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Nina [5.8K]
3 years ago
8

An object is dropped and is in free fall. Each second, the position of the object is marked. The distance between each mark is m

easured. Which of the following is correct? A. As the object falls, the distance between marks decreases because the object's speed is constant. B. As the object falls, the distance between marks decreases because the object's speed is increasing. C. As the object falls, the distance between marks increases because the object's speed is increasing. D. As the object falls, the distance between marks increases because the object's speed is constant.
Physics
1 answer:
navik [9.2K]3 years ago
8 0
C. hope this helps :)
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Please help me with questions 1, 2 and 3. <br> i need a step by step explanation
kifflom [539]

Answer:

1) d

2) 5 m/s

3) 100

Explanation:

The equation of position x for a constant acceleration a and an initial velocity v₀, initial position x₀, time t is:

(i) x=\frac{1}{2}at^2+v_0t+x_0

The equation for velocity v and a constant acceleration a is:

(ii) v=at+v_0

1) Solve equation (ii) for acceleration a and plug the result in equation (i)

(iii) a = \frac{v -v_0}{t}

(iv) x = \frac{v-v_0}{2t}t^2+v_0t + x_0

Simplify equation (iv) and use the given values v = 0, x₀ = 0:

(v) x=-\frac{v_0}{2}t + v_0t= \frac{v_0}{2}t

2) Given v₀= 3m/s, a=0.2m/s², t=10 s. Using equation (ii) to get the final velocity v:v=at+v_0=0.2\frac{m}{s^2} * 10s+3\frac{m}{s}=2\frac{m}{s}+3\frac{m}{s}=5\frac{m}{s}

3) Given v₀=0m/s, t₁=10s, t₂=1s and x₀=0. Looking for factor f = x(t₁)/x(t₂) using equation(i) to calculate x(t₁) and x(t₂):

f=\frac{x(t_1)}{x(t_2)}=\frac{\frac{1}{2}at_1^2 }{\frac{1}{2}at_2^2}=\frac{t_1^2}{t_2^2}=\frac{10^2}{1^2}=\frac{100}{1}

5 0
3 years ago
Mark and David are loading identical cement blocks onto David’s pickup truck. Mark lifts his block straight up from the ground t
Pepsi [2]

Answer:

b) true. The jobs are equal

Explanation:

The work on a body is the scalar product of the force applied by the distance traveled.

    W = F. d

Work is a scalar, the work equation can be developed

    W = F d cos θ

Where θ is the angle between force and displacement

Let's apply these conditions to the exercise

a) False, if we see the expression d cosT is the projection of the displacement in the direction of the force, so there may be several displacement, but its projection is always the same

b) true. The jobs are equal dx = d cosθ

c) False, because the force is equal and the projection of displacement is the same

d) False, knowledge of T is not necessary because the projection of displacement is always the same

e) False mass is not in the definition of work

5 0
3 years ago
Can someone answer this please.?? I really need help
svetlana [45]

Answer:

1. is the age group 35 and 44

2. is 2006 i think its 2006 i cant really tell in the picture but its the one before the last one!

6 0
2 years ago
The third floor of a house is 8m above street level. How much work is needed to move a 136kg refrigerator to the third floor?
jonny [76]

m = Mass of the refrigerator to be moved to third floor = 136 kg

g = Acceleration due to gravity by earth on the refrigerator being moved = 9.8 m/s²

h = Height to which the refrigerator is moved  = 8 m

W = Work done in lifting the object

Work done in lifting the object is same as the gravitational potential energy gained by the refrigerator. hence

Work done = Gravitation potential energy of refrigerator

W = m g h

inserting the values

W = (136) (9.8) (8)

W = 10662.4 J



8 0
2 years ago
A dog of mass 4 kg runs up a hill of height 8 m. How much gravitational potential energy does the dog gain?
Genrish500 [490]
A. 314 because when you use the formula for the GPE ; GPE=MGH or means mass times gravity time height (4x8x9.8) and thats equivalent to 313.6 which rounds up to 314. Hope it helps 
8 0
3 years ago
Read 2 more answers
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