Determine explicitly which is faster, 75 miles per hour or 75 m/s? Express all your results in MKS units and explain your conclusions from the numbers, and explain using words. Show all work.

Answers

Answer 1

Answer:

75 m/s is faster

Explanation:

MKS units stands for meter kilogram seconds

75 miles per hour = 75 mph

1 mile = 1609.34 meters

1 hour = 60×60 = 3600 seconds

1 mph = 1609.34/3600 = 0.44704 m/s

75 mph = 75×0.44707 = 33.52792 m/s

Comparing 75 mph = 33.52792 m/s with 75 m/s it can be seen that 75 m/s is faster. Even without calculating the values you can know the answer. 75 mph means that in 1 hour the object will move 75 miles. 75 m/s means that in one second the object will cover 75 meters multiply by 3600 and you will get 270000 m/h that is 270 km/h divide it by 1.6 and you can approximately get the value in mph that will be around 168 mph which is faster than 75 mph.


Related Questions

To practice Problem-Solving Strategy 11.1 for conservation of momentum problems.An 80-kg quarterback jumps straight up in the air right before throwing a 0.43-kg football horizontally at 15 m/s . How fast will he be moving backward just after releasing the ball?

Answers

Answer:

0.081 m /s

Explanation:

According to the conservation of momentum, the momentum of a system is conserved when no external force is applied on a body.

momentum of the system before throwing = momentum of the system after throwing

Let v be the velocity of quarterback after throwing the football.

80 x 0 + 0.43 x 0 = 80 x v + 0.43 x 15

0 = 80 v + 6.45

v = - 0.081 m /s

The negative sign shows that he is moving in backward direction.

Final answer:

The quarterback will be moving backward at approximately 0.080625 m/s just after releasing the football due to the conservation of momentum.

Explanation:

The student's question involves practicing Problem-Solving Strategy 11.1 for conservation of momentum problems. Specifically, the student is asked to calculate how fast an 80-kg quarterback will be moving backward just after releasing a football that has a mass of 0.43 kg and is thrown horizontally at a speed of 15 m/s. To solve this, we use the principle of conservation of momentum, which states that the total momentum of a system remains constant if no external forces act on it.

Before the throw, the total momentum of the system (quarterback and football) is zero because both are stationary with respect to horizontal motion. After the throw, the combined momentum must still be zero. We can set up the equation mQB × vQB + mball × vball = 0, where mQB is the mass of the quarterback, vQB is the quarterback's velocity after the throw, mball is the mass of the football, and vball is the velocity of the football. This simplifies to vQB = -(mball × vball) / mQB.

Plugging in the numbers gives us vQB = -(0.43 kg × 15 m/s) / 80 kg, which yields a velocity of vQB ≈ -0.080625 m/s. The negative sign indicates that the quarterback is moving in the opposite direction of the throw, which is backward.

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Does resistance in a circuit remain constant if the temperature changes?

Answers

Answer:

No the resistance of a given circuit does not remain constant if the temperature of the circuit changes.

Explanation:

The resistance of any resistor used in a circuit depends upon the temperature  of that resistor. This can be mathematically represented as follows

[tex]R(t)=R_{0}(1+\alpha \Delta t)[/tex]

Where,

R(t) is resistance of any resistor at temperature t

[tex]R_{o}[/tex] is the resistance of the resistor at time of fabrication

α is temperature coefficient of resistivity it's value is different for different materials

This change in the resistance is the cumulative effect of:

1) Variation of resistivity with temperature

2) Change in dimensions of the resistor with change in temperature

Final answer:

Resistance in a circuit changes with temperature due to increased atomic vibrations affecting electron movement.

Explanation:

Resistance in a circuit does not remain constant if the temperature changes. As temperature increases, the resistance of a conductor typically increases due to the atoms vibrating more rapidly, causing more collisions for the electrons passing through.

This change in resistance with temperature is a common phenomenon seen in various materials. It generally increases with increasing temperature due to more frequent electron collisions within the conductor.

Ohm's Law, which states that the current flowing through a conductor between two points is directly proportional to the voltage across the two points, assumes constant temperature.

In practical terms, if you were to graph resistance against temperature, for some materials, you would notice a linear increase in resistance for small temperature changes, while for large changes, the relationship can be nonlinear.

A projectile is fired at an upward angle of 29.7° from the top of a 108-m-high cliff with a speed of 130-m/s. What will be its speed (in m/s) when it strikes the ground below?

Answers

Answer:

79.2 m/s

Explanation:

θ = angle at which projectile is launched = 29.7 deg

a = initial speed of launch = 130 m/s

Consider the motion along the vertical direction

v₀ = initial velocity along the vertical direction = a Sinθ = 130 Sin29.7 = 64.4 m/s

y = vertical displacement = - 108 m

a = acceleration = - 9.8 m/s²

v = final speed as it strikes the ground

Using the kinematics equation

v² = v₀² + 2 a y

v² = 64.4² + 2 (-9.8) (-108)

v = 79.2 m/s

A 100-kg box is a rest on the floor. The coefficient of static friction between the box and the floor is 0.40, while the coefficient of kinetic friction is 0.20. You apply a rightward horizontal force of P = 300 N. Part B What is the normal force acting on the block?

Answers

Answer:

The normal force acting on the block is 980 N.

Explanation:

Mass of box = 100 kg

Coefficient of static friction = 0.40

Coefficient of kinetic friction = 0.20

We need to calculate the normal force

We know that,

When the object is placed on the flat surface then the normal force is equal to the weight of the object.

So. The normal force = mass x acceleration due to gravity

[tex]N=mg[/tex]

Where, N = Normal force

m = mass of object

g = acceleration due to gravity

[tex]N=100\times9.8[/tex]

[tex]N=980\ N[/tex]

Hence, The normal force acting on the block is 980 N.

Final answer:

The normal force acts perpendicular to the floor supporting the box and is equal to 980 N.

Explanation:

The question asks about the normal force acting on a 100-kg box at rest on the floor. To find the normal force on the box, we can use the equation that relates the normal force to the weight of the object, which is W = mg, where m is the mass and g is the acceleration due to gravity.

Given that the mass of the box is 100 kg and the acceleration due to gravity is 9.80 m/s², the normal force can be calculated as:

Normal Force (N) = Mass (m) × Acceleration due to gravity (g)

Normal Force = 100 kg × 9.80 m/s² = 980 N

The normal force acts perpendicular to the floor supporting the box and is equal to 980 N. It is often confused with the applied force P, but the normal force is related to the box's weight, not the horizontal force applied.

For what position of the object will a spherical concave mirror project on the screen an image smaller than the object? a. between focus and center
b. between the focus and the mirror
c. at the center of curvature
d. beyond the center of curvature

Answers

Answer:

option (d)

Explanation:

A concave mirror always forms a real and inverted image of an object except when the object placed between pole and focus of the mirror.

When the object is placed beyond the centre of curvature, it forms a image which is smaller than the object but it is real and inverted in nature.

Final answer:

The position for which a concave mirror projects a screen image smaller than the object is when the object is placed beyond the center of curvature.

Explanation:

For what position of the object will a spherical concave mirror project on the screen an image smaller than the object? The correct answer is d. beyond the center of curvature. When an object is placed beyond the center of curvature, the concave mirror forms a real, inverted image that is reduced in size, or smaller than the object itself.

This effect can be understood through ray diagrams where rays travelling parallel to the axis, striking the center of the mirror, and moving toward the focal point, all converge to form an image between the focal point and the center of curvature of the mirror. However, if the object is placed closer to the mirror, such as between the focus and the mirror, the produced image would be larger than the object.

An old millstone, used for grinding grain in a gristmill, is a solid cylindrical wheel that can rotate about its central axle with negligible friction. The radius of the wheel is 0.330 m. A constant tangential force of 200 N applied to its edge causes the wheel to have an angular acceleration of 0.936 rad/s2. (a) What is the moment of inertia of the wheel (in kg · m2)?

Answers

Answer:

The answer is I=70,513kgm^2

Explanation:

Here we will use the rotational mechanics equation T=Ia, where T is the Torque, I is the Moment of Inertia and a is the angular acceleration.

When we speak about Torque it´s basically a Tangencial Force applied over a cylindrical or circular edge. It causes a rotation. In this case, we will have that T=Ft*r, where Ft is the Tangencial Forge and r is the radius

Now we will find the Moment of Inertia this way:

[tex]Ft*r=I*a[/tex] -> [tex](Ft*r)/(a) = I[/tex]

Replacing we get that I is:

[tex]I=(200N*0,33m)/(0,936rad/s^2)[/tex]

Then [tex]I=70,513kgm^2[/tex]

In case you need to find extra information, keep in mind the Moment of Inertia for a solid cylindrical wheel is:  

[tex]I=(1/2)*(m*r^2)[/tex]

A solar heating system has a 25.0% conversion efficiency; the solar radiation incident on the panels is 1 000 W/m2. What is the increase in temperature of 30.0 kg of water in a 1.00-h period by a 4.00-m2-area collector? (cw = 4 186 J/kg×°C)

Answers

Final answer:

The increase in temperature of the water is 0.796 °C.

Explanation:

To calculate the increase in temperature of the water, we first need to calculate the total energy absorbed by the collector. The power incident on the collector can be calculated by multiplying the solar radiation intensity by the collector area:

Power incident on the collector = 1000 W/m² × 4 m² = 4000 W

The energy absorbed by the collector can be calculated by multiplying the power incident on the collector by the conversion efficiency:

Energy absorbed by the collector = 4000 W × 0.25 = 1000 J/s

Now we can calculate the increase in temperature of the water using the specific heat formula:

ΔT = Energy absorbed by the collector / (mass of water × specific heat of water)

ΔT = 1000 J/s / (30 kg × 4186 J/kg×°C) = 0.796 °C

A bullet is accelerated down the barrel of a gun by hot gases produced in the combustion of gun powder. What is the average force exerted on a 0.0300-kg bullet to accelerate it to a speed of 600 m/s in a time of 2.00 ms (milliseconds)?

Answers

Answer:

The average force exerted on the bullet are of F=9000 Newtons.

Explanation:

t= 2*10⁻³ s

m= 0.03 kg

V= 600 m/s

F*t= m*V

F= (m*V)/t

F= 9000 N

The average force exerted on the bullet to accelerate it to a speed of 600 m/s in a time of 2.00 ms is 9000 Newtons (N). when a bullet is accelerated down the barrel of a gun by hot gases produced in the combustion of gunpowder.

Given:

Mass of the bullet (m) = 0.0300 kg

The final velocity of the bullet (v) = 600 m/s

Time taken to reach the final velocity (t) = 2.00 ms = 2.00 × 10⁻³ s

acceleration (a) = (change in velocity) / (time taken)

a = (v - u) / t

a = (600 - 0 ) / (2.00 × 10⁻³)

Now, we can calculate the average force using Newton's second law:

force (F) = mass (m) × acceleration (a)

F = 0.0300 × [(600 ) / (2.00 × 10⁻³)]

F = 0.0300× (3.00 × 10⁵)

F = 9000 N

Therefore, the average force exerted on the bullet to accelerate it to a speed of 600 m/s in a time of 2.00 ms is 9000 Newtons (N).

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Be sure to answer all parts. Assume the diameter of a neutral helium atom is 1.40 × 102 pm. Suppose that we could line up helium atoms side by side in contact with one another. How many atoms would it take to make the distance 2.60 cm from end to end? × 10 He atoms (Enter your answer in scientific notation.)

Answers

Answer:

The number of atoms are [tex]1.86\times10^{8}[/tex].

Explanation:

Given that,

Diameter [tex]D = 1.40\times10^{2}\ pm[/tex]

[tex]D=1.40\times10^{2}\times10^{-12}\ m[/tex]

Distance = 2.60 cm

We calculate the number of atoms

Using formula of numbers of atoms

[tex]Number\ of\ atoms =\dfrac{2.60\times10^{-2}}{1.40\times10^{2}\times10^{-12}}[/tex]

[tex]Number\of\atoms =1.86\times10^{8}[/tex]

Hence, The number of atoms are [tex]1.86\times10^{8}[/tex].

A Raman line is observed at 4768.5 À, when acetylene was irradiated by 4358.3 A radiation, Calculate the equilibrium vibrational frequency that causes the shift.

Answers

Answer:

The equilibrium vibrational frequency that causes the shift is [tex]0.56\times10^{14}\ Hz[/tex]

Explanation:

Given that,

Wavelength of Raman line [tex]\lambda'=4768.5\ A[/tex]

Wavelength [tex]\lambda=4358.3\ A[/tex]

We need to calculate the frequency

Using formula of frequency

[tex]f =\dfrac{c}{\lambda}[/tex]

For 4748.5 A

The frequency is

[tex]f'=\dfrac{3\times10^{8}}{4748.5\times10^{-10}}[/tex]

[tex]f' =6.32\times10^{14}\ Hz[/tex]

For 4358.3 A

The frequency is

[tex]f=\dfrac{3\times10^{8}}{4358.3\times10^{-10}}[/tex]

[tex]f=6.88\times10^{14}\ Hz[/tex]

We need to calculate the shift

[tex]\Delta f=f-f'[/tex]

[tex]\Delta f=(6.88-6.32)\times10^{14}\ Hz[/tex]

[tex]\Delta f=0.56\times10^{14}\ Hz[/tex]

Hence, The equilibrium vibrational frequency that causes the shift is [tex]0.56\times10^{14}\ Hz[/tex]

(a) What is the acceleration of gravity on the surface of the Moon? The mass of the moon is 7.35 x 1022 kg and its radius is 1.74 x 106 m. (b) What is the acceleration of gravity on the surface (or outer limit) of Mercury? The mass of Mercury is 3.30 x 1023 kg and its radius is 2.44 x 106 m

Answers

Answer:

Part a)

a = 1.62 m/s/s

Part b)

a = 3.70 m/s/s

Explanation:

Part A)

Acceleration due to gravity on the surface of moon is given as

[tex]a = \frac{GM}{R^2}[/tex]

here we know that

[tex]M = 7.35 \times 10^{22} kg[/tex]

[tex]R = 1.74 \times 10^6 m[/tex]

now we have

[tex]a_g = \frac{(6.67 \times 10^{-11})(7.35 \times 10^{22})}{(1.74 \times 10^6)^2}[/tex]

[tex]a_g = 1.62 m/s^2[/tex]

Part B)

Acceleration due to gravity on surface of Mercury is given as

[tex]a = \frac{GM}{R^2}[/tex]

here we know that

[tex]M = 3.30 \times 10^{23} kg[/tex]

[tex]R = 2.44 \times 10^6 m[/tex]

now we have

[tex]a_g = \frac{(6.67 \times 10^{-11})(3.30 \times 10^{23})}{(2.44 \times 10^6)^2}[/tex]

[tex]a_g = 3.70 m/s^2[/tex]

A wire carries a steady current of 2.60 A. A straight section of the wire is 0.750 m long and lies along the x axis within a uniform magnetic field, vector B = 1.50 k T. If the current is in the positive x direction, what is the magnetic force on the section of wire?

Answers

Answer:

The magnetic force on the section of wire is [tex]-2.925\hat{j}\ N[/tex].

Explanation:

Given that,

Current [tex]I = 2.60\hat{i}\ A[/tex]

Length = 0.750 m

Magnetic field [tex]B = 1.50\hat{k}\ T[/tex]

We need to calculate the magnetic force on the section of wire

Using formula of magnetic force

[tex]\vec{F}=l\vec{I}\times\vec{B}[/tex]

[tex]\vec{F}=0.750\times2.60\hat{i}\times1.50\hat{k}[/tex]

Since, [tex]\hat{i}\times\hat{k}=-\hat{j}[/tex]

[tex]\vec{F}=-2.925\hat{j}\ N[/tex]

Hence, The magnetic force on the section of wire is [tex]-2.925\hat{j}\ N[/tex].

17. (a) What is the terminal voltage of a large 1.54-V carbon-zinc dry cell used in a physics lab to supply 2.00 A to a circuit, if the cell’s internal resistance is 0.100 Ω? (b) How much electrical power does the cell produce? (c) What power goes to its load?

Answers

Answer:

a) 1.34 Volts

b) 3.08 W

c) 2.68 W

Explanation:

Given:

Emf of the cell, E = 1.54 V

current, i = 2.0 A

internal resistance, r = 0.100Ω

(a) Terminal voltage (V) = E - v

where,

v is the potential difference across the resistance 'r'

now,

according to the Ohm's Law, we have

v = i × r  

substituting the values in the above equation we get

v = 2.0 × 0.100 = 0.2 Volts

thus,

Terminal voltage (V) = (1.54 - 0.2) = 1.34 V

(b) Now, the Total power (P) is given as

P = E × i = (1.54 × 2.0) = 3.08 W  

(c) Power into its load = [terminal voltage, v] * i  

= (1.34 × 2.0) = 2.68 W

Final answer:

Terminal voltage of a carbon-zinc cell is 1.34 V, producing 2.68 W of electrical power, with 2.68 W going to the load.

Explanation:

Terminal voltage: The formula to calculate terminal voltage is V = EMF - I * r, where V is the terminal voltage, EMF is the electromotive force, I is the current, and r is the internal resistance of the cell. So, V = 1.54 V - 2.00 A * 0.100 Ω = 1.34 V.

Electrical power: The electrical power produced by the cell is given by P = V * I, where P is power, V is voltage, and I is current. Substituting the values, P = 1.34 V * 2.00 A = 2.68 W.

Power to load: The power delivered to the load is equal to the voltage supplied to the load times the current flowing through it. Hence, the power to the load is P = V_load * I = 1.34 V * 2.00 A = 2.68 W.

A proton moves in a region of constant electric field. Does it follow that the proton’s velocity is parallel to the electric field? Does it follow that the proton’s acceleration is parallel to the electric field? Explain.

Answers

Answer:

Explanation:

A proton is positively charged in nature.

Let an electric field strength is E.

The force on a charged particle placed in an electric field is given by

F = q x E

Where q is the charge

Here, the proton is positively charged, so the direction of force acting on the proton is same as teh direction of electric field strength.

Thus, the motion of proton is parallel to the electric field and the motion is accelerated.

A rocket moves upward from rest with an acceleration of 40 m/s2 for 5 seconds. It then runs out of fuel and continues to move upward. Neglecting any air resistance, how high will it rise?

Answers

Answer:

Maximum height of rocket  = 2538.74 m

Explanation:

We have equation of motion s = ut + 0.5 at²

For first 5 seconds

          s = 0 x 5 + 0.5 x 40 x 5² = 500 m

Now let us find out time after 5 seconds rocket move upward.

We have the equation of motion v = u + at

After 5 seconds velocity of rocket

         v = 0 + 40 x 5 = 200 m/s

After 5 seconds the velocity reduces 9.8m/s per second due to gravity.

Time of flying after 5 seconds

          [tex]t=\frac{200}{9.81}=20.38s[/tex]

Distance traveled in this 20.38 s

          s = 200 x 20.38 - 0.5 x 9.81 x 20.38² = 2038.74 m

Maximum height of rocket = 500 +2038.74 = 2538.74 m

A charge Q is located inside a rectangular box The electric flux through each of the six surfaces of the box is 1 2060 Nm2 C2 1590 Nm2 C 3 1690 Nm2 C 4 3430 Nm2 C 5 1870 Nm2 C and 6 5760 Nm2 C What is Q

Answers

Answer:

[tex]Q = 1.45 \times 10^{-7} C[/tex]

Explanation:

Here flux passing through each surface is given

so total flux through whole cube is given by

[tex]\phi = 2060 + 1590 + 1690 + 3430 + 1870 + 5760[/tex]

[tex]\phi = 16400 Nm^2 C[/tex]

now we also know that total flux through a closed surface depends on the total charge enclosed in the surface

So we will have

[tex]\frac{Q}{\epsilon_0} = 16400[/tex]

[tex]Q = (8.85 \times 10^{-12})(16400)[/tex]

[tex]Q = 1.45 \times 10^{-7} C[/tex]

The total charge [tex](\( Q \))[/tex] enclosed by the rectangular box is approximately [tex]\( 2.544 \times 10^{-4} \, \text{C} \)[/tex].

The electric flux [tex](\( \Phi \))[/tex] through a closed surface is given by Gauss's Law:

[tex]\[ \Phi = \frac{Q}{\varepsilon_0} \][/tex]

where:

- [tex]\( \Phi \)[/tex] is the electric flux,

- [tex]\( Q \)[/tex] is the total charge enclosed by the closed surface,

- [tex]\( \varepsilon_0 \)[/tex] is the permittivity of free space [tex](\( \varepsilon_0 \approx 8.85 \times 10^{-12} \, \text{C}^2/\text{Nm}^2 \))[/tex].

For a closed box, the total electric flux [tex](\( \Phi_{\text{total}} \))[/tex] is the sum of the electric flux through each of its six surfaces. Let's denote the electric flux through each surface as [tex]\( \Phi_i \) where \( i = 1, 2, \ldots, 6 \)[/tex].

[tex]\[ \Phi_{\text{total}} = \sum_{i=1}^{6} \Phi_i \][/tex]

Now, we can set up an equation using the given values:

[tex]\[ \Phi_{\text{total}} = 12060 \, \text{Nm}^2/\text{C} + 1590 \, \text{Nm}^2/\text{C} + 1690 \, \text{Nm}^2/\text{C} + 3430 \, \text{Nm}^2/\text{C} + 1870 \, \text{Nm}^2/\text{C} + 5760 \, \text{Nm}^2/\text{C} \][/tex]

[tex]\[ \Phi_{\text{total}} = 28800 \, \text{Nm}^2/\text{C} \][/tex]

Now, use Gauss's Law to find the total charge [tex](\( Q \))[/tex] enclosed by the closed surface:

[tex]\[ Q = \Phi_{\text{total}} \cdot \varepsilon_0 \][/tex]

[tex]\[ Q = 28800 \, \text{Nm}^2/\text{C} \cdot 8.85 \times 10^{-12} \, \text{C}^2/\text{Nm}^2 \][/tex]

[tex]\[ Q \approx 2.544 \times 10^{-4} \, \text{C} \][/tex]

So, the total charge [tex](\( Q \))[/tex] enclosed by the rectangular box is approximately [tex]\( 2.544 \times 10^{-4} \, \text{C} \)[/tex].

If a material has an index of refraction of 1.61, Determine the speed of light through this medium

Answers

Answer:

1.86 x 10^8 m/s

Explanation:

n = 1.61

The formula for the refractive index is given by

n = speed of light in vacuum / speed of light in material

n = c / v

v = c / n

v = (3 x 10^8) / 1.61

v = 1.86 x 10^8 m/s

Final answer:

The speed of light in a material with an index of refraction of 1.61 is calculated as approximately 1.86 * 10^8 m/s, using the equation v = c/n where c is the speed of light in vacuum and n is the index of refraction.

Explanation:

The speed of light in a given material can be calculated using the index of refraction of the material, as defined by the equation n = c/v, where n is the index of refraction, c is the speed of light in a vacuum, and v is the speed of light in the material.

Given that the index of refraction for the material in question is 1.61, and the speed of light in vacuum, c = 3.00 * 10^8 m/s, the speed of light v in this medium would therefore be calculated by rearranging the equation to v = c/n.

 

By substituting the given values into the equation, v = 3.00 * 10^8 m/s / 1.61, we find that the speed of light in the material is approximately 1.86 * 10^8 m/s.

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The focal length of a planar-convex thin lens in air is 250.0 cm. The glass it is made of has an index of 1.530. Determine the radii of curvature of its surfaces. What would happen to the radii if n was reduced to 1.500?

Answers

Final answer:

The radii of curvature can be determined using the Lens Maker's Equation. For a planar-convex lens, we can consider one surface as flat and another as curved. If the refractive index decreases, the radius of curvature would increase.

Explanation:

To find the radii of curvature for the planar-convex thin lens in air, you can use the Lens Maker's Equation, which is 1/f = (n-1)(1/R1 - 1/R2). Here, f is the focal length, n is the refractive index of the glass, R1 and R2 are the radii of curvature for the two surfaces of the lens.

For a planar-convex lens, one surface is flat (which is the planar side) and another surface is curved (which is the convex side). So, we can consider R1 = ∞ for the flat surface and R2 = R (the required radius) for the convex surface. By substituting these values into the Lens Maker's Equation, we can solve for the radius of curvature of the convex surface.

If n was reduced to 1.500, the radius of curvature would increase because, according to the Lens Maker's Equation, radius of curvature is inversely proportional to (n-1). Thus, as n decreases, the radius of curvature increases.

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Suppose a particle moves along a straight line with velocity v(t)=t2e−2tv(t)=t2e−2t meters per second after t seconds. How many meters has it traveled during the first t seconds?

Answers

Final answer:

The distance traveled by a particle with velocity function

[tex]v(t)=t^2e^-2t[/tex]

over the first t seconds can be obtained by integrating that function from 0 to t using integration by parts.

Explanation:

The distance traveled by a particle can essentially be obtained by integrating the velocity function over a given timeframe. In this case, the velocity function is given by

[tex]v(t)=t^2e^-2t.[/tex]

To calculate the distance traveled over the first t seconds, we need to integrate this function from 0 to t. Therefore, the integral ∫v(t)dt from 0 to t will provide the required solution. Use the method of integration by parts, choosing u=t^2 and

[tex]dv=e^-2t dt,[/tex]

then complete the integration process following the standard procedure.

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A skier moving at 5.23 m/s encounters a long, rough, horizontal patch of snow having a coefficient of kinetic friction of 0.220 with her skis. How far does she travel on this patch before stopping?

Answers

im not that smart but maybe 2 x the mass of the wieght will give u the answer

Select True or False for the following statements about conductors in electrostatic equilibrium. All points of a conductor are at the same potential. Charges prefer to be uniformly distributed throughout the volume of a conductor. The electric field inside the conducting material is always zero. Just outside the surface of a conductor, the electric field is always zero.

Answers

Answers :

1. All points of a conductor are at the same potential. - True

2. Charges prefer to be uniformly distributed throughout the volume of a conductor. - False

3 The electric field inside the conducting material is always zero. -True

4.Just outside the surface of a conductor, the electric field is always zero. - False

a) True

b) False

c) True

d) False

What is a conductor ?

A conductor is a substance or material that allows electricity to flow through it.

a) All points of a conductor are at the same potential is True as charge distribution on the surface of the conductor is uniform

b) Charges prefer to be uniformly distributed throughout the volume of a conductor is False because all the charge comes on the surface and get distributed uniformly on the surface of the conductor and their is no charge inside the conductor

c) The electric field inside the conducting material is always zero is True

Since , all the charge is on the surface of the conductor so , there will not be any charges inside the conductor , this is why there will not be electric field .

d)Just outside the surface of a conductor, the electric field is always zero is False as due to charge on the surface there will be electric field outside the surface of conductor .

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A car is travelling at a constant speed of 26.5 m/s. Its tires have a radius of 72 cm. If the car slows down at a constant rate to 11.7 m/s over 5.2 s, what is the magnitude of the angular acceleration of the tires during that time? (in rad/s^2)

Answers

Answer:

Magnitude of angular acceleration = -3.95 rad/s²

Explanation:

Angular acceleration is the ratio of linear acceleration and radius.

That is

        [tex]\texttt{Angular acceleration}=\frac{\texttt{Linear acceleration}}{\texttt{Radius}}\\\\\alpha =\frac{a}{r}[/tex]

Radius = 72 cm = 0.72 m

Linear acceleration is rate of change of velocity.

[tex]a=\frac{11.7-26.5}{5.2}=-2.85m/s^2[/tex]

Angular acceleration

        [tex]\alpha =\frac{a}{r}=\frac{-2.85}{0.72}=-3.95rad/s^2[/tex]

Angular acceleration = -3.95 rad/s²  

Magnitude =  3.95 rad/s²     

16 A 20-pewton force daurected west and a 5 newton force directed north act concurently on a 5 kg object. Draw the resultant vector of the two forces, calculate its magnitude, and determine the acceleration of the object

Answers

Answer:

20.62 N

4.123 m/s^2

Explanation:

A = 20 N west

B = 5 N North

m = 5 kg

Both the forces acting at right angle

Use the formula of resultant of two vectors.

Let r be the magnitude of resultant of two vectors.

[tex]R = \sqrt{A^{2} + B^{2} + 2 A B Cos\theta}[/tex]

[tex]R = \sqrt{20^{2} + 5^{2} + 2 \times 20 \times 5 \times Cos90}[/tex]

R = 20.62 N

Let a be the acceleeration.

a = Net force / mass = R / m = 20.62 / 5

a = 4.123 m/s^2

A solid block of mass m2 = 1.14 kg, at rest on a horizontal frictionless surface, is connected to a relaxed spring (with spring constant k = 125 N/m whose other end is fixed. Another solid block of mass m1 = 2.27 kg and speed v1 = 2.00 m/s collides with the 1.14 kg block. If the blocks stick together, what is their speed immediately after the collision?

Answers

Answer:

v = 1 m/s

Explanation:

from the principle of conservation of momentum, we have following relation

initial momentum = final momentum

[tex]m_{1}v_{1}+m_{2}v_{2} = (m_{1}+m_{2})v^{2}[/tex]

where

m1 = 1.14 kg

v1 = 2.0 m/s

m2 = 1.14 kg

v2  = 0 m/s

putting all value in the above equation

[tex]1.14 *2.0+ 0 =(1.14+1.14)v^{2}[/tex]

[tex]v =\frac{1.14*2.0}{1.14+1.14}[/tex]

v = 1 m/s

Final answer:

By applying the conservation of momentum principle, the speed of the two blocks immediately after the collision is found by equating the momentum before and after the collision.

Explanation:

In this physics problem, we need to determine the speed of the two blocks attached by a spring immediately after the collision. The principle of conservation of momentum can be applied to this. Before the collision, only block m1 is moving, so the total momentum is m1*v1. After the collision, the two blocks move together at speed v', resulting in total momentum of (m1 + m2) * v'. According to the principle of conservation of momentum, the momentum before collision equals to momentum after collision. Hence, m1*v1 = (m1 + m2)*v'.

Solving this equation will yield the value of v'. Substituting the values of m1 = 2.27 kg, m2 = 1.14 kg, and v1 = 2.00 m/s will result in v' = (2.27 kg * 2.00 m/s) / (2.27 kg + 1.14 kg), which gives the speed of the blocks immediately after collision.

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A tank contains gas at 13.0°C pressurized to 10.0 atm. The temperature of the gas is increased to 95.0°C, and half the gas is removed from the tank. What is the pressure of the remaining gas in the tank?

Answers

Answer:

The pressure of the remaining gas in the tank is 6.4 atm.

Explanation:

Given that,

Temperature T = 13+273=286 K

Pressure = 10.0 atm

We need to calculate the pressure of the remaining gas

Using equation of ideal gas

[tex]PV=nRT[/tex]

For a gas

[tex]P_{1}V_{1}=nRT_{1}[/tex]

Where, P = pressure

V = volume

T = temperature

Put the value in the equation

[tex]10\times V=nR\times286[/tex]....(I)

When the temperature of the gas is increased

Then,

[tex]P_{2}V_{2}=\dfrac{n}{2}RT_{2}[/tex]....(II)

Divided equation (I) by equation (II)

[tex]\dfrac{P_{1}V}{P_{2}V}=\dfrac{nRT_{1}}{\dfrac{n}{2}RT_{2}}[/tex]

[tex]\dfrac{10\times V}{P_{2}V}=\dfrac{nR\times286}{\dfrac{n}{2}R368}[/tex]

[tex]P_{2}=\dfrac{10\times368}{2\times286}[/tex]

[tex]P_{2}= 6.433\ atm[/tex]

[tex]P_{2}=6.4\ atm[/tex]

Hence, The pressure of the remaining gas in the tank is 6.4 atm.

Which object is hotter? Multiple Choice -22:49 O Object 1 at T = 0°C Object 3 at T = OK Object 2 at T = 0°F

Answers

Answer:

Object 1 is hotter.

Explanation:

Object 1 T=[tex]0^{\circ}C[/tex]  

Object 2 T=0 K

Object 3 T=[tex]0^{\circ}F[/tex]

Relation between Celcius ,Kelvin and Fahrenheit

[tex]\dfrac{C-0}{100}=\dfrac{K-273}{100}=\dfrac{F-32}{180}[/tex]

K=C+273,  [tex]K=\dfrac{5}{9}(F-32)+273[/tex].

So now we will convert all in one unit.

Object 1 T=273 K

Object 2 T=0 K

Object 3 T=255.22 K

From above we can say that Object 2 is coolest and object 1 is hottest.

So  Object 1 is hotter.

Alex throws a 0.15-kg rubber ball down onto the floor. The ball's speed just before impact is 6.5 m/s and just after is 3.5 m/s. What is the change in the magnitude of the ball's momentum? A. 0.09 kg*m/s B.1.5 kg*m/s C. 4.3 kg*m/s D. 4.3 kg*m/s

Answers

Answer:

Change in ball's momentum is 1.5 kg-m/s.

Explanation:

It is given that,

Mass of the ball, m = 0.15 kg

Speed before the impact, u = 6.5 m/s

Speed after the impact, v = -3.5 m/s (as it will rebound)

We need to find the change in the magnitude of the ball's momentum. It is given by :

[tex]\Delta p=m(v-u)[/tex]

[tex]\Delta p=0.15\ kg(-3.5\ m/s-6.5\ m/s)[/tex]

[tex]\Delta p=-1.5\ kg-m/s[/tex]

So, the change in the ball's momentum is 1.5 kg-m/s. Hence, this is the required solution.

Final answer:

The change in the magnitude of the ball's momentum is 1.125 kg·m/s.

Explanation:

The change in the magnitude of an object's momentum can be calculated by subtracting its initial momentum from its final momentum. In this case, the initial momentum of the ball is calculated by multiplying its mass (0.15 kg) by its initial speed (6.5 m/s), and the final momentum is calculated by multiplying its mass by its final speed (3.5 m/s).

So, the change in the magnitude of the ball's momentum is calculated as:

|pf| - |pi| = (0.15 kg)(3.5 m/s) - (0.15 kg)(6.5 m/s) = -0.15 kg·m/s - 0.975 kg·m/s = -1.125 kg·m/s.

Since momentum is a vector quantity, the change in its magnitude is given by its absolute value, which is 1.125 kg·m/s. Therefore, the correct answer is A. 1.125 kg·m/s.

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A black hole is a ________

Answers

Answer:

Answer to the question:

Explanation:

A black hole is a finite region of space within which there is a mass concentration high and dense enough to generate a gravitational field such that no material particle, not even light, can escape it.

Does the initial velocity of an object have anything to do with its acceleration? For example, compared to dropping an object, if you throw it downward would the acceleration be different after you released it?

Answers

Answer:

Explanation:

This is an excellent question to get an answer for. It teaches you much about the nature of physics.

The answer is no.

The distance will be quite different. The time might be different in getting to the distance.  But the acceleration will be the same in either case.

How do you know? Look at one of the formulas, say

d = vi * t + 1/2*a * t^2

What does vi do? vi will alter both t and d. if vi = 0 then both d and/or t will be found. But what will "a" do? Is there anything else acting in the up or down line of action? You should answer no.

If vi is not zero, t will be less and d will take less time to get where it is going.

Final answer:

Yes, the initial velocity of an object affects its acceleration. If you drop an object, the initial velocity is zero, resulting in a constant acceleration solely due to gravity. If you throw the object downward, the initial velocity adds to the acceleration due to gravity, leading to a higher overall acceleration.

Explanation:

Yes, the initial velocity of an object can affect its acceleration. When you drop an object, its initial velocity is zero, so the acceleration is solely due to gravity and is constant at approximately 9.8 m/s2 (assuming no air resistance). However, if you throw an object downward, it already has an initial downward velocity, which adds to the acceleration due to gravity. So, the acceleration of the object after you release it will be greater than if you had dropped it.

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You measure the pressure at 10 feet in a column of water and determine it to be 43.36 lbs/in. What would this measurement be in N/m2? 18.2 N/m2 299,034 N/m2 ) 187,334 N/m2 19.9 N/m 2

Answers

Answer:

299034 N/m²

Explanation:

1 lbs = 4.448 N

1 in = 0.0254 m

1 in² = 0.254² m²

thus,

[tex]1\frac{lbs}{in^2} = \frac{4.448N}{0.0254^2m^2}=6894.413N/m^2[/tex]

therefore,

43.36lbs/in² in N/m² will be

= 43.36 × 6894.413

= 298941.77  N/m² ≈ 299034 N/m²

so the correct option is 299034 N/m²

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