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3 years ago

An object is dropped from a 45 m high building. At the same time, another object is thrown

Physics

1 answer:

Dafna11 [192]3 years ago

4 0

Answer:

-92.33 (meaning the objects will not meet above the ground).

Explanation:

We can use the kinematic equation <em>displacement = initial velocity*time + 1/2*acceleration*time^2.</em>

We can plug in the known values of the 2 objects into the equation, where t is the time and x is the displacement:

x = 0*t + 1/2*(-9.8)*t^2+45

x = 8.5*t + 1/2*(-9.8)*t^2

We need to first solve for t to solve for x. Since both equations are equal to x, we can set them equal to each other and solve for t:

0*t + 1/2*(-9.8)*t^2+45 = 8.5*t + 1/2*(-9.8)*t^2

-4.9*t^2 +45 = 8.5*t + -4.9*t^2

45 = 8.5*t

t = 45/8.5 ≈5.294

Now, we can plug t as 5.294 into any of the equations above to solve for x:

x = 0*5.294 + 1/2*-9.8*(5.294)^2+45 ≈ -92.33

That means, the objects will not meet above the ground.

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An object of mass 2kg is on an incline where there is an applied force of 15N. The

seraphim [82]

a) See free-body diagram in attachment

b) The acceleration is $2.46 m/s^2$

Explanation:

The free-body diagram of an object is a diagram representing all the forces acting on the object. Each force is represented by a vector of length proportional to the magnitude of the force, pointing in the same direction as the force.

The free-body diagram for this object is shown in the figure in attachment.

There are three forces acting on the object:

The weight of the object, labelled as $mg$ (where m is the mass of the object and g is the acceleration of gravity), acting downward
The applied force, $F_a$ , acting up along the plane
The force of friction, $F_f$ , acting down along the plane

In order to find the acceleration of the object, we need to write the equation of the forces acting along the direction parallel to the incline. We have:

$F_a - F_f - mg sin \theta = ma$

where:

$F_a = 15 N$ is the applied force, pushing forward

$F_f = 5 N$ is the frictional force, acting backward

$mg sin \theta$ is the component of the weight parallel to the incline, acting backward, where

m = 2 kg is the mass of the object

$g=9.8 m/s^2$ is the acceleration of gravity

$\theta=15^{\circ}$ is the angle between the horizontal and the incline (it is not given in the problem, so I assumed this value)

a is the acceleration

Solving for a, we find:

$a=\frac{F_a - F_f - mg sin \theta}{m}=\frac{15-5-(2)(9.8)(sin 15^{\circ})}{2}=2.46 m/s^2$

Learn more about inclined planes:

brainly.com/question/5884009

#LearnwithBrainly

8 0

3 years ago

Why are citizens obligated to respond to such documents?

Zina [86]

To protect the constitutional right to confront ones accused

5 0

3 years ago

The distance from Abdullah's house to his school is 2.4km. Abdulla takes 0.6h to go to school on his cycle but takes only 0.4h t

vladimir1956 [14]

Answer:

The average speed can be calculated as the quotient between the distance travelled and the time needed to travel that distance.

To go to the school, he travels 2.4 km in 0.6 hours, then here the average speed is:

s = (2.4km)/(0.6 hours) = 4 km/h

To return to his home, he travels 2.4km again, this time in only 0.4 hours, then here the average speed is:

s' = (2.4 km)/(0.4 hours) = 6 km/h.

Now, if we want the total average speed (of going and returning) we have that the total distance traveled is two times the distance between his home and school, and the total time is 0.6 hours plus 0.4 hours, then the average speed is:

S = (2*2.4 km)/(0.6 hours + 0.4 hours)

S = (4.8km)/(1 h) = 4.8 km/h

5 0

2 years ago

When the voltage is at a maximum positive value, the current is at a value that is_________.

alukav5142 [94]

Answer:

When the voltage is at a maximum positive value, the the current is at a value that is maximum and positive

Explanation:

We know that the relation between the Voltage and the current is given using the Ohm's law, which states that the voltage (V) is directly proportional to the current (I)

Mathematically,

V ∝ I

Hence,

When the voltage is at a maximum positive value, the the current is at a value that is maximum and positive

4 0

3 years ago

If the primary coil of a transformer has 200 turns and is supplied with 120 v ac power, how many turns must the secondary coil h

Oduvanchick [21]

The ratio of the turns to the voltage should be equal
i.e: 200/120 = t/12
so the secondary coil should have 20 turns

5 0

3 years ago