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3 years ago

X=3t^2-12t+5 calculate time t=1 and time when velocity zero

1 answer:

4 0

You plug your 1 into all spots where the variable is T. Then after you’ve done this, you simply simplify the problem. Then after simplifying you should have an equation where you are simply adding and subtracting which will give you the final answer of 2.

X=3T^2-12T+5
X=3(1)^2-12(1)+5
X=(3)^2-12+5
X=9-12+5

X=2

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Two geological field teams are working in a remote area. A global positioning system (GPS) tracker at their base camp shows the

Masja [62]

Answer:

distance of 2nd team from 1st team will be: 58.2

Direction of 2nd team from 1st team will be: 14.90 deg North of east

Explanation:

ASSUME Vector is R and makes angle A with +x-axis,

therefore component of vector R is

$R_x = Rcos A$

$R_y = Rsin A$

From above relation

Assuming base camp as the origin, location of 1st team is

$R_1 = 37 km$ away at 21 deg North of west (North of west is in 2nd quadrant, So x is -ve and y is positive)

$R_{1x} = -R_1*cos A_1 = -37*cos 21 deg = -34.54 km$

$R_{1y} = R_1*sin A_1 = 37*sin 21 deg = 13.25 km$

location of 2nd team is at

$R_2 = 32 km$ , at 38 deg East of North = 32 km, at 58 deg North of east (North of east is in 1st quadrant, So x and y both are +ve)

$R_{2x} = R_2*cos A_2 = 32*cos 58 deg = 16.95 km$

$R_{2y} = R_2*sin A_2 = 32*sin 58 deg = 27.13 km$

Now position of 2nd team with respect to 1st team will be given by:

$R_3 = R_2 - R_1$

$R_3 = (R_{2x} - R_{1x}) i + (R_{2y} - R_{1y}) j$

Using above values:

$R_3 = (16.95 - (-34.54)) i + (27.13 - 13.42) j$

$R_3 = 51.49 i + 13.71 j$

distance of 2nd team from 1st team will be:

$\left | R_3 \right | = \sqrt (51.49^2 +13.71^2)$

$\left | R_3 \right | = 53.28 km = 58.2 km$

Direction of 2nd team from 1st team will be:

$Direction = tan^{-1} \frac{R_{3y}}{R_{3x}} = tan^{-1}[ \frac{13.71}{51.49}]$

Direction = 14.90 deg North of east

6 0

2 years ago

(b) 360 days into seconds.

Angelina_Jolie [31]

Explanation:

(b) We know that,

1 day = 24 hours

1 hour = 3600 s

So, we found that, 1 day = 86400 s

We need to find the 360 days into seconds. So,

1 day = 86400 s

360 days = 86400×360

360 days = 31104000 seconds

(d) Weight of a body, W = 600 N

Acceleration due to gravity on mars is 3.7 m/s²

Weight, W = mg

m is mass of body

$m=\dfrac{W}{g}\\\\m=\dfrac{600}{3.7}\\\\m=162.16\ kg$

(e) Mass of body, m = 100 kg

Acceleration due to gravity on the moon, 1.6 m/s²

Weight, W = 100 × 1.6

W = 160 N

8 0

3 years ago

Temperature is a measure of the average ____________ energy of an object's particles. light mechanical potential kinetic

Zielflug [23.3K]

Temperature is a measure of the average kinetic energy of the particles of a substance.
Hope this helps! :)

6 0

3 years ago

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You are pushing on a heavy desk with a force of 65 N the desk does not slide the force of friction between the desk and the floo

astra-53 [7]

When an object does not move even on pushing , static frictional force acts on in opposite direction of the applied force to stop the object from moving. static frictional force is a self adjusting force and it adjust its value according to the applied force if the applied force is smaller than the maximum value of static frictional force. The object starts moving once the applied force on it becomes greater than the maximum static frictional force. hence the statement is true.

7 0

3 years ago

11. A 3.8 kg object is lifted 12 meters. Approximately how much work is performed during the lifting?

Mademuasel [1]

Hope this answers your question!! Ask any help at anytime

4 0

3 years ago