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Leto [7]
3 years ago
5

If we decrease the amount of force applied to an object, and all other factors remain the same, the amount of work completed wil

l
A decrease.
B increase.
C increase, then decrease.
D not change.
Physics
2 answers:
mina [271]3 years ago
6 0
The answer is A.) decrease.
Take pushing a box for example. You push your hardest but then give out, still trying to push. You're doing less work than what you started with.
Dominik [7]3 years ago
6 0

Answer:

Option A is the correct answer.

Explanation:

We know that the expression for work

          Work = Force x Displacement

So work is directly proportional to force.

Here force is decreasing, so work also decreases.

Option A is the correct answer.

You might be interested in
5. bending the electric wire a.change in size <br>b.change in shape <br>c.change in texture​
il63 [147K]

Answer:

change in shape

hope this will help you

8 0
3 years ago
Darren drives to school in rush hour traffic and averages 28 mph. He returns home in mid-afternoon when there is less traffic an
Gala2k [10]

Answer:

distance between school and home is 21 miles

Explanation:

given data

in rush hour speed  s1 = 28 mph

less traffic speed s2 = 42 mph

time t = 1 hr 15 min = 1.25 hr

to find out

distance  d

solution

we consider here distance home to school is d and t1 time to reach at school

we get here distance equation when we go home to school that is

distance = 28 × t1    .......................1

and when we go school to home distance will be

distance = 42 × ( t - t1 )

distance = 42 × ( 1.25 - t1 )     ...................2

so from equation 1 and 2

28 × t1 = 42 × ( 1.25 - t1 )

t1 = 0.75

so

from equation 1

distance = 28 × t1

distance = 28 × 0.75

distance = 21 miles

4 0
3 years ago
During a summer with little rainfall, your house on a hill slope experiences an interval during which your well runs dry. You ha
irina1246 [14]

Answer and Explanation:

You don't have water because of two possible reasons:

  1. Because of the summer and the little rain, the underwater supply goes low.
  2. The slope in the hill you live makes the underground water goes down by the effect of gravity. Imagine the underground water like a small tank, when the water is reduced for any reason the bottom of the tank will have the remaining water, while the top part will be "empty".

6 0
3 years ago
a liquid reactant is pumped through a horizontal, cylindrical, catalytic bed. The catalyst particles are spherical, 2mm in diame
natulia [17]

Answer:

The upper limit on the flow rate = 39.46 ft³/hr

Explanation:

Using Ergun Equation to calculate the pressure drop across packed bed;

we have:

\frac{\delta P}{L}= \frac{150 \mu_oU(1- \epsilon )^2}{d^2p \epsilon^3} + \frac{1.75 \rho U^2(1-\epsilon)}{dp \epsilon^3}

where;

L = length of the bed

\mu = viscosity

U = superficial velocity

\epsilon = void fraction

dp = equivalent spherical diameter of bed material (m)

\rho = liquid density (kg/m³)

However, since U ∝ Q and all parameters are constant ; we can write our equation to be :

ΔP = AQ + BQ²

where;

ΔP = pressure drop

Q = flow rate

Given that:

9.6 = A12 + B12²

Then

12A + 144B = 9.6       --------------   equation (1)

24A + 576B = 24.1    ---------------  equation (2)

Using elimination methos; from equation (1); we first multiply it by 2 and then subtract it from equation 2 afterwards ; So

288 B = 4.9

       B = 0.017014

From equation (1)

12A + 144B  = 9.6

12A + 144(0.017014) = 9.6

12 A = 9.6 - 144(0.017014)

A = \frac{9.6 -144(0.017014}{12}

A = 0.5958

Thus;

ΔP = AQ + BQ²

Given that ΔP = 50 psi

Then

50 = 0.5958 Q + 0.017014 Q²

Dividing by the smallest value and then rearranging to a form of quadratic equation; we have;

Q² + 35.02Q - 2938.8 = 0

Solving the quadratic equation and taking consideration of the positive value for the upper limit of the flow rate ;

Q = 39.46 ft³/hr

3 0
3 years ago
Momentum for a system can be conserved in one direction while not being conserved in another. What is the anglebetween the direc
katrin [286]

Answer:

90°

Explanation:

The angle will be 90° when momentum for a system can be conserved in one direction while not being conserved in another.

The example can be

If we apply force on an object horizontally in west direction, then as in other direction south or north we cannot apply the principal of momentum conservation.

5 0
3 years ago
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