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2 years ago

10

susan and joe push a table 18 meters down a hall by exerting a force of 200 newtons (n). what is the total amount of work done o

n the table? 1600 joules 1800 joules 2000 joules 3600 joules

1 answer:

prisoha [69]2 years ago

3 0

Answer:

3600joules

Explanation:

formula :W=FS

W=work done (J)

F=force (N)

S=displacement moved in the direction of force (m)

200N×18m

=3600J

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A 75-kilogram hockey player is skating across the ice at a speed of 6.0 meters per second. What is the magnitude of the average

saveliy_v [14]

Answer:

(3) 690 N

Explanation:

Force: This is the product of mass and acceleration of a body. The S.I unit of force is Newton(N).

The formula for force is given as,

F = ma..................... Equation 1

Where F = force, m = mass, a = acceleration.

Also,

a = (v-u)/t................... Equation 2

Where v = Final velocity, u = initial velocity, t = time.

Given: u = 6.0 m/s, v = 0 m/s (bring to stop), t = 0.65 s.

Substitute into equation 2

a = (0-6)/0.65

a = -6/0.65

a = -9.23 m/s²

Also given: m = 75 kg

Substitute again into equation 1

F = 75(-9.23)

F = -692.25 N

The negative sign tells that the force oppose the motion of the player

F ≈ 690 N

Hence the right option is (3) 690 N

3 0

4 years ago

On an alien planet, a simple pendulum of length 0.300 m has oscillation frequency 0.709 Hz. Find the period of the pendulum. Fin

sweet [91]

Answer:

1.41s

5.95m/s

0.2746m

Explanation:

The time period

T = 1/f

= 1/0.709s

= 1.41 seconds

We have

T = 2π√l/g

T² = 4π²l/g

g = 4π²l/T²

g = 4x3.14²x0.300/1.41²

g = 5.95m/s² this is the acceleration due to gravity.

Then the time period of the glide

T2 = 2π√m/k

Length of pendulum = l

Time period T

T2 = 2π√l/g

Then T1 = T2

2π√m/k = 2π√l/g

M/k = l/g

L = g.m/k

L = 5.95x0.450/9.75

L = 0.2746

This must be the length of the simple pendulum

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3 years ago

Which of the following devices would you expect to consume the most energy for each hour that it operates?

Nat2105 [25]

B. An electric space heater

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3 years ago

Use the diagram to answer each question.

valentina_108 [34]

Figure A shows cross section of a land form or rock. In Figure B, compression stress is applied on it. When compression stresses are applied on a rock, it squeezes the rock cause fold or fracture. The fault formed by compression stress is called thrust fault. If the compression stresses/ force continue to act on a rock it will converge and form thrust fault. In Figure C, tension stresses is applied on the rock. When a tension stress applied on a rock it deforms/ lengthen. There are three type of deformations occur due to tension stresses. One is elastic deformation, in which, rock retains it original shape when force/stresses are removed. Second is plastic deformation, in which rock lengthen and change occur permanently. Third type of deformation is result into fracture or breaking of rock. In Figure C, shear stresses are applied on rock. Shear stresses are applied with equal magnitude but in opposite direction. It cause breaking of rock.

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4 years ago

Read 2 more answers

A car starts from rest with an acceleration of 6

mrs_skeptik [129]

Answer:

$t=16.67s$

Explanation:

From the question, acceleration from $6m/s^2 \ to \ 0m/s^2$ in 10 seconds.

Acceleration function can be written as:-

$a_t=6-0.6t$

From the acceleration equation, we can obtain the velocity equation:

$v(t)=\int\limits^t_b {0} \, dv=\int\limits^t_0 {a(t)} \, dt\\v(t)=\int\limits^t_0 {(6-06t)} \, dt=6t-0.3t^2\\*dv=a(t)dt$

We calculate velocity after 10sec from the above v(t) as $30m/s$ .

To obtain distance travelled after 10 seconds:-

$S=\int\limits^{10}_0 {6t-0.3t^2} \, dt=|3t^2-0.1t^3| \ *lims(10,0)\\ S=200m$

Therefore 200m is for 10seconds, and next 200m at 30m/s

Total time= $10+\frac{200}{30}=16.67s$

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4 years ago