How long will it take to go 150,000 m traveling at 50 km/hr

A box has a mass of 35kg.he pulls the rope horizontally with a force of 175 N. Find the horizontal acceleration of the box as it

Cloud [144]

Answer:

3.43 m/s^2

Explanation:

Force is equal to mass times acceleration. (F=ma). You can use inverse operations to get the formula for acceleration, which is acceleration is equal to force divided by mass. (a=F/m). Since there are two forces here, the force friction (55 N), and the force applied (175 N), we must solve for the net force. To solve for the net force, you take the applied force (175 N) and subtract the frictional force from it (55 N). Thus, the net force is 120 N. With this done, we can now solve for our acceleration.

Using the equation for acceleration, we take the force and divide it by mass.

120/35

Answer: 3.43* m/s^2**

*Note: This is rounded to the nearest hundredth, the full answer is: 3.42857143

**Note: In case you're confused, this is meters per second squared.

8 0

2 years ago

Which of these galaxies would you most likely find at the center of a large cluster of galaxies?

Arada [10]

Answer:

b. a large elliptical galaxy

Explanation:

In elliptical galaxies the stars are grouped in an elliptical shape, it has a low quantity of gas and dust in comparison to spiral galaxies, and its stars belong to an old population, there is not new stellar formation in it.

The stars orbit in a messy way which made to believe that they form from the merger of galaxies.

They are also really massive (around $10^{4}$ solar masses).

The most massive and luminous can be found in the center of cluster of galaxies.

4 0

2 years ago

Two forces F1 and F2 act on an object at a point P in the indicated directions. The magnitude of F1 is 15 lb and the magnitude o

adoni [48]

Answer

given,

F₁ = 15 lb

F₂ = 8 lb

θ₁ = 45°

θ₂ = 25°

Assuming the question's diagram is attached below.

now,

computing the horizontal component of the forces.

F_h = F₁ cos θ₁ - F₂ cos θ₂

F_h = 15 cos 45° - 8 cos 25°

F_h = 3.36 lb

now, vertical component of the forces

F_v = F₁ sin θ₁ + F₂ sin θ₂

F_v = 15 sin 45° + 8 sin 25°

F_v = 13.98 lb

resultant force would be equal to

$F = \sqrt{F_h^2+F_v^2}$

$F = \sqrt{3.36^2+13.98^2}$

F = 14.38 lb

the magnitude of resultant force is equal to 14.38 lb

direction of forces

$\theta =tan^{-1}(\dfrac{F_v}{F_h})$

$\theta =tan^{-1}(\dfrac{13.98}{3.36})$

θ = 76.48°

3 0

2 years ago

You are in downtown Chicago (where streets run N-S and E-W). You started from 600 N. Michigan Avenue, and walked 3 blocks toward

pickupchik [31]

Answer:

Displacement is 565.69 m at 45° west of north

Explanation:

Let north represent positive y axis and east represent positive x axis.

We have journey started from 600 N. Michigan Avenue, and walked 3 blocks toward north, 4 blocks toward west, and 1 block toward north to a train station.

3 blocks toward north = 300 j m

4 blocks toward west = -400 i m

1 blocks toward north = 100 j m

Total displacement = -400 i + 400 j m

Magnitude

$s=\sqrt{(-400)^2+400^2}=565.69m$

Direction,

$\theta =tan^{-1}\left ( \frac{400}{-400}\right )=135^0$

Direction is 45° west of north.

Displacement is 565.69 m at 45° west of north

4 0

2 years ago

A shot-putter projects the shot at 42.00˚ to the horizontal from a height of 2.100 m. It lands 17.00 m away horizontally. Next,

Tamiku [17]

Answer:

Explanation:

Let 100 m/s be the velocity of projection.

So horizontal component

= 100 cos42

= 74.31 m /s

Vertical component = - 100 sin 42 . in upward direction

66.91 m/s

Net displacement = 2.1 downwards ( + ve )

Using s = ut + 1/2 gt²

2.1 = - 66.91 t + .5 x 9.8 x t²

4.9 t² - 66.91 t - 2.1 = 0

t = 13.685 s

Horizontal distance covered

= 13.685 x 74.31

= 1016.93 m

If angle of projction is 40°

So horizontal component

= 100 cos40

= 76.60 m /s

Vertical component = - 100 sin 42 . in upward direction

64.27 m/s

Net displacement = 2.1 downwards ( + ve )

Using s = ut + 1/2 gt²

2.1 = -76.60 t + .5 x 9.8 x t²

4.9 t² - 76.60 t - 2.1 = 0

t = 15.659 s

Horizontal distance covered

= 15.659 x 76.60

= 1199.49 m

So horizontal range is increased , if angle of projection is increased .

8 0

2 years ago