A carbon resistor is to be used as a thermometer. On a winter day when the temperature is 3°C, the resistance of the carbon resistor is 217.6 2 Ω. What is the temperature on a spring day when the resistance is 215.3 2 Ω? (Take the temperature coefficient of resistivity for carbon to be a-5.00 x 10^-4°C^-1) (Give your answer in decimal using "degreeC" °C) as unit)

Answers

Answer 1

Answer:

24.14 ⁰C

Explanation:

T₀ = initial temperature = 3 °C

R₀ = initial resistance of thermometer at initial temperature = 217.62 Ω

R = Final resistance of thermometer at final temperature = 215.32 Ω

T = final temperature = ?

α = temperature coefficient of resistivity for carbon = - 5.00 x 10⁻⁴ C⁻¹

Final resistance of thermometer at final temperature is given as

R = R₀ (1 + α (T - T₀ ))

Inserting the values

215.32 = 217.62 (1 + (- 5.00 x 10⁻⁴) (T - 3))

T = 24.14 ⁰C


Related Questions

What is the final velocity of a 28 kg object that has an initial velocity of 5 m/s and has a force of 232 N exerted on in for 5.2 seconds? Do not leave a space between the units and the answer. No decimals-round to nearest whole number.

Answers

Answer:

48 m/s

Explanation:

m = mass of the object = 28 kg

F = magnitude of net force acting on the object = 232 N

acceleration of the object is given as

a = F/m

a = 232/28

a = 8.3 m/s²

v₀ = initial velocity of the object = 5 m/s

v = final velocity of the object

t = time interval = 5.2 s

using the kinematics equation

v = v₀ + a t

v = 5 + (8.3) (5.2)

v = 48 m/s

A force does 210 J of work when it acts on a moving object and its direction is in the same direction as the object’s displacement. How much work does this force do when the angle between it and the object’s displacement is 56°

Answers

Answer:

When the angle is 56° the work done is 117.43 J

Explanation:

Work = F . s = Fscosθ

We have

   W1 = 210 J, θ = 0°

Substituting

      210 = F x s x cos 0 = Fs

Now we have to find W2 when angle θ = 56°

Substituting

      W2 = F x s x cos 56 = 210 cos56= 117.43 J

When the angle is 56° the work done is 117.43 J

A wave travels through a medium at 251 m/s and has a wavelength of 5.10 cm. What is its frequency? What is its angular frequency? What is its period of oscillation?

Answers

Final answer:

The frequency of the wave is 4921.57 Hz. The angular frequency is 30937.48 rad/s. The period of oscillation is 0.000203 s.

Explanation:

To find the frequency of the wave, we can use the formula:

v = λ × f

Plugging in the values given, we have:

251 = 0.051 × f

solving for f, we get:

f = 251 / 0.051 = 4921.57 Hz

The angular frequency (represented as ω) can be calculated using the formula:

ω = 2π × f

Substituting the value of f we found, we get:

ω = 2×3.1416×4921.57 = 30937.48 rad/s

The period of oscillation (represented as T) is the reciprocal of the frequency.

T = 1 / f = 1 / 4921.57 = 0.000203 s

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A football is kicked and leaves the ground at an angle 37.0° with a velocity of 20.0 ms^-1. Calculate the maximum height, the time of travel before the football hits the ground and how far away it hits the ground. Assume the ball leaves the foot at ground level, and ignore air resistance and rotation of the ball.

Answers

Maximum height is approximately 10.4 meters.Time of travel is around 3.22 seconds.Horizontal distance is about 54.3 meters.

1. Maximum Height:

The maximum height (H) can be calculated using the vertical motion equation for projectile motion:

[tex]\[ H = \dfrac{v_{0y}^2}{2g}, \][/tex]

Here:

[tex]\(v_{0y}\)[/tex] is the initial vertical component of velocity (initial velocity [tex]\(v_0\)[/tex] multiplied by the sine of the launch angle),

g is the acceleration due to gravity ([tex]\(9.81 \, \text{m/s}^2\)[/tex] near the surface of the Earth).

Substitute the given values:

[tex]\[ H = \dfrac{(20.0 \, \text{m/s} \cdot \sin(37.0^\circ))^2}{2 \cdot 9.81 \, \text{m/s}^2}\\\\ \approx 10.4 \, \text{m}. \][/tex]

2. Time of Travel:

The total time of flight ([tex]\(t_{\text{total}}\)[/tex]) can be found using the equation for vertical motion:

[tex]\[ t_{\text{total}} = \dfrac{2v_{0y}}{g}. \][/tex]

Substitute the given values:

[tex]\[ t_{\text{total}} = \dfrac{2 \cdot 20.0 \, \text{m/s} \cdot \sin(37.0^\circ)}{9.81 \, \text{m/s}^2}\\\\ \approx 3.22 \, \text{s}. \][/tex]

3. Horizontal Distance:

The horizontal distance (d) can be calculated using the horizontal motion equation:

[tex]\[ d = v_{0x} \cdot t_{\text{total}}, \][/tex]

Here:

[tex]\(v_{0x}\)[/tex] is the initial horizontal component of velocity (initial velocity [tex]\(v_0\)[/tex] multiplied by the cosine of the launch angle),

[tex]\(t_{\text{total}}\)[/tex] is the total time of flight calculated in the previous step.

Substitute the given values:

[tex]\[ d = 20.0 \, \text{m/s} \cdot \cos(37.0^\circ) \cdot 3.22 \, \text{s}\\\\ \approx 54.3 \, \text{m}. \][/tex]

Thus, Maximum height is approximately 10.4 meters, time of travel is around 3.22 seconds, and horizontal distance is about 54.3 meters.

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Final answer:

The football kicked at an angle of 37.0° with a velocity of 20.0 m/s will reach a maximum height of 7.32 meters. It will stay in the air for a total of 2.44 seconds and land 39 meters away.

Explanation:

The subject of your question is projectile motion, which is a part of Physics. Given the initial angle of 37.0° and a velocity of 20.0 m/s, we can calculate the maximum height, time of travel, and horizontal distance traveled using the equations of motion.

First, we need to find the components of the initial velocity. The vertical component (Voy) is found using Voy = Vo*sin(θ) = 20.0 m/s * sin(37.0°) = 12.01 m/s. The horizontal component (Vox) is Vox = Vo*cos(θ) = 20.0 m/s * cos(37.0°) = 15.97 m/s.

The maximum height (H), also known as the apex of the trajectory, can be calculated using the formula H = Voy^2 / (2*g) where g is the acceleration due to gravity (9.81 m/s²). So, H = (12.01 m/s)^2 / (2*9.81 m/s²) = 7.32 meters.

The total time of travel (t), from launch to when the ball returns back to the ground, is t = 2*Voy/g = 2*(12.01 m/s) / (9.81 m/s²) = 2.44 seconds.

Finally, the horizontal distance traveled, known as the range (R), is R = Vox * t = 15.97 m/s * 2.44 seconds = 39 meters.

So the maximum height the football reaches is 7.32 m, the time it stays in the air is 2.44 seconds, and it hits the ground 39 meters away.

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A ball is thrown with a velocity of 3.0 m/s at an angle of 30° above horizontal. Its vertical velocity is zero when: The ball is at its maximum height. Just before the ball hits the ground. When the ball changes direction. Never during the ball's flight.

Answers

Answer:

The ball is at its maximum height.

Explanation:

As we know that when ball is thrown at some angle with the horizontal then the component of its velocity is given as

[tex]v_x = vcos\theta[/tex]

[tex]v_y = vsin\theta[/tex]

now here vertical velocity is the velocity in y direction

so it is given as

[tex]v_y = 3.0 sin30 = 1.5 m/s[/tex]

now as the velocity of ball in vertical direction becomes zero

then in that case

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

[tex]0 - 1.5 = (-9.8)t[/tex]

[tex]t = 0.15 s[/tex]

since at this position the vertical component of the velocity is zero

so this is the position of ball when its height is maximum

Final answer:

The vertical velocity of a ball thrown at an angle becomes zero at its maximum height. This is the point in its trajectory where gravity has completely counteracted its initial upward velocity. At no other time during its flight will the ball's vertical velocity be zero.

Explanation:

The motion of a projectile launched at an angle involves two components of motion – horizontal and vertical. When a ball is thrown with a velocity of 3.0 m/s at an angle of 30° above horizontal, it will have both horizontal and vertical components of velocity. The vertical velocity of the ball becomes zero when the ball reaches its maximum height, as there is no upward velocity to counteract gravity at this point.

To find the vertical component of initial velocity (Vy), we use the formula Vy = V × sin(θ), where V is the initial velocity and θ is the launch angle. So in this case, the vertical velocity component is Vy = 3.0 m/s × sin(30°).

At maximum height, gravity has slowed the vertical velocity to zero. This is the only point during the ball's trajectory where the vertical component of velocity is zero. Just before the ball hits the ground, its vertical velocity is not zero but is equal in magnitude and opposite in direction to its vertical velocity on launch. When the ball changes direction horizontally, this does not affect the vertical velocity component. Therefore, the ball's vertical velocity is zero only when it reaches its highest point.

g You are standing on a skateboard, initially at rest. A friend throws a very heavy ball towards you. You can either catch the object or deflect the object back towards your friend (such that it moves away from you with the same speed as it was originally thrown). What should you do in order to MINIMIZE your speed on the skateboard?

Answers

Answer:

catch the ball

Explanation:

if you push the ball back, that equal amount of force will applied to you. however if you catch it, you absorb less of the directional energy

The star nearest to our sun is Proxima Centauri, at a distance of 4.3 light-years from the sun. How far away, in km, is Proxima Centauri from the sun?

Answers

Answer: [tex]4.068(10)^{13} km[/tex]

Explanation:

A light year is a unit of length and is defined as "the distance a photon would travel in vacuum during a Julian year at the speed of light at an infinite distance from any gravitational field or magnetic field. "

In other words: It is the distance that the light travels in a year.

This unit is equivalent to [tex]9.461(10)^{12}km[/tex], which mathematically is expressed as:

[tex]1Ly=9.461(10)^{12}km[/tex]

Doing the conversion:

[tex]4.3Ly.\frac{9.461(10)^{12}km}{1Ly}=4.068(10)^{13}km[/tex]

The Answer: The star closest to our sun is 4.2 light years away from the sun.

Explanation:

Suppose that the centripetal force acting on an object in circular motion were doubled to a new value, and the object remained in a circular path with the same radius. How would the motion be affected?

Answers

Explanation:

Centripetal force is mass times centripetal acceleration:

F = m v² / r

If force is doubled while mass and radius are held constant, then velocity will increase.

2F = m u² / r

2 m v² / r = m u² / r

2 v² = u²

u = v√2

So the velocity increases by a factor of √2.

Doubling the centripetal force acting on an object in circular motion, while maintaining the same radius, necessitates an increase in the object's velocity, leading to faster circular motion.

The question explores how doubling the centripetal force acting on an object in uniform circular motion, while maintaining the same radius, affects its motion. According to Newton's second law, the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. The formula for centripetal force is given as Fc = m(v2/r), where m is mass, v is velocity, and r is the radius of the circular path. Doubling the centripetal force while keeping the radius constant means that for the force to remain balanced and the object to stay in circular motion, the velocity of the object must increase. This is because the square of the velocity (v2) is directly proportional to the force applied. Therefore, the object's speed around the circular path will increase, resulting in a faster circular motion.

Consider two identical small steel balls dropped into two identical containers, one filled with water and the other with oil. Which ball will reach the bottom of the container first? Why?

Answers

Answer:

Explanation:

When a body is moving in a liquid, it experiences a backward dragging force. this backward dragging force is given by the formula

F = 6 π η r v

where, η is called the coefficient of viscosity, r be the radius, v be the velocity of object.

As we know that the coefficient of friction for oil is more than water so the dragging force in oil is more than water.

So, the balls in water comes first.

Assume that in interstellar space the distance between two electrons is about 0.56 cm. The electric force between the two electrons is 1. Attractive. 2. Repulsive.

Answers

Answer: 2. Repulsive

The electric force is a force that is inversely proportional to the square of the distance. This can be proved by Coulomb's Law, which states:  

"The electrostatic force [tex]F_{E}[/tex] between two point charges [tex]q_{1}[/tex] and [tex]q_{2}[/tex] is proportional to the product of the charges and inversely proportional to the square of the distance [tex]d[/tex] that separates them, and has the direction of the line that joins them"  

Mathematically this law is written as:  

[tex]F_{E}= K\frac{q_{1}.q_{2}}{d^{2}}[/tex]

Where [tex]K[/tex] is a proportionality constant.

So, if we have two electrons, this means we have two charges with the same sign, hence the electric force between them will be repulsive.

Final answer:

Explaining the attractive electrostatic force and gravitational force between two electrons at atomic scales.

Explanation:

The electrostatic force between two electrons separated by 10^-10 m is attractive. The magnitude of this force can be calculated using Coulomb's law, and for two electrons with charge e = 1.6 x 10^-19 C, the force is strong at atomic distances.

The gravitational force between the electrons is always attractive and weaker than the electrostatic force, with a magnitude of 5.54 x 10^-51 N. This force is much smaller compared to the electrostatic force at atomic scales.

The balance between attractive and repulsive forces is crucial in understanding the stability of atomic structures and the concept of bond length and bond energy in molecules.

Find the amount of Heat conducted per second through a bar of aluminum if the cross sectional area is 30 cm the length of the bar is 1.5 m and one of the ends has a temperature of 25°C and the other has a temperature of 300°C. Thermal conductivity of Aluminum is 1.76 x 10^4 Cal cm / m^2 h°c Convert the units as needed

Answers

Answer:

9680 cal

Explanation:

A = cross-sectional area of the bar = 30 cm² = 30 x 10⁻⁴ m²

L = length of the bar = 1.5 m

T₁ = Temperature at one end of the bar = 25 °C

T₂ = Temperature at other end of the bar = 300 °C

k = Thermal conductivity of Aluminum = 1.76 x 10⁴ Cal cm /(m² ⁰C)

Q = amount of heat conducted per second

Amount of heat conducted per second is given as

[tex]Q = \frac{k A (T_{2} - T_{1})}{L}[/tex]

Q = (1.76 x 10⁴) (30 x 10⁻⁴) (300 - 25)/1.5

Q = 9680 cal

what is the radius of the circular path ofa charged particle in a magnetic field?

Answers

Answer:

The radius of a circular path of a charged particle orbit depends on the charge and velocity of the particle.

Explanation:

A uniform disk is constrained to rotate about an axis passing through its center and perpendicular to the plane of the disk. If the disk starts with an angular velocity of 7.0 rad/s and is subject to a constant angular acceleration of 3.0 rad/s2, find the angular displacement of a point on the rim of the disk as it rotates under these conditions for 15 s. (Assume the positive direction is in the initial direction of the rotation of the disk. Indicate the direction with the sign of your answer.)

Answers

Answer:

442.5 rad

Explanation:

w₀ = initial angular velocity of the disk = 7.0 rad/s

α = Constant angular acceleration = 3.0 rad/s²

t = time period of rotation of the disk = 15 s

θ = angular displacement of the point on the rim

Angular displacement of the point on the rim is given as

θ = w₀ t + (0.5) α t²

inserting the values

θ = (7.0) (15) + (0.5) (3.0) (15)²

θ = 442.5 rad

A 2.0 kg wooden block is slid along a concrete surface (μk = 0.21) with an initial speed of 15 m/s. How far will the block slide until it stops?

Answers

Answer:

The distance is 54.6 m

Explanation:

Given that,

Mass = 2.0 kg

Frictional coefficient = 0.21

Initial velocity = 15 m/s

We need to calculate the acceleration

Using formula of frictional force  

[tex]F = \mu mg[/tex]

[tex]F=0.21\times2.0\times9.8[/tex]

[tex]F = 4.12\ N[/tex]

We need to calculate the acceleration

[tex]F = ma[/tex]

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

[tex]a =\dfrac{4.12}{2.0}[/tex]

[tex]a=2.06\ m/s^2[/tex]

We need to calculate the initial velocity

Using equation of motion

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

Put the value  

[tex]0=15^2-2\times2.06\times s[/tex]

[tex]s = \dfrac{15^2}{2\times2.06}[/tex]

[tex]s=54.6\ m[/tex]

Hence, The distance will be 54.6 m.

A proton starting from rest travels through a potential of 1.0 kV and then moves into a uniform 0.040-T magnetic field. What is the radius of the proton's resulting orbit? (mproton = 1.67 × 10-27 kg, e = 1.60 × 10-19 C)

Answers

Answer:

Radius = 0.11 m

Explanation:

To find the speed of the proton we know that

[tex]KE = PE[/tex]

here we have

[tex]\frac{1}{2}mv^2 = qV[/tex]

now we have

[tex]v = \sqrt{\frac{2qV}{m}}[/tex]

now we have

[tex]v = \sqrt{\frac{2(1.60 \times 10^{-19})(1000)}{(1.67\times 10^{-27})}}[/tex]

[tex]v = 4.38 \times 10^5 m/s[/tex]

Now for the radius of the circular motion of charge we know

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

[tex]R = \frac{mv}{qB}[/tex]

[tex]R = \frac{(1.67\times 10^{-27})(4.38 \times 10^5)}{(1.60\times 10^{-19})(0.040)}[/tex]

[tex]R = 0.11 m[/tex]

The radius of the proton orbit is 0.114m

kinetic energy:

When the proton travels through the potential V it gains kinetic energy given below:

[tex]\frac{1}{2}mv^2=qV\\\\v=\sqrt[]{\frac{2qV}{m} }\\\\v=\sqrt{\frac{2\times (1.6\times10^{-19})\times10^3}{1.67\times10^{-27}}[/tex]

[tex]v=4.37\times10^5m/s[/tex]

magnetic force:

Now, a moving charge under magnetic field B undergoes circular motion due to the magnetic force being perpendicular to the velocity of the charge, given by:

[tex]\frac{mv^2}{r}=qvB[/tex]

[tex]r=\frac{mv}{qB} \\\\r=\frac{1.67\times10^{-27}\times4.37\times10^5}{1.6\times10^{-19}\times0.040} m\\\\r=0.114m[/tex]

the radius of the orbit is 0.114m

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What is the wavelength of an earthquake wave if it has a speed of 8 km/s and a frequency of 3 Hz?

Answers

Answer:

2666.67 m

Explanation:

Changing the 8km/s into SI units we 8000m/s.

Speed of a wave = λf where λ is the wavelength and f he frequency of the wave.

v=λf

Therefore making wavelength the subject of the formula we get

λ= v/f

Substituting with the values in the question we get:

λ= (8000m/s)/3Hz

=2666.667 m

The answer is 266666

You can report sexual harassment to:
Your Supervisor Or Human Resources Department
The New York State Division Of Human Rights
The Equal Employment Opportunity Commission
All Of The Above

Answers

You can report sexual harassment to all of the above.

A small bulb is rated at 7.5 W when operated at 125 V. Its resistance (in ohms) is : (a) 17 (b) 7.5 (c) 940 (d) 2100 (e) 0.45

Answers

Answer:

Resistance of the bulb is 2100 watts.

Explanation:

Given that,

Power of the bulb, P = 7.5 watts

Voltage, V = 125 volts

We have to find the resistance of the bulb. The power of an electrical appliance is given by the following formula as :

[tex]P=\dfrac{V^2}{R}[/tex]

[tex]R=\dfrac{V^2}{P}[/tex]

[tex]R=\dfrac{(125\ V)^2}{7.5\ W}[/tex]

R = 2083.34 ohms

or

R = 2100 ohms

Hence, the correct option is (d) "2100 ohms"

A baseball approaches home plate at a speed of 40.0 m/s, moving horizontally just before being hit by a bat. The batter hits a pop-up such that after hitting the bat, the baseball is moving at 57.0 m/s straight up. The ball has a mass of 145 g and is in contact with the bat for 2.10 ms. What is the average vector force the ball exerts on the bat during their interaction? (Let the +x-direction be in the initial direction of motion, and the +y-direction be up.) F

Answers

Answer:

[tex]F = (3.9\times 10^3)\hat j - 2.8 \times 10^3\hat i[/tex]

Explanation:

Initial momentum of the ball is given as

[tex]P_i = mv_i[/tex]

[tex]P_i = 0.145 (40) = 5.8 kg m/s \hat i[/tex]

now final momentum of the ball is given as

[tex]P_f = 0.145(57) = 8.3 kg m/s \hat j[/tex]

now by the formula of force we have

[tex]F = \frac{P_f - P_i}{\Delta t}[/tex]

now we have

[tex]F = \frac{8.3 \hat j - 5.8 \hat i}{2.10 \times 10^{-3}}[/tex]

[tex]F = (3.9\times 10^3)\hat j - 2.8 \times 10^3\hat i[/tex]

The average vector force the ball exerts on the bat during their interaction is

[tex]\rm F = 3.9\times10^3\;\hat{j}-2.8\times10^3\;\hat{i}[/tex]

Given :

Initial Speed = 40 m/sec

Final Speed = 57 m/sec

Mass = 0.145 Kg

Time = 0.0021 sec

Solution :

Initial Momentum is,

[tex]\rm P_i = mv_i[/tex]

[tex]\rm P_i = 0.145\times 40 = 5.8\;Kg .m/sec\;\hat{i}[/tex]

Final momentum is,

[tex]\rm P_f = mv_f = 0.145\times57=8.3\; Kg.m/sec\; \hat{j}[/tex]

Now,

[tex]\rm F=\dfrac{P_f-P_i}{\Delta t}= \dfrac{8.3\hat{j}-5.8\hat{i}}{2.10\times10^-^3}[/tex]

[tex]\rm F = 3.9\times10^3\;\hat{j}-2.8\times10^3\;\hat{i}[/tex]

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A uniform conducting rod of length 22 cm has a potential difference across its ends equal to 41 mV (millivolts). What is the magnitude of the electric field inside the conductor in units of N/C?

Answers

Answer:

0.186 N/C

Explanation:

The relationship between electric field strength and potential difference is:

[tex]E=\frac{\Delta V}{d}[/tex]

where

E is the electric field strength

[tex]\Delta V[/tex] is the potential difference

d is the distance

Here we have

[tex]\Delta V=41 mV=0.041 V[/tex]

d = 22 cm = 0.22 m

So the electric field magnitude is

[tex]E=\frac{0.041 V}{0.22 m}=0.186 N/C[/tex]

Final answer:

To find the electric field inside a conductor, use the formula E = V/d. With a potential difference of 41 mV and length of 22 cm, the electric field is 0.186 N/C.

Explanation:

The magnitude of the electric field inside a conductor is calculated using the formula E = V/d, where V is the potential difference and d is the length across which the potential difference is applied.

In this case, the conducing rod has a length of 22 cm (0.22 meters) and a potential difference across its ends of 41 mV (0.041 volts). Therefore, the magnitude of the electric field E in the conductor can be calculated as:

E = V/d = 0.041 V / 0.22 m = 0.18636... V/m

Since 1 V/m is equivalent to 1 N/C, the magnitude of the electric field in the conductor is approximately 0.186 N/C.

The protons in a nucleus are approximately 2 ✕ 10^−15 m apart. Consider the case where the protons are a distance d = 1.93 ✕ 10^−15 m apart. Calculate the magnitude of the electric force (in N) between two protons at this distance.

Answers

Answer:

61.8 N

Explanation:

Given data

Charge of the protons (q): 1.60 × 10⁻¹⁹ CoulombDistance between the protons (d): 1.93 × 10⁻¹⁵ metersCoulomb's constant (k): 8.99 × 10⁹ N.m².C⁻²

We can find the magnitude of the electric force (F) between the two protons using Coulomb's law.

[tex]F=k.\frac{q_{1}q_{2}}{d^{2} } \\F=8.99 \times 10^{9} N.m^{2} .C^{-2} .\frac{(1.60 \times 10^{-19}C)^{2} }{(1.93 \times 10^{-15}m)^{2} } \\F=61.8N[/tex]

The magnitude of the electric force is 61.8 N.

The magnitude of the electric force between two protons at this distance is mathematically given as

F=61.8N

What is the magnitude of the electric force (in N) between two protons at this distance?

Question Parameter(s):

The protons in a nucleus are approximately 2 ✕ 10^−15 m apart.

a distance d = 1.93 ✕ 10^−15 m apart.

Generally, the equation for the electric force (F)   is mathematically given as

[tex]F=k.\frac{q_{1}q_{2}}{d^{2} }[/tex]

Therefore

[tex]F=8.99* 10^{9} N.m^{2} .C^{-2} .\frac{(1.60 * 10^{-19}C)^{2} }{(1.93* 10^{-15}m)^{2} }[/tex]

F=61.8N

In conclusion, The force is

F=61.8N

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Consider a car that is initially traveling along a straight stretch of highway at 15 m/s. At t=0 the car begins to accelerate at 2.0 m/s2 in order to pass a truck. The final velocity is vx=25m/s. Calculate how far the car travels during its 5.0 s of acceleration.

Answers

Answer:

The car travels 100 meter during its 5.0 s of acceleration

Explanation:

We have the equation of motion v²=u²+2as, where v is the final velocity, u is the initial velocity, a is the acceleration ans s is the displacement.

v = 25 m/s

a = 2 m/s²

u = 15 m/s

Substituting

      25²=15²+2 x 2 x s

       s = 100 m

The car travels 100 meter during its 5.0 s of acceleration

Red light has a wavelength of 500 nm. What is the frequency of red light?

Answers

Answer:

[tex]0.6 \times 10 {}^{15} [/tex]

Explanation:

By using relation,

Speed of light = frequency × wavelength (in m)

Car goes 60min/hr. What is speed in m/s? 1 mile = 1.6 km

Answers

Answer:

26.7 m/s

Explanation:

The speed of the car is

v = 60 mi/h

We know that

1 mile = 1.6 km = 1600 m

1 h = 60 min = 3600 s

So we can convert the speed from mi/h into m/s by multiplying by the following factor:

[tex]v = 60 \frac{mi}{h} \cdot \frac{1600 m/mi}{3600 s/h}=26.7 m/s[/tex]

A 53.4 kg ice skater is at rest when she throws a snowball east at 14.8 m/s. The skater recoils west at 0.100 m/s. What is the mass of the snowball?

Answers

Answer:

0.36 kg

Explanation:

We can solve the problem by using the law of conservation of momentum: the total momentum at the beginning must be equal to the total momentum after the skater has thrown the ball.

Before the launch, the skater and the snowball are at rest, so the initial total momentum is zero:

[tex]p_i = 0[/tex]

After the launch, the total momentum is:

[tex]p_f = M V + m v[/tex]

where

M = 53.4 kg is the mass of the ice skater

v = -0.100 m/s is the velocity of the ice skater (here we assumed that east is the positive direction)

m is the mass of the snowball

v = +14.8 m/s is the velocity of the snowball

Since momentum must be conserved,

[tex]p_i = p_f\\0 = MV +mv[/tex]

so we can find m:

[tex]m=-\frac{MV}{v}=-\frac{(53.4 kg)(-0.100 m/s)}{14.8 m/s}=0.36 kg[/tex]

Final answer:

The question pertains to the principle of conservation of momentum. The mass of the snowball is found by equating the momentum of an ice skater with the snowball, resulting in a mass of about 0.36 kg for the snowball.

Explanation:

This question represents a physics concept known as conservation of momentum, stating that the total initial momentum of a system equals the total final momentum in the absence of external forces. In this scenario, the initial momentum of the system is zero because both the ice skater and the snowball are initially at rest.

In the final state, the 53.4 kg ice skater moves west at 0.100 m/s, and the snowball moves east at 14.8 m/s. The eastward momentum equates to the westward momentum.

To find the mass of the snowball, we can set the two momenta to be equal, resulting in the equation: (Mass of Skater) * (Velocity of Skater) = (Mass of Snowball) * (Velocity of Snowball), which simplifies to (53.4 kg) * (0.100 m/s) = (Mass of Snowball) * (14.8 m/s). Solving for the mass of the snowball gives approximately 0.36 kg.

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Two children of mass 27 kg and 51 kg sit balanced on a seesaw with the pivot point located at the center of the seesaw. If the children are separated by a distance of 4 m, at what distance from the pivot point is the small child sitting in order to maintain the balance?

Answers

Answer:

2.62 m

Explanation:

Let the small child sit at a distance x from the pivot.

The distance of big child from the pivot is 4 - x .

By using the concept of moments.

Clockwise moments = anticlockwise moments

27 x = 51 ( 4 - x )

27 x = 204 - 51 x

78 x = 204

x = 2.62 m

Which would have the highest frequency of vibration? (Prove mathematically.) Pendulum A with a 200 g mass on a 1.0 m string Pendulum B with a 400 g mass on a 0.5 string

Answers

Answer:

Pendulum B

Explanation:

The time period of a pendulum is given by :

[tex]T=2\pi\sqrt{\dfrac{L}{g}}[/tex]

Case 1.

Mass, m = 200 g = 0.2 kg

Length of string, l = 1 m

Time, [tex]T_1=2\pi\sqrt{\dfrac{1\ m}{9.8\ m/s^2}}[/tex]

T₁ = 2.007 Seconds

Since, [tex]f=\dfrac{1}{T_1}[/tex]

[tex]f_1=\dfrac{1}{2.007}[/tex]

f₁ = 0.49 Hz

Case 2.

Mass, m = 400 g = 0.4 kg

Length of string, l = 0.5 m

Time, [tex]T_2=2\pi\sqrt{\dfrac{0.5\ m}{9.8\ m/s^2}}[/tex]

T₂ = 1.41 seconds

[tex]f₂=\dfrac{1}{T_2}[/tex]

[tex]f₂=\dfrac{1}{1.41}[/tex]

f₂ = 0.709 seconds

Hence, pendulum B have highest frequency of vibration.

Calculate the power output of a 1.5 g fly as it walks straight up a windowpane at 2.4 cm/s .

Express your answer using two significant figures. (Watts)

Answers

Final answer:

The power output of a 1.5 g fly moving upwards at a speed of 2.4 cm/s can be calculated by applying the physics formula for power. The resultant power output is ~0.036 Watts.

Explanation:

The power output of any moving body can be calculated using the formula P = mgh/t where 'P' is power, 'm' is mass, 'g' is acceleration due to gravity, 'h' is height and 't' is time. In order to calculate the power output of the fly, we need to convert the variables to the appropriate units.

So first, convert the mass of the fly to kg, which gives 0.0015 kg. Then, convert the speed from cm/s to m/s, giving us 0.024 m/s. The acceleration due to gravity is approximately 9.8 m/s2. We want to find the power as the fly walks 1 meter.

Substituting the values into the power equation, we get P = (0.0015 kg * 9.8 m/s2 * 1 m) / (1 m / 0.024 m/s) = 0.036 W, which when rounded off to two significant figures is 0.036 Watts.

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Final answer:

To calculate the power output of a fly walking up a windowpane, we need to use the equation Power = force x velocity. By converting the mass of the fly into kilograms, calculating the force exerted by the fly using the equation Force = mass x gravity, and multiplying the force by the velocity, we find that the power output of the fly is approximately 0.00004 Watts or 4 x 10^-5 Watts.

Explanation:

To calculate the power output of the fly, we can use the equation: Power = force x velocity.

First, let's convert the mass of the fly from grams to kilograms. 1.5 g = 0.0015 kg. The force exerted by the fly is equal to its weight, which is given by the equation: Force = mass x gravity.

Assuming the acceleration due to gravity is 9.8 m/s^2, the force exerted by the fly is: Force = 0.0015 kg x 9.8 m/s^2. Finally, we can calculate the power output of the fly by multiplying the force by the velocity: Power = (0.0015 kg x 9.8 m/s^2) x 0.024 m/s. This gives us a power output of approximately 0.00004 Watts or 4 x 10^-5 Watts.

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The moment of inertia for a 5500 kg solid disc is 12100 kg-m^2. Find the radius of the disc? (a) 2.111 m (b) 2.579 m (c) 1.679 m (d) 2.574 m (e) 2.098 m (f) 2.457 m

Answers

Answer:

The radius of the disc is 2.098 m.

(e) is correct option.

Explanation:

Given that,

Moment of inertia I = 12100 kg-m²

Mass of disc m = 5500 kg

Moment of inertia :

The moment of inertia is equal to the product of the mass and square of the radius.

The moment of inertia of the disc is given by

[tex]I=\dfrac{mr^2}{2}[/tex]

Where, m = mass of disc

r = radius of the disc

Put the value into the formula

[tex]12100=\dfrac{5500\times r^2}{2}[/tex]

[tex]r=\sqrt{\dfrac{12100\times2}{5500}}[/tex]

[tex]r= 2.098\ m[/tex]

Hence, The radius of the disc is 2.098 m.

The actual size of a gold nucleus is approximately 7 fm. Knowing this, calculate the kinetic energy (KE, or KEa, however you may choose to label the measurement) that an alpha particle would need to just touch the outside of the nucleus. Does this seem like a reasonable number?

Hint - the LHC is currently the most powerful particle accelerator on Earth, and i?t operates at 13 TeV (TeV = 1012 eV).

Answers

Answer:

(bruh moment)

Explanation:

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