Two point charges are located on the y axis as follows: charge q1 = -2.30 nC at y1 = -0.600 m , and charge q2 = 2.80 nC at the origin (y = 0). What is the magnitude of the net force exerted by these two charges on a third charge q3 = 7.50 nC located at y3 = -0.300 m ?

Explanation:

Given that,

Initial speed of the sports car, u = 80 km/h = 22.22 m/s

Final speed of the runner, v = 0

Distance covered by the sports car, d = 80 km = 80000 m

Let a is the acceleration of the sports car.  It can be calculated using third equation of motion as :

[tex]v^2-u^2=2ad[/tex]

[tex]a=\dfrac{v^2-u^2}{2d}[/tex]

[tex]a=\dfrac{0-(22.22)^2}{2\times 80000}[/tex]

[tex]a=-0.00308\ m/s^2[/tex]

Value of g, [tex]g=9.8\ m/s^2[/tex]

[tex]a=\dfrac{-0.00308}{9.8}\ m/s^2[/tex]

[tex]a=(-0.000314)\ g\ m/s^2[/tex]

Hence, this is required solution.

Answer:

10 years

Explanation:

As you can understand from the question it is given that the planet is already filled to half of its capacity. Also the population doubles in 10 years. To fill up the planet completely the population needs to double only once. To do that only 10 years are required.

As it is mentioned there are no other factors affecting the growth rate, in 10years the planet will be filled to its carrying capacity.

Answer:[tex]0.0686 m^3[/tex]

Explanation:

Given

No of moles=2

[tex]T=25^{\circ}\approx 298 K[/tex]

[tex]P_1=210 KPa[/tex]

[tex]T_2=48 ^{\circ}\approx 321 k[/tex]

[tex]P_2=77.7 KPa[/tex]

PV=nRT

Substitute

[tex]77.7\times V_2=2\times 8.314\times 321[/tex]

[tex]V_2=0.0686 m^3[/tex]

Answer:

Magnitude of resultant force is  1190.314 N

Direction of force is 13.352°

Explanation:

given data:

thrust force = 725 N

Angle = 32.4 degree

Let x is consider as positive direction

Resultant force in x direction is  

Rx = 725 + 513cos32.4 = 1158.14 N

and Resultant force perpendicular to x direction is:

Ry = 513sin32.4 = 274.88 N

Magnitude of resultant force is

[tex]R=\sqrt{R_x^2+R_y^2} = 1190.314N[/tex]

and resultant force direction is  

[tex]\theta=tan^{-1}\frac{R_y}{R_x} = 13.352\degree[/tex]

Answer:0.835 s

Explanation:

Given

Red ball initial velocity([tex]u_r[/tex])=1.1 m/s

height of building(h)=28 m

after 0.5 sec blue ball is thrown with a velocity([tex]u_b[/tex])=24.3 m/s

Height of blue after 2 sec red ball is thrown

i.e. height of blue ball at t=1.5 sec after blue ball is thrown upward

[tex]h=24.3\times 1.5-\frac{9.81\times 1.5^2}{2}=25.414 m[/tex]

therefore blue ball is at height of 25.414+0.9=26.314 from ground

moment after the two ball is at same height

for red ball

[tex]14=1.1\times \left ( t+0.5\right )+\frac{9.81\times \left ( t+0.5\right )^2}{2}-----1[/tex]

for blue ball

[tex]13.1=24.3\times t-\frac{9.81\times t^2}{2}-----2[/tex]

add 1 & 2

we get

[tex]27.1=1.1t+0.55+24.3t+\frac{g\left ( t+0.25\right )}{2}[/tex]

27.1=25.4t+0.55+4.905t+1.226

t=0.835 s

Answer:

Explanation:

The weight w its:

[tex]w \ = \ m \ a[/tex]

where m is the mass and a is the acceleration.

Local acceleration of gravity = [tex]9.779 \frac{m}{s^2}[/tex]

weight :

[tex]w = \ 80\ kg \ * \ 9.779 \frac{m}{s^2}[/tex]

[tex]w = \ 782.32 \ N [/tex]

Local acceleration of gravity = [tex]9.796 \frac{m}{s^2}[/tex]

weight :

[tex]w = \ 80\ kg \ * \ 9.796 \frac{m}{s^2}[/tex]

[tex]w = \ 783.68 \ N [/tex]

Local acceleration of gravity = [tex]9.798 \frac{m}{s^2}[/tex]

weight :

[tex]w = \ 80\ kg \ * \ 9.798 \frac{m}{s^2}[/tex]

[tex]w = \ 783.84 \ N [/tex]

Local acceleration of gravity = [tex]9.803 \frac{m}{s^2}[/tex]

weight :

[tex]w = \ 80\ kg \ * \ 9.803 \frac{m}{s^2}[/tex]

[tex]w = \ 784.24 \ N [/tex]

Local acceleration of gravity = [tex]9.815 \frac{m}{s^2}[/tex]

weight :

[tex]w = \ 80\ kg \ * \ 9.815 \frac{m}{s^2}[/tex]

[tex]w = \ 785.2 \ N [/tex]

Answer

mass of a person is 80 kg

to calculate weight of the person = ?

a) Mexico City, Mexico

acceleration due to gravity in mexico = 9.779 m/s²

   weight = 80× 9.779 = 782.32 N

b) Cape Town, South Africa

acceleration due to gravity in cape town = 9.796 m/s²

   weight = 80× 9.796 = 783.68 N

c) Tokyo, Japan

acceleration due to gravity in Tokyo = 9.798 m/s²

   weight = 80× 9.798 = 783.84 N

d) Chicago, IL

acceleration due to gravity in Chicago = 9.80 m/s²

   weight = 80× 9.80 = 784 N

e) Copenhagen, Denmark

acceleration due to gravity in Copenhagen = 9.81 m/s²

   weight = 80× 9.81 = 784.8 N

Answer:

Explanation:

This problem can be solved easily if we represent velocity in the form of vector.

The velocity of 351 was towards easterly direction so

V₁ = 351 i

The velocity of 351 was towards south west making - 48° with east or + ve x direction.

V₂ = 351 Cos 48 i - 351 sin 48 j

V₂ = 234.86 i - 260.84 j

Change in velocity

= V₂ - V₁ = 234.86 i - 260.84 j - 351 i

= -116.14 i - 260.84 j

acceleration

= change in velocity / time

(-116.14 i - 260.84 j )/ 1

= -116.14 i - 260.84 j

magnitude = 285.53 ms⁻²

Direction

Tan θ = 260.84 / 116.14 = 2.246

θ = 66 degree south of west .

Answer:

Part a)

[tex]T = 59.5 s[/tex]

Part b)

v = 22.8 m/s

Explanation:

Part a)

Time taken by the taxi to reach the maximum speed is given as

[tex]v_f - v_i = at[/tex]

[tex]27 - 0 = 2(t)[/tex]

[tex]t = 13.5 s[/tex]

now it moves with constant speed for next 41 s and then finally comes to rest in next 5 s

so total time for which it will move is given as

[tex]T = 13.5 s + 41 s + 5 s[/tex]

[tex]T = 59.5 s[/tex]

Part b)

Distance covered by the taxi while it accelerate to its maximum speed is given as

[tex]d_1 = \frac{v_f + v_i}{2} t[/tex]

[tex]d_1 = \frac{27 m/s + 0}{2} (13.5)[/tex]

[tex]d_1 = 182.25[/tex]

Now it moves for constant speed for next 41 s so distance moved is given as

[tex]d_2 = (27)(41)[/tex]

[tex]d_2 = 1107 m[/tex]

Finally it comes to rest in next 5 s so distance moved in next 5 s

[tex]d_3 = \frac{v_f + v_i}{2} t[/tex]

[tex]d_3 = \frac{27 + 0}{2}(5)[/tex]

[tex]d_3 = 67.5 m[/tex]

so here we have total distance moved by it is given as

[tex]d = 182.25 + 1107 + 67.5 [/tex]

[tex]d = 1356.75 m[/tex]

average speed is given as

[tex]v = \frac{1356.75}{59.5}[/tex]

[tex]v = 22.8 m/s[/tex]

Final answer:

The taxi is in motion for 59.5 seconds and its average velocity is 19.38 m/s.

Explanation:

(a) To find the total time the taxi is in motion, we need to calculate the time it takes to accelerate to 27.0 m/s, the time it travels at constant speed, and the time it takes to stop. The time to accelerate is given by the equation:

t = (vf - vi) / a = (27.0 m/s - 0 m/s) / 2.00 m/s² = 13.5 s

The time traveling at constant speed is given as 41.0 s, and the time to stop is given as 5.00 s. Therefore, the total time the taxi is in motion is:

Total time = Time to accelerate + Time at constant speed + Time to stop = 13.5 s + 41.0 s + 5.00 s = 59.5 s

(b) The average velocity is given by the total displacement divided by the total time. The displacement during acceleration is given by:

Displacement = vi × t + 0.5 × a × t² = 0 m/s × 13.5 s + 0.5 × 2.00 m/s² × (13.5 s)² = 182.25 m

The displacement during constant speed is equal to the constant speed multiplied by the time:

Displacement = constant speed × time = 27.0 m/s × 41.0 s = 1,107 m

The displacement during deceleration is given by:

Displacement = vf × t + 0.5 × (-a) × t² = 27.0 m/s × 5.00 s + 0.5 × (-2.00 m/s²) × (5.00 s)² = -137.5 m

Therefore, the total displacement is 182.25 m + 1,107 m - 137.5 m = 1,152.75 m.

The average velocity is then:

Average velocity = Total displacement / Total time = 1,152.75 m / 59.5 s = 19.38 m/s

Answer:

51.94 ft/s²

257.63 ft/s

Explanation:

t = Time taken = 4 s

u = Initial velocity = 34 mi/h

v = Final velocity

s = Displacement = 615 ft

a = Acceleration

Converting velocity to ft/s

[tex]34\ mi/h=\frac{34\times 5280}{3600}=49.87\ ft/s[/tex]

Equation of motion

[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow a=2\frac{s-ut}{t^2}\\\Rightarrow a=2\left(\frac{615-49.87\times 4}{4^2}\right)\\\Rightarrow a=51.94\ ft/s^2[/tex]

Acceleration is 51.94 ft/s²

[tex]v=u+at\\\Rightarrow v=49.87+51.94\times 4\\\Rightarrow v=257.63\ ft/s[/tex]

Final velocity at this time is 257.63 ft/s

Answer:

The blue boat traveled 6.1 km farther west than the green boat

The green boat traveled 10.7 km farther south than the blue boat

Explanation:

The equation for linear uniform speed movement is

X(t) = X0 + v * t

Since we have two coordinates (X, Y) we use

X(t) = X0 + vx * t

Y(t) = Y0 + vy * t

The dock will be the origin of coordinates so X0 and Y0 will be zero. The X axis will be pointing west and the Y axis south.

The blue boat moves with a direction 24° south of west, so it will have speeds:

vxb = 104 * cos(24) = 95 km/h

vyb = 104 * sin(24) = 42.3 km/h

And the green boat:

vxg = 104 * cos(37.7) = 82.3 km/h

vyg = 104 * sin(37.7) = 63.6 km/h

After half hour the boats will have arrived at positions

Xb = 95 * 0.5 = 47.5 km

Yb = 42.3 * 0.5 = 21.1 km

And

Xg = 82.3 * 0.5 = 41.4 km

Yg = 63.6 * 0.5 = 31.8 km

The difference in positions of the boats

47.5 - 41.4 = 6.1 km

31.8 - 21.1 = 10.7 km

Answer:

The height of the cliff is 39.655 m

Given:

Height at which the arrow was shot, h = 1.3 m

Velocity of the arrow, u = 34 m/s

Angle, [tex]\theta' = 58.1^{\circ}[/tex]

Time of the fight, t = 4.1 s

Solution:

Let the Height of the cliff be H

Since, the motion of the object is projectile motion and the direction of motion  is vertical at some angle

Therefore, we consider the vertical component of velocity, [tex]u_{y} = usin\theta[/tex].

Now,

The Height of the cliff is given by applying the second equation of motion in the projectile:

Thus

[tex]s = u_{y}t - \frac{1}{2}gt^{2}[/tex]

[tex]s = 34sin60^{\circ}\times 4.1 - \frac{1}{2}\times 9.8\times 4.1^{2}[/tex]

s = 38.355 m

Now, the height of the cliff, H:

H = s + h = 38.355 + 1.3 = 39.655 m

Answer:

[tex]C_{eq}=1.97\ \mu F[/tex]

Explanation:

Given that,

Capacitance 1, [tex]C_1=0.5\ \mu F[/tex]

Capacitance 2, [tex]C_2=11\ \mu F[/tex]

Capacitance 3, [tex]C_3=1.5\ \mu F[/tex]

C₁ and C₂ are connected in series. Their equivalent is given by :

[tex]\dfrac{1}{C'}=\dfrac{1}{C_1}+\dfrac{1}{C_2}[/tex]

[tex]\dfrac{1}{C'}=\dfrac{1}{0.5}+\dfrac{1}{11}[/tex]

[tex]C'=0.47\ \mu F[/tex]

Now C' and C₃ are connected in parallel. So, the final equivalent capacitance is given by :

[tex]C_{eq}=C'+C_3[/tex]

[tex]C_{eq}=0.47+1.5[/tex]

[tex]C_{eq}=1.97\ \mu F[/tex]

So, the equivalent capacitance of the combination is 1.97 micro farad. Hence, this is the required solution.

Answer with Explanation:

The general wave equation is given by

[tex]\frac{\partial ^2u}{\partial x^2}+\frac{\partial ^2u}{\partial y^2}=\frac{1}{c^2}\frac{\partial ^2u}{\partial t^2}[/tex]

where

'c' is the velocity of the wave

Comparing with the given equation

[tex]\frac{\partial ^2h}{\partial x^2}+\frac{\partial ^2h}{\partial y^2}=\frac{1}{gd}\frac{\partial ^2h}{\partial t^2}[/tex]

We can see that

[tex]c^2=gd\\\\\therefore c=\sqrt{gd}[/tex]

Thus the velocity of  wave is given by [tex]v=\sqrt{gd}[/tex]

Answer:

The equivalent capacitance will be [tex]15\mu F[/tex]  

Explanation:

We have given two capacitance [tex]C_=10\mu F\ and\ C_2=20\mu F[/tex]

They are connected in parallel

So equivalent capacitance [tex]C=C_1+C_2=10+20=30\mu F[/tex]

This equivalent capacitance is now connected in series with [tex]30\mu F[/tex]

In series combination of capacitors the equivalent capacitance is given by [tex]\frac{1}{C}=\frac{1}{30}+\frac{1}{30}[/tex]

[tex]C=\frac{30}{2}=15\mu F[/tex]

So the equivalent capacitance will be [tex]15\mu F[/tex]  

The equivalent capacitance of the arrangement where capacitors of 10 uF and 20 uF are connected in parallel, and in series with a 30 uF capacitor, is 15 uF.

To determine the equivalent capacitance of a combination in which capacitors are connected both in parallel and in series, one must consider the rules for each type of connection. For capacitors connected in parallel, the equivalent capacitance is simply the sum of their individual capacitances. When capacitors are connected in series, the reciprocal of the equivalent capacitance is the sum of the reciprocals of the individual capacitances. In your case, the 10 uF and 20 uF capacitors are connected in parallel, so their equivalent capacitance is 10 uF + 20 uF = 30 uF. This combined capacitance is then in series with the 30 uF capacitor.

The next step is to calculate the equivalent capacitance for the series connection using the formula:

1/Cseries = 1/C1 + 1/C2

Where C1 is the equivalent parallel capacitance (30 uF) and C2 is the capacitance of the third capacitor (30 uF). Substituting the values:

1/Cseries = 1/30 uF + 1/30 uF = 1/15 uF

So, Cseries = 15 uF. This is the total equivalent capacitance of the combination of capacitors.

The coefficient of friction in this scenario is calculated as the ratio of the frictional force to the normal force. After calculating the frictional force to be 50N and the normal force to be 300N, the coefficient of friction is 0.167.

Dana is pulling a box on a level surface with a rope that makes an angle of 30 degrees. The box has a weight of 300 N and the tension in the rope is 100 N. We first must understand that the force of friction is balancing the force that Dana is applying to move the box.

The vertical component of the tension (Tsin30) should be equal to the force of friction since they are balancing each other. Therefore, Tsin30 = Friction. Substituting the given values (100N*sin30 = Friction), we can calculate friction as 50 N.

The coefficient of friction can be calculated as the ratio of the frictional force to the normal force. As there is no vertical acceleration, the normal force is equal to the weight of the box which is 300N. Therefore, the coefficient of friction (mu) equals friction/normal force is equal to 50N/300N which is 0.167.

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The coefficient of friction is approximately 0.247.

The coefficient of friction between the box and the surface is given by the ratio of the frictional force to the normal force. The frictional force is the horizontal component of the tension in the rope, and the normal force is the vertical component of the tension in the rope plus the weight of the box acting downward.

First, let's calculate the horizontal and vertical components of the tension in the rope. The horizontal component can be found using the cosine of the angle, and the vertical component can be found using the sine of the angle:

[tex]\[ T_{\text{horizontal}} = T \cdot \cos(\theta) \] \[ T_{\text{vertical}} = T \cdot \sin(\theta) \][/tex]

Given that the tension in the rope (T) is 100 N and the angle [tex](\(\theta\))[/tex] is 30 degrees, we have:

[tex]\[ T_{\text{horizontal}} = 100 \text{ N} \cdot \cos(30^\circ) \] \\T_{\text{vertical}} = 100 \text{ N} \cdot \sin(30^\circ) \][/tex]

Using the values of cosine and sine for 30 degrees:

[tex]\[ T_{\text{horizontal}} = 100 \text{ N} \cdot \frac{\sqrt{3}}{2} \approx 86.6 \text{ N} \]\\T_{\text{vertical}} = 100 \text{ N} \cdot \frac{1}{2} = 50 \text{ N} \][/tex]

The normal force (N) is the sum of the weight of the box (W) and the vertical component of the tension:

[tex]\[ N = W + T_{\text{vertical}} \] \[ N = 300 \text{ N} + 50 \text{ N} \] \[ N = 350 \text{ N} \][/tex]

The frictional force (f) is equal to the horizontal component of the tension since the box is moving at constant velocity (implying that the net force in the horizontal direction is zero):

[tex]\[ f = T_{\text{horizontal}} \] \[ f = 86.6 \text{ N} \][/tex]

Now, the coefficient of friction is the ratio of the frictional force to the normal force:

[tex]\[ \mu = \frac{f}{N} \] \[ \mu = \frac{86.6 \text{ N}}{350 \text{ N}} \] \[ \mu \approx 0.247 \][/tex]

Answer:

The answer is: 0.00563 mg = 5.63 μg

Explanation:

Significant figures of a number are the figures or digits that carry meaning and contributes to the precision of the given number.

The given quantity 0.00563 mg, has three significant figures as all the leading zeros are not significant.

Now to represent 0.00563 mg between 0.1 and 1000, the given unit milligram (mg) is converted to microgram (μg).

As, 1 mg = 1000 mg

Therefore, 0.00563 mg = 5.63 μg, has three significant figures as non-zeros figures are significant.

Therefore, 5.63 μg lies between between 0.1 and 1000 and has significant figures.

The magnitude of the tension on the ends of the clothesline is approximately [tex]\( 41.99 \, \text{N} \).[/tex]

To find the tension on the ends of the clothesline, we can use the principle of equilibrium. When the mass is at rest in the middle of the clothesline, the tension in the clothesline must balance the gravitational force acting on the mass.

Let's denote:

- [tex]\( T \)[/tex] as the tension in the clothesline.

- [tex]\( m \)[/tex] as the mass tied to the clothesline.

- [tex]\( g \)[/tex] as the gravitational acceleration.

The gravitational force acting on the mass is given by [tex]\( F_{\text{gravity}} = m \cdot g \).[/tex]

When the mass is at rest in the middle of the clothesline, the horizontal components of the tensions on either side cancel each other out, leaving only the vertical components to support the weight.

Since the clothesline sags a distance of 3 meters in the middle, each side of the clothesline forms a right triangle with the sagging distance as one leg and half of the distance between the poles as the other leg. Therefore, each side of the clothesline forms a right triangle with legs of 8 meters and 3 meters.

Using the Pythagorean theorem, we can find the length of the clothesline (the hypotenuse of each triangle), which is the same as the length of each segment between the midpoint and the poles:

[tex]\[ L = \sqrt{(8 \, \text{m})^2 + (3 \, \text{m})^2} \]\[ L = \sqrt{64 + 9} \]\[ L = \sqrt{73} \][/tex]

Now, we can set up an equation for the vertical equilibrium:

[tex]\[ 2T \sin(\theta) = F_{\text{gravity}} \][/tex]

Where [tex]\( \theta \)[/tex] is the angle the clothesline makes with the horizontal.

We can find [tex]\( \sin(\theta) \)[/tex] using trigonometry:

[tex]\[ \sin(\theta) = \frac{3}{L} \][/tex]

Now, substitute this expression for [tex]\( \sin(\theta) \)[/tex] into the equation for vertical equilibrium:

[tex]\[ 2T \left(\frac{3}{L}\right) = m \cdot g \]\[ 2T \left(\frac{3}{\sqrt{73}}\right) = 3 \times 9.8 \]\[ T = \frac{3 \times 9.8 \times \sqrt{73}}{2 \times 3} \]\[ T = \frac{9.8 \times \sqrt{73}}{2} \][/tex]

Now, we can calculate the tension:

[tex]\[ T \approx \frac{9.8 \times \sqrt{73}}{2} \]\[ T \approx \frac{9.8 \times 8.544}{2} \]\[ T \approx \frac{83.9712}{2} \]\[ T \approx 41.9856 \][/tex]

So, the magnitude of the tension on the ends of the clothesline is approximately [tex]\( 41.99 \, \text{N} \).[/tex]

The magnitude of the tension at the ends of the clothesline is approximately [tex]\[T = 41.87 \, \text{N}\][/tex]

To solve for the tension in the clothesline, we first consider the forces and the geometry involved.

Given:

Distance between poles, [tex]\( L = 16 \) meters[/tex]

Sagging distance,[tex]\( h = 3 \) meters[/tex]

Mass, [tex]\( m = 3 \) kilograms[/tex]

Gravitational acceleration, [tex]\( g = 9.8 \) m/s\(^2\)[/tex]

Determine the Weight of the Mass

The weight [tex]\( W \)[/tex] of the mass is given by:

[tex]\[W = mg = 3 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 29.4 \, \text{N}\][/tex]

Analyzing the Geometry

[tex]\[\text{Half the distance between poles} = \frac{L}{2} = \frac{16}{2} = 8 \, \text{meters}\][/tex]

[tex]\[\text{Sagging distance} = h = 3 \, \text{meters}\][/tex]

Calculate the Length of the Hypotenuse

[tex]\[\sqrt{(8 \, \text{m})^2 + (3 \, \text{m})^2} = \sqrt{64 + 9} = \sqrt{73} \, \text{meters}\][/tex]

Analyze Forces in Vertical and Horizontal Directions

[tex]\[2T_v = W \quad = \quad T_v = \frac{W}{2} = \frac{29.4 \, \text{N}}{2} = 14.7 \, \text{N}\][/tex]

The vertical component of the tension [tex]\( T \)[/tex] in each half of the clothesline is:

[tex]\[T_v = T \sin \theta\][/tex]

where \( \theta \) is the angle the clothesline makes with the horizontal. From the triangle, we know:

[tex]\[\sin \theta = \frac{h}{\sqrt{(L/2)^2 + h^2}} = \frac{3}{\sqrt{73}}\][/tex]

[tex]\[T \sin \theta = 14.7 \, \text{N}\][/tex]

Solve for the Tension[tex]\( T \)[/tex]

[tex]\[T \sin \theta = 14.7 \, \text{N} \quad = \quad T \cdot \frac{3}{\sqrt{73}} = 14.7 \, \text{N}\][/tex]

[tex]\[T = \frac{14.7 \times \sqrt{73}}{3}\][/tex]

[tex]\[T = 14.7 \times \frac{\sqrt{73}}{3}\][/tex]

[tex]\[T = 14.7 \times \frac{8.544}{3}\][/tex]

[tex]\[T = 14.7 \times 2.848 = 41.87 \, \text{N}\][/tex]

Answer:

1) Distance traveled equals 23.1 meters.

2) Final velocity equals 21.658 m/s.

Explanation:

The problem can be solved using second equation of kinematics as

[tex]s=ut+\frac{1}{2}at^{2}[/tex]

where

s is the distance covered

u is the initial speed of the ball

a is the acceleration the ball is under

t is time of travel

Applying the given values in the above equation we get

[tex]s=4.0\times 1.8+\frac{1}{2}\times 9.81\times 1.8^{2}\\\\s=23.1meters[/tex]

Part 2)

The velocity after 't' time can be obtained using first equation of kinematics.

[tex]v=u+at[/tex]

Applying the given values we get

[tex]v=4+9.81\times 1.80\\\\\therefore v=21.658m/s[/tex]

Final answer:

The ball travels 23.07 meters in 1.80 seconds, and the velocity of the ball at that time is 21.66 m/s downward.

Explanation:

The student wants to know how far a ball projected vertically downward at a speed of 4.00 m/s will travel in 1.80 s and the ball's velocity at that time. To solve these questions, we use the kinematic equations that govern linear motion with constant acceleration due to gravity, ignoring air resistance.

The distance traveled by the ball can be found using the equation:

s = ut +  [tex]rac{1}{2}at^2[/tex]

Where:

s is the distance traveled,

u is the initial velocity (4.00 m/s),

t is the time (1.80 s),

a is the acceleration due to gravity (9.81 m/s^2).

Plugging in the values, we have:

s = 4.00 m/s  imes 1.80 s + 0.5  imes 9.81 [tex]m/s^2[/tex] imes [tex](1.80 s)^2[/tex]

s = 7.20 m + 15.87 m

s = 23.07 m

The ball travels 23.07 meters in 1.80 seconds.

The velocity of the ball after 1.80 seconds can be calculated using the equation:

v = u + at

Where:

v is the final velocity,

u is the initial velocity,

a is the acceleration,

t is the time.

So, v = 4.00 m/s + 9.81 [tex]m/s^2[/tex] imes 1.80 s

v = 4.00 m/s + 17.66 m/s

v = 21.66 m/s

The velocity of the ball after 1.80 seconds is 21.66 m/s downward.

Answer:

Q = - 256 X 10⁻⁷ C .

Explanation:

Electric field due to a charge Q at a distance d from the center is given by the expression

E = k Q /d² Where k is a constant and it is equal to 9 x 10⁹

Put the given value in the equation

9 x 10² = [tex]\frac{9\times10^9\times Q}{(14+2)^2\times10^{-4}}[/tex]

Q = [tex]\frac{9\times16^2\times10^{-2}}{9\times10^9}[/tex]

Q = - 256 X 10⁻⁷ C .

It will be negative in nature as the field is directed towards the center.

Final answer:

To find the charge on the surface of the copper ball, we use Coulomb's law and the provided electric field magnitude, resulting in an approximate charge of 2.56 x 10⁻⁹ C or 2.56 nC, indicating a positive charge.

Explanation:

The question involves calculating the amount of charge on the surface of a copper ball using the given electric field intensity. The electric field due to a charged sphere at a distance from its surface can be calculated using Coulomb's law and the principle of superposition, which for a sphere of charge translates to E = kQ/r², where E is the electric field, k is Coulomb's constant (8.99 x 10⁹ Nm²/C²), Q is the charge, and r is the distance from the center of the sphere to the point of interest (in this case, 14 cm from the surface or 16 cm from the center considering the radius of the copper ball is 2 cm).

Substituting the given values into the equation, we have 9.0 x 10² = (8.99 x 10⁹)Q/(0.16)². Solving for Q gives us Q = 9.0 x 10² x (0.16)² / (8.99 x 10⁹) C, which yields approximately 2.56 x 10⁻⁹ C or 2.56 nC. The charge is positive, as the electric field is directed towards the center of the ball, indicating a positive source.

Answer:

Work done,[tex]w=5.12\times10^{6}\ \rm J[/tex]

change in internal Energy ,[tex]\Delta U=5.12\times10^6\ \rm J[/tex]

Explanation:

Given:

Since It is given that the process is adiabatic process it means that there is no exchange of heat between the system and surroundings

[tex]T_iV_i^{\gamma -1}=T_fV_f^{\gamma -1}\\\\286\times V_i^{\gamma -1}=T_f \left( \dfrac{V_i}{5} \right )^{\gamma -1}\\T_f=840.76\ \rm K[/tex]

Let n be the number of moles of Helium given by

[tex]n=\dfrac{m}{M}\\n=\dfrac{3\times10^3}{4}\\n=0.75\times10^3[/tex]

Work done in Adiabatic process

Let W be the work done

[tex]W=\dfrac{nR(T_1-T_2)}{\gamma-1}\\W=\dfrac{0.75\times10^3\times8.314(286-840.76)}{1.67-1}\\W=-5.12\times 10^6\ \rm J[/tex]

The Internal Energy change in any Process is given by

Let [tex]\Delta U[/tex] be the change in internal Energy

[tex]\Delta U=nC_p\Delta T\\\Delta U=0.75\times10^3\times1.5R\times(840.76-286)\\\Delta U=5.12\times10^6\ \rm J[/tex]

Answer:

The final velocity of the ball just hits the ground is [tex]\sqrt{2gh}[/tex].

Explanation:

A ball is dropped from the height h. As the ball is dropped means there is no initial velocity given to the ball.

GIVEN:

The distance traveled by the ball is equal to h.

Initial velocity of the ball is zero.

Concept:

Two type of force comes into play in dropping of ball.

1. Gravitational force

Due to gravitational force ball forced to fall downward towards the surface.

2. Air resistance

Air develops drag force on the ball in opposite direction of the gravitational force.

Assumption:

There is no air resistance in falling of ball. So, there will be no drag force on the ball.

Calculation:

It is given that a ball is dropped from a building of height h.

The Newton’s equations of motion are as follows:

[tex]V^{2}-u^{2}=2as[/tex]

[tex]s=ut+\frac{1}{2}at^{2}[/tex]

[tex]v=u+at[/tex]

STEP 1

Newton’s equation of motion is expressed as follows:

[tex]V^{2}-u^{2}=2as[/tex]

Here, V is final velocity of the ball just before the ball hits the ground, a is acceleration due to gravity, u is the initial velocity of the ball and s is the distance traveled by the ball.

Here, initial velocity is zero. So, u = 0 m/s.

Vertical distance traveled by the ball is h. So, s= h.

Only gravitational force is responsible for motion as well as for acceleration. Thus, the acceleration due to gravity acts on the ball. Acceleration due to gravity is denoted by small g.

Acceleration due to gravity is constant for different planate. The acceleration due to gravity for the earth is 9.8 m/s². Here, no planet is defined. So the acceleration due to gravity is a = g.

Substitute 0 for m/s, g for a and h for s in above equation as follows:

[tex]V^{2}-u^{2}=2as[/tex]

[tex]V^{2}-(0)^{2}=2gh[/tex]

[tex]V^{2}=2gh[/tex]

[tex]v=\sqrt{2gh}[/tex]

Thus, the final velocity of the ball just hits the ground is [tex]\sqrt{2gh}[/tex].

Answer:

a) 2034 J

b) -1471 J

c) -509 J

Explanation:

The trunk has a mass of 50 kg, so its weight is

f = m * a

f = 50 * 9.81 = 490 N

If the incline is of 30 degrees, the force tangential to the incline is:

ft = f * sin(a)

ft = 490 * sin(30) = 245 N

And the normal force is:

fn = f * cos(a)

fn = 490 * cos(30) = 424 N

The frictional force is:

ff  = μ * fn

ff = 0.2 * 424 = 84.8 N

To push the trunk up one must apply a force slightly greater than the opposing forces, the opposing are the tangential component of the weight and the friction force

fp = fn + ff

fp = 245 + 84.8 = 339 N

The work of the applied force is:

L = f * d

Lp = fp * d

Lp = 339 * 6 = 2034 J

The work of the weight is done by the tangential component:

Lw = -245 * 6 = -1471 J (it is negative because the weight was opposed to the direction of movement)

The work of the friction force is

Lf = -84.8 * 6 = -509 J

Answer:

22.02 lb

Explanation:

The weight of astronaut Earth (w) = 588 N

We know that,

[tex]4.45N = 1lb[/tex]

Thus,

[tex]588N = \frac{1}{4.45}\times 588lb[/tex]

588N = 132.13 lb

Acceleration due to gravity on Earth (g) = 9.8 m/s²

Acceleration due to gravity on Moon = g'

[tex]g'=\frac{g}{6}[/tex]

We know that weight of an object on Earth is,

[tex]w = m\times g[/tex]

[tex]m = \frac{w}{g}[/tex]

Similarly, weight on Moon will be

[tex]w' = m\times g'[/tex]

[tex]w' = \frac{w}{g}\times\frac{g}{6}[/tex]

[tex]w' = \frac{132.13}{6}[/tex]

[tex]w' = 22.02[/tex]

Thus the astronaut will weigh 22.02 lb on Moon.

Therefore, the stone takes 3.06 seconds to reach the bottom, has a speed of 29.8 m/s downwards just before hitting, and travels a total distance of 150.0 meters.

Here's how to solve the problems about the falling stone:

1. Time to reach the bottom:

First half of the journey (upward):

Use the equation v = u - gt, where v is the final velocity (0 m/s at the top), u is the initial velocity (15.0 m/s), g is the acceleration due to gravity (9.81 m/s^2), and t is the time.

Solving for t, we get t = u/g = 15.0 m/s / 9.81 m/s^2 = 1.53 seconds.

Second half of the journey (downward):

The stone falls freely again with an initial velocity of 0 m/s. The time taken will be the same as the upward journey, 1.53 seconds.

Total time to reach the bottom:

Add the times for both halves: 1.53 seconds + 1.53 seconds = 3.06 seconds.

2. Speed just before hitting:

Use the same equation v = u - gt, but this time u is 0 m/s (at the top) and t is the total time (3.06 seconds).

Solving for v, we get v = 0 m/s - 9.81 m/s^2 * 3.06 seconds = -29.8 m/s (negative sign indicates downward direction).

Therefore, the stone's speed just before hitting is 29.8 m/s downwards.

3. Total distance traveled:

The stone travels twice the height of the cliff: once going up and once going down.

Total distance = 2 * cliff height = 2 * 75.0 m = 150.0 m.

Therefore, the stone takes 3.06 seconds to reach the bottom, has a speed of 29.8 m/s downwards just before hitting, and travels a total distance of 150.0 meters.

An equipotential line is like the contour line of a map that had lines of equal altitude. In this case the "altitude"is the electrical potential or voltage. Equipotential lines are always perpendicular to the electric field. In three dimensions these lines form equipotential surfaces. The movement along an equipotential surface does not perform work, because that movement is always perpendicular to the electric field.

Answer:

(a ) vf= 188.12m/s  : Final speed at 8.75 s

(b) d= 823.04 m   : Distance the rocket sled traveled

Explanation:

Rocket sled kinematics :The rocket sled moves with a uniformly accelerated movement, then we apply the following formulas:

d =vi*t+1/2a*t² Formula (1)

vf= vi+at            Formula(2)

Where:

vi: initial speed =0

a: acceleration=21.5 m/s²

t: time=8.75 s

vf: final speed in m/s

d:displacement in meters(m)

Calculation of displacement (d) and final speed (vf)

We replace data in formulas (1) and (2):

d= 0+1/2*21.5*8.75²

d= 823.04 m

vf= 0+21.5*8.75

vf= 188.12m/s

To calculate the final velocity of someone running with constant acceleration, we first find the acceleration using the distance and time, and then apply it to determine the final velocity. The final velocity after 10 seconds is 8 m/s.

To find the final velocity of the person running with constant acceleration, we use the kinematic equation v = u + at, where v is final velocity, u is initial velocity, a is acceleration, and t is time. The person starts from rest, so u = 0. We need to find a first, and we can do so using the equation s = ut + (1/2)at², where s is the distance traveled.

Insert the given values: 40m = 0 + (1/2)a(10s)², resulting in a = 0.8 m/s². Now apply the acceleration to the initial equation: v = 0 + 0.8 m/s² * 10s, which gives us v = 8 m/s. Therefore, the person's final velocity after 10 seconds is 8 m/s.

Answer:

I think D is the answer.

D. The development of reading and writing in young children

Answer:

7.15 m/s

Explanation:

We use a frame of reference in which the origin is at the point where the trucck passed the car and that moment is t=0. The X axis of the frame of reference is in the direction the vehicles move.

The truck moves at constant speed, we can use the equation for position under constant speed:

Xt = X0 + v*t

The car is accelerating with constant acceleration, we can use this equation

Xc = X0 + V0*t + 1/2*a*t^2

We know that both vehicles will meet again at x = 578

Replacing this in the equation of the truck:

578 = 24 * t

We get the time when the car passes the truck

t = 578 / 24 = 24.08 s

Before replacing the values on the car equation, we rearrange it:

Xc = X0 + V0*t + 1/2*a*t^2

V0*t = Xc - 1/2*a*t^2

V0 = (Xc - 1/2*a*t^2)/t

Now we replace

V0 = (578 - 1/2*1.4*24.08^2) / 24.08 = 7.15 m/s

Answer:

speed when the block had slid 3.40 m is 2.68 m/s

Explanation:

given data

distance = 6.80 m

speed = 3.80 m/s

to find out

speed when the block had slid 3.40 m

solution

we will apply here equation of motion that is

v²-u² = 2×a×s   ..............1

here s is distance, a is acceleration and v is speed and u is initial speed that is 0

so put here all value in equation 1 to get a

v²-u² = 2×a×s

3.80²-0 = 2×a×6.80

a = 1.06 m/s²

so

speed when distance 3.40 m

from equation 1 put value

v²-u² = 2×a×s

v²-0 = 2×1.06×3.40

v² = 7.208

v = 2.68

so speed when the block had slid 3.40 m is 2.68 m/s