An oil bath maintained at 50.5°C loses heat to its surroundings at the rate of 4.68 kJ/min. Its temperature is maintained by an electrically heated coil with a resistance of 60 operated from a 110 V line. A thermoregulator switches the current on and off. What fraction of the time will the current be turned on?

Explanation:

Given that,

Weight of the astronaut on the surface of Earth, W = 450 N

We know that the acceleration due to gravity on the earth is, [tex]g=9.8\ m/s^2[/tex]

Weight of an object is given by, W = mg

[tex]m=\dfrac{W}{g}[/tex]

[tex]m=\dfrac{450\ N}{9.8\ m/s^2}[/tex]

m = 45.91 kg

Also, the acceleration due to gravity on the surface of moon is one -sixth of the acceleration due to gravity on the surface of Earth, [tex]g'=1.62\ m/s^2[/tex]

As mass remains constant. So, the weight on the moon is :

W' = mg'

[tex]W'=45.91\times 1.62[/tex]

W' = 74.37 N

Hence, this is the required solution.

Answer:Sometime decreases and sometime increases.

Explanation:

Given

Projectile is launched at [tex]35 ^{\circ}[/tex]

so horizontal velocity=ucos35

vertical velocity=usin35

since gravity acts in the vertical direction, therefore, it affects only the vertical component of the projectile.

So Vertical velocity first decrease up to maximum height and then starts to increase.

Answer:

- 196.2 J

Explanation:

Given:

Mass of the ball = 2.00 kg

Height through which ball drops = 10.0 m

Since it is mentioned that the ball comes to a stop in the mud, therefore the velocity of the ball will be zero i.e v = 0 m/s

Thus.

the kinetic energy = [tex]\frac{1}{2}mv^2[/tex]  

= [tex]\frac{1}{2}m\times0^2[/tex]  = 0 J

Now,

The potential energy = - mgh

here, g is the acceleration due to gravity

and, the negative sign depicts that the ball is falling down

thus,

The potential energy = - 2 × 9.81 × 10 =  - 196.2 J

Hence, the total energy = - 196.2 + 0 = - 196.2 J

Final answer:

The sum of the gravitational potential and kinetic energy of the 2.00-kg ball after it comes to a stop in the mud at the bottom of a 10.0-m deep well is 0 Joules, since both potential and kinetic energy are zero at rest.

Explanation:

When the 2.00-kg ball is dropped into a 10.0-m deep well, it initially has a certain amount of gravitational potential energy which is converted entirely to kinetic energy just before it hits the bottom, assuming there is no air resistance. The sum of its potential and kinetic energy at the top is equal to the gravitational potential energy it has due to its position. This can be calculated using the formula for gravitational potential energy (GPE), which is GPE = mgh, where m is the mass, g is the acceleration due to gravity (approximately 9.81 m/s² on Earth), and h is the height above the reference point. In this case, when the ball comes to a stop in the mud at the bottom of the well, all its potential energy has been converted to kinetic energy during the fall, and then to other forms of energy (like heat or sound) upon impact. The ball's potential energy at the bottom of the well is zero and the kinetic energy is also zero since the ball is at rest, so the sum of its potential and kinetic energy after coming to a stop is:

mgh = 2*9.81*10 = -196.2 J (as the ball is falling down)

Sum = -196.2J

According to the conservation of energy principle, energy is neither created nor destroyed, only transformed from one form to another. Therefore, initially, the ball had potential energy which got converted to kinetic energy during the fall and finally to other forms of energy when it came to rest.

Answer:

(a) 10 s

(b) 236.5 m

(c) Kathy's speed = 47.3 m/s

    Stan's speed = 42.9 m/s

Explanation:

Given:

Assumptions:

Part (a):

 Kathy will catch Stan only if the distances traveled by each of them are equal at the same instant.

[tex]\therefore s_s=s_k\\\Rightarrow u_st_s+\dfrac{1}{2}a_st_s^2=u_kt_k+\dfrac{1}{2}a_kt_k^2\\ \Rightarrow (0)(t+1)+\dfrac{1}{2}(3.9)(t+1)^2=(0)(t)+\dfrac{1}{2}(4.73)t^2\\ \Rightarrow \dfrac{1}{2}(3.9)(t+1)^2=\dfrac{1}{2}(4.73)t^2\\\Rightarrow (3.9)(t+1)^2=(4.73)t^2\\\Rightarrow \dfrac{(t+1)^2}{t^2}=\dfrac{4.73}{3.9}\\\textrm{Taking square root in both sides}\\\dfrac{t+1}{t}= 1.1\\\Rightarrow t+1=1.1t\\\Rightarrow 0.1t = 1\\\Rightarrow t = 10\\[/tex]

Hence, Kathy catches Stan after 11 s from the Stan's starting times.

Part (b):

Distance traveled by Kathy to catch Stan will be distance the distance traveled by her in 10 s.

[tex]s_s = u_kt_k+\dfrac{1}{2}a_kt_k^2\\\Rightarrow s_s= (0)(t)+\dfrac{1}{2}(4.73)t^2\\\Rightarrow s_s= \dfrac{1}{2}(4.73)(10)^2\\\Rightarrow s_s= 236.5[/tex]

Hence, Kathy traveled a distance of 236.5 m to overtake Stan.

Part (c):

[tex]v_k = u_k+a_kt_k\\\Rightarrow v_k = 0+(4.73)(t)\\\Rightarrow v_k = (4.73)(10)\\\Rightarrow v_k =47.3[/tex]

The speed of Kathy at the instant she catches Stan is 47.3 m/s.

[tex]v_s = u_s+a_st_s\\\Rightarrow v_s = 0+(3.9)(t+1)\\\Rightarrow v_s = (3.9)(10+1)\\\Rightarrow v_s =42.9[/tex]

The speed of Stan at the instant Kathy catches him is 42.9 m/s.

Answer:

E=[8.1X-9.63Y]*10^{3}N/m

Explanation:

Field in the point is the sum of the point charge electric field and the field of the infinite line.

First, we calculate the point charge field:

[tex]E_{Charge}=\frac{1}{4\pi \epsilon_0}  *\frac{Q}{||r_p -r||^2} *Unitary vector\\||r_p -r||^2=(x_p-x)^2+(y_p -y)^2=1.25 m^2\\Unitary Vector=\frac{(r_p -r)}{||r_p -r||}=\frac{(x_p-x)X+(y_p -y)Y}{||r_p -r||}\\=\frac{2\sqrt{5} }{5}X- \frac{\sqrt{5} }{5}Y \\E_{Charge}=K*\frac{3\mu C}{1.25m^2}*(\frac{2\sqrt{5} }{5}X- \frac{\sqrt{5} }{5}Y)[/tex]

It is vectorial, where X and Y represent unitary vectors in X and Y. we recall the Coulomb constant k=[tex]\frac{1}{4\pi \epsilon_0}[/tex] and not replace it yet. Now we compute the line field as follows:

[tex]E_{Line}=\frac{\lambda}{2\pi \epsilon_0 distance} *Unitary Vector\\Unitary Vector=X[/tex] (The field is only in the perpendicular direction to the wire, which is X)

[tex]E_{Line}=\frac{-3\mu C/m*2}{2*2\pi \epsilon_0 5*m}X=K*\frac{6\mu C/m}{ 5*m}(-X)[/tex]

We multiplied by 2/2 in order to obtain Coulomb constant and express it that way. Finally, we proceed to sum the fields.

[tex]E=K*\frac{3\mu C}{1.25m^2}*(\frac{2\sqrt{5} }{5}X- \frac{\sqrt{5} }{5}Y)+K*\frac{6\mu C}{ 5*m^2}(-X)\\E=K*[2.15-1.2]X-K*[1.07]Y \mu N/m\\E=K*[0.9X-1.07Y] \mu C/m^2\\E=[8.1X-9.63Y]*10^{3}N/m[/tex]

Answer:

1.875 cm from 60 microcoulomb charge.

Explanation:

Let the third charge be Q. Let it be put at x distance from 60 micro coulomb charge for balance.

Force on this charge due to first charge

= [tex]\frac{k\times60\times10^{-6}Q}{x^2}[/tex]

Force on this charge due to second charge

= [tex]\frac{k\times6.66\times10^{-6}Q}{(2.5-x)^2}[/tex]

Since both these forces are equal [tex]\frac{k\times60\times10^{-6}Q}{x^2}=\frac{k\times6.66\times10^{-6}Q}{(2.5-x)^2}[/tex]

[tex]\frac{60}{6.66} = \frac{x^2}{(2.5-x)^2}[/tex][tex]\frac{x}{2.5 -x} = \frac{3}{1}[/tex]

x = 1.875

1.875 cm from 60 microcoulomb charge.

Answer:

False.

Explanation:

Yes the magnitude of a vector is always positive , but a vector consists of

when two vectors are added their direction may be opposite to each other For example-

[tex]a=3i+3j[/tex]

[tex]b=3i-6j[/tex] ,

then their resultant

[tex]r=b+a\\r=3i+3j+3i-6j\\r=6i-3j[/tex]

This resultant vector's x and y component equal to y and x component of vector b so its magnitude will be equal to magnitude of vector b.

Therefore, the resultant magnitude not necessary equal to the magnitude of either vector.

Answer:

option (c) 4.71 seconds

Explanation:

Given:

Distance to be covered = 1 km = 1000 m

Acceleration, a = 90 m/s²

From Newton's equation of motion

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

where,  

s is the distance

u is the initial speed  = 0 m/s ( since it starts from rest )

a is the acceleration

t is the time

on substituting the respective values, we get

[tex]1000=0\times t+\frac{1}{2}\times90\times t^2[/tex]

or

45t² = 1000

or

t² = 22.22

or

t = 4.71 seconds

Hence, the correct answer is option (c) 4.71

Answer:

The column compressed is 57.142 Pa.

Explanation:

Given that,

Height = 3.50 m

Diameter = 50.0 cm

Radius = 25.0 cm

Load [tex]F=5.00\times10^{5}\ kg[/tex]

We need to calculate the column compressed

Using formula of compressed

[tex]P = \dfrac{F}{A}[/tex]

[tex]P=\dfrac{F}{l\times r}[/tex]

Where, F = load

l = length

r = radius

Put the value into the formula

[tex]P=\dfrac{5.00\times10^{5}}{3.50\times25\times10^{2}}[/tex]

[tex]P=57.142\ Pa[/tex]

Hence, The column compressed is 57.142 Pa.

Answer:

A)Ep=-81.3N/C  :Electric field at the point x = +0.2

Ep Magnitude =81.3N/C  

Direction of the electric field ( Ep): -x

B)Graphic attached

Explanation:

Conceptual analysis

The electric field at a point P due to a point charge is calculated as follows:

E = k*q/d²

E: Electric field in N/C

q: charge in Newtons (N)

k: electric constant in N*m²/C²

d: distance from charge q to point P in meters (m

Graphic attached

The attached graph shows the field due to the charges:

The electric field at a point P due to several point charges is the vector sum of the electric field due to individual charges.

E₁: Electric Field at point  Xp=0.2 m due to charge q₁. As the charge q1 is positive (q₁+) ,the field leaves the charge.

E₁: Electric Field at point  Xp=0.2 m due to charge q₂. As the charge q1 is positive (q₂+) ,the field leaves the charge

E₃: Electric Field at point Xp=0.2 m due to charge q₃. As the charge q₃ is negative (q₃-), the field enters the charge.

Equivalence

+1.0μC=1* 10⁶C

Data

q₁=+1.0μC=1* 10⁶C

q₂=+1.0μC=1* 10⁶C

q₃=-2.0μC=-2* 10⁶C

Xp=0.2m

x₁=0

x₂=0.01 m

x₃=0.02m

Calculation of the distances of the charges to the point P

d= Xp-x

d₁=Xp-x₁= 0.2-0= 0.2m

d₂=Xp-x₂=0.2-0.01= 0.19m

d₃=Xp-x₃=0.2-002= 0.18m

Calculation of electric fields due to charges q1, q2 and q3 at point P

E₁=k*q₁/d₁²=9*10⁹*1*10⁻⁶/0.2²=225*10³N/C

E₂=k*q₂/d₂²=9*10⁹*1*10⁻⁶/0.19²=249.3*10³N/C

E₃=-k*q₃/d₃²=9*10⁹*2*10⁻⁶/0.18²=-555.6*10³N/C

Calculation of electric field at point P due to charges q₁, q₂ and q₃

To calculate Ep, the electric fields E₁,E₂ and E₃ are added algebraically:

Ep=E₁+E₂ +E₃

Ep=(225*10³+249.3*10³ -555.6*10³)N/C

Ep=-81.3N/C  

Ep Magnitude =81.3N/C  in -x direction

If[tex]\( q_1 = 0 \), then \( q_2 = 11 \)[/tex] which is positive.

Thus, [tex]\( q_1 = 0 \, \text{C} \) and \( q_2 = 11 \, \text{C} \)[/tex] are the valid solutions.

To find the charges [tex]\( q_1 \) and \( q_2 \),[/tex] we can use Coulomb's law, which states that the magnitude of the force between two point charges is given by:

[tex]\[ F = \frac{k \cdot |q_1 \cdot q_2|}{r^2} \][/tex]

Given that the force exerted by one charge on the other has a magnitude of 8 mN (milli-Newtons) when they are separated by 4.5 m, we can set up the equation as follows:

[tex]\[ 8 \times 10^{-3} \, \text{N} = \frac{(8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2) \cdot |q_1 \cdot q_2|}{(4.5 \, \text{m})^2} \][/tex]

Solving for[tex]\( |q_1 \cdot q_2| \[/tex]), we get:

[tex]\[ |q_1 \cdot q_2| = \frac{8 \times 10^{-3} \, \text{N} \cdot (4.5 \, \text{m})^2}{8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2} \][/tex]

[tex]\[ |q_1 \cdot q_2| = \frac{8 \times 10^{-3} \times 20.25}{8.99 \times 10^9} \, \text{C}^2 \][/tex]

[tex]\[ |q_1 \cdot q_2| = \frac{0.162}{8.99 \times 10^9} \, \text{C}^2 \][/tex]

[tex]\[ |q_1 \cdot q_2| = 1.802 \times 10^{-11} \, \text{C}^2 \][/tex]

Now, since both charges are positive and repel each other, we know that [tex]\( q_1 \)[/tex]is smaller than [tex]\( q_2 \)[/tex]. So, let's assume [tex]\( q_1 = x \) and \( q_2 = 11 - x \)[/tex]. Substituting these values into the equation for [tex]\( |q_1 \cdot q_2| \)[/tex], we get:

[tex]\[ x \cdot (11 - x) = 1.802 \times 10^{-11} \, \text{C}^2 \][/tex]

[tex]\[ 11x - x^2 = 1.802 \times 10^{-11} \, \text{C}^2 \][/tex]

Now, solve this quadratic equation to find  x , which represents [tex]\( q_1 \).[/tex] Once we find x , we can find [tex]\( q_2 = 11 - x \).[/tex] Then, we verify that both charges are positive and satisfy the given conditions.

To solve the quadratic equation [tex]\( 11x - x^2 = 1.802 \times 10^{-11} \, \text{C}^2 \)[/tex], we first rearrange it to the standard form:

[tex]\[ x^2 - 11x + 1.802 \times 10^{-11} = 0 \][/tex]

We can use the quadratic formula to find[tex]\( x \):[/tex]

[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]

where [tex]\( a = 1 \), \( b = -11 \), and \( c = 1.802 \times 10^{-11} \).[/tex]

Plugging in these values:

[tex]\[ x = \frac{-(-11) \pm \sqrt{(-11)^2 - 4(1)(1.802 \times 10^{-11})}}{2(1)} \][/tex]

[tex]\[ x = \frac{11 \pm \sqrt{121 - 7.208 \times 10^{-11}}}{2} \][/tex]

[tex]\[ x = \frac{11 \pm \sqrt{121 - 7.208 \times 10^{-11}}}{2} \][/tex]

[tex]\[ x = \frac{11 \pm \sqrt{121 - 7.208 \times 10^{-11}}}{2} \][/tex]

[tex]\[ x = \frac{11 \pm \sqrt{121 - 7.208 \times 10^{-11}}}{2} \][/tex]

[tex]\[ x = \frac{11 \pm \sqrt{121 - 7.208 \times 10^{-11}}}{2} \][/tex]

[tex]\[ x \approx \frac{11 \pm \sqrt{121}}{2} \][/tex]

[tex]\[ x \approx \frac{11 \pm 11}{2} \][/tex]

[tex]\[ x_1 \approx \frac{11 + 11}{2} \][/tex]

[tex]\[ x_1 \approx \frac{22}{2} \][/tex]

[tex]\[ x_1 \approx 11 \][/tex]

[tex]\[ x_2 \approx \frac{11 - 11}{2} \][/tex]

[tex]\[ x_2 \approx \frac{0}{2} \][/tex]

[tex]\[ x_2 \approx 0 \][/tex]

So, we have [tex]\( x_1 = 11 \) and \( x_2 = 0 \).[/tex]

Since [tex]\( q_1 \)[/tex]  must be smaller than [tex]\( q_2 \),[/tex] we take [tex]\( x_1 = 11 \) as \( q_1 \) and \( x_2 = 0 \) as \( q_2 \).[/tex]

Therefore,[tex]\( q_1 = 11 \, \text{C} \) and \( q_2 = 0 \, \text{C} \)[/tex]. But we need both charges to be positive, so this solution is not valid.

Let's check the other root [tex]\( x_2 = 0 \):[/tex]

If[tex]\( q_1 = 0 \), then \( q_2 = 11 \)[/tex] which is positive.

Thus, [tex]\( q_1 = 0 \, \text{C} \) and \( q_2 = 11 \, \text{C} \)[/tex] are the valid solutions.

The average velocity between the first two signs is 9.1 m/s, between the second and third signs is 10.4 m/s, and the car's acceleration is approximately 0.26 m/s².

To solve for the average velocity and acceleration of the car passing three equally spaced traffic signs, we can use the kinematic equations for uniformly accelerated motion.

Part a: average velocity between the first two signs

The average velocity is calculated by dividing the displacement (distance between the signs) by the time interval: vavg1,2 = d /
[tex](t_2 - t_1)[/tex]. Substituting the given values, we have vavg1,2 = 25 m / (3.9 s - 1.3 s), resulting in an average velocity of 9.1 m/s.

Part b: average velocity between the second and third signs

Similarly, for the average velocity between the second and third signs: vavg2,3 = d / ([tex]t_3 - t_2[/tex]). Using the given times, we calculate vavg2,3 = 25 m / (5.5 s - 3.9 s), giving an average velocity of 10.4 m/s.

Part c: magnitude of the acceleration

To find the acceleration, we can use the velocities we just calculated, and the time between the signs. We can set up an equation for motion for the time intervals. As initial velocity v0 for the second interval is the final velocity of the first interval, we get v0 = vavg1,2 and final velocity v = vavg2,3. Then we can use the equation v = v0 + a[tex](t_3 - t_2)[/tex] to find the acceleration. Rearranging for a, we have a = [tex](v - v_0) / (t_3 - t_2)[/tex]. After calculating, we find that the acceleration is approximately 0.26 m/s².

Answer:

Explanation:

Given

Velocity of women relative to the car is u

Let v be the velocity of car after women jump off

Therefore women velocity relative to the ground is v-u

Conserving momentum

Mv+Nm(v-u)=0

Mv+Nmv=Nmu

[tex]v=\frac{Nm}{M+Nm}[/tex]

(b)Let [tex]v_r[/tex] be the velocity of car after r women has jumped and [tex]v_N[/tex] be the final velocity of the car after N women have jumped.

After r women jumped and N-r on cart momentum is given by

[tex]P_r=\left [ M+\left ( N-r\right )m\right ]v_r[/tex]

After next woman jumps car has velocity of [tex]v_{r+1}[/tex] while the woman has velocity is [tex]v_{r+1}-u[/tex](relative to the ground)

Total momentum

[tex]P_{r+1}=\left [ M+\left ( N-r-1\right )m\right ]v_{r+1}+m\left ( v_{r+1}-u\right )[/tex]

Since total momentum in horizontal direction is conserved then

[tex]P_{r+1}=P_r[/tex]

[tex]v_{r+1}-v_r=\frac{mu}{\left ( M+\left ( N-r\right )m\right )}[/tex]

Summing the above expression from r=0 to r=N-1 we get

[tex]\sum_{j=1}^{j=N}\frac{mu}{\left ( M+jm\right )}[/tex]

(c)Answer in part b is greater because j\leq N[/tex]

Thus Velocity in part b is greater.

Intuitively if you jump after an another person you will impart extra momentum to the cart compared to when all persons jumped off simultaneously.

When all women jump off the flatcar at the same time, the final velocity of the flatcar is zero. However, when the women jump off one at a time, the flatcar gains momentum and moves in the opposite direction, resulting in a non-zero final velocity.

To answer the given question, we can apply the principle of conservation of momentum. When all the women jump off the flatcar at the same time, the total momentum of the system must remain constant. Since the women are jumping off the flatcar with equal and opposite velocities, the total momentum of the system before and after their jumps will be zero. Therefore, the final velocity of the flatcar will also be zero.

On the other hand, when the women jump off one at a time, the total momentum of the system does not remain constant. Initially, the flatcar and the first woman have a total momentum of zero. But as the first woman jumps off with a velocity u, the flatcar gains an equal and opposite momentum. When the second woman jumps off, the flatcar gains momentum again, and this process continues until all women have jumped off. As a result, the flatcar gains momentum with each jump and moves in the opposite direction of the women. Therefore, the final velocity of the flatcar when the women jump off one at a time will be greater than zero.

From a physical and intuitive perspective, the answer to (b) is greater because as each woman jumps off, the flatcar gains momentum in the opposite direction. This momentum accumulates with each jump, resulting in a higher final velocity for the flatcar compared to when all the women jump off simultaneously.

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Answer:

[tex]f_{beat}=f_2-f_1=353Hz -350Hz=3Hz[/tex]

Explanation:

When there are two waves with similar frequency, the superposition of these waves create a new wave with a particular beat. The beat frequency from this constructive superposition is equal to the value of the difference in frequency of the two waves.

[tex]f_{beat}=f_2-f_1=353Hz -350Hz=3Hz[/tex]

Answer:

Part a)

[tex]a = 4.86 m/s^2[/tex]

Part b)

[tex]d = 155.52 m[/tex]

Part c)

[tex]v_f = 48.6 m/s[/tex]

Explanation:

As we know that car start from rest and reach to final speed of 87 mph

so we have

[tex]v_f = 87 mph = 38.88 m/s[/tex]

now we have

Part a)

acceleration is rate of change in velocity

[tex]a = \frac{v_f - v_i}{t}[/tex]

[tex]a = \frac{38.88 - 0}{8}[/tex]

[tex]a = 4.86 m/s^2[/tex]

Part b)

distance moved by car with uniform acceleration is given as

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

[tex]d = \frac{38.88 + 0}{2} 8[/tex]

[tex]d = 155.52 m[/tex]

Part c)

As we know that the car start from rest

so final speed after t = 10 s

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

[tex]v_f = 0 + (4.86)10[/tex]

[tex]v_f = 48.6 m/s[/tex]

Answer:

The image height is 0.00283 m or 0.283 cm and is inverted due to the negative sign

Explanation:

u = Object distance =  9 m

v = Image distance = 1.7 cm (as the image is forming on the retina)

[tex]h_u[/tex]= Object height = 1.5 m

Magnification

[tex]m=-\frac{v}{u}=\frac{h_v}{h_u}\\\Rightarrow -\frac{v}{u}=\frac{h_v}{h_u}\\\Rightarrow -\frac{0.017}{9}=\frac{h_v}{1.5}\\\Rightarrow h_v=-0.00283\ m[/tex]

The image height is 0.00283 m or 0.283 cm and is inverted due to the negative sign

The image formed on the retina is 1.8 cm behind the lens.

The image formed on the retina can be calculated using the lens formula: 1/f = 1/do + 1/di, where f is the focal length of the lens, do is the object distance from the lens, and di is the image distance.

In this case, the lens-to-retina distance is given as 2.00 cm, so di = -2.00 cm. The object distance do can be calculated as the difference between the person's height and the distance at which they are standing: do = 9.0 m - 1.5 m = 7.5 m. Substituting the values into the lens formula, we get 1/2.00 cm = 1/7.5 m + 1/di. Solving for di, we find that the image is formed 1.8 cm behind the lens.

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Answer:

Slope of position time is velocity

Explanation:

The position time graph means a relation between the position of the object and the time which is represent on a graph.

The graph line shows that how the position of an object changes with respect to time.

The slope of the position time graph shows the rate of change of position of the object with respect to time.

The rate of change of position with respect to time is called velocity.

thus, the slope of position time graph gives the velocity of the object.

Answer:

Explanation:

Relative velocity = 100 - 15 = 85 m /s

Distance to be covered up with this relative speed = 1.3km

= 1300m

If t be the time taken

85 x t = 1300

t = 1300/ 85

= 15.29

Time of fall of the bomb t₁ by 1 km

1000 = 1/2 g t₁²

t₁² = 2000/9.8

t₁ = 14.28 s

Since t₁≠ t₂

It will not score a direct hit.

Answer:

The energy of the capacitors connected in parallel is 0.27 J

Given:

C = [tex]2.0\micro F = 2.0\times 10^{- 6} F[/tex]

C' = [tex]4.0\micro F = 4.0\times 10^{- 6} F[/tex]

Potential difference, V = 300 V

Solution:

Now, we know that the equivalent capacitance of the two parallel connected capacitors is given by:

[tex]C_{eq} = C + C' = 2.0 + 4.0 = 6.0\micro F = 6.0\times 10^{- 6} F[/tex]

The energy of the capacitor, E is given by;

[tex]E = \frac{1}{2}C_{eq}V^{2}[/tex]

[tex]E = \frac{1}{2}\times 6.0\times 10^{- 6}\times 300^{2} = 0.27 J[/tex]

Answer:

Velocity of the truck, v = 4 m/s                                                            

Explanation:

It is given that,

Distance covered by the truck, d = 100 m

Time taken by the truck, t = 25 s

Let v is the velocity of the truck. We know that the velocity is a vector quantity. Mathematically, it is given by :

[tex]v=\dfrac{d}{t}[/tex]

[tex]v=\dfrac{100\ m}{25\ s}[/tex]

v = 4 m/s

So, the velocity of the truck is 4 m/s. Hence, this is the required solution.

Answer:

it increases-

Explanation:

When the mass of a rocket decreases as it burns through its fuel and the force ( thrust) is constant then by newtons second law of motion

F= ma  here F is constant this means that   ma= constant

⇒ m= F /a    this implies that mass is inversely proportional to  acceleration.

its means when the mass decreases the acceleration must increase. hence the acceleration increases

The charge on each ball is approximately [tex]\( 8.08 \times 10^{-9} \) C[/tex].

Step 1

To determine the charge on each ball, let's analyze the forces acting on one of the charged balls when they are in equilibrium. The forces are:

1. Gravitational force [tex](\( F_g \))[/tex] downward.

2. Tension [tex](\( T \))[/tex] in the thread.

3. Electrostatic force [tex](\( F_e \))[/tex] between the two charges.

Step 2

Given:

Length of the thread [tex](\( L \)) = 0.26 m[/tex]

Mass of each ball [tex](\( m \)) = \( 7.75 \times 10^{-4} \) kg[/tex]

Angle between the threads = 35°, so each thread makes an angle of 17.5° with the vertical.

Step 3

First, calculate the gravitational force:

[tex]\[ F_g = mg = 7.75 \times 10^{-4} \, \text{kg} \times 9.8 \, \text{m/s}^2 = 7.595 \times 10^{-3} \, \text{N} \][/tex]

The electrostatic force must balance the horizontal component of the tension:

[tex]\[ F_e = T \sin(17.5^\circ) \][/tex]

The vertical component of the tension balances the gravitational force:

[tex]\[ T \cos(17.5^\circ) = F_g \][/tex]

[tex]Solving for \( T \):[/tex]

[tex]\[ T = \frac{F_g}{\cos(17.5^\circ)} = \frac{7.595 \times 10^{-3} \, \text{N}}{\cos(17.5^\circ)} \approx 7.99 \times 10^{-3} \, \text{N} \][/tex]

Step 4

Now calculate the horizontal component of the tension:

[tex]\[ T \sin(17.5^\circ) = 7.99 \times 10^{-3} \, \text{N} \times \sin(17.5^\circ) \approx 2.41 \times 10^{-3} \, \text{N} \][/tex]

Step 5

This is the electrostatic force:

[tex]\[ F_e = \frac{k q^2}{r^2} \][/tex]

where [tex]\( r \)[/tex] is the distance between the balls.

Step 6

To find [tex]\( r \)[/tex]:

[tex]\[ r = 2L \sin(17.5^\circ) = 2 \times 0.26 \, \text{m} \times \sin(17.5^\circ) \approx 0.156 \, \text{m} \][/tex]

Using Coulomb's law:

[tex]\[ 2.41 \times 10^{-3} \, \text{N} = \frac{(8.99 \times 10^9 \, \text{N m}^2/\text{C}^2) q^2}{(0.156 \, \text{m})^2} \][/tex]

Step 7

Solving for [tex]\( q \)[/tex]:

[tex]\[ q^2 = \frac{2.41 \times 10^{-3} \, \text{N} \times (0.156 \, \text{m})^2}{8.99 \times 10^9 \, \text{N m}^2/\text{C}^2} \approx 6.53 \times 10^{-17} \, \text{C}^2 \][/tex]

[tex]\[ q \approx \sqrt{6.53 \times 10^{-17}} \approx 8.08 \times 10^{-9} \, \text{C} \][/tex]

Answer:

Electric Field at a distance d from one end of the wire is [tex]E=\dfrac{Q}{4\pi \epsilon_0(L+d)d}[/tex]

Electric Field when d is much grater than length of the wire =[tex]\dfrac{Q}{4\pi \epsilon_0\ d^2}[/tex]

Explanation:

Given:

Let dE be the Electric Field due to the small elemental charge on the wire at a distance x from the one end of the wire and let [tex]\lambda[/tex] be the charge density of the wire

[tex]E=\dfrac{dq}{4\pi \epsilon_0x^2}[/tex]

Now integrating it over the entire length varying x from x=d to x=d+L we have and replacing [tex]\lambda=\dfrac{Q}{L}[/tex] we have

[tex]E=\int\dfrac{\lambda dx}{4\pi \epsilon_0x^2}\\E=\dfrac{Q}{4\pi \epsilon_0 (L+d)(d)}[/tex]

When d is much greater than the length of the wire then we have

1+\dfrac{L]{d}≈1

So the Magnitude of the Electric Field at point P = [tex]\dfrac{Q}{4\pi \epsilon_0\ d^2}[/tex]

Answer:

It takes you 32.27 seconds to travel 121 m using the speed ramp

Explanation:

Lets explain how to solve the problem

- The speed ramp has a length of 121 m and is moving at a speed of

 2.2 m/s relative to the ground

- That means the speed of the ramp is 2.2 m/s

- You can cover the same distance in 78 seconds when walking on

  the ground

Lets find your speed on the ground

Speed = Distance ÷ Time

The distance is 121 meters

The time is 78 seconds

Your speed on the ground = 121 ÷ 78 = 1.55 m/s

If you walk at the same rate with respect to the speed ramp that

you walk on the ground

That means you walk with speed 1.55 m/s and the ramp moves by

speed 2.2 m/s

So your speed using the ramp = 2.2 + 1.55 = 3.75 m/s

Now we want to find the time you will take to travel 121 meters using

the speed ramp

Time = Distance ÷ speed

Distance = 121 meters

Speed 3.75 m/s

Time = 121 ÷ 3.75 = 32.27 seconds

It takes you 32.27 seconds to travel 121 m using the speed ramp

Answer:

5.4 m

Explanation:

mass, m = 34 kg

initial velocity, u = 9 m/s

final velocity, v = 0 m/s

coefficient of friction, μ = 0.6

Angle of inclination, θ = 10°

Let teh distance traveled before to come into rest is d.

According to the work energy theorem, the work done by all the forces is equal to the change in kinetic energy of the block.

Work done by the gravitational force = W1 = -  mg Sinθ x d  

Work done by the frictional force = W2 = - μ N = - μ mg Cosθ x d

negative sign shows that the direction of force and the direction of displacement is opposite to each other.

Total work done W = W1 + w2

W = - 34 x 9.8 x Sin 10 x d - 0.6 x 34 x 9.8 x cos 10 x d

W = - 254.86 d

Change in kinetic energy = 0.5 x m (v^2 - u^2)

                                         = 0.5 x 34 (0 - 81) = - 1377

So, W = change in KE

- 254.86 d = - 1377

d = 5.4 m

Answer:

T₁ = 19.39N   with an angle of 52° with the horizontal.

T₂ = 15.64N  with an angle of 40° with the horizontal.

Explanation:

1)We calculate the weight of holiday decoration

W = m×g

W: holiday decoration weight in Newtons (N)

m: holiday decoration weight = 3 kg

g: acceleration due to gravity = 9.8 m/s²

W = m×g = 3 kg × 9.8 m/s² = 29.4 N

2)We apply Newton's first law to the system in equilibrium:

ΣFx = 0

T₁Cos52° - T₂Cos40°=0

T₁Cos52° = T₂Cos40°

T₁ = T₂(Cos40° / T₁Cos52°)

T₁ = 1,24×T₂  Equation (1)

ΣFy = 0

T₁Sin52° + T₂Sin40° - 29.4 = 0 Equation (2)

We replace T₁ of  Equation (1) in the Equation (2)

1.24×T₂×Sin52° + T₂Sin40° - 29.4 = 0

1.88*T₂ = 29.4

T₂ = 29.4/1.88

T₂ = 15.64N

We replace T₂ = 15.64N in the Equation (1)  to calculate T₁:

T₁ = 1.24×15.64N

T₁ = 19.39N

Answer:

0.224 psi

Explanation:

The pressure using a differential manometer is calculated with the delta H.

Delta H = 293 - 275 = 18 mm

The formula for the pressure is:

P = rho * g * h,

where rho : density of the fluid inside the manometer

g : gravitational acceleration

h : delta H inside the manometer.

It is importar the use of units.

8.75 g/cm3 = 8750 kg/m3

g = 9.8 m/s2

h = 18 mm = 0.018 m

P = 1543,5 Pa ;  1 psi = 6894.8 Pa

P = 1543,5/6894,8 = 0.224 psi

Answer:

Explanation:

Force due to charges 1.75 and 5 nC is given below

F =K Q₁Q₂ / d²

F₁ = [tex]\frac{9\times10^9\times1.75\times10^{-9}\times5\times 10^{-9}}{(2.7\times10^{-2})^2}[/tex]

F₁ = 10.8 X 10⁻⁵ N . It will at in x direction.

Force due to other charge placed at origin

F₂ = [tex]\frac{9\times10^9\times1.75\times10^{-9}\times5\times 10^{-9}}{22.5\times10^{-4}}[/tex]

F₂ = 3.5 x 10⁻⁵ N.

Its x component

= F₂ Cos θ

= 3.5 x 10⁻⁵ x 3.9/ 4.74

= 2.88 x 10⁻⁵ N

Its y component

F₂ sin θ

= 3.5 x 10⁻⁵ x 2.7/4.743

= 1.99 x 10⁻⁵ N

Total x  component

=  10.8 X 10⁻⁵ +2.88 x 10⁻⁵

= 13.68 x 10⁻⁵ N.

Magnitude of total force  F

F²  = (13.68 x 10⁻⁵)² + (1.99 x 10⁻⁵ )²

F = 13.82 X 10⁻⁵ N

Direction θ with x axis .

Tanθ = 1.99/ 13.68

θ = 8 °

Answer:

time required after impact for a puck is 2.18 seconds

Explanation:

given data

mass = 30 g = 0.03 kg

diameter = 100 mm = 0.1 m

thick = 0.1 mm = 1 ×[tex]10^{-4}[/tex] m

dynamic viscosity = 1.75 ×[tex]10^{-5}[/tex] Ns/m²

air temperature = 15°C

to find out

time required after impact for a puck to lose 10%

solution

we know velocity varies here 0 to v

we consider here initial velocity = v

so final velocity = 0.9v

so change in velocity is du = v

and clearance dy = h

and shear stress acting on surface is here express as

= µ [tex]\frac{du}{dy}[/tex]

so

= µ  [tex]\frac{v}{h}[/tex]   ............1

put here value

= 1.75×[tex]10^{-5}[/tex] × [tex]\frac{v}{10^{-4}}[/tex]

= 0.175 v

and

area between air and puck is given by

Area = [tex]\frac{\pi }{4} d^{2}[/tex]

area  =  [tex]\frac{\pi }{4} 0.1^{2}[/tex]

area = 7.85 × [tex]\frac{v}{10^{-3}}[/tex] m²

so

force on puck is express as

Force = × area

force = 0.175 v × 7.85 × [tex]10^{-3}[/tex]

force = 1.374 × [tex]10^{-3}[/tex] v    

and now apply newton second law

force = mass × acceleration

- force = [tex]mass \frac{dv}{dt}[/tex]

- 1.374 × [tex]10^{-3}[/tex] v = [tex]0.03 \frac{0.9v - v }{t}[/tex]

t =  [tex] \frac{0.1 v * 0.03}{1.37*10^{-3} v}[/tex]

time = 2.18

so time required after impact for a puck is 2.18 seconds

To calculate the time required for the puck to lose 10% of its initial speed, you can use the equation for deceleration. First, find the final velocity of the puck using the given 10% decrease. Then, calculate the acceleration of the puck using the equation for acceleration. Finally, substitute the acceleration back into the equation for time to find the answer.

To calculate the time required for the puck to lose 10% of its initial speed, we need to find the deceleration of the puck. We can use the equation for deceleration, which is a = (v_f - v_i) / t, where a is the acceleration, v_f is the final velocity, v_i is the initial velocity, and t is the time. In this case, since the puck is losing speed, we can use the negative value of the acceleration.

Given that the initial speed of the puck is v_i = (2 * distance) / t, we can calculate the final velocity v_f = 0.9 * v_i, where 0.9 represents the 10% decrease. Substituting these values into the deceleration equation, we can solve for t as follows:

t = (v_f - v_i) / a = (0.9 * v_i - v_i) / (-a) = 0.1 * v_i / a.

Now we can find the acceleration by using the equation a = (6 * π * η * r) / (m * v_i), where η is the dynamic viscosity of air, r is the radius of the puck (which is half the diameter), m is the mass of the puck, and v_i is the initial velocity. Substituting the given values, we can calculate the acceleration. Finally, substituting the acceleration back into the equation for t, we can calculate the time required for the puck to lose 10% of its initial speed.

Answer:

1.89 seconds

Explanation:

t = Time taken

u = Initial velocity = 9.6 m/s

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

[tex]v=u+at\\\Rightarrow 0=9.6-9.8\times t\\\Rightarrow \frac{-9.6}{-9.8}=t\\\Rightarrow t=0.97 \s[/tex]

Time taken to reach maximum height is 0.97 seconds

[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow s=9.6\times 0.97+\frac{1}{2}\times -9.8\times 0.97^2\\\Rightarrow s=4.7\ m[/tex]

So, the stone would travel 4.7 m up

So, total height ball would fall is 4.7+12.8 = 17.5 m

[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow 17.5=0t+\frac{1}{2}\times 9.8\times t^2\\\Rightarrow t=\sqrt{\frac{17.5\times 2}{9.8}}\\\Rightarrow t=1.89\ s[/tex]

Time taken by the stone to travel 17.5 m is 1.89 seconds