A 50-kg object is accelerating to the right at 10 m/s². What is the magnitude of the rightward net force acting

To prevent collisions and violations at intersections that have traffic signals, use the _____ to ensure the intersection is cle

Ann [662]

The answer should be:
To prevent collisions and violations at intersections that have traffic signals, use the delayed acceleration technique to ensure the intersection is clear before you enter it.
Delayed acceleration technique refers to waiting to go through an intersection until you have a chance to scan for other vehicles.

3 0

2 years ago

A 21200 kg sailboat experiences an eastward force 42700 N due to the tide pushing its hull while the wind pushes the sails with

My name is Ann [436]

Answer:

$2.95 m/s^{2}$

Explanation:

The resultant force F

$F=\sqrt {F_{1}^{2}+F_{2}^{2}+2F_{1}F_{2}cos135^{o}}$ where $F_{1}$ is eastward force, $F_{2}$ is force directed towards the North

$F=\sqrt {42700^{2}+85000^{2}+(2*42700*85000)cos135^{o}}=62573.17217 N$

F=62573.2 N

The magnitude of acceleration of sailboat is given by

$a=\frac {F}{m}=\frac {62573.2}{21200}=2.95 m/s^{2}$

7 0

2 years ago

Suppose an X-ray binary is found in which the visible star is a 12 M⦿ red giant, the orbital period is 3.65 days, and the semima

nataly862011 [7]

Answer:

If the mass of a star is greater than 3 solar masses, it will create a black hole. If its mass is less, it will create a neutron star.

Explanation:

If a star's gravity is high enough, when it condenses on itself, it will form a black hole. Otherwise, it will create a large amount of highly dense matter, such as a neutron star. It can be said that if the mass of a star is greater than 3 solar masses, it will create a black hole. If its mass is less, it will create a neutron star.

8 0

2 years ago

If the outer conductor of a coaxial cable has radius 2.6 mm , what should be the radius of the inner conductor so that the induc

mel-nik [20]

Answer:

Inner radius = 2 mm

Explanation:

In a coaxial cable, series inductance per unit length is given by the formula;

L' = (µ/(2π))•ln(R/r)

Where R is outer radius and r is inner radius.

We are given;

L' = 50 nH/m = 50 × 10^(-9) H/m

R = 2.6mm = 2.6 × 10^(-3) m

Meanwhile µ is magnetic constant and has a value of µ = µ_o = 4π × 10^(−7) H/m

Plugging in the relevant values, we have;

50 × 10^(-9) = (4π × 10^(−7))/(2π)) × ln(2.6 × 10^(-3)/r)

Rearranging, we have;

(50 × 10^(-9))/(2 × 10^(−7)) = ln((2.6 × 10^(-3))/r)

0.25 = ln((2.6 × 10^(-3))/r)

So,

e^(0.25) = (2.6 × 10^(-3))/r)

1.284 = (2.6 × 10^(-3))/r)

Cross multiply to give;

r = (2.6 × 10^(-3))/1.284)

r = 0.002 m or 2 mm

5 0

2 years ago

Racing greyhounds are capable of rounding corners at very high speeds. A typical greyhound track has turns that are 45-m-diamete

Margarita [4]

Explanation:

It is given that,

Diameter of the semicircle, d = 45 m

Radius of the semicircle, r = 22.5 m

Speed of greyhound, v = 15 m/s

The greyhound is moving under the action of centripetal acceleration. Its formula is given by :

$a=\dfrac{v^2}{r}$

$a=\dfrac{(15)^2}{22.5}$

$a=10\ m/s^2$

We know that, $g=9.8\ m/s^2$

$a=\dfrac{10\times g}{9.8}$

$a=1.02\ g$

Hence, this is the required solution.

5 0

2 years ago