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chubhunter [2.5K]
3 years ago
12

An object undergoes two successive displacements:

Physics
1 answer:
Gre4nikov [31]3 years ago
8 0

The magnitude of the net displacement is 95.3 m

Explanation:

To find the magnitude of the net displacement, we have to resolve each of the two displacements into the horizontal and vertical direction first.

1st displacement is:

d_1=79 m at 16.9^{\circ}

So its components are

d_{1x}=(79)(cos 16.9^{\circ})=75.6 m\\d_{1y}=(79)(sin 16.9^{\circ})=23.0 m

2nd displacement is:

d_2=16.7 m at 31.1^{\circ}

So its components are

d_{2x}=(16.7)(cos 31.1^{\circ})=14.3 m\\d_{2y}=(16.7)(sin 31.1^{\circ})=8.6 m

Therefore, the x- and y-components of the net displacement are:

d_x=d_{1x}+d_{2x}=75.6+14.3=89.9 m\\d_y=d_{1y}+d_{2y}=23.0+8.6=31.6 m

Therefore, the magnitude of the final displacement is:

d=\sqrt{d_x^2+d_y^2}=\sqrt{(89.9)^2+(31.6)^2}=95.3 m

Learn more about displacement:

brainly.com/question/3969582

#LearnwithBrainly

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gizmo_the_mogwai [7]

Answer:

10s

Explanation:

Acceleration is a measure of a rate of change of velocity, or in other words, a measure of how quickly the velocity is changing.

If acceleration is constant, then the velocity is changing by a constant amount.

With an acceleration of 100 m/s^2, starting from the launching pad (and thus, an initial velocity of zero), we can calculate how long it will take to reach a final velocity of 1000m/s with the following formula:

v=at+v_o where "v" is the final velocity at some later time "t", "a" is the constant acceleration, and "v" sub-zero is the initial velocity.

v=at+v_o

(1000\text{ [m/s]})=(100 \text{ } [\text{m/s}^2] )t+(0\text{ [m/s]})

1000\text{ [m/s]}=100 \text{ } [\text{m/s}^2] *t

\dfrac{1000\text{ [m/s]}}{100 \text{ } [\text{m/s}^2]}=\dfrac{100 \text{ } [\text{m/s}^2] *t}{100 \text{ } [\text{m/s}^2]}

10\text{ [s]}=t

So, it will take 10 seconds for the rocket to reach 1000m/s when starting from the launching pad, with a constant velocity of 100m/s^2.

<u>Verification:</u>

In this situation, it is quick to verify that 10 seconds is correct by looking at what the velocities will be each second.

Recognizing that the acceleration is a=\dfrac{100 [\frac{m}{s}]}{1[s]}, the velocity increases by 100 units [m/s] every second.

At time 0[s], the velocity is 0[m/s]

At time 1[s], the velocity is 100[m/s]

At time 2[s], the velocity is 200[m/s]

At time 3[s], the velocity is 300[m/s]

At time 4[s], the velocity is 400[m/s]

At time 5[s], the velocity is 500[m/s]

At time 6[s], the velocity is 600[m/s]

At time 7[s], the velocity is 700[m/s]

At time 8[s], the velocity is 800[m/s]

At time 9[s], the velocity is 900[m/s]

At time 10[s], the velocity is 1000[m/s]

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