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S_A_V [24]
3 years ago
15

Through what vertical height is a 50 N object moved if 250 J of work is done lifting it against the gravitational field of Earth

?
Physics
1 answer:
zimovet [89]3 years ago
6 0

5m

Explanation:

Given parameters:

Weight of object = 50N

Work done in lifting object = 250J

Unknown:

Vertical height = ?

Solution:

The work done on an object is the force applied to lift a body in a specific direction.

   Work done = force x distance

  Weight is a force in the presence of gravity;

  Work done = weight x height of lifting

Height of lifting = \frac{work done }{weight}

 Height of lifting = \frac{250}{50} = 5m

The vertical height through which the object was lifted is 5m

learn more:

Work done brainly.com/question/9100769

#learnwithBrainly

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1. A 100-kg crate is pulled across a warehouse floor using a rope with a force of 250 N at an angle of 45o from the horizontal.
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
I have a question, it concerns hydrostatic buoyancy and Archimedes' law.
saw5 [17]

Answer: the lvl wud remain the same

Explanation: as per Archimedes Principle, the weight of the water displaced by the object is equal to the weight of the object. When the ship initially went into the pool, it wud hv displaced some water. When the anchor is dropped, the level does not change coz the anchor was already in the ship and no extra weight has been added, so the weight of the anchor has already been accounted for in the first place when the ship was first placed in the pool

4 0
3 years ago
TWO QUESTIONS!
solong [7]

The correct answer to the question is : 1) Converting energy and 2) Within about twenty years, natural gas forms.

EXPLANATION:

In coal fired power plants, the water is heated by burning coals. The heat produced in this process will convert water into steam. The steam is allowed to rotate the turbine. The turbine is attached to a generator which converts the kinetic energy of the steam into electric energy due to electromagnetic induction.

Hence, turbine helps in converting energy.

As per the second question, Walt listed the steps that occur during the formation of natural gas.

The natural gas is produced due to natural phenomenon. For this, millions of years are passed. So, it is a long range process in which the fossil of animals and planets buried beneath the surface are converted into natural gas under high pressure.

Hence, the mistake of Walt is  that the natural gas is produced withing twenty years.

7 0
3 years ago
Read 2 more answers
James threw a ball vertically upward with a velocity of 41.67ms-1 and after 2 second David threw a ball vertically upward with a
Reptile [31]

Answer:

When have passed 3.9[s], since James threw the ball.

Explanation:

First, we analyze the ball thrown by James and we will find the final height and velocity by the time two seconds have passed.

We'll use the kinematics equations to find these two unknowns.

y=y_{0} +v_{0} *t+\frac{1}{2} *g*t^{2} \\where:\\y= elevation [m]\\y_{0}=initial height [m]\\v_{0}= initial velocity [m/s] =41.67[m/s]\\t = time passed [s]\\g= gravity [m/s^2]=9.81[m/s^2]\\Now replacing:\\y=0+41.67 *(2)-\frac{1}{2} *(9.81)*(2)^{2} \\\\y=63.72[m]\\

Note: The sign for the gravity is minus because it is acting against the movement.

Now we can find the velocity after 2 seconds.

v_{f} =v_{o} +g*t\\replacing:\\v_{f} =41.67-(9.81)*(2)\\\\v_{f}=22.05[m/s]

Note: The sign for the gravity is minus because it is acting against the movement.

Now we can take these values calculated as initial values, taking into account that two seconds have already passed. In this way, we can find the time, through the equations of kinematics.

y=y_{o} +v_{o} *t-\frac{1}{2} *g*t^{2} \\y=63.72 +22.05 *t-\frac{1}{2} *(9.81)*t^{2} \\\\y=63.72 +22.05 *t-4.905*t^{2} \\

As we can see the equation is based on Time (t).

Now we can establish with the conditions of the ball launched by David a new equation for y (elevation) in function of t, then we match these equations and find time t

y=y_{o} +v_{o} *t+\frac{1}{2} *g*t^{2} \\where:\\v_{o} =55.56[m/s] = initial velocity\\y_{o} =0[m]\\now replacing\\63.72 +22.05 *t-(4.905)*t^{2} =0 +55.56 *t-(4.905)*t^{2} \\63.72 +22.05 *t =0 +55.56 *t\\63.72 = 33.51*t\\t=1.9[s]

Then the time when both balls are going to be the same height will be when 2 [s] plus 1.9 [s] have passed after David throws the ball.

Time = 2 + 1.9 = 3.9[s]

4 0
3 years ago
A Boeing 777 aircraft has a mass of 300,000 kg. At a certain instant during its landing, its speed is 27.0 m/s. If the braking f
GarryVolchara [31]

Answer:

Speed of the airplane 10.0 s later = 12.2 m/s

Explanation:

Mass of Boeing 777 aircraft = 300,000 kg

Braking force = 445,000 N

Deceleration

            a=\frac{445000}{300000}=1.48m/s^2

Initial velocity, u = 27 m/s

Time , t = 10 s

We have equation of motion, v =u +at

            v = 27 + (-1.48) x 10 = 27 - 14.8 = 12.2 m/s

Speed of the airplane 10.0 s later = 12.2 m/s

6 0
3 years ago
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