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seropon [69]
1 year ago
14

(a) What is the sum of the following four vectors in unit-vector notation? For that sum, what are (b) the magnitude, (c) the ang

le in degrees, and (d) the angle in radians? Positive angles are counterclockwise from the positive direction of the x axis; negative angles are clockwise.
: 6.00 m at + 0.900 rad
: 5.00 m at - 75.0°
: 4.00 m at + 1.20 rad
: 6.00 m at - 210°
Physics
1 answer:
avanturin [10]1 year ago
7 0

6m at 0.9 radian means 6m at 51.57^0

Since positive angles are counter clockwise

6m at 51.57^0 can be written as 6*cos51.57^0i+6*sin51.57^0j = 3.73i+4.70j

5m at -75^0 can be written as 5*cos(-75)^0i+5*sin(-75)^0j = 1.294i-4.83j

4m at 1.2 radian means 4m at 68.75^0

4m at 68.75^0 can be written as 4*cos68.75^0i+4*sin68.75^0j = 1.45i+3.73j

6m at -210^0 can be written as 6*cos(-210)^0i+6*sin(-210)^0j = -5.20i-3j

a) So sum of the vector = 1.274i+0.6j

b) Magnitude = \sqrt{1.274^2+0.6^2} = 1.98 m

c) Angle,  tan\theta = \frac{0.6}{1.274} \\ \\ \theta = 25.22^0

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Answer:

The force applied on the big piston is 1306.67 N

Explanation:

Given;

force applied on small piston, F₁ = 200 N

diameter of the small piston, d₁ = 4.37 cm

radius of the small piston, r₁ = d₁/2 = 2.185 cm

Area of the small piston, A₁ = πr₁² = π(2.185 cm)² = 15 cm²

Area of the big piston, A₂ = 98 cm²

The pressure of the piston is given by;

P = \frac{F}{A} \\\\\frac{F_1}{A_1} = \frac{F_2}{A_2}\\\\ F_2 = \frac{F_1A_2}{A_1}

Where;

F₂ is the force on big piston

F_2 = \frac{200*98}{15} \\\\F_2 = 1306.67 \ N

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1 year ago
A student measures the pH of a solution to be 6.8. Which should the student add if she wants to decrease the pH of the solution?
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7 0
1 year ago
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"The predictions of Einstein’s Theory of General Relativity were tested on a double pulsar system in January of 2004. His equati
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Answer:

99.95%

Explanation:

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8 0
2 years ago
A 2530-kg test rocket is launched vertically from the launch pad. Its fuel (of negligible mass) provides a thrust force so that
gavmur [86]

Answer:

A = 1.4 m/s²

B = -0.10493 m/s³

a = 1.29507 m/s²

T = 28095.8271 N

T = 1.13198 W

Explanation:

t = Time taken

g = Acceleration due to gravity = 9.81 m/s²

The equation

v(t)=At+Bt^2

Differentiating with respect to time

\frac{dv}{dt}=\frac{d(At+Bt^2)}{dt}\\\Rightarrow 1.4=A+2Bt

At t = 0

1.4=A

Hence, A = 1.4 m/s²

B=\frac{v-At}{t^2}\\\Rightarrow B=\frac{2.18-1.4\times 1.8}{1.8^2}\\\Rightarrow B=-0.10493\ m/s^3

B = -0.10493 m/s³

At t = 5 seconds

a=1.4+2\times -0.010493\times 5=1.29507\ m/s^2

a = 1.29507 m/s²

T=m(a+g)\\\Rightarrow T=2530(1.29507+9.81)\\\Rightarrow T=28095.8271\ N

T = 28095.8271 N

Weight of rocket

W=2530\times 9.81=24819.9\ N

\frac{T}{W}=\frac{28095.8271}{24819.9}\\\Rightarrow \frac{T}{W}=1.13198\\\Rightarrow T=1.13198W

T = 1.13198 W

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