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Delicious77 [7]
3 years ago
10

If you drop an object from rest, the distance it falls is given by (1/2)at2, where a is the acceleration of the object and t is

the amount of time fallen. Suppose you drop an object and measure the distance it falls. Suppose you drop it again, but it falls for three times the amount of time as before. How far did it fall on the second drop?
Select one:
a. It will fall twice the distance compared to the first time dropped.
b. It will fall three times the distance compared to the first time dropped.
c. It will fall nine times the distance compared to the first time dropped.
d. It depends on the planet you are on when you drop it.
Physics
1 answer:
klio [65]3 years ago
7 0

If,

d \:  =  \:   \frac{1}{2} a {t}^{2}

then, with 3x time t, (suppose, new distance is h)

h \:  =  \:   \frac{1}{2} a {(3t)}^{2}

=  \frac{1}{2} a9 {t}^{2}

= 9 \:  \frac{1}{2} a{t}^{2}

= 9d

Therefore, new distance h will be 9 times bigger than distance d.

answer: c

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harkovskaia [24]

Answer:

(a) The net force is 80.394 N

    The acceleration of the crate is 0.804 m/s²

(b) the final velocity of the crate is 5.02 m/s

Explanation:

Given;

mass of the crate, m = 100 kg

applied force, F = 250 N

angle of inclination, θ = 45°

coefficient of friction, μ = 0.12

Applied force in y-direction, F_y = Fsin \theta = 250sin45 = 176.78 \ N

Applied force in x-direction, F_x = Fcos \theta = 250cos45 = 176.78 \ N

The normal force is calculated as;

N + Fy -W = 0

N = W - Fy

N = (100 x 9.8) - 176.78

N = 980 - 176.78 = 803.22 N

The frictional force is given by;

Fk = μN

Fk = 0.12 x 803.22

Fk = 96.386 N

(a) The net force is given by;

F_{net} = F_x - F_k\\\\F_{net} = 176.78-96.386\\\\F_{net} = 80.394 \ N

Apply Newton's second law of  motion;

F = ma

a = \frac{F_{net}}{m}\\\\ a = \frac{80.394}{100}\\\\ a = 0.804 \ m/s^2

(b) the velocity of the crate after 5.0 s

F = ma= \frac{m(v-u)}{t} \\\\Ft =m(v-u)\\\\v-u = \frac{Ft}{m}\\\\ v = \frac{Ft}{m} + u\\\\v = \frac{F_{net}*t}{m} + u\\\\v = \frac{80.394*5}{100} + 1\\\\v = 5.02 \ m/s

7 0
3 years ago
A bag of sugar weighs 3.50 lb on Earth. What would it weigh in newtons on the Moon, where the free-fall acceleration isone-sixth
Alex73 [517]

Answer:

F_{Earth}= 15.57 N

F_{Moon}= 2.60 N

F_{Uranus}= 16.98 N

The mass of the bag is the same on the three planets. m=1.59 kg

Explanation:

The weight of the sugar bag on Earth is:

g=9.81 m/s²

m=3.50 lb=1.59 kg

F_{Earth}=m·g=1.59 kg×9.81 m/s²= 15.57 N

The weight of the sugar bag on the Moon is:

g=9.81 m/s²÷6= 1.635 m/s²

F_{Moon}=m·g=1.59 kg× 1.635 m/s²= 2.60 N

The weight of the sugar bag on the Uranus is:

g=9.81 m/s²×1.09=10.69 m/s²

F_{Uranus}=m·g=1.59 kg×10.69 m/s²= 16.98 N

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Answer:

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Given:

Time taken to make one oscillation, T = 0.25 s

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Frequency, f of oscillation is given as the reciprocal of time taken for one oscillation and is given by:

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The period of oscillation can be defined as the time taken by the suspended mass for completion of one oscillation.

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