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maxonik [38]
3 years ago
14

A student walks 350 m [S], then 400 m [E20°N], and finally 550 m [N10°W]. Using the component method, find the resultant (total)

displacement). Round your answer to the appropriate significant figures. Round your angle to the nearest degree.
Physics
1 answer:
levacccp [35]3 years ago
5 0

In component form, the displacement vectors become

• 350 m [S]   ==>   (0, -350) m

• 400 m [E 20° N]   ==>   (400 cos(20°), 400 sin(20°)) m

(which I interpret to mean 20° north of east]

• 550 m [N 10° W]   ==>   (550 cos(100°), 550 sin(100°)) m

Then the student's total displacement is the sum of these:

(0 + 400 cos(20°) + 550 cos(100°), -350 + 400 sin(20°) + 550 sin(100°)) m

≈ (280.371, 328.452) m

which leaves the student a distance of about 431.8 m from their starting point in a direction of around arctan(328.452/280.371) ≈ 50° from the horizontal, i.e. approximately 431.8 m [E 50° N].

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Answer:

After 1 sec = 4.9 m

After 2 sec = 19.6 m

After 3 sec = 44.1 m

After 4 sec =  78.4 m

After 5 sec = 122.5 m

Explanation:

After 1 sec:

<em>u=0m/s   t=1 s  a=9.8m/s²</em>

s = ut + (1/2)at²

=0(1) + (1/2)(9.8)(1²) = 4.9m

After 2 sec:

<em>u=0m/s   t=2 s  a=9.8m/s²</em>

s = ut + (1/2)at²

=0(2) + (1/2)(9.8)(2²) = 19.6m

After 3 sec:

<em>u=0m/s   t=3 s  a=9.8m/s²</em>

s = ut + (1/2)at²

=0(3) + (1/2)(9.8)(3²) = 44.1m

After 4 sec:

<em>u=0m/s   t=4 s  a=9.8m/s²</em>

s = ut + (1/2)at²

=0(4) + (1/2)(9.8)(4²) = 78.4m

After 5 sec:

<em>u=0m/s   t=5 s  a=9.8m/s²</em>

s = ut + (1/2)at²

=0(5) + (1/2)(9.8)(5²) = 122.5m

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What is the advantage of constantly performing endurance training such as Brisk walking or jogging, Yard work, Dancing, Swimming
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Exercise or Getting grader Stomata
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31. Draw a free body diagram for a 15.5N box that is being pushed to the right with a 18. N force while experiencing 4.30 N of r
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Answer:

See answers below

Explanation:

a.

F = mg,

15.5 N = m(9.8 m/s²)

m = 1.58 kg

b.

Fnet = Applied force - resistance,

Fnet = 18 N - 4.30 N,

Fnet = 13.70 N

Fnet = ma

13.70 N = (1.58 kg)a

a = 8.67 m/s²

For the free body diagram, draw a box with an upward arrow labeled 15.5 N, a downward label labeled 15.5 N, a right label labeled 18 N, and a left label labeled 4.30 N.

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3 years ago
The pressure in an automobile tire depends on the temperature of the air in the tire. When the air temperature is 25°C, the pres
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Answer:0.0704 kg

Explanation:

Given

initial Absolute pressure(P_1)=210+101.325=311.325

T_1=25^{\circ}\approx 298 K

V=0.025 m^3

T_2=50^{\circ}\approx 323 K

as the volume remains constant therefore

\frac{P_1}{T_1}=\frac{P_2}{T_2}

\frac{311.325}{298}=\frac{P_2}{323}

P_2=337.44 KPa

therefore Gauge pressure is 337.44-101.325=236.117 KPa

Initial mass m_1=\frac{P_1V}{RT_1}=\frac{311.325\times 0.025}{0.0287\times 298}

m_1=0.91 kg

Final mass m_2=\frac{P_2V}{RT_2}=\frac{311.325\times 0.025}{0.0287\times 323}

m_2=0.839

Therefore m_1-m_2=0.91-0.839=0.0704 kg of air needs to be removed to get initial pressure back

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3 years ago
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Explanation:

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Focal length of the concave mirror, f = -13.5 cm

Image distance, v = -37.5 cm (in front of mirror)

Let u is the object distance. It can be calculated using the mirror's formula as :

\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}

\dfrac{1}{u}=\dfrac{1}{f}-\dfrac{1}{v}

\dfrac{1}{u}=\dfrac{1}{(-13.5)}-\dfrac{1}{(-37.5)}

u = -21.09 cm

The magnification of the mirror is given by :

m=\dfrac{-v}{u}

m=\dfrac{-(-37.5)}{(-21.09)}

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So, the magnification produced by the mirror is (-1.77). Hence, this is the required solution.

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