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iogann1982 [59]
3 years ago
15

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

Physics
1 answer:
seraphim [82]3 years ago
8 0

a) See free-body diagram in attachment

b) The acceleration is 2.46 m/s^2

Explanation:

a)

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

b)

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

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3 0
3 years ago
These are equal and opposite forces that do not cause a change in position or motion. True or False
Contact [7]

Answer:

TRUE

Explanation:

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This force is comprised of two most important properties namely the strength and direction. When any of the two forces is higher then it result in the motion of the object.

Thus, the above given statement is TRUE.

3 0
4 years ago
it took 3.5 hours for a train to travel the distance between two cities at a velocity of 120 miles per hour. How many miles lie
Art [367]

Given:

Time: 3.5 hrs

Velocity: 120 miles/hr

Now Distance=  Speed × Time

Now Velocity and speed have the same magnitude. Velocity being a vector quantity has a definite direction. Whereas speed is a scalar quantity,it indicates only the magnitude an doesn't define any direction.

Hence Distance = Velocity x time

Distance = 3.5 × 120 = 420 miles

7 0
3 years ago
What is the voltage across a semiconductor bar if the current through it is 0.17 A? The electron concentration in the bar is 2.7
Anastaziya [24]

Answer:

The voltage across a semiconductor bar is 0.068 V.

Explanation:

Given that,

Current = 0.17 A

Electron concentration n= 2.7\times10^{18}\ cm^{-3}

Electron mobility \mu=1000 cm^2/Vs

Length = 0.1 mm

Area = 500 μm²

We need to calculate the resistivity

Using formula of resistivity

\sigma=n\times q\times \mu

\rho=\dfrac{1}{\sigma}

Put the value into the formula

\rho=\dfrac{1}{2.7\times10^{18}\times10^{6}\times1.6\times10^{-19}\times1000\times10^{-4}}

\rho=2\ \mu \Omega m

We need to calculate the resistance

Using formula of resistance

R=\dfrac{\rho l}{A}

R=\dfrac{2\times10^{-6}\times0.1\times10^{-3}}{500\times(10^{-6})^2}

R=0.4\ \Omega

We need to calculate the voltage

Using formula of voltage

V= IR

Put the value into the formula

V=0.17\times0.4

V=0.068\ V

Hence, The voltage across a semiconductor bar is 0.068 V.

6 0
3 years ago
To calibrate the calorimeter electrically, a constant voltage of 3.6 V is applied and a current of 2.6 A flows for a period of 3
iren [92.7K]

Answer:

372.3 J/^{\circ}C

Explanation:

First of all, we need to calculate the total energy supplied to the calorimeter.

We know that:

V = 3.6 V is the voltage applied

I = 2.6 A is the current

So, the power delivered is

P=VI=(3.6)(2.6)=9.36 W

Then, this power is delivered for a time of

t = 350 s

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Finally, the change in temperature of an object is related to the energy supplied by

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E = 3276 J is the energy supplied

C is the heat capacity of the object

\Delta T =29.1^{\circ}-20.3^{\circ}=8.8^{\circ}C is the change in temperature

Solving for C, we find:

C=\frac{E}{\Delta T}=\frac{3276}{8.8}=372.3 J/^{\circ}C

5 0
3 years ago
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