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faltersainse [42]
2 years ago
12

Pls help in these 2 questions

Physics
1 answer:
Nadya [2.5K]2 years ago
4 0

Answer:

Taking forces along the plane

F cos θ - M g sin θ -100 = M a       net of forces along the plane

F = (M a + M g * .5 + 100) / .866     solving for F

F = (80 * 1.5 + 80 * 9.8 * .5 + 100) / .866 = 707 N

F = 707 N acting along the plane

Fn = F sin θ + M g cos θ       forces acting perpendicular to plane

Fn = 707 * 1/2 + 80 * 9.8 * .866 = 1030 Newtons   forces normal to plane

(this would give a coefficient of friction of 100 / 1030 = .097 = Fn)

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What can happen overnight to soil?
UkoKoshka [18]
(GABS) Overnight, all of the particles settled down to the bottom , and the larger particles were on the bottom and the smaller particles were on the top. Therefore, clay was on top, hummus was in the middle, and soil was on the bottom.
Particles dissolve is an unique way
7 0
3 years ago
During which stage of the cell cycle does the cell condense chromosomes and break down the nuclear envelope? Group of answer cho
Natalka [10]

Answer:

Prophase

Explanation:

During the Prophase

6 0
3 years ago
A tennis player swings her 1000 g racket with a speed of 10 m/s. She hits a 60 g tennis ball that was approaching her at a speed
Mashutka [201]

(a)

consider the motion of the tennis ball. lets assume the velocity of the tennis ball going towards the racket as positive and velocity of tennis ball going away from the racket as negative.

m = mass of the tennis ball = 60 g = 0.060 kg

v₀ = initial velocity of the tennis ball before being hit by racket = 20 m/s

v = final velocity of the tennis ball after being hit by racket = - 39 m/s

ΔP = change in momentum of the ball

change in momentum of the ball is given as

ΔP = m (v - v₀)

inserting the above values

ΔP = (0.060) (- 39 - 20)

ΔP = - 3.54 kgm/s

hence , magnitude of change in momentum : 3.54 kgm/s

7 0
3 years ago
65. The weight of a body when totally immersed in a liquid is 4.2N if he weight of the liquid displaced is 2.5N. Find the weight
Anna35 [415]

Answer:

Given, Apparent weight(W₂)=4.2N

          Weight of liquid displaced (u)=2.5N

          Let weight of body in air = W₁

Solution,

             U=W₁-W₂

              W₁=4.2=2.5=6.7N

∴Weight of body in air is 6.7N

5 0
2 years ago
A 6 cm object is 15 cm from a convex lens that has a focal length of 5 cm. What is the distance of the image from the lens, to t
Vesna [10]

Answer:

7.50 cm

Explanation:

The formula

1/v + 1/u = 1/f

Is used.

where.

u is the object distance.

v is the image distance.

f is the focal length of the lens.

1/v + 1/15 = 1/5

1/v = 1/5 - 1/15

1/v = (3-1)/15

1/v = 2/15

2v = 15

V = 15/2

V = 7.5 cm

For focal length, f in lens is always taken as negative for concave and positive for convex. ... And for image distance, V in lens it is taken as positive in Convex lens since image is formed on +X side. It is taken as negative in Concave lens since image is formed in -X side of the Cartesian.

4 0
3 years ago
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