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3 years ago

Classify the following situations as involving balanced or unbalanced forces.

Physics

1 answer:

tester [92]3 years ago

5 0

Answer:

1) Unbalanced

2) Balance

3) Balanced

4) Unbalanced

5) Unbalanced

Explanation:

For 1 and 5, since the objects are not at a constant speed/velocity, their forces must be unbalanced.

For 2 and 3, their speeds are constant so that means the force is balanced. If the bycicle or box started accelerating, that would indicate an unbalanced force

For 4, the speed is constant, but it's direction is not indicating another unbalanced outside force causing it to turn. The force is unbalanced.

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Reading from left to right, describe the spacing between the dots in lines 1, 2, and 3.

Ierofanga [76]

Answer:

Line 1: the spacing is even the whole time

Line 2: the spacing increases over time

Line 3: the spacing decreases over time

hope this helps!

5 0

3 years ago

Force F=2.0N i - 3.0N k acts on a pebble with position vectorr=0.50m j - 2.0m k relative to the origin. In unit vector notation,

klasskru [66]

Answer with Explanation:

We are given that

Force acts on a pebble= $2\hat{i}-3\hat{k}$ N

Position vector= $r=0.5\hat{j}-2\hat{k}$ m

a.We have to find the resulting torque on the pebble about origin.

Torque= $r\times F$

Substitute the values then we get

$Torque= (0.5j-2k)\times (2i-3k)$

Torque= $-k-1.5i-4j$ N-m

By using $i\times j=k,j\times k=i,k\times i=j,j\times i=-k,k\times j=-i,i\times k=-j,i\times i=j\times j=k\times k=0$

b. $r=2i-3k$

$r-r_1=(0.5j-2k)-(2i-3k)=-2i+0.5j+k$

Torque about point (2,0,-3)

$\tau=(-2i+0.5j+k)\times (2i-3k)$

$\tau=-6j-k-1.5i+2j=-1.5i-4j-k$ N-m

6 0

3 years ago

Most landfills contain very little air and moisture, which many microorganisms need to survive. How do you think these condition

Orlov [11]

Since substances break down much faster under aerobic conditions, as oxygen helps break the molecules apart, in a process called oxidation, biodegradation that actually takes place does so slowly that effects may not be visible in decades after a product was buried, if even then.

7 0

4 years ago

1. A particle is projected vertically upwards with a velocity of 30 ms from a point 0. Find (a) the maximum height reached(b) th

Bogdan [553]

Assuming the particle is in free fall once it is shot up, its vertical velocity v at time t is

v = 30 m/s - g t

where g = 9.8 m/s² is the magnitude of the acceleration due to gravity, and its height y is given by

y = (30 m/s) t - 1/2 g t ²

(a) At its maximum height, the particle has 0 velocity, which occurs for

0 = 30 m/s - g t

t = (30 m/s) / g ≈ 3.06 s

at which point the particle's maximum height would be

y = (30 m/s) (3.06 s) - 1/2 g (3.06 s)² ≈ 45.9184 m ≈ 46 m

(b) It takes twice the time found in part (a) to return to 0 height, t ≈ 6.1 s.

(c) The particle falls 35 m below its starting point when

-35 m = (30 m/s) t - 1/2 g t ²

Solve for t to get a time of about t ≈ 7.1 s

7 0

3 years ago

Calculate the force, when a object acelerattes to 30 m/s^2 and have a mass of 40 kg

faust18 [17]

Hello!

Use the second law of Newton:

F = ma

Replacing:

F = 40 kg * 30 m/s^2

Resolving:

F = 1200 N

The force is 1200 Newtons.

3 0

3 years ago

Read 2 more answers