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Fudgin [204]
3 years ago
14

John drives a distance of 90 Km to the right at a steady speed of V1=25.00 m/s. Then he stops at the gas station for 10 minutes.

Then it returns and drives back ( to the left) for half an hour to the left at speed of V2 = 20.0 m/s.Speed =total distance/total timeVelocity=Displacement /total time1 km= 1000 m, 1 minute = 60 seconds, 1 hour= 60 minutesa. What is the difference between speed and Velocity?b. Calculate the total time of his round trip. c. Calculate the total distance.d. Calculate the total displacement.e. Calculate the average speed. f. Calculate the average velocity.
Physics
1 answer:
aleksley [76]3 years ago
4 0

a. Speed is defined as rate of change of distance per unit time whereas velocity is defined as rate of change of displacement per unit time.

b. t=6000\ s is the total time taken in the trip

c. d=126000\ m is the total distance

d. s=54000\ m towards right from the starting point.

e. v_a=21\ m.s^{-1}

f. \vec v_a=9\ m.s^{-1} towards right.

Explanation:

a.

Speed is a scalar quantity while velocity is a vector quantity.

Speed is defined as rate of change of distance per unit time whereas velocity is defined as rate of change of displacement per unit time.

Speed is a directionless quantity while velocity constitutes direction.

b.

<em>Total time of round trip when we're given:</em>

  • distance travelled to the right, d_r=90000\ m
  • speed while travelling to the right, v_r=25\ m.s^{-1}
  • time spent at gas station, t_g=600\ s
  • time spent while travelling back towards the left, t_l=30\times 60=1800\ s
  • speed while travelling to the left, v_{_l}=20\ m.s^{-1}

<em>Now time taken for travelling towards right:</em>

t_r=\frac{d_r}{v_r}

t_r=\frac{90000}{25}

t_r=3600\ s

<u>Therefore total time taken in the round trip:</u>

t=t_r+t_l+t_g

t=3600+600+1800

t=6000\ s

c.

<em>Now, distance travelled towards left:</em>

d_l=v_{_l}\times t_l

d_l=20\times1800

d_l=36000\ m

<u>Therefore total distance:</u>

d=d_l+d_r

d=36000+90000

d=126000\ m

d.

Now, total displacement:

s=d_r-d_l

s=90000-36000

s=54000\ m towards right from the starting point.

e.

<u>Average speed:</u>

v_a=\frac{d}{t}

v_a=\frac{126000}{6000}

v_a=21\ m.s^{-1}

f.

<u>Average velocity:</u>

\vec v_a=\frac{s}{t}

\vec v_a=\frac{54000}{6000}

\vec v_a=9\ m.s^{-1} towards right.

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anyanavicka [17]

Answer:

1. E x 4πr² = ( Q x r³) / ( R³ x ε₀ )

Explanation:

According to the problem, Q is the charge on the non conducting sphere of radius R. Let ρ be the volume charge density of the non conducting sphere.

As shown in the figure, let r be the radius of the sphere inside the bigger non conducting sphere. Hence, the charge on the sphere of radius r is :

Q₁ = ∫ ρ dV

Here dV is the volume element of sphere of radius r.

Q₁ = ρ x 4π x ∫ r² dr

The limit of integration is from 0 to r as r is less than R.

Q₁ = (4π x ρ x r³ )/3

But volume charge density, ρ = \frac{3Q}{4\pi R^{3} }

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Applying Gauss law of electrostatics ;

∫ E ds = Q₁/ε₀

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Substitute the value of Q₁ in the above equation. Hence,

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3 years ago
One horse is pulling a 755 kg sled straight ahead applying a force of 1988 N. If the acceleration of the sled is 1.36 m/s2, what
Inessa [10]

Answer:

The coefficient of kinetic friction is 0.13

Explanation:

Newton's second law states that the acceleration of an object is proportional to the net force on it, the factor of proportionality is the mass. So, we can express that law mathematically as:

\sum\overrightarrow{F}=m\overrightarrow{a} (1)

With F the net force, m the mass and a the acceleration of the object. In our case we're interested on what's happening to the sled, then we have to analyze the forces on it, those forces are the weight and the normal force on the vertical direction and the pulling force and frictional force in the horizontal direction. So, because (1) is a vector equation we can express that in their vertical (y) and horizontal (x) components:

F_y=ma_y (2)

F_x=ma_x (3)

On y we have that the acceleration is zero because the sled is not moving upward or downward, remember that the net force on y is the weight (W) pointing downward and the normal force pointing upward:

F_y=W+n=0

Following the convention that positive is upward and negative downward, W=mg=(755)(-9.81):

F_y=(755)(-9.81)+n=0

n=7406.55 N (4)

Now on the x direction we have the sum of the forces is the pulling force (T) and friction force (f)

F_x=F+f=ma_x

Choosing the direction where the horse is pulling F=1988N and the acceleration should be positive too, then:

1988+f=m(1.36)

f=(755)(1.36)-1988=-961.2 N

The negative sign means it's in the opposite direction the horse is pulling

The frictional force is related with the coefficient of kinetic friction in the next way:

|f|=\mu_k n

with μk the coefficient of kinetic friction, and n the normal force that we already found on (4), so we simply solve the last equation for μk:

\mu_k=\frac{|f|}{n}=\frac{961.2}{7406.55}=0.13

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