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2 years ago

14

What is the direction of the sum of these two vectors?

2 answers:

gayaneshka [121]2 years ago

7 0

Answer:

what is the direction of the sum of these two vectors?

jolli1 [7]2 years ago

6 0

Answer:

69.6

Explanation:

Ax= 0

Ay= 63.5

Bx= 101 cos 57= 55.009

By= 101 sin 57= 84.706

Tan-1 = (84.706/55.009)

= 69.6

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A 15-kN tensile load will be applied to a 50-m length of steel wire with E = 200 GPa. Determine the smallest diameter wire that

sasho [114]

Answer:

d=13.81 mm

Explanation:

Given that

P = 15 KN ,L = 50 m

E= 200 GPa

ΔL = 25 mm

σ = 150 MPa

Lets take d=Diameter

There are we have two criteria to find out the diameter of the wire

Case I :

According to Stress ,σ = 150 MPa

P = σ A

$A=\dfrac{P}{\sigma}$

$d=\sqrt{\dfrac{4P}{\pi \sigma}}$

By putting the values

$d=\sqrt{\dfrac{4\times 15000}{\pi \times 150}}$

d= 11.28 mm

Case II:

According to elongation ,ΔL = 25 mm

$\Delta L=\dfrac{PL}{AE}$

$A=\dfrac{PL}{E\Delta L}$

$A=\dfrac{4PL}{\pi E\Delta L}$

$d=\sqrt{\dfrac{4\times 15000\times 50000}{\pi \times 200\times 1000\times 25}}$

d=13.81 mm

Therefore the answer will be 13.81 mm .Because it satisfy both the conditions.

3 0

3 years ago

Liquid pools of methane are found on the surface of Titan, one of Saturn's moons. The temperature on the surface of Titan is -18

Rina8888 [55]

Answer:

-292 degrees F

Explanation:

C = 5/9( F -32)

-180 = 5/9 (F-32)

F = -292 degrees

7 0

2 years ago

The maximum distance from the Earth to the Sun (at aphelion) is 1.521 1011 m, and the distance of closest approach (at perihelio

LUCKY_DIMON [66]

Answer:

29274.93096 m/s

$2.73966\times 10^{33}\ J$

$-5.39323\times 10^{33}\ J$

$2.56249\times 10^{33}\ J$

$-5.21594\times 10^{33}$

Explanation:

$r_p$ = Distance at perihelion = $1.471\times 10^{11}\ m$

$r_a$ = Distance at aphelion = $1.521\times 10^{11}\ m$

$v_p$ = Velocity at perihelion = $3.027\times 10^{4}\ m/s$

$v_a$ = Velocity at aphelion

m = Mass of the Earth = 5.98 × 10²⁴ kg

M = Mass of Sun = $1.9889\times 10^{30}\ kg$

Here, the angular momentum is conserved

$L_p=L_a\\\Rightarrow r_pv_p=r_av_a\\\Rightarrow v_a=\frac{r_pv_p}{r_a}\\\Rightarrow v_a=\frac{1.471\times 10^{11}\times 3.027\times 10^{4}}{1.521\times 10^{11}}\\\Rightarrow v_a=29274.93096\ m/s$

Earth's orbital speed at aphelion is 29274.93096 m/s

Kinetic energy is given by

$K=\frac{1}{2}mv_p^2\\\Rightarrow K=\frac{1}{2}\times 5.98\times 10^{24}(3.027\times 10^{4})^2\\\Rightarrow K=2.73966\times 10^{33}\ J$

Kinetic energy at perihelion is $2.73966\times 10^{33}\ J$

Potential energy is given by

$P=-\frac{GMm}{r_p}\\\Rightarrow P=-\frac{6.67\times 10^{-11}\times 1.989\times 10^{30}\times 5.98\times 10^{24}}{1.471\times 10^{11}}\\\Rightarrow P=-5.39323\times 10^{33}$

Potential energy at perihelion is $-5.39323\times 10^{33}\ J$

$K=\frac{1}{2}mv_a^2\\\Rightarrow K=\frac{1}{2}\times 5.98\times 10^{24}(29274.93096)^2\\\Rightarrow K=2.56249\times 10^{33}\ J$

Kinetic energy at aphelion is $2.56249\times 10^{33}\ J$

Potential energy is given by

$P=-\frac{GMm}{r_a}\\\Rightarrow P=-\frac{6.67\times 10^{-11}\times 1.989\times 10^{30}\times 5.98\times 10^{24}}{1.521\times 10^{11}}\\\Rightarrow P=-5.21594\times 10^{33}$

Potential energy at aphelion is $-5.21594\times 10^{33}\ J$

6 0

2 years ago

I meed help with these 2 questions plz

ludmilkaskok [199]

-- 30N

-- the sum of all the forces

4 0

3 years ago

What is the relationship between distance of the objects and the gravitational force between them? (Another way of asking: What

Nadya [2.5K]

Explanation:

increase in the force of gravity

6 0

2 years ago