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.
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
solar masses).
The most massive and luminous can be found in the center of cluster of galaxies.
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 = 14.38 lb
the magnitude of resultant force is equal to 14.38 lb
direction of forces


θ = 76.48°
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

Direction,
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Direction is 45° west of north.
Displacement is 565.69 m at 45° west of north
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 .