One degree Celsius indicates the same temperature change as: A) one kelvin B) one degree Fahrenheit C) 9/5 kelvin. D) 5/9 degree Fahrenheit.

Initial momentum of the Play-doh: 0.0600 x 2.60 = 0.156 kg/m/s

Total mass of the block and play-doh: 0.38 + 0.0600 = 0.44 kg.

Final momentum is mass x velocity = 0.44v

V = Initial momentum / mass

V = 0.156 / 0.44 = 0.3545 m/s

Work done by spring is equal to the Kinetic enrgy.

Work Done by spring = 1/2 *28.0 * distance^2 = 14 * d^2

KE = 1/2 * 0.44* 0.3545^2

set to equal each other:

14 * d^2 = 0.22 *0.12567

Solve for d:

d = √(0.22*0.12567)/14

d = 0.44 meters = 4.4cm

Answer:

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Answer:

Initial velocity of ball 2 = 38.21 m/s = 125.36 ft/s    

At the time of collision velocity of ball one is descending.

Explanation:

Velocity of ball 1 = 146 ft/sec = 44.50m/s

The balls are to collide at an altitude of 234 ft

H = 234 ft = 71.32 m

We have equation of motion

         v² = u² + 2as

         v² = 44.50² + 2 x (-9.81) x 71.32

         v = ±24.10 m/s.

Time for each velocity can be calculated using equation of motion

        v = u + at

         24.10 = 44.50 - 9.81 t , t = 2.07 s

         -24.10 = 44.50 - 9.81 t , t = 6.99 s      

Since the second ball throws after 2.3 seconds  we can avoid case with t = 2.07 s.

So at the time of collision velocity of ball one is descending.

The collision occurs at t = 6.99 s.

Time of flight of ball 2 = 6.99 - 2.3 = 4.69 seconds.

Height traveled by ball 2 = 71.32 m

We need to find velocity

We have

           s = ut + 0.5 at²

           71.32 = u x 4.69 - 0.5 x 9.81 x 4.69²

           u = 38.21 m/s = 125.36 ft/s

Initial velocity of ball 2 = 38.21 m/s = 125.36 ft/s    

Answer:

v2=139 ft

Explanation:

First we just look at the motion of the first particle. It is moving vertically in a gravitational field so is decelerating with rate g = 9.81 m/s^2 = 32.18 ft/s^2. We can write it's vertical position as a function of time.  

h_1=v_1*t-(a*t/2)

We set this equal to 234 ft to find when the body is passing that point, a solve the quadratic equation for t.  

t_1,2=v_1±(√v_1^2-4*a/2*h_1)/a=2.57 s, 7.44 s

Since we know the second ball was launched after 2.3 seconds, we know that the time we are looking for is the second one, when the first ball is descending. The second ball will have 2.3 seconds less so the time we further use is t_c = 7.44 - 2.3 = 5.14 s. With this the speed of the second ball needed for collision at given height, can be found.

Solving a similar equation, but this time for v2 to obtain the result.

h_2=234 ft=v2*t_c-(a*t_c^2/2)--->v2=139 ft

Answer:

25000 V

Explanation:

The formula for potential is

V = Kq/r

Potential at B due to the charge placed at origin O

V1 = K q / OB

[tex]V_{1}= \frac{9 \times 10^{9} \times 30 \times 10^{-6}}{27}[/tex]

V1 = 10000 V

Potential at B due to the charge placed at A

V2 = K q / AB

[tex]V_{2}= \frac{9 \times 10^{9} \times 30 \times 10^{-6}}{18}[/tex]

V2 = 15000 V

Total potential at B

V = V1 + V2 = 10000 + 15000 = 25000 V

Answer:

[tex]\frac{dQ}{dt} = 966 W[/tex]

Explanation:

As we know that the rate of heat transfer due to temperature difference is given by the formula

[tex]\frac{dQ}{dt} = \frac{KA(\Delta T)}{L}[/tex]

here we know that

[tex]K = 0.69 W/m-K[/tex]

A = 4 m x 7 m

thickness = 30 cm

temperature difference is given as

[tex]\Delta T = 20 - 5 = 15 ^oC[/tex]

now we have

[tex]\frac{dQ}{dt} = \frac{(0.69W/m-K)(28 m^2)(15)}{0.30}[/tex]

[tex]\frac{dQ}{dt} = 966 W[/tex]

Answer:

[tex]F = 56.75 N[/tex]

Explanation:

As per the free body diagram of the box we can say that the applied force must be greater than or equal to the sum of friction force and component of the weight of the box along the hill

So we can say that

[tex]F_{net} = F cos35 - (mgsin20 + F_f)[/tex]

on the other side the force perpendicular to the plane must be balanced so that it remains in equilibrium in that direction

so we can say that

[tex]F_n + Fsin35 = mgcos20[/tex]

now we will have

[tex]F_f = \mu F_n[/tex]

[tex]F_f = (0.20)(mg cos20 - Fsin35)[/tex]

now we have

[tex]0 = Fcos35 - mg sin20 - (0.20)(mg cos20 - Fsin35)[/tex]

[tex]F(cos35 + 0.20 sin35) = mg sin20 + 0.20 mgcos20[/tex]

[tex]F = \frac{mg sin20 + 0.20 mgcos20}{(cos35 + 0.20 sin35)}[/tex]

[tex]F = \frac{100(sin20 + 0.20 cos20)}{(cos35 + 0.20 sin35)}[/tex]

[tex]F = 56.75 N[/tex]

Answer:

Power dissipated in 3 ohms resistor is 5.32 watts                

Explanation:

Resistor 1, R₁ = 3 ohms

Resistor 2, R₂ = 6 ohms

Resistor 3, R₃ = 4 ohms

Voltage source, V = 12 V

We need to find the power dissipated in the 3 ohms resistor. Firstly, we will find the equivalent resistance of R₁ and R₂.

[tex]\dfrac{1}{R'}=\dfrac{1}{R_1}+\dfrac{1}{R_2}[/tex]

[tex]\dfrac{1}{R'}=\dfrac{1}{3}+\dfrac{1}{6}[/tex]

R' = 2 ohms

Now R' is connected in series with R₃. Their equivalent is given by :

[tex]R_{eq}=R'+R_3[/tex]

[tex]R_{eq}=2+4[/tex]

[tex]R_{eq}=6\ ohms[/tex]

Total current flowing through the circuit, [tex]I=\dfrac{12}{6}=2\ A[/tex]

Voltage across R', [tex]V'=IR'=2\times 2=4\ V[/tex]

The voltage across R₁ and R₂ is 4 volts as they are connected in parallel. So, current across 3 ohm resistor is,

[tex]I=\dfrac{4}{3}=1.33\ A[/tex]

Power dissipated is given by, P = I × V

[tex]P=1.33\ A\times 4\ \Omega[/tex]

P = 5.32 watts

So, 5.32 watt of power is dissipated  in 3 ohms resistor. Hence, this is the required solution.

Final answer:

The required force is approximately 21.63 newtons.

Explanation:

The question involves determining the average force that a shooter must exert to keep a gun from moving while it fires bullets. To find this force, we can use the concept of impulse, which relates force, time, and change in momentum.

The gun fires 115 bullets per minute, which is 115/60 bullets per second. Multiplying the number of bullets per second by the mass and speed of each bullet gives the total momentum per second, which is the same as the force when mass is in kg and speed is in m/s.

First, we find the momentum of each bullet:

Bullet speed (v) = 342 m/s

F ≈ 21.63 N

The shooter must exert an average force of approximately 21.63 newtons to keep the gun from moving.

Answer:

8.3×10² W/m²

Explanation:

I₀ = Intensity of unpolarized light

θ = angle between the filters = 25°

I = Intensity of the polarized light after passing through three filters = 280 W/m²

[tex]I=\frac{I_0}{2}cos^2{25}cos^2{25}\\\Rightarrrow I_0=\frac{2I}{cos^4{25}}\\\Rightarrow I_0=\frac{2\times 280}{cos^4{25}}\\\Rightarrow I_0=830.01\ W/m^2[/tex]

∴ Intensity of unpolarized light is 8.3×10² W/m²

To find the intensity of the incident beam of light passing through three polarizing filters, we can use the formula for the intensity of polarized light after passing through a filter. The incident light is initially unpolarized, so we need to calculate the intensity after passing through each filter. Given the intensity after passing through the third filter and the angle between the filters, we can find the intensity of the incident beam.

To find the intensity of the incident beam of light, we need to use the formula for the intensity of polarized light after passing through a polarizing filter: I = Io * cos^2(theta). Since the incident light is initially unpolarized, the intensity of the incident beam is twice the intensity of the light after passing through the first filter. Let's call this intensity I1. The intensity after passing through the second filter is I2 = I1 * cos^2(theta), and the intensity after passing through the third filter is I3 = I2 * cos^2(theta). Given that I3 = 280 W/m^2 and theta increases by 25 degrees from one filter to the next, we can calculate the intensity of the incident beam, Io, using these formulas.

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The question is about determining the initial speed of a car involved in an elastic collision with a parked truck, using the laws of conservation of momentum.

The subject of this question is Physics, and it involves the principles of conservation of momentum and elastic collisions. We are given masses for two vehicles (a car and a truck) and the post-collision speed of one of the vehicles (the truck). To find the car's initial speed, we use the equations for an elastic collision, which ensures that both momentum and kinetic energy are conserved.

For an elastic collision, the conservation of momentum is given by:

m1 * v1 + m2 * v2 = m1 * v1' + m2 * v2'

where m1 and m2 are the masses of the car and truck respectively; v1 and v2 are the initial velocities of the car and truck respectively (since the truck is initially parked, v2 = 0); v1' and v2' are the final velocities of the car and truck respectively.

To find the final velocity v1' of the car, we must use the conservation of kinetic energy along with the above equation. However, we only need the conservation of momentum equation solved for v1 (initial speed of the car) since v2' (final velocity of the truck) is given.

After plugging in the given values (car's mass 1689 kg, truck's mass 2000 kg, truck's final velocity 17 km/h), the equation is solved to find the car's initial velocity.

Answer:

122241.02 J

Explanation:

Work Done = 230000 J

[tex]T_H=\text {High Temperature Reservoir}=830\ K\\T_L=\text {Low Temperature Reservoir}=288\ K\\\text{In case of Carnot Cycle}\\\text{Efficiency}=\eta\\\eta=1-\frac{T_L}{T_H}\\\Rightarrow \eta =1-\frac{288}{830}\\\Rightarrow \eta=0.65\\[/tex]

[tex]\eta=\frac{\text{Work}}{\text{Heat}}\\\Rightarrow \eta=\frac{W}{Q_H}\\\\\Rightarrow 0.65=\frac{230000}{Q_H}\\\Rightarrow Q_H=\frac{230000}{0.65}\\\Rightarrow Q_H=352214.02[/tex]

[tex]Q_L=Q_H-W\\\Rightarrow Q_L=352214.02-230000\\\Rightarrow Q_L=122241.02\ J[/tex]

∴Heat exhausted into the air to produce 230,000 J of work is 122241.02 Joule

Correct Answer:

122000 J

Answer:

[tex]\tau = - 30 \hat k[/tex]

Explanation:

Position vector of the point of application of point of application of force is given as

[tex]\vec r = 2\hat i + 2\hat j[/tex]

now we have have force

[tex]\vec F = 5 \hat i - 10\hat j[/tex]

now the moment of force is given as

[tex]\tau = \vec r \times \vec F[/tex]

[tex]\tau = (2\hat i + 2\hat j) \times (5\hat i - 10\hat j)[/tex]

[tex]\tau = -20\hat k - 10 \hat k[/tex]

[tex]\tau = - 30 \hat k[/tex]

The moment of the force about the coordinate origin is 5 Nm.

To find the moment of a force about the coordinate origin, we need to calculate the cross product of the force vector and the position vector from the origin to the point where the force is applied. The position vector is given by: r = 2i + 2j. The cross product is obtained by taking the determinant of a matrix that includes the unit vectors for i, j, and k. The moment is then the magnitude of the cross product.

So, the cross product is: (i, j, k) ⨯ (5i, -10j, 0k) = (0, 0, 5)

Since the moment is the magnitude of the cross product, the moment of the force about the coordinate origin is 5 Nm.

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Answer:

1.27 m

Explanation:

Distance = 192 m

number of rotations = 48

Distance traveled in one rotation = 2 x π x r

Where, r be the radius of wheel.

so, distance traveled in 48 rotations = 48 x 2 x 3.14 x r

It is equal to the distance traveled.

192 = 48 x 2 x 3.14 x r

r = 0.637 m

diameter of wheel = 2 x radius of wheel = 2 x 0.637 = 1.27 m

Answer:

1.29 meters

Explanation:

In order to calculate this you first have to calculate the circumference of the wheel, as you know to travel 192 meters the wheel did 48 revolutions, that means that it took 48 circumferences of the wheel to cover 192 meters, we first have to divide the 192 meters by the circumference:

Circumference:192 m/48=4m

So to know the diameter we just need to divide the circumference by pi:

diameter=circumference/pi

Diameter=4 meters/3.14

Diameter= 1.2861 meters.

Answer:

[tex]I(t)=4[1-e^{-2t}][/tex]

Explanation:

For a LR circuit as shown the current at any time t in the circuit is given by

[tex]I(t)=\frac{V}{R}[1-e^{\frac{-Rt}{L}}][/tex]

where

'V' is the voltage

'R' is resistance in the circuit

'L' is the inductance of the circuit

't' is time after circuit is turned on

Applying the given values we get

[tex]I(t)=\frac{32}{8}[1-e^{\frac{-8t}{4}}]\\\\I(t)=4[1-e^{-2t}][/tex]

Answer:

Part a)

i = 0

Part b)

i = 1 A

Explanation:

Part a)

As per Lenz law we know that inductor in series circuit opposes the sudden change in current

And if the flux in the circuit will change then it will induce back EMF to induce opposite current in it

Now when we close the switch at t = 0

then initially it will induce the opposite EMF in such a way that net EMF of the circuit will be ZERO

so current = 0

Part b)

After long time the induced EMF in the circuit will be zero as the flux will become constant

so here we can say

[tex]EMF = i(R_1 + R_2)[/tex]

[tex]12 = i (8 + 4)[/tex]

[tex]i = 1 A[/tex]

The initial current through the resistor in a series circuit with an inductor and resistor can be calculated using Ohm's Law. After a very long time, the current through the resistor becomes zero.

In a series circuit with an inductor and a resistor, the initial current through the resistor can be calculated using Ohm's Law. The total resistance in the circuit is the sum of the resistance of the inductor and the resistor. Using the formula I = V / R, where V is the voltage from the battery and R is the total resistance, we can calculate the initial current.

After a very long time, the inductor reaches a state of equilibrium with no change in current. So the current through the resistor a very long time after the switch is closed would be zero.

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Answer:

Option B is the correct answer.

Explanation:

Period of simple pendulum is given by the expression [tex]T=2\pi \sqrt{\frac{l}{g}}[/tex], where l is the length of pendulum and g is the acceleration due to gravity value.

So if we have a simple pendulum with length l we can find its period. Using the above formula we can calculate  acceleration due to gravity value of that place.

So using simple pendulum we can measure acceleration due to gravity value.

Option B is the correct answer.

Answer:

Elastic collision is that collision in which two objects are collides and separate out without and deformation in shape and size

example: if a block connected with ideal spring will collide with another block in which spring will compress during the collision and then again come to its natural state.

Inelastic collision in that in which two objects will collide and then stick with each other after collision and there is permanent deformation in the objects.

example : if two cars will collide and stuck together after collision and move together with same speed

So here in both examples and both type of collision there is no external force during collision of two objects so here total momentum of the system will remains conserved.

But since the shape will change in inelastic collision so kinetic energy is not conserved in inelastic collision while in elastic collision kinetic energy is conserved

Final answer:

An example of an elastic collision is two colliding billiard balls, with both momentum and kinetic energy conserved, while an inelastic collision, such as a car crash, conserves momentum but not kinetic energy.

Explanation:

An example of an elastic collision is a collision between two billiard balls where they bounce off of each other without any deformation or generation of heat. In this case, both momentum and kinetic energy are conserved. Momentum conservation means the total momentum before the collision is equal to the total momentum after the collision. Similarly, kinetic energy conservation implies that the total kinetic energy before the collision is the same as the total kinetic energy after the collision.

In contrast, an example of an inelastic collision is a car crash, where the cars collide and deform, often sticking together. In this type of collision, while again momentum is conserved, kinetic energy is not conserved because some of it has been transformed into other forms of energy, such as heat or sound. The reduction in internal kinetic energy is a hallmark of inelastic collisions.

The conservation of momentum and kinetic energy together allow us to calculate the final velocities of objects in one-dimensional, two-body collisions. However, in inelastic collisions, energy is not conserved due to the conversion of kinetic energy to other energy forms.

Answer:

The coordinates of the point is (0,0.55).

Explanation:

Given that,

First charge [tex]q_{1}=9\times10^{-6}\ C[/tex] at origin

Second charge [tex]q_{2}=6\times10^{-6}\ C[/tex]

Second charge at point P = (0,1)

We assume that,

The net electric field between the charges is zero at mid point.

Using formula of electric field

[tex]E=\dfrac{kq}{r^2}[/tex]

[tex]0=\dfrac{k\times9\times10^{-6}}{d^2}+\dfrac{k\times6\times10^{-6}}{(1-d)^2}[/tex]

[tex]\dfrac{(1-d)}{d}=\sqrt{\dfrac{6}{9}}[/tex]

[tex]\dfrac{1}{d}=\dfrac{\sqrt{6}}{3}+1[/tex]

[tex]\dfrac{1}{d}=1.82[/tex]

[tex]d=\dfrac{1}{1.82}[/tex]

[tex]d=0.55\ m[/tex]

Hence, The coordinates of the point is (0,0.55).

It is a physical field occupied by a charged particle on another particle in its surrounding. The coordinates of the point where the net electric field strength due to these charges is zero will be (0,0.55).

It is a physical field occupied by a charged particle on another particle in its surrounding.

The following data are given as

q₁ is the first charge  at the origin

q₂ is the Second charge  

point of the Second charge is P = (0,1)

As we know the net electric field between the charges is zero at the midpoint.

The relation of the electric field with the distance between the charged particle is given by the formula

[tex]\rm E=\frac{Kq}{d^2} \\\\0=\frac{K\times9\times10^{-6}}{d^2} +\frac{K\times6\times10^{-6}}{(1-d)^2} \\\\\frac{1-d}{d} =\sqrt{\frac{6}{9} } \\\\\frac{1}{d} =\sqrt{\frac{{6} }{9} } +1\\\\\rm\frac{1}{d} =1.82\\\\\rm d=0.55m[/tex]

Hence the coordinates of the point where the net electric field strength due to these charges is zero will be (0,0.55).

To learn more about the electric force refer to the link;

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Answer:

it will go up along the inclined plane by d = 9.62 m

Explanation:

As we know that puck is moving upward along the slide

then the net force opposite to its speed is given as

[tex]F_{net} = - mgsin\theta - \mu mgcos\theta[/tex]

so here deceleration is given as

[tex]a = -g(sin\theta + \mu cos\theta)[/tex]

now plug in all values in it

[tex]a = -9.81(sin20 + 0.2cos20)[/tex]

[tex]a = -5.2 m/s^2[/tex]

now the distance covered by the puck along the plane is given as

[tex]v_f^2 - v_i^2 = 2 a d[/tex]

[tex]0 - 10^2 = 2(-5.2)d[/tex]

[tex]d = 9.62 m[/tex]

Answer:

(a) 45 micro coulomb

(b) 6 micro Coulomb

Explanation:

C = 3 micro Farad = 3 x 10^-6 Farad

V = 15 V

(a) q = C x V

where, q be the charge.

q = 3 x 10^-6 x 15 = 45 x 10^-6 C = 45 micro coulomb

(b)

V = 2 V, C = 3 micro Farad = 3 x 10^-6 Farad

q = C x V

where, q be the charge.

q = 3 x 10^-6 x 2 = 6 x 10^-6 C = 6 micro coulomb

Answer:

Distance ran by basketball player = 4.5 m

Explanation:

We have equation of motion, v = u+at

Substituting

           6 = 0 + a x 1.5

           a = 4m/s²

Now we need to find distance traveled, we have the equation of motion

           v² = u² + 2as

Substituting

           6² = 0² + 2 x 4 x s

           8 s = 36

              s = 4.5 m

Distance ran by basketball player = 4.5 m

Answer:

The object's maximum speed remains unchanged.

Explanation:

The speed of a particle in SHM is given by :

[tex]v(t)=A\omega\ sin(\omega t)[/tex]

Maximum speed is, [tex]v_{max}=A\omega[/tex]

If A' = 2A and T' = 2T

[tex]v'_{max}=(2A)\dfrac{2\pi}{2T}[/tex]

[tex]v'_{max}=(A)\dfrac{2\pi}{T}[/tex]

[tex]v'_{max}=v_{max}[/tex]

So, the maximum speed of the object remains the same i.e. it remains unchanged. Hence, this is the required solution.

If the amplitude and period of an object's simple harmonic motion are both doubled, the object's maximum speed will be halved.

When an object moves with simple harmonic motion, doubling the amplitude and period will affect various properties of the motion. In this case, if the amplitude and the period are both doubled, the object's maximum speed will be halved. This means that the object will reach its maximum velocity at a slower rate compared to its original motion.

To understand why this happens, it's important to know that in simple harmonic motion, the maximum speed occurs when the object passes through equilibrium. Doubling the period means that the object will take twice as long to complete one full cycle, which results in a decrease in its maximum speed by a factor of 2.

Answer:

The speed of wave in the second string is 55.3 m/s.

Explanation:

Given that,

Speed of wave in first string= 58 m/s

We need to calculate the wave speed

Using formula of speed for first string

[tex]v_{1}=\sqrt{\dfrac{T}{\mu_{1}}}[/tex]...(I)

For second string

[tex]v_{2}=\sqrt{\dfrac{T}{\mu_{2}}}[/tex]...(II)

Divided equation (II) by equation (I)

[tex]\dfrac{v_{2}}{v_{1}}=\sqrt{\dfrac{\dfrac{T}{\mu_{2}}}{\dfrac{T}{\mu_{1}}}}[/tex]

Here, Tension is same in both string

So,

[tex]\dfrac{v_{2}}{v_{1}}=\sqrt{\dfrac{\mu_{1}}{\mu_{2}}}[/tex]

The linear density of the second string

[tex]\mu_{2}=\mu_{1}+\dfrac{10}{100}\mu_{1}[/tex]

[tex]\mu_{2}=\dfrac{110}{100}\mu_{1}[/tex]

[tex]\mu_{2}=1.1\mu_{1}[/tex]

Now, Put the value of linear density of second string

[tex]\dfrac{v_{2}}{v_{1}}=\sqrt{\dfrac{\mu_{1}}{1.1\mu_{1}}}[/tex]

[tex]v_{2}=v_{1}\times\sqrt{\dfrac{1}{1.1}}[/tex]

[tex]v_{2}=58\times\sqrt{\dfrac{1}{1.1}}[/tex]

[tex]v_{2}=55.3\ m/s[/tex]

Hence, The speed of wave in the second string is 55.3 m/s.

Final answer:

The resulting wave speed in the second string with a 10% greater linear density and the same tension will be the same as the wave speed in the first string.

Explanation:

The wave speed in a string can be determined by the tension and linear mass density of the string. In this case, the wave speed in the first string is 58 m/s. To find the resulting wave speed in the second string with a 10% greater linear density and the same tension, we can use the formula for wave speed:

v = √(T/μ)

Let's assume the linear mass density of the first string is μ and the linear mass density of the second string is 1.1μ. Since the tension is the same in both strings, we have:

v1 = v2

√(T/μ) = √(T/(1.1μ))

Cross multiplying and simplifying, we get:

Tμ = 1.1T

μ = 1.1

So, the resulting wave speed in the second string will be 58 m/s. Therefore, the wave speed remains the same.

Answer:

c).  [tex]a = 0.60 m/s^2[/tex]

Explanation:

As we know that weight of the automobile is given here

so weight = mass times gravity

[tex]W = mg[/tex]

[tex]9800 = m(9.8)[/tex]

[tex]m = 1000 kg[/tex]

now from Newton's law

[tex]F = ma[/tex]

[tex]600 = 1000 a[/tex]

[tex]a = \frac{600}{1000}[/tex]

[tex]a = 0.60 m/s^2[/tex]

The question is about finding the acceleration of a car with a given weight and applied force. To do this, the weight of the car is first converted to mass, then using Newton's Second Law (F = ma), we can determine the acceleration. The calculated acceleration of the car is 0.6 m/s^2.

The subject matter of this question is related to physics, specifically Newton's Second Law of Motion. Noting that the force, mass, and acceleration of the automobile are connected by the formula F = ma (force equals mass times acceleration), the given problem can be solved. First, we need to convert the weight of the automobile (9800 N) into mass, as weight is the product of mass and gravity (9.8 m /s^2). Therefore, the mass of the car is 1000 kg.

Using this converted mass and the applied force (600 N) in the formula, we can calculate the acceleration. The calculation goes as follows: 600N = 1000 kg * a, where a is the unknown acceleration we're trying to find. When you solve for 'a', you get a = 0.6 m/s^2. Hence, the acceleration of the 9800-N automobile when a forward force of 600N is applied, neglecting friction, is option c) 0.6 m/s^2.

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Answer:

- 1 m/s, 20 m

Explanation:

u = 9 m/s, a = - 2 m/s^2, t = 5 sec

Let s be the displacement and v be the velocity after 5 seconds

Use first equation of motion.

v = u + a t

v = 9 - 2 x 5 = 9 - 10 = - 1 m/s

Use second equation of motion

s = u t + 1/2 a t^2

s = 9 x 5 - 1/2 x 2 x 5 x 5

s = 45 - 25 = 20 m

Final answer:

The runner's displacement in the first 5 seconds after crossing the finish line is 22.5 meters. Her velocity 5 seconds after crossing the finish line is 1 m/s in the positive direction. These calculations are based on the formulas for displacement and velocity under constant acceleration.

Explanation:

As a runner crosses the finish line, she decelerates from a velocity of 9 m/s at a rate of 2 m/s2. To find her displacement during the first 5 seconds after crossing the finish line, we use the formula for displacement under constant acceleration, δx = v0t + ½at2, where v0 = 9 m/s, a = -2 m/s2 (deceleration means acceleration is in the opposite direction to velocity), and t = 5 s. Substituting these values gives us a displacement of 22.5 meters. To find her velocity 5 seconds after crossing the finish line, we use the formula v = v0 + at, which yields a final velocity of 1 m/s in the positive direction.

This result makes sense as the runner is slowing down but not yet stopped 5 seconds after crossing the finish line. The calculated displacement of 22.5 meters is the total distance covered during these 5 seconds of deceleration.

Answer:

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

Explanation:

As per Newton's 2nd law we know that net force on an object of mass "m" is given by the formula

[tex]F = ma[/tex]

so net force is the product of mass and acceleration

here we know that

mass = 5 kg

net force on the particle is 4.5 N

so from above equation we have

[tex]4.5 = 5 \times a[/tex]

so the acceleration is given as

[tex]a = \frac{4.5}{5} = 0.9 m/s^2[/tex]

Answer:

v = 2.43 m/s

Explanation:

As we know that block is placed at rest at distance

d = 2 m

so here the centripetal force on the block to move in the circle is due to static friction force

now when block is just going to slide over the disc

then the friction force is maximum static friction which is given as

[tex]\mu_s mg = \frac{mv^2}{R}[/tex]

now we have

[tex]v = \sqrt{\mu_s Rg}[/tex]

now plug in the values in the above equation

[tex]v = \sqrt{0.3(2)(9.81)}[/tex]

[tex]v = 2.43 m/s[/tex]

To determine the maximum speed at which the block can attain before it begins to slip, calculate the force of static friction using the given coefficient and weight of the block. Then, use the equation fs = m(a+g) to determine the maximum speed.

To determine the maximum speed at which the block can attain before it begins to slip, we need to calculate the force of static friction. The formula for static friction is fs ≤ μsN, where μs is the coefficient of static friction and N is the normal force. In this case, the normal force is the weight of the block, which is mg. The maximum force of static friction is therefore μs(mg). Since the block is on a horizontal surface, the normal force is equal to the weight of the block, and the maximum force of static friction is μs(mg). We can calculate μs using the given coefficient of static friction and determine the maximum speed using the equation fs = m(a+g), where fs is the force of static friction, m is the mass of the block, a is the acceleration, and g is the acceleration due to gravity.

Plug in the given values: ms = 0.3, m = (mass of the block), a = 2 m/s², and g = 9.8 m/s².

Then, solve for the maximum speed by rearranging the equation to isolate v: v = sqrt((fs - m*g)/m).

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Answer:

The ball remain in air for 3.06 seconds.

Maximum height reached = 22.94 m

Explanation:

We have equation of motion v = u + at

At maximum height, final velocity, v =0 m/s

       Initial velocity = 15 m/s

       acceleration = -9.81 m/s²

Substituting

       0 = 15 -9.81 t

        t = 1.53 s

Time of flight = 2 x 1.53 = 3.06 s

The ball remain in air for 3.06 seconds.

We also have equation of motion v² = u² + 2as

At maximum height, final velocity, v =0 m/s

       Initial velocity = 15 m/s

       acceleration = -9.81 m/s²

Substituting

       0² = 15² - 2 x 9.81 x s

        s = 22.94 m

Maximum height reached = 22.94 m

Answer:

acceleration is 9.58 × [tex]10^{7}[/tex] (- 14 ĵ + 51 k  ) m/s

Explanation:

given data

uniform electric field E  = 54.0 ĵ V/m

uniform magnetic field B = (0.200 î + 0.300 ĵ + 0.400 k) T

velocity v  = 170 î m/s.

to find out

acceleration

solution

we know magnetic force for proton is

i.e = e (velocity × uniform magnetic field)

magnetic force = e (170 î × (0.200 î + 0.300 ĵ + 0.400 k) )

magnetic force = e (- 68 ĵ + 51 k  )   ..................1

and now for electric force for proton i.e

= uniform electric field × e ĵ

electric force = e (54 ĵ )   ............2

so net force will be  add magnetic force + electric force

from equation 1 and 2

e (- 68 ĵ + 51 k  )  + e (54 ĵ )

e (- 14 ĵ + 51 k  )

so the acceleration (a)  for proton will be

net force = mass × acceleration

a = e (- 14 ĵ + 51 k  )  / 1.6 [tex]10^{-19}[/tex] / 1.67 × [tex]10^{-27}[/tex]

acceleration = 9.58 × [tex]10^{7}[/tex] (- 14 ĵ + 51 k  ) m/s

The acceleration of the proton can be determined using the equation F = qvBsinθ. Plug in the given values to calculate the acceleration.

The acceleration of the proton can be determined using the equation F = qvBsinθ, where F is the force, q is the charge of the proton, v is the velocity, B is the magnetic field strength, and θ is the angle between the velocity and the magnetic field. Since the question does not provide the angle, we assume that θ = 0. Therefore, the acceleration of the proton is given by:

a = (q * v * B) / m

where m is the mass of the proton. Plugging in the given values, we can calculate the acceleration.

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Answer:

Final kinetic energy, [tex]KE_f=5\ J[/tex]  

Explanation:

It is given that,

Kinetic energy of toy car, [tex]KE_i = 8\ J[/tex]

Frictional force, F = 0.5 N

Distance, d = 6 m

We need to find the kinetic energy after this frictional force has acted on it. We know that frictional force is an opposing force. The work done by the toy car is given by :

[tex]W=F\times d[/tex]

[tex]W=-0.5\ N\times 6\ m[/tex]

W = -3 J

We know that the work done is equal to the change in kinetic energy. Let [tex]KE_f[/tex] is the final kinetic energy.

[tex]W=KE_f-KE_i[/tex]

[tex]-3=KE_f-8[/tex]

[tex]KE_f=5\ J[/tex]

So, the final kinetic energy is the toy car is 5 J. Hence, this is the required solution.

The period of the vertical block-spring system on the moon is approximately 2.45 seconds.

The period of a vertical block-spring system is determined by the square root of the ratio of the mass of the block to the spring constant. Since the mass and spring constant remain the same, the period will only be affected by the acceleration due to gravity.

On the moon, where the acceleration due to gravity is about 1/6 that of Earth, the period of the system will be sqrt(6), or approximately 2.45 seconds.

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