A pole-vaulter just clears the bar at 5.53 m and falls back to the ground. The change in the vaulter's potential energy during the fall is -3200 J. What is his weight?

At 30°C, the rms speed of a sample of oxygen with a molar mass of 16 g/mol is approximately 482.34 m/s.

The root mean square (rms) speed of a gas molecule is a measure of the average speed of the gas particles in a sample. It can be calculated using the formula:

vrms = √(3kT/m)

Where:

vrms is the rms speed

k is the Boltzmann constant (1.38 x 10^-23 J/K)

T is the temperature in Kelvin

m is the molar mass of the gas in kilograms

To calculate the rms speed of oxygen at 30°C (303 Kelvin) with a molar mass of 16 g/mol, we need to convert the molar mass to kilograms by dividing it by 1000:

m = 16 g/mol = 0.016 kg/mol

Substituting the values into the formula, we have:

vrms = √((3 * 1.38 x 10^-23 J/K * 303 K) / (0.016 kg/mol))

Calculating this expression yields the rms speed of the oxygen sample:

vrms ≈ 482.34 m/s

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So the right answer is 0.98 watt.

look at the attached picture

Hope it will help you

Explanation:

41.3 kilocalories = 172 799.2 joules

\(hope \: it \: will \: help\)

Answer:

Inertia; Newton's First Law

Explanation:

When you hit you're breaks, the force you feel that pulls you forward is due to inertia.

Newton's first law says that "every object will remain at rest or in uniform motion in a straight line unless compelled to change its state by the action of an external force." The question is a perfect example of this.

It is due to the inertia of motion of the objects that they fall on the floor bed. This is also called Newton's first law of motion.

Newton's first law of motion is known as the law of inertia.

Given is a man who is driving a car at 35 mph hits his brakes suddenly in order to avoid hitting a dog. During the incident, everything in the front seat of his car slides forward and into the floorboard.

The property of body by virtue of which it resist any change in its state of rest or uniform motion is called inertia of body. When the car was moving at 35 mph, every object on front seat was also moving with the same speed. However, when the brakes were applied, they were applied on the car and not on the objects in the front seat. Hence, they keep moving forward and stop only when they strike and fall down. This is called inertia of motion. This is Newton's first law of motion.

Therefore, it is due to the inertia of motion of the objects that they fall on the floor bed. This is also called Newton's first law of motion.

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The skater's final angular velocity is approximately 9.86 rad/s.

The skater's final angular velocity can be calculated using the principle of conservation of angular momentum. The equation for angular momentum is given by:

L = Iω

where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.

Initially, the skater has an angular momentum of:

L_initial = I_initial * ω_initial

Substituting the given values:

L_initial = 2.12 kg m² * 3.25 rad/s

The skater's final angular momentum remains the same, as angular momentum is conserved:

L_final = L_initial

The final moment of inertia is given as 0.699 kg m². Therefore, the final angular velocity can be calculated as:

L_final = I_final * ω_final

0.699 kg m² * ω_final = 2.12 kg m² * 3.25 rad/s

Solving for ω_final:

ω_final = (2.12 kg m² * 3.25 rad/s) / 0.699 kg m²

Hence, the skater's final angular velocity is approximately 9.86 rad/s.

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part a.  the wavelength of this wave  is 26 cm

part b. The amplitude (A) of a wave is 9 cm

The wavelength (λ) of a wave is described as the distance between two consecutive points on the wave that are in phase.

In this scenario, the distance between two corresponding points on the wave will be equal to the horizontal distance from crest to trough, which is 26 cm.

Hence, the wavelength of the wave is: λ = 26 cm

The amplitude (A) of a wave is described as the maximum displacement of a particle from its rest position as the wave passes through it.

In this scenario, the vertical distance from crest to trough is 18 cm, which is equal to twice the amplitude.

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Answer:

B

Explanation:

Answer:

Refer to the step-by-step Explanation.

Step-by-step Explanation:

Simplify the equation with given substitutions,

Given Equation:

\(mgh+(1/2)mv^2+(1/2)I \omega^2=(1/2)mv_{_{0}}^2+(1/2)I \omega_{_{0}}^2\)

Given Substitutions:

\(\omega=v/R\\\\ \omega_{_{0}}=v_{_{0}}/R\\\\\ I=(2/5)mR^2\)\(\hrulefill\)

Start by substituting in the appropriate values: \(mgh+(1/2)mv^2+(1/2)I \omega^2=(1/2)mv_{_{0}}^2+(1/2)I \omega_{_{0}}^2 \\\\\\\\\Longrightarrow mgh+(1/2)mv^2+(1/2)\bold{[(2/5)mR^2]} \bold{[v/R]}^2=(1/2)mv_{_{0}}^2+(1/2)\bold{[(2/5)mR^2]}\bold{[v_{_{0}}/R]}^2\)

Adjusting the equation so it easier to work with.\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2=\dfrac12mv_{_{0}}^2+\dfrac12\Big[\dfrac25mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\)

\(\hrulefill\)

Simplifying the left-hand side of the equation:

\(mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\)

Simplifying the third term.

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2}\cdot \dfrac{2}{5} \Big[mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\)

\(\\ \boxed{\left\begin{array}{ccc}\text{\Underline{Power of a Fraction Rule:}}\\\\\Big(\dfrac{a}{b}\Big)^2=\dfrac{a^2}{b^2} \end{array}\right }\)

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2 \cdot\dfrac{v^2}{R^2} \Big]\)

"R²'s" cancel, we are left with:

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5}mv^2\)

We have like terms, combine them.

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{7}{10} mv^2\)

Each term has an "m" in common, factor it out.

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)\)

Now we have the following equation:

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)=\dfrac12mv_{_{0}}^2+\dfrac12\Big[\dfrac25mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\)

\(\hrulefill\)

Simplifying the right-hand side of the equation:

\(\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac12\cdot\dfrac25\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}^2}{R^2}\Big]\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\cdot\dfrac{v_{_{0}}^2}{R^2}\Big]\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15mv_{_{0}}^2\Big\\\\\\\\\)

\(\Longrightarrow \dfrac{7}{10}mv_{_{0}}^2\)

Now we have the equation:

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)=\dfrac{7}{10}mv_{_{0}}^2\)

\(\hrulefill\)

Now solving the equation for the variable "v":

\(m(gh+\dfrac{7}{10}v^2)=\dfrac{7}{10}mv_{_{0}}^2\)

Dividing each side by "m," this will cancel the "m" variable on each side.

\(\Longrightarrow gh+\dfrac{7}{10}v^2=\dfrac{7}{10}v_{_{0}}^2\)

Subtract the term "gh" from either side of the equation.

\(\Longrightarrow \dfrac{7}{10}v^2=\dfrac{7}{10}v_{_{0}}^2-gh\)

Multiply each side of the equation by "10/7."

\(\Longrightarrow v^2=\dfrac{10}{7}\cdot\dfrac{7}{10}v_{_{0}}^2-\dfrac{10}{7}gh\\\\\\\\\Longrightarrow v^2=v_{_{0}}^2-\dfrac{10}{7}gh\)

Now squaring both sides.

\(\Longrightarrow \boxed{\boxed{v=\sqrt{v_{_{0}}^2-\dfrac{10}{7}gh}}}\)

Thus, the simplified equation above matches the simplified equation that was given.  

Answer:

The power output of this engine is  \(P = 17.5 W\)

The  the maximum (Carnot) efficiency is  \(\eta_c = 0.7424\)

The  actual efficiency of this engine is  \(\eta _a = 0.46\)

Explanation:

From the question we are told that

    The temperature of the hot reservoir is  \(T_h = 1250 \ K\)

      The temperature of the cold reservoir  is  \(T_c = 322 \ K\)

     The energy absorbed from the hot reservoir is \(E_h = 1.37 *10^{5} \ J\)

       The energy exhausts into  cold reservoir is  \(E_c = 7.4 *10^{4} J\)

The power output is mathematically represented as

      \(P = \frac{W}{t}\)

Where t is the time taken which we will assume to be 1 hour =  3600 s  

W is the workdone which is mathematically represented as

      \(W = E_h -E_c\)

substituting values

       \(W = 63000 J\)

So

    \(P = \frac{63000}{3600}\)

    \(P = 17.5 W\)

The Carnot efficiency is mathematically represented as

          \(\eta_c = 1 - \frac{T_c}{T_h}\)

         \(\eta_c = 1 - \frac{322}{1250}\)

         \(\eta_c = 0.7424\)

The actual efficiency is mathematically represented as

        \(\eta _a = \frac{W}{E_h}\)

substituting values

         \(\eta _a = \frac{63000}{1.37*10^{5}}\)

         \(\eta _a = 0.46\)

If the applied force is 20 N and the displacement is from X=0m to X=5.0m, the work done is 100 Joules.

To determine the work done by a force, you need to know the displacement and the component of the force parallel to the displacement.

In this case, if the force applied is 20 N and the displacement is from X=0m to X=5.0m, you need to determine if the force is parallel to the displacement.

If the force is parallel to the displacement, then the work done is simply the product of the force and the displacement, which is:

Work = Force x Displacement

Work = 20 N x (5.0 m - 0 m)

Work = 100 Joules

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The complete question is below:

What is the work done from X =0m to 5.0m? if the applied force is 20 N

it would be the 3rd one. so C

Answer:

10 kg

Explanation:

The mass of an object does not change even if the amount of gravtiy changes.

Answer:

11,000 kg

(a) 11.2 m/s

(b) 1.6 m/s

Explanation:

Momentum is conserved.

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

(2200 kg) (60.0 km/h) + m (0 km/h) = (2200 kg) (10 km/h) + m (10 km/h)

132,000 kg km/h = 22,000 kg km/h + m (10 km/h)

110,000 kg km/h = m (10 km/h)

m = 11,000 kg

Momentum is conserved.

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

(m) (-v) + (2m) (5v) = (m) (v₁) + (2m) (v₂)

-mv + 10mv = m v₁ + 2m v₂

9mv = m (v₁ + 2 v₂)

9v = v₁ + 2 v₂

Since the collision is elastic, kinetic energy is also conserved.

½ m₁u₁² + ½ m₂u₂² = ½ m₁v₁² + ½ m₂v₂²

m₁u₁² + m₂u₂² = m₁v₁² + m₂v₂²

(m) (-v)² + (2m) (5v)² = m v₁² + (2m) v₂²

mv² + 50mv² = m v₁² + 2m v₂²

51mv² = m (v₁² + 2 v₂²)

51v² = v₁² + 2 v₂²

We know v = 1.60 m/s.  So the two equations are:

14.4 = v₁ + 2 v₂

130.56 = v₁² + 2 v₂²

Solve the system of equations using substitution.

130.56 = (14.4 − 2 v₂)² + 2 v₂²

130.56 = 207.36 − 57.6 v₂ + 4 v₂² + 2 v₂²

0 = 6 v₂² − 57.6 v₂ + 76.8

0 = v₂² − 9.6 v₂ + 12.8

v₂ = [ 9.6 ± √(9.6² − 4(1)(12.8)) ] / 2(1)

v₂ = 1.6 or 8

If v₂ = 1.6 m/s, then v₁ = 14.4 − 2 v₂ = 11.2 m/s.

If v₂ = 8 m/s, then v₁ = 14.4 − 2 v₂ = -1.6 m/s.

We know v₁ can't be -1.6 m/s, since that would mean puck A didn't change speeds after the collision.  Therefore, v₁ = 11.2 m/s and v₂ = 1.6 m/s.

Answer:

Explanation:

given:

radius (r) = 1.88 km

velocity (v) = 136.3 km/hr

required:

banking angle ∡ ?

first:

convert 1.88 km to m = 1.88km * 1000m / 1km

r = 1880 m

convert velocity v = 136.3 km/hr to m/s = 136.3 km/hr * (1000 m/ 3600s)

v = 37.86 m/s

now.. calculate the angle

Ф = inv tan (v² / r * g)            we know that gravity = 9.8 m/s²

Ф = inv tan (37.86² / (1880 * 9.8))

Ф = 4.4°

Explanation:

Friction acts between two surfaces in contact only and opposes the movement of one body with respect to another. Therefore friction is a force which opposes the relative motion between two bodies. If a body is moving it is slowed down by frictional force applied on it by the surface on which it is moving.

This question involves the concepts of Newton's Law of Gravitation and mass.

The force on Procyon A from Procyon B will be "equal" to the force on Procyon B from Procyon A, which has a value of "3.75 x 10²⁶ N".

Applying Newton's Law of Gravitation, we can find the force on Procyon A from Procyon B, which is equal to the force on Procyon B from Procyon A:

\(F=\frac{Gm_1m_2}{r^2}\)

where,

F = force = ?

G = universal gravitational constant = 6.67 x 10⁻¹¹ N.m²/kg²

m₁ = mass of Procyon A = 3 x 10³⁰ kg

m₂ = mass of Procyon B = (2.5)(3 x 10³⁰ kg) = 7.5 x 10³⁰ kg

r = distance between them = 2 x 10¹² m

Therefore,

\(F=\frac{(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(3\ x\ 10^{30}\ kg)(7.5\ x\ 10^{30}\ kg)}{(2\ x\ 10^{12}\ m)^2}\)

F = 3.75 x 10²⁶ N

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Answer:

How to write a powerful design brief in 9 easy steps

Start with an overview of the business.

Cover the scope.

Define the audience.

Understand the competition.

Set specific goals.

Take inventory of what you already have.

Set the schedule.

Determine the budget.

Explanation:

Answer:C, Atoms of iron are different from atoms of nickel.

Explanation:All the atoms that make up each type of element are alike, and they are different from the atoms that make up every other type of element. So the element iron is different from the element nickel because the atoms that make up the iron are different from the atoms that make up the nickel.

Answer:

E_total = 26.97 N/C

Electric field points towards the positive charge

Explanation:

We are given;

Charge 1; q1 = 6 nC = 6 × 10^(-9) C

Charge 2; q2 = 3 nC = 3 × 10^(-9) C

Distance between both charges; R_o = 2 m

Since we want to find electric field midway, the distance midway is r = 2/2 = 1 m

Using coulumbs law;

E = kq/r²

Where k is a constant with a value of 8.99 × 10^(9) N.m/C²

Thus;

E1 = kq1/r²

E1 = (8.99 × 10^(9) × 6 × 10^(-9))/1²

E1 = 53.94 N/C

Similarly;

E2 = kq2/r²

E2 = (8.99 × 10^(9) × 3 × 10^(-9))/1²

E2 = 26.97 N/C

Since both electric fields are positive, it means that they are both moving towards the midpoint of the distance between both charges.

This implies they will have opposite directions.

Thus, total electric field at the midway point is;

E_total = E1 - E2

E_total = 53.94 - 26.97

E_total = 26.97 N/C

The tax rate you will pay is displayed in tax brackets for each category of taxable income.

Thus, For instance, in 2022, the first $10,275 of your taxable income is subject to the lowest tax rate of 10% if you are single.

Up until the maximum amount of your taxable income, the following portion of your income is taxed at a rate of 12%.

As taxable income rises, the tax rate rises under the progressive tax system. Overall, this has the result that taxpayers with higher incomes often pay a greater rate of income tax than taxpayers with lower incomes.

Thus, The tax rate you will pay is displayed in tax brackets for each category of taxable income.

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Answer:the ballon and the comb have the same electrical charge

Explanation:because I got the answer right

The comb is able to repel the balloon is that the balloon and the comb possess the same electrical charge. Therefore, option (B) is correct.

Electric charge is the physical property of matter that offers it to experience a force when keep in an electric or magnetic field. An electric charge is related to an electric field, and the moving electric charge produced a magnetic field.

The interaction of the charges creates an electromagnetic force and the combination of an electric field and magnetic field is called an electromagnetic field.

When the students rub a balloon and comb against his hair several times static charges are built up on the balloon and the comb.  When the students place the comb near the balloon and the comb repels the balloon means that both the comb and the balloon possess the same type of electrical charges.

Therefore, the balloon and the comb acquired the same electrical charge.

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Answer: C

Explanation: 60 divided by 4 =15

1.Unit vector in the direction BA: BA/|BA| = (2/3, 2/3, 1/3)

2.The angular velocity (ω) of the body is given as 3 radians per second.

3.Without the position of point P given, it is not possible to write the position vector of P.

4.Left-handed rotation refers to the direction of rotation where the rotation follows the left-hand rule.

5.Position vector of point A: (4, 1, 1)

Position vector of point B: (2, -1, 0)

The direction vector BA = (-2, -2, -1)

1.To determine the unit vector pointing in the direction BA, we subtract the coordinates of point B from the coordinates of point A and normalize the resulting vector.

The direction vector BA is given by:

BA = (4 - 2, 1 - (-1), 1 - 0) = (2, 2, 1)

To obtain the unit vector in the direction of BA, we divide the direction vector by its magnitude:

|BA| = √(2^2 + 2^2 + 1^2) = √(4 + 4 + 1) = √9 = 3

Unit vector in the direction BA: BA/|BA| = (2/3, 2/3, 1/3)

2.The angular velocity (ω) of the body is given as 3 radians per second.

3.Without the position of point P given, it is not possible to write the position vector of P. Please provide the position of point P to proceed with the calculation.

4.Left-handed rotation refers to the direction of rotation where the rotation follows the left-hand rule. In a left-handed coordinate system, if you curl the fingers of your left hand in the direction of rotation, your thumb will point in the direction of the rotation axis. It is the opposite direction to a right-handed rotation.

5.The position vectors of points A and B are:

Position vector of point A: (4, 1, 1)

Position vector of point B: (2, -1, 0)

The direction vector BA can be obtained by subtracting the coordinates of point A from the coordinates of point B:

BA = (2 - 4, -1 - 1, 0 - 1) = (-2, -2, -1)

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Taking the positive direction to be to the right,  the total momentum of this system is - 14kg-m/s

Option B is correct.

momentum is described as the product of the mass and velocity of an object and a vector quantity possessing a magnitude and a direction.

momentum = mass x velocity

momentum 1  = 5 x 8kg = 40 kg-m/s

momentum 2 = 4 x 15 kg = 60 kg-m/s

momentum 3 = 2 x 3kg = 6 kg-m/s

Taking the positive direction to be to the right,  the total momentum of this system is momentum 1  - momentum 2 + momentum 3

total momentum =  40 kg-m/s - 60 kg-m/s + 6 kg-m/s

total momentum  = -20kg-m/ + 6 kg-m/s

total momentum =  - 14 kg-m/s

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Answer: B im sorry if its wrong

Explanation:

Answer:

1 Newton (N) 1 kg*m/s²

Acceleration is measured in m/s², so combine your starting figures in such a way as to get m/s² (you have to divide away the kg):

9N/2kg = 4.5 m/s²

Explanation:

Answer:

Force

Explanation:

You can't "make" things move but you can apply enough force on to an object for it to move, pls brainliest!

Radius of wire is not used to determine the magnetic force on a current-carrying wire in a magnetic field

A magnetic field is a region in space where magnetic forces can be detected. It is created by the motion of electric charges, such as the movement of electrons in an electric current or the spinning of electrons in an atom.

A magnetic field is represented by lines of force that can be visualized using magnetic field lines. These lines indicate the direction of the magnetic field at each point in space and the strength of the magnetic field is indicated by the density of the lines.

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This question is not complete, the complete question is;

A railroad handcar is moving along straight, frictionless tracks with negligible air resistance.

In the following cases, the car initially has a total mass (car and contents) of 170 kg and is traveling east with a velocity of magnitude 4.60 m/s.

Find the final velocity of the car in each case, assuming that the handcar does not leave the tracks.

Part A

An object with a mass of 22.0 kg is thrown sideways out of the car with a speed of 2.50 m/s relative to the car's initial velocity.

Part B

An object with a mass of 22.0 kg is thrown backward out of the car with a velocity of 4.60 m/s relative to the initial motion of the car.

Answer:  

Part A) the final velocity of the car is  4.6 m/s

Part B) the final velocity of the car is 5.28 m/s

Explanation:

Given the data in the question;

Total mass (m₁+m₂) = 170 kg

velocity of magnitude Vx = 4.60 m/s

PART A)

An object with a mass of 22.0 kg is thrown sideways out of the car with a speed of 2.50 m/s relative to the car's initial velocity,

i.e

m₂ = 22.0 kg

so m₁ = 170 - 22 = 148 kg

so, we apply conservation of momentum

since the object thrown out of the car, it has nothing to do with the car's velocity.

(m₁+m₂)Vx = m₁Vx₁ + m₂Vx₂

we substitute

(170)4.60 = 148Vx₁ + 22(4.60)

782 = 148Vx₁ + 101.2

148Vx₁ = 782 - 101.2

148Vx₁ = 680.8

Vx₁ = 680.8 / 148

Vx₁ = 4.6 m/s

Therefore, the final velocity of the car is  4.6 m/s

Part B)

An object with a mass of 22.0 kg is thrown backward out of the car with a velocity of 4.60 m/s relative to the initial motion of the car.

Vx = V(m₁+m₂) / ((m₁+m₂) - m₁)

we substitute

Vx = 4.60(170) / ((170) - 22)

Vx = 782 / 148

Vx = 5.28 m/s

Therefore, the final velocity of the car is 5.28 m/s