In which direction will the object accelerate?

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o-na [289]

Answer:

<u>(B) wheat</u>

Explanation:

Wheat is an <u>anemophilous</u> plant (form of pollination by which wind distributes pollen)
They lack petals and sepals (outer whorls of the flower)
The female plants consist of an ovary (female reproductive part) from which two styles (connecting stigma and ovary) arise
The feathery stigma catches the pollen and sends it to the ovary for fertilization

6 0

2 years ago

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Force due to gravity. While JWST is in orbit, the Earth will be at a distance of 1.494 x 109 m from the telescope, and the Sun w

Alona [7]

Answer:

the gravitational force JWST will feel from the Sun is 37.785 Newton { leftward }

Explanation:

Given that;

distance of earth from the telescope r1 = 1.494 × 10⁹ m

and the Sun will be 1.49598 × 10¹¹ m further away

so r2 = r1 + 1.49598 × 10¹¹

r2 = 1.494 × 10⁹ m + 1.49598 × 10¹¹ m = 1.51092 × 10¹¹ m

mass of sun Ms = 1.9884 × 10³⁰ kg

mass of earth Me = 5.945× 10²⁴ kg

mass of JWST Mj = 6500 kg

What is the gravitational force JWST will feel from the Sun (strength and direction)?

the gravitational force of the sun will be attractive based on Newton law of gravitational force; so

Fjs = GMjMs / r2²

constant G = 6.674 × 10⁻¹¹ Nm²/kg²

Force on the JWST by the sun will be;

Fjs = GMjMs / r2² { leftward}

we substitute

Fjs = [(6.674 × 10⁻¹¹ Nm²/kg²)(6500 kg )(1.9884 × 10³⁰ kg)] / (1.51092 × 10¹¹ m)²

= (8.62587804 × 10²³) / ( 2.28287925 × 10²² )

= 37.785 Newton { leftward }

Therefore, the gravitational force JWST will feel from the Sun is 37.785 Newton { leftward }

7 0

3 years ago

The water flowing through a 2.0 cm (inside diameter) pipe flows out through three 1.3 cm pipes. (a) If the flow rates in the thr

Verdich [7]

Answer:

a)54L/min

b)0.845

Explanation:

a) A x V= $A_1V_1+ A_2V_2+A_3V_3$

where suffix 1,2,3 refers to the three pipes.

=27L/min+16L/min+11 L/min

=54L/min

b) A x V=54L/min => $\frac{\pi }{4} d^2$ x v

d= 2 cm

$\frac{\pi }{4} d^2$ x v = 54

v= $\frac{4}{\pi }$ x $\frac{54}{2^2}$

-> $A_1$ x $V_1$ =27L/min => $\frac{\pi }{4} d_1^2$ x $v_1$

$d_1$ = 1.3cm

$\frac{\pi }{4} d^2$ x $v_1$ = 27

$v_1$ = $\frac{4}{\pi }$ x $\frac{27}{1.3^2}$

Next is to find the ratio of speed i.e $\frac{v}{v_1}$

$\frac{4}{\pi }$ x $\frac{54}{2^2}$ / $\frac{4}{\pi }$ x $\frac{27}{1.3^2}$ => $\frac{54}{27}$ $\frac{1.3^2}{2^2}$

$\frac{v}{v_1}$ = 0.845

8 0

3 years ago

There is a seasaw that's holding two men. The seesaw has a length of 18m that can pivot from a point at its center. Man 1 has a

Alex

Answer:

Distance= 2.3864m

Explanation:

So that the balance is in equilibrium parallel to the floor, we must match the moment each man makes with respect to the pivot point.

In many cases the point of application of force does not coincide with the point of application in the body. In this case the force acts on the object and its structure at a certain distance, by means of an element that transfers that action of this force to the object.

This combination of force applied by the distance to the point of the structure where it is applied is called the moment of force F with respect to the point. The moment will attempt a rotation shift or rotation of the object. The distance from the force to the point of application is called the arm.

Mathematically it is calculated by expression:

M= F×d

The moment caused by the first man is:

M1= 75kg × (9.81m/s²) × 1.75m= 1287.5625 N×m

The moment caused by the second man must be equal to that caused by the first by which:

M2= 1287.5625 N×m= 55kg × (9.81m/s²) × distance ⇒

⇒distance= (1287.5625 N×m)/( (55kg × (9.81m/s²) )= 2.3864m

At this distance from the pivot point, the second should sit down so that the balance is balanced parallel to the ground.

3 0

3 years ago

A roller of radius 12.5 cm turns at 14 revolutions per second. What is the linear velocity of the roller in meters per second?

Firdavs [7]

12.5 times 14 and convert to meters its 1.75 meters per second

7 0

3 years ago

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