A player kicks a ball with an initial vertical velocity of 12 m/s and horizontal velocity of 16 m/s. (a) At what speed does the ball hit the ground? (b) For how long does the ball remain in the air? (c)What maximum height is attained by the ball?

Answer:

(a) Area = [tex]83.022m^2[/tex] (b)  Area = [tex]9.07m^2[/tex]

Explanation:

We have given radius of the circle r = 5.142 m and r = 1.7 m

We have to find the area of the circle

We know that area of the circle is given  by

[tex]A=\pi r^2[/tex], here r is the radius of the circle

(a) Radius = 5.142 m

So area [tex]A=\pi r^2=3.14\times 5.142^2=83.022m^2[/tex]

(b) Radius = 1.7 m

So area [tex]A=\pi r^2=3.14\times 1.7^2=9.07m^2[/tex]

Explanation:

(a) Mass of the cars, m₁ = m₂ = 1160 kg

Distance between cars, d = 35 m

The gravitational force between two cars is given by :

[tex]F=G\dfrac{m_1m_2}{d^2}[/tex]

[tex]F=6.67\times 10^{-11}\times \dfrac{(1160)^2}{(35)^2}[/tex]

[tex]F=7.32\times 10^{-8}\ N[/tex]

So, the force of gravity between the cars is [tex]7.32\times 10^{-8}\ N[/tex].

(b)The weight of a car is given by :

[tex]W=mg[/tex]

[tex]W=1160\times 9.8[/tex]

W = 11368 N

On comparing,

[tex]\dfrac{W}{F}=\dfrac{11368}{7.32\times 10^{-8}}[/tex]

[tex]\dfrac{W}{F}=1.55\times 10^{11}[/tex]

[tex]W=1.55\times 10^{11}\times F[/tex]

So, the weight of the car is [tex]1.55\times 10^{11}[/tex] times the force of gravitation between the cars.

Answer:

a) 1200 kN/m²

b) 1,200,000 kg/ms²

c) 1.2 × 10⁹ kg/km.s²

Explanation:

Given:

Pressure = 1200 kPa

a) 1 Pa = 1 N/m²

thus,

1000 N = 1 kN

1200 kPa = 1200 kN/m²

b)  1 Pa = 1 N/m² =  1 kg/ms²

Thus,

1200 kPa = 1200000 Pa

or

1200000 Pa = 1200000 × 1 kg/ms²

or

= 1,200,000 kg/ms²

c) 1 km = 1000 m

or

1 m = 0.001 Km

thus,

1,200,000 kg/ms² = [tex]\frac{1,200,000}{0.001}\frac{kg}{km.s^2}[/tex]

or

= 1.2 × 10⁹ kg/km.s²

(a) In kN and m units, the pressure is [tex]$\boxed{1200 \text{ kPa}}$[/tex].  (b) In kg, m, and s units, the pressure is [tex]$\boxed{1200 \text{ kg/(m·s}^2\text{)}}.[/tex] (c) In kg, km, and s units, the pressure is [tex]\boxed{1200 \text{ kg/(km·s}^2\text{)}}[/tex].

Explanation and logic of the

To convert pressure from kPa to the required units, we need to understand the relationship between pressure, force, and area. Pressure (P) is defined as the force (F) applied per unit area (A), given by P = F/A.

(a) In the International System of Units (SI), pressure is measured in pascals (Pa), where 1 Pa = 1 N/m². Since 1 kPa = 1000 Pa, and 1 kN = 1000 N, the pressure in kN/m² is numerically the same as in kPa. Therefore, the pressure in the tank is 1200 kN/m².

(b) To express the pressure in base SI units of kg, m, and s, we need to consider that 1 N = 1 kg·m/s². Since 1 kPa = 1 kN/m², and 1 kN = 1000 N, we have:

[tex]\[ 1200 \text{ kPa} = 1200 \times \frac{1000 \text{ kg\·m/s}^2}{1 \text{ m}^2} = 1200 \times 1000 \text{ kg/(m\s}^2\text{)} \][/tex]

(c) To express the pressure in kg, km, and s, we convert the area from m² to km²:

[tex]\[ 1 \text{ m}^2 = \frac{1}{1000 \times 1000} \text{ km}^2 = \frac{1}{1000000} \text{ km}^2 \][/tex]

Thus, the pressure in kg/(km.s²) is:

[tex]\[ 1200 \text{ kPa} = 1200 \times \frac{1000 \text{ kg\·m/s}^2}{\frac{1}{1000000} \text{ km}^2} = 1200 \times 1000000 \text{ kg/(km\·s}^2\text{)} \][/tex]

However, since 1 kPa is equivalent to 1 kN/m², and 1 kN = 1000 N, we can simplify the expression by recognizing that the conversion from m² to km² results in a factor of 1/1000000, which cancels out the factor of 1000 from the conversion of kN to N, leaving us with:

[tex]\[ 1200 \text{ kPa} = 1200 \text{ kg/(km\s}^2\text{)} \][/tex]

Therefore, the pressure in the tank, when expressed in kg/(km·s²), is 1200 kg/(km·s²).

Answer:

The correct answer is  option 'a': It decreases with increase in altitude

Explanation:

Acceleration due to gravity is the acceleration that a  body is subjected to when it is freely dropped from a height from surface of any planet, ignoring the resistance that the object may face in it's motion such as drag due to any fluid.. The acceleration due to gravity is same for all the objects and is independent of their masses, it only depends on the mass of the planet and the radius of the planet on which the object is dropped. it's values varies with:

1) Depth from surface of planet.

2)Height from surface of planet.

3) Latitude of the object.

Hence it neither is a fundamental quantity nor an universal constant.

The variation of acceleration due to gravity with height can be mathematically written as:

[tex]g(h)=g_{surface}(1-\frac{2h}{R_{planet}})[/tex]

where,

R is the radius of the planet

[tex]g_{surface}[/tex] is value of acceleration due to gravity at surface.

hence we can see that upon increase in altitude the value of 'g' goes on decreasing.

A girl drops a pebble from a high cliff into a lake far below. She sees the splash of the pebble hitting the water [tex]2.00\ s[/tex] later. The pebble was going at a speed of [tex]19.62\ m/s[/tex] when it hit the water.

Use kinematic equations to solve this problem. The pebble is dropped from rest, so its initial velocity is 0 m/s. The time it takes for the pebble to hit the water is given as 2.00 seconds.

The speed of the pebble when it hits the water:

Given:

Initial velocity [tex](v_o) = 0 m/s[/tex]

Time [tex](t) = 2.00\ seconds[/tex]

Acceleration due to gravity [tex](a) = -9.81\ m/s^2[/tex]

Use the kinematic equation:

[tex]v = v_o + a \times t\\v = 0 + (-9.81) \times (2.00)[/tex]

Calculate the numerical value of v:

[tex]v = -19.62\ m/s[/tex]

So, the pebble was going at a speed of [tex]19.62\ m/s[/tex] when it hit the water.

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The pebble was traveling at a speed of 19.6 m/s when it hit the water.

In order to determine the speed at which the pebble hit the water, we can use the equations of motion. Assuming the acceleration due to gravity is 9.8 m/s^2 and neglecting air resistance, we can use the equation:

s = ut + (1/2)at^2

Where s is the distance traveled, u is the initial velocity, t is the time, and a is the acceleration. In this case, the distance traveled is the height of the cliff, the initial velocity is 0 (since the pebble is dropped), t is 2.00 s, and the acceleration is 9.8 m/s^2. Plugging in these values, we can solve for the initial velocity:

s = 0 * (2.00) + (1/2) * 9.8 * (2.00)^2

s = 19.6 m

Therefore, the pebble was traveling at a speed of 19.6 m/s when it hit the water.

Answer:

Distance traveled by car in 25 L IS 396.15 km

Explanation:

We have given average millage of car = 60 km /gal

Means car travel 60 km in 1 gallon

The amount of gasoline = 25 L

We know that 1 L = 0.2641 gallon

So [tex]25L=25\times 0.2641=6.6025gallon[/tex]

As the car travels 60 km in 1 gallon

So traveled distance by car in 6.6025 gallon of gasoline = 60×6.6025 = 396.15 km

Answer:

[tex]V_{2}=\frac{V_{1}}{2}[/tex]    

Explanation:

The potential of a conducting sphere is the same as a punctual charge,

First sphere:

[tex]V_{1}=k*q_{1}/r_{1}[/tex]    (1)

Second sphere:

[tex]V_{2}=k*q_{2}/r_{2}[/tex]        (2)

But, second sphere's radius is twice first sphere radius, and their charges the same:

[tex]q_{1}=q_{2}[/tex]

[tex]r_{2}=2*r_{1}[/tex]

If we divide the equations (1) and (2), to solve V2:

[tex]V_{2}=V_{1}*r_{1}/r_{2}=V_{1}/2[/tex]    

Answer:

[tex]D_{B}=1173.98m\\D_{C}=675.29m[/tex]

Explanation:

If we express all of the cordinates in their rectangular form we get:

A = (1404.77 , 655.06) m

[tex]B = A + ( -D_{B} *sin(41) , -D_{B} * cos(41) )[/tex]

[tex]C = A + B + ( -D_{C} *cos(20) , D_{C} * sin(20) )[/tex]

Since we need C to be (0,0) we stablish that:

[tex]C = (0,0) = A + B + ( -D_{C} *cos(20) , D_{C} * sin(20) )[/tex]

That way we make an equation system from both X and Y coordinates:

[tex]A_{x} + B_{x} + C_{x} = 0[/tex]

[tex]A_{y} + B_{y} + C_{y} = 0[/tex]

Replacing values:

[tex]1404.77 - D_{B}*sin(41) - D_{C}*cos(20) = 0[/tex]

[tex]655.06 - D_{B}*cos(41) + D_{C}*sin(20) = 0[/tex]

With this system we can solve for both Db and Dc and get the answers to the question:

[tex]D_{B}=1173.98m[/tex]

[tex]D_{C}=675.29m[/tex]

Answer:

Explanation:

For elestic collision

v₁ = [tex]\frac{(m_1-m_2)u_1}{m_1+m_2} +\frac{2m_2u_2}{m_1+m_2}[/tex]

[tex]v_2 = [tex]\frac{(m_2-m_1)u_2}{m_1+m_2} +\frac{2m_1u_1}{m_1+m_2}[/tex][/tex]

Here u₁ = 0 , u₂ = 22 m/s , m₁ = 77 kg , m₂ = .15 kg ,  v₁ and v₂ are velocity of goalie and puck after the collision.

v₁ = 0 + ( 2 x .15 x22 )/ 77.15  

= .085 m / s

Velocity of goalie will be .085 m/s in the direction of original velocity of ball before collision.

v₂ = (.15 - 77)x 22 / 77.15 +0

= - 21.91 m /s

=Velocity of puck will be - 21.91 m /s  in the direction opposite  to original velocity of ball before collision.

Answer:

a)8.59 m/s

b)-0.8818 m/s²

Explanation:

a) Given the van moved 40 m in 7.70 seconds to a final velocity of 1.80 m/s

Apply the equation for motion;

[tex]d=(\frac{V_i+V_f}{2} )*t[/tex]

where

t=time the object moved

d=displacement of the object

Vi=initial velocity

Vf=final velocity

Given

t=7.70s

Vf=1.80 m/s

d=40m

Vi=?

Substitute values in equation

[tex]40=(\frac{V_i+1.80}{2} )7.70\\\\\\40=\frac{7.70V_i+13.86}{2} \\\\80=7.70V_i+13.86\\\\80-13.86=7.70V_i\\\\66.14=7.70V_i\\\\\frac{66.14}{7.70} =\frac{7.70V_i}{7.70} \\8.59=V_i[/tex]

b)Acceleration is the rate of change in velocity

Apply the formula

Vf=Vi+at

where;

Vf=final velocity of object

Vi=Initial velocity of the object

a=acceleration

t=time the object moved

Substitute values in equation

Given;

Vf=1.80 m/s

Vi=8.59 m/s

t=7.70 s

a=?

Vf=Vi+at

1.80=8.59+7.70a

1.80-8.59=7.70a

-6.79=7.70a

-6.79/7.70=7.70a/7.70

-0.8818=a

The van was slowing down.

When the brakes are applied to a moving van, it travels a distance of 40.0 m in 7.70 s with a final velocity of 1.80 m/s.

a) The initial speed of the van just before the brakes were applied was 8.59 m/s.  

b) The acceleration of the van while the brakes were applied was -0.88 m/s².  

a) The initial speed of the van can be calculated as follows:

[tex] v_{f} = v_{i} + at [/tex]

Where:  

[tex] v_{f}[/tex]: is the final velocity = 1.80 m/s  

[tex] v_{i}[/tex]: is the initial velocity =?

a: is the acceleration

t: is the time = 7.70 s

By solving the above equation for a we have:

[tex] a = \frac{v_{f} - v_{i}}{t} [/tex]  (1)

Now, we need to use other kinematic equation to find the initial velocity.

[tex] v_{i}^{2} = v_{f}^{2} - 2ad [/tex]   (2)

By entering equation (1) into (2) we have:

[tex] v_{i}^{2} = v_{f}^{2} - 2d(\frac{v_{f} - v_{i}}{t}) = (1.80 m/s)^{2} - 2*40.0 m(\frac{1.80 m/s - v_{i}}{7.70 s}) [/tex]

After solving the above equation for [tex]v_{i}[/tex] we get:

[tex] v_{i} = 8.59 m/s [/tex]

Hence, the initial velocity is 8.59 m/s.

b) The acceleration can be calculated with equation (1):

[tex] a = \frac{v_{f} - v_{i}}{t} = \frac{1.80 m/s - 8.59 m/s}{7.70 s} = -0.88 m/s^{2} [/tex]

Then, the acceleration is -0.88 m/s². The minus sign is because the van is decelerating.      

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Answer:

Option B

Solution:

As per the question:

Heat produced at the rate of 10 W  

The resistor R and 2R are in series.

Also, in series, same current, I' passes through each element in the circuit.

Therefore, current is constant in series.

Also,

Power,[tex] P' = I'^{2}R[/tex]

When current, I' is constant, then

P' ∝ R

Thus

[tex]\frac{P'}{2R} = \frac{10}{R}[/tex]

P' = 20 W

The power dissipated in the resistor 2R is B. 20 W.

When resistors are connected in series, the current through each resistor is the same. Given two resistors, R and 2R, connected in series across a battery, we know that the power dissipated by resistor R is 10 W.

In a series circuit, the voltage across each resistor is different, but the current is the same.

The power dissipated by a resistor in a circuit is given by the formula:

P = I²R

Since the resistors are in series, the same current (I) flows through both resistors. Let's denote the current flowing through the series circuit as I. Using the given power dissipation for resistor R, we get:

10 W = I²R

To find the current:

I = √(10/R)

Next, we calculate the power dissipated by the resistor 2R:

P = I²(2R)

Substituting the value of I:

P = (10/R) × (2R) = 20 W

Thus, the power dissipated in resistor 2R is 20 W.

Answer:

a =  1.02834008 m/s2

Explanation:

given  data:

initial velocity u = -1.03 m/s

time t = 2.47 s later

final velocity v = 1.51 m/s

average acceleration is given as a

[tex]a  = \frac{(v - u)}{t}[/tex]

putting all value to get required value of acceleration:

   [tex]= \frac{(1.51 -(- 1.03))}{2.47}[/tex]

   [tex]= \frac{1.51+1.03}{2.47}[/tex]  

   = 1.02834008 m/s2

Answer:same

Explanation:

Given

ball A initial velocity=3 m/s(upward)

Ball B initial velocity=3 m/s (downward)

Acceleration on both the balls will be acceleration due to gravity which will be downward in direction

Both acceleration is equal

For ball A

maximum height reached is [tex]h_1=\frac{3^2}{2g}[/tex]

After that it starts to move downwards

thus ball have to travel a distance of h_1+h(building height)

so ball A final velocity when it reaches the ground is

[tex]v_a^2=2g\left ( h_1+h\right )[/tex]

[tex]v_a^2=2g\left ( 0.458+h\right )[/tex]

[tex]v_a=\sqrt{2g\left ( 0.458+h\right )}[/tex]

For ball b

[tex]v_b^2-\left ( 3\right )^2=2g\left ( h\right )[/tex]

[tex]v_b^2=2g\left ( \frac{3^2}{2g}+h\right )[/tex]

[tex]v_b=\sqrt{2g\left ( 0.458+h\right )}[/tex]

thus [tex]v_a=v_b[/tex]

Final answer:

A. The magnitudes of the accelerations of both balls are equal, but they have opposite directions. B. The final speeds of both balls when they reach the ground will be the same.

Explanation:

A. The acceleration of both balls will be the same in magnitude but in opposite directions. Since the acceleration due to gravity acts downward, ball A will have a negative acceleration while ball B will have a positive acceleration. Therefore, the magnitudes of their accelerations will be equal.

B. When both balls reach the ground, their final speeds will also be the same. This is because the vertical motion of the balls is independent of their initial speeds when air resistance is ignored. The time it takes for them to reach the ground will be the same, and hence their final velocities will also be equal.

The solution for the distance x from [tex]\(q_1\)[/tex] along the line connecting the charges where the total electric field is zero is given by: [tex]\[x = \frac{s \cdot (q_1 + q_1\sqrt{q_2})}{q_1 + q_2}\][/tex].

Let's go through the complete solution step by step.

Given:

- Charges: [tex]\(q_1\)[/tex] and [tex]q_2[/tex]

- Distance between charges: s

- Coulomb's constant: [tex]\(k = \frac{1}{4\pi\epsilon_0}\)[/tex]

We want to find the distance (x) from charge [tex]\(q_1\)[/tex] along the line connecting the charges where the total electric field is zero.

The electric field (E) due to a point charge (q) at a distance (r) is given by Coulomb's law:

[tex]\[E = \dfrac{k \cdot q}{r^2}\][/tex]

At a distance x from charge [tex]\(q_1\)[/tex], the electric fields due to [tex]\(q_1\) (\(E_1\))[/tex] and [tex]\(q_2\) (\(E_2\))[/tex] will have magnitudes given by:

[tex]\[E_1 = \dfrac{k \cdot q_1}{x^2}\][/tex]

[tex]\[E_2 = \dfrac{k \cdot q_2}{(s - x)^2}\][/tex]

For the total electric field to be zero, [tex]\(E_1\)[/tex] and [tex]\(E_2\)[/tex] must cancel each other out:

[tex]\[E_1 + E_2 = 0\][/tex]

Substitute the expressions for [tex]\(E_1\)[/tex] and [tex]\(E_2\)[/tex]:

[tex]\[\dfrac{k \cdot q_1}{x^2} + \dfrac{k \cdot q_2}{(s - x)^2} = 0\][/tex]

Cross-multiply and simplify:

[tex]\[q_1 \cdot (s - x)^2 = -q_2 \cdot x^2\][/tex]

Expand and rearrange:

[tex]\[q_1 \cdot (s^2 - 2sx + x^2) = -q_2 \cdot x^2\][/tex]

Solve for x:

[tex]\[q_1 \cdot s^2 - 2q_1 \cdot sx + q_1 \cdot x^2 + q_2 \cdot x^2 = 0\][/tex]

[tex]\[(q_1 + q_2) \cdot x^2 - 2q_1 \cdot sx + q_1 \cdot s^2 = 0\][/tex]

This is a quadratic equation in terms of x. Solve for x using the quadratic formula:

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

Where:

[tex]\(a = q_1 + q_2\)[/tex]

[tex]\(b = -2sq_1\)[/tex]

[tex]\(c = q_1s^2\)[/tex]

Calculate x using the quadratic formula:

[tex]\[x = \dfrac{-(-2sq_1) \pm \sqrt{(-2sq_1)^2 - 4(q_1 + q_2)(q_1s^2)}}{2(q_1 + q_2)}\][/tex]

Simplify the expression inside the square root:

[tex]\[x = \dfrac{2sq_1 \pm \sqrt{4s^2q_1^2 - 4(q_1 + q_2)(q_1s^2)}}{2(q_1 + q_2)}\][/tex]

[tex]\[x = \dfrac{2sq_1 \pm \sqrt{4s^2q_1^2 - 4q_1^2s^2 - 4q_2q_1s^2}}{2(q_1 + q_2)}\][/tex]

[tex]\[x = \dfrac{2sq_1 \pm \sqrt{-4q_2q_1s^2}}{2(q_1 + q_2)}\][/tex]

[tex]\[x = \dfrac{2sq_1 \pm 2q_1s\sqrt{-q_2}}{2(q_1 + q_2)}\][/tex]

[tex]\[x = \dfrac{s \cdot (q_1 \pm q_1\sqrt{-q_2})}{q_1 + q_2}\][/tex]

Since [tex]\(q_1\)[/tex] and [tex]\(q_2\)[/tex] are both positive charges, x will be a positive value.

Thus, the solution for the distance x from [tex]\(q_1\)[/tex] is given by [tex]\[x = \dfrac{s \cdot (q_1 + q_1\sqrt{q_2})}{q_1 + q_2}\][/tex].

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Answer:

A) The car will hit the barrier

b) 19.82 m/s

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

Distance = Speed × Time

⇒Distance = (75/3.6)×0.6

⇒Distance = 12.5 m

Distance traveled by the car during the reaction time is 12.5 m

Equation of motion

[tex]v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-\frac{75}{3.6}^2}{2\times -8.5}\\\Rightarrow s=25.53\ m[/tex]

Total distance traveled is 12.5+25.53 = 38.03 m

So, the car will hit the barrier

Distance = Speed × Time

⇒d = u0.6

[tex]v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-u^2}{2\times -8.5}\\\Rightarrow s=\frac{u^2}{17}[/tex]

d + s = 35

[tex]\\\Rightarrow u0.6+\frac{u^2}{17}=35\\\Rightarrow \frac{u^2}{17}+0.6u-35=0[/tex]

[tex]10u^2+102u-5950=0\\\Rightarrow u=\frac{-51+\sqrt{62101}}{10},\:u=-\frac{51+\sqrt{62101}}{10}\\\Rightarrow u=19.82, -30.02[/tex]

Hence, the maximum velocity by which the car could be moving and not hit the barrier 35 meters ahead is 19.82 m/s.

Answer:

-600 J

Explanation:

F₁ = 8i +29 j + 32k

F₂ = 48 i - 59 j - 22 k

F = F₁ +F₂ = 8i +29 j + 32k +48 i - 59 j - 22 k

F = 56i - 30 j + 10 k

displacement d = ( 0 - 20 )i + ( 0 - 15 )j + ( 7 -0) k

d = - 20 i - 15 j + 7 k  

Work Done = F dot product d

F . d = - 56 x 20 - 30 x - 15 + 10 x 7

=  - 1120 +450 + 70

= -600 J

Answer:

B. Work one on one with children in need

Explanation:

Whole group time allows early childhood teachers to introduce new material. It ensures that each student is presented and reviewed with uniform key concepts. It also sets expectations in planning and developing the lessons because it provides baseline assessments.The only thing that seems impossible in this activity is that teacher can give individual or one-to-one attention to each student.

hence the correction will be option B Work one on one with children in need

Answer:

Part a)

[tex]d = 21\sqrt2 = 29.7 m[/tex]

Part b)

Direction is 45 degree North of West

Part c)

[tex]v_{avg} = 1.41 m/s[/tex]

Part d)

direction of velocity will be 45 Degree North of West

Part e)

[tex]a = 0.875 m/s^2[/tex]

Part f)

[tex]\theta = 45 degree[/tex] North of East

Explanation:

Initial position of the cyclist is given as

[tex]r_1 = 21.0 m[/tex] due East

final position of the cyclist after t = 21.0 s

[tex]r_2 = 21.0 m[/tex] due North

Part a)

for displacement we can find the change in the position of the cyclist

so we have

[tex]d = r_2 - r_1[/tex]

[tex]d = 21\hat j - 21\hat i[/tex]

so magnitude of the displacement is given as

[tex]d = 21\sqrt2 = 29.7 m[/tex]

Part b)

direction of the displacement is given as

[tex]\theta = tan^{-1}\frac{y}{x}[/tex]

[tex]\theta = tan^{-1}\frac{21}{-21}[/tex]

so it is 45 degree North of West

Part c)

For average velocity we know that it is defined as the ratio of displacement and time

so here the magnitude of average velocity is defined as

[tex]v_{avg} = \frac{\Delta x}{t}[/tex]

[tex]v_{avg} = \frac{29.7}{21}[/tex]

[tex]v_{avg} = 1.41 m/s[/tex]

Part d)

As we know that average velocity direction is always same as that of average displacement direction

so here direction of displacement will be 45 Degree North of West

Part e)

Here we also know that initial velocity of the cyclist is 13 m/s due South while after t = 21 s its velocity is 13 m/s due East

So we have

change in velocity of the cyclist is given as

[tex]\Delta v = v_f - v_i[/tex]

[tex]\Delta v = 13\hat i - (-13\hat j)[/tex]

now average acceleration is given as

[tex]a = \frac{\Delta v}{\Delta t}[/tex]

[tex]a = \frac{13\hat i + 13\hat j}{21}[/tex]

so the magnitude of acceleration is given as

[tex]a = \frac{13\sqrt2}{21} = 0.875 m/s^2[/tex]

Part f)

direction of acceleration is given as

[tex]\theta = tan^{-1}\frac{y}{x}[/tex]

[tex]\theta = tan^{-1}\frac{13}{13}[/tex]

[tex]\theta = 45 degree[/tex] North of East

Answer:

Explanation:

Combined mass = 815 + 60 = 875 kg

Net weight acting downwards

= 875 x 9.8

= 8575 N

Tension in the string acting upwards

= 9410

Net upward force = 9410 - 8575

= 835 N

Acceleration

Force / mass

a = 835 / 875

a = .954 ms⁻²

Since the elevator is going up with acceleration a

Total reaction force on the woman from the ground

= m ( g + a )

60 ( 9.8 + .954)

= 645.25 N.

Reading of scale = 645.25 N

Answer:

Answer:

6.68 x 10^16 m/s^2

Explanation:

Electric field, E = 3.8 x 10^5 N/C

charge of electron, q = 1.6 x 10^-19 C

mass of electron, m = 9.1 x 10^-31 kg

Let a be the acceleration of the electron.

The force due to electric field on electron is

F = q E

where q be the charge of electron and E be the electric field

F = 1.6 x 10^-19 x 3.8 x 10^5

F = 6.08 x 10^-14 N

According to Newton's second law

Force  = mass x acceleration

6.08 x 10^-14 = 9.1 x 10^-31 x a

a = 6.68 x 10^16 m/s^2

Explanation:

Answer:

Wave number, [tex]k=20\ m^{-1}[/tex]

Explanation:

The given equation of wave is :

[tex]y=0.02\ sin(20x-400t)[/tex]............(1)

The general equation of the wave is given by :

[tex]y=A\ sin(kx-\omega t)[/tex]..............(2)

k is the wave number of the wave

[tex]\omega[/tex] is the angular frequency

On comparing equation (1) and (2) :

[tex]k=20\ m^{-1}[/tex]

[tex]\omega=400\ rad[/tex]

So, the wave number of the wave is [tex]20\ m^{-1}[/tex]. Hence, this is the required solution.

Answer:

[tex]1.232\times 10^{-5}\ T.[/tex]

Explanation:

Given:

According to Ampere's circuital law, the line integral of magnetic field over a closed loop, called Amperian loop, is equal to [tex]\mu_o[/tex] times the net current threading the loop.

[tex]\oint \vec B \cdot d\vec l=\mu_o I.[/tex]

In case of a circular loop, the directions of magnetic field and the line element [tex]d\vec l[/tex], both are along the tangent of the loop at that point, therefore, [tex]\vec B\cdot d\vec l = B\ dl[/tex].

[tex]\oint \vec B \cdot d\vec l = \oint B\ dl = B\oint dl.\\[/tex]

[tex]\oint dl[/tex] is the circumference of the Amperian loop = [tex] 2\pi r[/tex]

Therefore,

[tex]B\ 2\pi r=\mu_o I\\B=\dfrac{\mu_o I}{2\pi r}.[/tex]

It is the magnetic field due to a current carrying wire at a distance r from it.

For the first wire, passing through the origin:

Consider an Amperian loop of radius 3.510 m, concentric with the axis of the wire, such that it passes through the point where magnetic field is to be found, therefore, [tex]r_1 = 3.510\ m.[/tex]

The magnetic field at the given point due to this wire is given by:

[tex]B_1 = \dfrac{\mu_o I_1}{2\pi r_1}\\=\dfrac{4\pi \times 10^{-7}\times 250}{2\pi \times 3.510}=1.42\times 10^{-5}\ T.[/tex]

For the first wire, passing through the y-axis:

Consider an Amperian loop of radius (3.510+1.8) m = 5.310 m,  concentric with the axis of the wire, such that it passes through the point where magnetic field is to be found, therefore, [tex]r_2 = 5.310\ m.[/tex]

The magnetic field at the given point due to this wire is given by:

[tex]B_2 = \dfrac{\mu_o I_2}{2\pi r_2}\\=\dfrac{4\pi \times 10^{-7}\times 50}{2\pi \times 5.310}=1.88\times 10^{-6}\ T.[/tex]

The directions of current in both the wires are opposite therefore, the directions of the magnetic field due to both the wires are also opposite to that of each other.

Thus, the net magnetic field at [tex]r_r=-3.510\ m[/tex] is given by

[tex]B=B_1-B_2 = 1.42\times 10^{-5}-1.88\times 10^{-6}=1.232\times 10^{-5}\ T.[/tex]

Answer:

the bearing of the line AC will be 154° 51' 48"

Explanation:

given,

bearing of the line AB = 234° 51' 48"

an anticlockwise measure of an angle to the point C measured 80°

to calculate the bearing of AC.

As the bearing of line AB is calculated clockwise from North direction.

the angle is moved anticlockwise now the bearing of AC will be calculated by

                  =  bearing of line AB - 80°

                  =  234° 51' 48"  - 80°

                 =  154° 51' 48".

                  = 154.5148

so, the bearing of the line AC will be 154° 51' 48"

Answer:

1. 24375 N/C

2. 2925 V

Explanation:

d = 12 cm = 0.12 m

F = 3.9 x 10^-15 N

q = 1.6 x 10^-19 C

1. The relation between the electric field and the charge is given by

F = q E

So, [tex]E=\frac{F}{q}[/tex]

[tex]E=\frac{3.9 \times 10^{-15}}{1.6 \times 10^{-19}}[/tex]

E = 24375 N/C

2. The potential difference and the electric field is related by the given relation.

V = E x d

where, V be the potential difference, E be the electric field strength and d be the distance between the electrodes.

By substituting the values, we get

V = 24375 x 0.12 = 2925 Volt

Final answer:

The electric field at the position of the electron is approximately 2.43 x 10^4 N/C, and the potential difference between the plates is about 2916 V.

Explanation:

The electric field (E) within two charged parallel plates is given by the expression E = F/q, where F is the force experienced by a charge q. Since an electron has a charge of approximately 1.602 x 10^-19 C, and it is experiencing a force 3.9 x 10^-15 N, we can calculate the electric field as E = (3.9 x 10^-15 N) / (1.602 x 10^-19 C) ≈ 2.43 x 10^4 N/C. The direction of the electric field is from the positively charged plate to the negatively charged plate.

The potential difference (V) between the plates can be found using the relationship V = E × d, where d is the distance between the plates. Given that the plates are 12 cm apart, this distance in meters is 0.12 m, so the potential difference is V = (2.43 x 10^4 N/C) × 0.12 m ≈ 2916 V.

Answer with explanation:

The given vectors in are reduced to their componednt form as shown

For vector A it can be written as

[tex]\overrightarrow{v}_{a}=16cos(44^{o})\widehat{i}+16sin(44^{o})\widehat{j}[/tex]

Similarly vector B can be written as

[tex]\overrightarrow{v}_{b}=-13\widehat{i}[/tex]

Hence The sum and difference is calculated as

[tex]\overrightarrow{v}_{a}+\overrightarrow{v}_{b}=16cos(44^{o})\widehat{i}+16sin(44^{o})\widehat{j}+(-13\widehat{i})\\\\\overrightarrow{v}_{a}+\overrightarrow{v}_{b}=(16cos(44^{o})-13)\widehat{i}+16sin(44^{o})\widehat{j}\\\\\therefore \overrightarrow{v}_{a}+\overrightarrow{v}_{b}=-1.49\widehat{i}+11.11\widehat{j}\\\\\therefore |\overrightarrow{v}_{a}+\overrightarrow{v}_{b}|=\sqrt{(-1.49)^{2}+11.11^{2}}=11.21m[/tex]

The direction is given by

[tex]\theta =tan^{-1}\frac{r_{y}}{r_{x}}\\\\\theta =tan^{-1}\frac{11.11}{-1.49}=97.64^{o}[/tex]with positive x axis.

Similarly

[tex]\overrightarrow{v}_{a}-\overrightarrow{v}_{b}=16cos(44^{o})\widehat{i}+16sin(44^{o})\widehat{j}-(-13\widehat{i})\\\\\overrightarrow{v}_{a}-\overrightarrow{v}_{b}=(16cos(44^{o})+13)\widehat{i}+16sin(44^{o})\widehat{j}\\\\\therefore \overrightarrow{v}_{a}-\overrightarrow{v}_{b}=24.51\widehat{i}+11.11\widehat{j}\\\\\therefore |\overrightarrow{v}_{a}-\overrightarrow{v}_{b}|=\sqrt{(24.51)^{2}+11.11^{2}}=26.91m[/tex]

The direction is given by

[tex]\theta =tan^{-1}\frac{r_{y}}{r_{x}}\\\\\theta =tan^{-1}\frac{11.11}{24.51}=24.38^{o}[/tex]with positive x axis.

Answer:

(a). The electric force is 0.9 N.

(b). The acceleration is 0.9 m/s²

Explanation:

Given that,

Mass = 1.0 kg

Distance = 1.0 m

Charge = 10 μC

(a). We need to calculate the electric force

Using formula of electric force

[tex]F = \dfrac{kq_{1}q_{2}}{r^2}[/tex]

Put the value into the formula

[tex]F=\dfrac{9\times10^{9}\times10\times10^{-6}\times10\times10^{-6}}{1.0^2}[/tex]

[tex]F=0.9\ N[/tex]

(b). We need to calculate the acceleration

Using newton's second law

[tex]F = ma[/tex]

[tex]a =\dfrac{F}{m}[/tex]

Put the value into the formula

[tex]a=\dfrac{0.9}{1.0}[/tex]

[tex]a=0.9\ m/s^2[/tex]

Hence, (a). The electric force is 0.9 N.

(b). The acceleration is 0.9 m/s²

By applying Coulomb's law and Newton's second law, we find that the magnitude of the electric force on one of the masses is 0.0899 N and the initial acceleration of one of the masses is 0.0899 m/s^2.

This question pertains to Coulomb's law, which deals with the force between two charges. Coulomb's law states that the force between two charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

The equation used for this law is F = k(q1*q2)/r², where F represents the force, q1 and q2 represent the charges, r represents the distance between the charges and k is Coulomb's constant (8.99 × 10⁹ N×m²/C²).

a) To find the magnitude of the electric force on one of the masses, we would use the Coulomb's law equation: F = k(q1×q2)/r², where q1 and q2 are both +10μC (which need to be converted to C by multiplying by 10⁻⁶), and r is 1.0m. This gives us F = (8.99 × 10⁹ N×m²/C²) × ((10 × 10⁻⁶ C)²) / (1.0m)² = 0.0899 N.

b) To find the initial acceleration of one of the masses, we would use Newton's second law (F = ma), where F is the force (0.0899 N from the previous calculation), m is the mass (1.0 kg), and a is acceleration. So a = F/m = 0.0899 N / 1.0 kg = 0.0899 m/s².

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Answer:

A) 240 W

B) 864000 J

Explanation:

Hi!

We can easily calculate the required power multiplying the voltage of the process times the circulating current:

[tex]P=12V\times 20A=12(\frac{J}{coulomb})\times20(\frac{coulomb}{s})\\P = 240\frac{J}{s}=240W[/tex]

To calculate the total energy provided we need to multiply the power times the total time:

[tex]E=P\times t=240\frac{J}{s} \times 1 hr =240\frac{J}{s} \times 3600s\\ E=864 000J[/tex]

Hope this helps

Have a nice day!

Final answer:

The power required to charge the battery is 240 watts, and the energy provided during the charging process is 864,000 joules.

Explanation:

A student has asked how much power is required to charge a 12 V storage battery by a current of 20 A for 1 hour, and how much energy has been provided during the process.

A) To find the power required to charge the battery, we use the formula:

Power (P) = Voltage (V)  imes Current (I)

P = 12 V  imes 20 A

P = 240 W

Therefore, the power required to charge the battery at this rate is 240 watts.

B) The energy provided during the charging process can be calculated using the formula:

Energy (E) = Power (P)  imes Time (t)

Since power is given in watts and time in hours, we must convert the time to seconds for consistency in units (1 hour = 3600 seconds).

E = 240 W  imes 1 hour  imes 3600 seconds/hour

E = 240 W  imes 3600 s

E = 864,000 J

The energy provided during the charging process is 864,000 joules (or 864 kJ).

Answer:

The atmospheric pressure is [tex]9.845 \times 10^4\ Pa[/tex]

Explanation:

There are two ways of solving this exercise:

1)

In physics you can find that mmHg is a unit of pressure.

Pressure = Force/area.

If you consider the weight of mercury as your force (mass* acceleration of gravity = density*volume* acceleration of gravity ), then

[tex]P =\frac{\rho Vg}{A} = \frac{\rho Ahg}{A} =\rho g h[/tex]

where h is the height of the mercury column and rho its density.

[tex]\Delta P= P_{atm} - P_{hur} = 21.4\ mm Hg[/tex].

if normal atmospheric pressure is [tex]1.013 \times 10^5\ Pa = 759.81\ mmHg[/tex]

then the pressure in the presence of the hurricane is

[tex]P_{hur} = 759.81 - 21.4 = 738.41\ mmHg = 9.845 \times 10^4\ Pa[/tex]

2)

Considering the definition of pressure

[tex]\Delta P = \rho g h[/tex]

where [tex]\rho = 13.6\ g/cm^3[/tex], [tex]g = 9.8\ m/s^2 =980\ cm/s^2[/tex] and [tex]h = 21.4\ mm = 2.14\ cm[/tex].

[tex]\Delta P = P_{atm} - P_{hur} = 28521,92\ g/cms^2 = 2852,192\ kg/ms^2[/tex], where [tex]kg/ms^2 = Pa[/tex].

if

[tex]P_{atm} = 1.013 \times 10^5\ Pa[/tex],

then

[tex]P_{hur} = 1.013 \times 10^5 - 2852,192 =9.845 \times 10^4\ Pa[/tex].

Answer:

water

Explanation:

The specific heat of an object denotes the amount of energy which is required to raise the objects temperature by 1 unit of temperature and mass of the object is 1 unit.

If the same amount of heat is added to both water and iron the temperature difference in their final and initial temperatures will be different.

The difference in the initial and final temperature of iron will be higher than that of water. This means that the amount of energy which is required to raise the objects temperature by 1 unit of temperature for iron is lower than water.

So, water has higher specific heat

Also

Q = mcΔT

where

Q = Heat

m = Mass

c = Specific heat

ΔT = Change in temperature

[tex]c=\frac{Q}{m\Delta T}[/tex]

This means specific heat is inversely proportional to change in temperature.

So, for water the change in temperature will be lower than iron which means that water will have a higher specific heat.

The stone impacts the ground with a speed of approximately [tex]\(17.32 \, \text{m/s}\)[/tex].

To find the impact speed of the stone, we can use the kinematic equation that relates initial velocity [tex](\(v_0\))[/tex], final velocity [tex](\(v\))[/tex], acceleration [tex](\(g\))[/tex], and displacement [tex](\(s\))[/tex]:

[tex]\[v^2 = v_0^2 + 2gs\][/tex]

Where:

- [tex]\(v\)[/tex] is the final velocity (impact speed),

- [tex]\(v_0\)[/tex] is the initial velocity (throwing speed),

- [tex]\(g\)[/tex] is the acceleration due to gravity (9.81 m/s²),

- [tex]\(s\)[/tex] is the displacement (height the stone falls, -18.1 m, as it falls downward).

Substituting the known values:

[tex]\[v^2 = (7.61 \, \text{m/s})^2 + 2 \times (9.81 \, \text{m/s}^2) \times (-18.1 \, \text{m})\][/tex]

[tex]\[v^2 = 58.0321 - 2 \times 9.81 \times 18.1\][/tex]

[tex]\[v^2 \approx 58.0321 - 357.801\][/tex]

[tex]\[v^2 \approx -299.7689\][/tex]

Since the stone is impacting the ground, we consider the positive root:

[tex]\[v \approx \sqrt{299.7689}\][/tex]

[tex]\[v \approx 17.32 \, \text{m/s}\][/tex]

Therefore, the stone impacts the ground with a speed of approximately [tex]\(17.32 \, \text{m/s}\)[/tex].