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3 years ago

9

Which vector has an x-component with a length of 4?

1 answer:

matrenka [14]3 years ago

5 0

Answer:the scaler ax vertex

Explanation:

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Two cars are traveling at the same speed of 27 m/s on a curve thathas a radius of 120 m. Car A has a mass of 1100 kg and car B h

Lera25 [3.4K]

Answer:

$a_{cA} = 6.075$ m/s²

$a_{cB} = 6.075$ m/s²

$F_{cA} = 6682.5 N$

$F_{cB} = 9720 N$

Explanation:

Normal or centripetal acceleration measures change in speed direction over time. Its expression is given by:

$a_{c} = \frac{v^{2} }{r}$ Formula 1

Where:

$a_{c}$ : Is the normal or centripetal acceleration of the body ( m/s²)

v: It is the magnitude of the tangential velocity of the body at the given point

.(m/s)

r: It is the radius of curvature. (m)

Newton's second law:

∑F = m*a Formula ( 2)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

Data

$v_{A} = 27 \frac{m}{s}$

$v_{B} = 27 \frac{m}{s}$

$m_{A} = 1100 kg$

$m_{B} = 1600 kg$

r= 120 m

Problem development

We replace data in formula (1) to calculate centripetal acceleration:

$a_{cA} = \frac{(27)^{2} }{120}$

$a_{cA} = 6.075$ m/s²

$a_{cB} = \frac{(27)^{2} }{120}$

$a_{cB} = 6.075$ m/s²

We replace data in formula (2) to calculate centripetal force Fc) :

$F_{cA} = m_{A} *a_{cA} = 1100kg*6.075\frac{m}{s^{2} }$

$F_{cA} = 6682.5 N$

$F_{cB} = m_{B} *a_{cB} = 1600kg*6.075\frac{m}{s^{2} }$

$F_{cB} = 9720 N$

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4 years ago

Calculate the average induced voltage between the tips of the wings of a Boeing 747 flying at 940 km/hr above East Lansing. The

ANTONII [103]

Answer:

option (b)

Explanation:

B = 0.7 x 10^-4 T

v = 940 km/h = 940 x 5 / 18 = 261.11 m /s

l = 60 m

Th emotional emf is given by

e = B v l

e = 0.7 x 10^-4 x 261.11 x 60 = 1.096 V

e = 1.1 V

5 0

3 years ago

An earthquake produces longitudinal P waves that travel outward at 8000 m/s and transverse S waves that move at 4500 m/s. A seis

vivado [14]

Answer:

1234285.7 m or 1234.3 km

Explanation:

Let the distance be $d$ , the time taken by P waves be $t_P$ and the time taken by the S waves be $t_S$ .

$\text{Velocity}\dfrac{\text{Distance}}{\text{Time}}$

$\text{Time}\dfrac{\text{Distance}}{\text{Velocity}}$

For the P waves,

$t_P=\dfrac{d}{8000}$

$d=8000t_P$

For the S waves,

$t_S=\dfrac{d}{4500}$

$d=4500t_S$

Equating the $d$ ,

$8000t_P=4500t_S$

Divide both sides of the equation by 500 to reduce the terms.

$16t_P=9t_S$

Since S waves arrive 2 minutes (= 120 seconds) after P waves,

$t_S-t_P=120$

$t_S=120+t_P$

Substitute this in the equation of the distance.

$16t_P=9(t_P+120)$

$16t_P=9t_P+1080$

$7t_P=1080$

$t_P=\dfrac{1080}{7}$

Substitute this in the equation for $d$ involving $t_P$ .

$d=8000t_P$

$d=8000\times\dfrac{1080}{7}$

$d=1234285.7 \text{ m }= 1234.3 \text{ km}$

4 0

3 years ago