c$

From the law of conservation of linear momentum, sum of initial and final momentum equals each other hence

$1515v_a-(1125*67.4666668 ft/s)= 2640v_c$

The acceleration of two cars $a=-\mug=-0.75*32.17=-24.1275 ft/s^{2}$

From kinematic equation

$v^{2}=u^{2}+2as$ hence

$v^{2}-u^{2}=2as$

$0^{2}-(v_c)^{2}=2*-24.1275*19.5$

$v_c=\sqrt{2*24.1275*19.5}=30.67 ft/s$

Substituting the value of $v_c$ in equation $1515v_a-(1125*67.4666668 ft/s)= 2640v_c$

$1515v_a-(1125*67.4666668 ft/s)= 2640*30.67$

$1515v_a=(1125*67.4666668 ft/s)+2640*30.67$

$v_a=\frac {156868.8}{1515}=103.5438 ft/s$

$\frac {103.5438}{1.46667}=70.59787 mph\approx 70.60 mph$

3 0

3 years ago

An object travels in a circular path of radius 5.0 meters at a uniform speed of 10. m/s. What is the magnitude of the object's a

vladimir1956 [14]

I think F= mv²/r
And F=ma
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a = v²/r
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a = 20 m/s

8 0

3 years ago

Read 2 more answers

A solid cube of side 5cm has a mass of 250g. What is its density in kgm ??

Mariulka [41]

Answer:

5kgm

Explanation:

convert cm to m and g to kg

250/1000=0.25kg

5/1000=0.05m

then find the density

density=mass/volume

=0.25kg/0.05m

=5kgm

7 0

3 years ago