A long, straight wire with 2 A current flowing through it produces magnetic field strength 1 T at its surface. If the wire has a radius R, where within the wire is the field strength equal to 84 % of the field strength at the surface of the wire? Assume that the current density is uniform throughout the wire. (μ 0 = 4π × 10-7 T · m/A)

Answer:

The magnetic field in the center of the solenoid is [tex]7.8\times10^{-3}\ T[/tex].

Explanation:

Given that,

Length of solenoid = 0.425 m

Number of turns N = 950

Current I = 2.75 A

The magnetic field in the center of the solenoid is the product of the current , number of turns per unit length and permeability.

In mathematical form,

[tex]B = \mu_{0}nI[/tex]

Where, [tex]n = \dfrac{N}{l}[/tex]

N = number of turns

L = length

I = current

Now, The magnetic field

[tex]B = \dfrac{\mu_{0}NI}{l}[/tex]

Put the value into the formula

[tex]B=\dfrac{4\pi\times10^{-7}\times950\times2.75}{0.425}[/tex]

[tex]B=\dfrac{4\times3.14\times10^{-7}\times950\times2.75}{0.425}[/tex]

[tex]B=7.8\times10^{-3}\ T[/tex]

Hence, The magnetic field in the center of the solenoid is [tex]7.8\times10^{-3}\ T[/tex].

Answer:

The frequency of square coil is 47 Hz.

(A) is correct option.

Explanation:

Given that,

Radius =10 cm

Side = 20 cm

Magnetic field = 1.5 T

Frequency = 60 Hz

We need to calculate the the maximum induced voltage

[tex]V_{m}=BAN\omega[/tex]

Where, B = magnetic field

A = area of cross section

N = number of turns

[tex]\omega=2\pi f[/tex]= angular frequency

Put the value into the formula

[tex]V_{m}=1.5\times3.14\times(10\times10^{-2})^2\times1\times2\times3.14\times60[/tex]

[tex]V_{m}=17.75\ V[/tex]

If the square coil have the same induced voltage.

Area of square of coil [tex]A =(20\times10^{-2})^2[/tex]

[tex]A=0.04\ m^2[/tex]

Now, The angular velocity of square coil

[tex]\omega=\dfrac{V_{m}}{AB}[/tex]

[tex]\omega=\dfrac{17.75}{0.04\times1.5}[/tex]

[tex]\omega=295.8\ \dfrac{rad}{s}[/tex]

Now, frequency of rotation

[tex]f = \dfrac{\omega}{2\pi}[/tex]

Put the value into the formula of frequency

[tex]f=\dfrac{295.8}{2\times3.14}[/tex]

[tex]f=47\ Hz[/tex]

Hence, The frequency of square coil is 47 Hz.

Let [tex]\theta[/tex] be the direction the swimmer must swim relative to east. Then her velocity relative to the water is

[tex]\vec v_{S/W}=\left(4.0\dfrac{\rm km}{\rm h}\right)(\cos\theta\,\vec\imath+\sin\theta\,\vec\jmath)[/tex]

The current has velocity vector (relative to the Earth)

[tex]\vec v_{W/E}=\left(3.0\dfrac{\rm km}{\rm h}\right)\,\vec\imath[/tex]

The swimmer's resultant velocity (her velocity relative to the Earth) is then

[tex]\vec v_{S/E}=\vec v_{S/W}+\vec v_{W/E}[/tex]

[tex]\vec v_{S/E}=\left(\left(4.0\dfrac{\rm km}{\rm h}\right)\cos\theta+3.0\dfrac{\rm km}{\rm h}\right)\,\vec\imath+\left(4.0\dfrac{\rm km}{\rm h}\right)\sin\theta\,\vec\jmath[/tex]

We want the resultant vector to be pointing straight north, which means its horizontal component must be 0:

[tex]\left(4.0\dfrac{\rm km}{\rm h}\right)\cos\theta+3.0\dfrac{\rm km}{\rm h}=0\implies\cos\theta=-\dfrac{3.0}{4.0}\implies\theta\approx138.59^\circ[/tex]

which is approximately 41º west of north.

The swimmer should swim approximately 36.87 degrees in the north-west direction to counteract the eastward ocean current and reach Victoria, British Columbia directly from Port Angeles, Washington. This is derived by considering the velocity vectors of the swimmer and the current.

This question is about the interaction of velocity vectors which is a concept from physics. The swimmer's speed and the ocean current each represent a vector with both a magnitude (speed) and direction. We must factor in these two vectors to work out where the swimmer should aim to swim.

Assuming the north direction as +y axis and east as +x axis, the swimmer intends to swim north with velocity 4.0 km/h and the water's current is moving towards east with velocity 3.0 km/h. In order for the swimmer to move directly north, the swimmer needs to swim against the current, which means westwards. By utilizing the Pythagorean Theorem and trigonometric principles, we can calculate the angle of her direction which is arctan(3/4) in the north-west direction or around 36.87 degrees from north.

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D. The chemical potential energy of real battery decreases, it develops internal resistance and therefore the potential difference across its terminals decreases if its current increases.

Answer:

d) 1000 times

Explanation:

As we know that difference of sound level is given as

[tex]L_2 - L_1 = 10 Log \frac{I_2}{I_1}[/tex]

so here we need to find the ratio of two intensity

it is given as

[tex]Log\frac{I_2}{I_1} = \frac{(L_2- L_1)}{10}[/tex]

[tex]Log\frac{I_2}{I_1} = \frac{60 - 30}{10}[/tex]

[tex]Log\frac{I_2}{I_1} = 3[/tex]

now we have

[tex]\frac{I_2}{I_1} = 10^3[/tex]

so it is

d) 1000 times

Answer:

2.17 s

Explanation:

Here, w0 = 33.3 rpm = 33.3 /60 rps

= 2 × pi × 33.3 / 60 rad/ s

= 3.4854 rad / s

w = 78 rpm = 78 / 60 rps

= 2 × pi × 78 / 60 = 8.164 rad / s

a = 2.15 rad/s^2

Use first equation of motion for rotational motion

w = w0 + a t

8.164 - 3.485 = 2.15 × t

t = 2.17 s

Answer:

B) Its velocity is perpendicular to the acceleration.

Explanation:

For general projectile motion, the horizontal acceleration is 0 and the vertical acceleration is -g.  This is true for all points on the trajectory.

At the highest point, the vertical velocity is 0.  So you have only a horizontal velocity as well as a vertical acceleration.  So the two are perpendicular.

Answer:

(e) 98,1 KJ

Explanation:

The engine produces 19%; it means, it rejects 81% of energy. ⇒ 81/19=4.26 times.

The engine produces 23 kJ; it means it rejects 23 * 4.26 = 98.05263 kJ

Answer:

true

Explanation:

the sun hits it ig idek

The surfaces of equal potential around an infinite charged plane are parallel to the plane: True.

According to the law of electrostatic forces, the surfaces of equal potential around an infinite charged plane are always parallel to the plane because they cannot intersect the charged plane.

Since this conducting sphere of charge is centered at the origin (0), the quantity of energy that would be required to move a point charge from the surface of this sphere (r = R) to the center of the sphere is given by:

E = qV

E = q × 0

E = 0 Joules.

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Answer:

the answer is that u need to divide the 26 by the 22 and then u add seven to that number and then u have to multiply by 2. And you will get your answer

Explanation:

Answer:

Angle, θ = 29.35

Explanation:

It is given that,

Radius of the curve, r = 80 m

Velocity of the car, v = 21 m/s

The highway curve is to be banked so that the car will not skid sideways in the presence of friction. We need to find the angle should the curve be banked. Let the angle is θ. It is given by :

[tex]tan{\theta}=\dfrac{v^2}{rg}[/tex]

Where

g = acceleration due to gravity

[tex]tan\ {\theta}=\dfrac{(21\ m/s)^2}{80\ m\times 9.8\ m/s^2}[/tex]

[tex]\theta=tan^{-1}(0.5625)[/tex]

θ = 29.35°

Hence, this is the required solution.

To find the normal component of acceleration, differentiate the velocity and acceleration functions.

In order to find the normal component of the acceleration, we need to first find the velocity and acceleration vectors as functions of time. Using the given position function, we can differentiate it to find the velocity function:

v(t) = dr/dt = -3(4sin(3t)i + 4cos(3t)ĵ)

Next, we can differentiate the velocity function to find the acceleration function:

a(t) = dv/dt = -3(3cos(3t)i - 3sin(3t)ĵ)

Since the tangential component of the acceleration is constant, we can set the coefficient of the tangential component equal to a constant value:

-3(3sin(3t)) = Constant

Solving this equation will give us the value of the constant. Once we know the value of the constant, we can substitute it back into the acceleration function to find the normal component of the acceleration as a function of time.

Answer:

1.41 m/s^2

Explanation:

First of all, let's convert the two speeds from km/h to m/s:

[tex]u = 94.0 km/h \cdot \frac{1000 m/km}{3600 s/h} = 26.1 m/s[/tex]

[tex]v=46.0 km/h \cdot \frac{1000 m/km}{3600 s/h}=12.8 m/s[/tex]

Now we find the centripetal acceleration which is given by

[tex]a_c=\frac{v^2}{r}[/tex]

where

v = 12.8 m/s is the speed

r = 140 m is the radius of the curve

Substituting values, we find

[tex]a_c=\frac{(12.8 m/s)^2}{140 m}=1.17 m/s^2[/tex]

we also have a tangential acceleration, which is given by

[tex]a_t = \frac{v-u}{t}[/tex]

where

t = 17.0 s

Substituting values,

[tex]a_t=\frac{12.8 m/s-26.1 m/s}{17.0 s}=-0.78 m/s^2[/tex]

The two components of the acceleration are perpendicular to each other, so we can find the resultant acceleration by using Pythagorean theorem:

[tex]a=\sqrt{a_c^2+a_t^2}=\sqrt{(1.17 m/s^2)+(-0.78 m/s^2)}=1.41 m/s^2[/tex]

Answer:

a = 1.406 m/s²

Explanation:

We are told that, the speed of the train decreases from 94.0 km/h to 46.0 km/h

Let's convert both to m/s.

Thus,

v1 =94 km/h =(94 x 10)/36 =26.11 m/s

v2=46 km/h =(46 x 10)/36=12.78 m/s

The formula to calculate the tangential acceleration is given by;

a_t = -dv/dt

Where;

dv is change in velocity

dt is time difference

dv is calculated as; dv = v1 - v2

Thus, dv = 26.11 - 12.78 = 13.33 m/s

We are given that, t = 17 seconds

Thus;

a_t = -13.33/17 = -0.784 m/s²

The negative sign implies that the acceleration is inwards.

Now, let's calculate the radial acceleration;

a_r = v²/r

Where;

r is the radius of the path = 140m

v is the velocity at the instant given

a_r is radial acceleration.

Thus,

a_r = 12.78²/140 = 1.167 m/s²

Now, the tangential and radial components of acceleration are perpendicular to each other. Thus, we can use using Pythagoreas theorem to find the resultant acceleration;

Thus;

a² = (a_t)² + (a_r)²

Plugging in the relevant values, we have;

a² = (-0.784)² + (1.167)²

a² = 1.976545

a = √1.976545

a = 1.406 m/s²

Answer: [tex]1.1052g/cm^{3}[/tex]

Explanation:

Density [tex]D[/tex] is a characteristic property of a material and is defined as the relationship between the mass [tex]m[/tex] and volume [tex]V[/tex] of a specific substance or material. So, the density of the asteroid is given by the following equation:

[tex]D=\frac{m}{V}[/tex]   (1)

On the other hand, we know the asteroid has a mass [tex]m=1000kg[/tex] and is spherical. This means its volume is given by the following formula:

[tex]V=\frac{4}{3}}\pi r^{3}[/tex]   (2)

Where [tex]r=\frac{d}{2}=\frac{1.2m}{2}=0.6m[/tex]  is the radius of the sphere and is half its diameter [tex]d[/tex].

Knowing this, we can calculate the volume:

[tex]V=\frac{4}{3}}\pi (0.6m)^{3}[/tex]   (3)

[tex]V=0.904m^{3}[/tex]   (4)

Substituting (4) in (1):

[tex]D=\frac{1000kg}{0.904m^{3}}=1105.242\frac{kg}{m^{3}}[/tex]   (5) This is the density of the asteroid, but we were asked to find it in [tex]\frac{g}{cm^{3}}[/tex]. This means we have to make the conversion:

[tex]D=1105.242\frac{kg}{m^{3}}.\frac{1000g}{1kg}.\frac{1m^{3}}{(100cm)^{3}}[/tex]

Finally:

[tex]D=1.1052\frac{g}{cm^{3}}[/tex]

Answer:

Spring constant, k = 47 N/m

Explanation:

It is given that,

Force applied to a spring, F = 34 N

A very light ideal spring moves 0.73 m from equilibrium position i.e. x = 0.73 m

We have to find the force constant or spring constant of the spring. It can be calculated using Hooke's law. According to him, the force acting on the spring when it compresses or stretches is given by :

[tex]F=-kx[/tex]  (-ve sign shows opposite direction)

[tex]k=\dfrac{F}{x}[/tex]

[tex]k=\dfrac{34\ N}{0.73\ m}[/tex]

k = 46.5 N/m

or

k = 47 N/m

Hence, the spring constant of the spring is 47 N/m.

Utilizing Hooke's law, which posits that the force needed to compress or extend a spring is proportional to the distance, the spring constant can be calculated to be approximately 46.6 N/m by dividing the force (34N) by the displacement (0.73m). So, the closest option amongst the provided ones is (A) 47N/m.

In the situation described where a force of 34N stretches a very light ideal spring 0.73 m from equilibrium, you would want to find the spring's force constant, often denoted as k. This involves a principle in physics known as Hooke's law, which states that the force needed to extend or compress a spring by some distance is proportional to that distance. The equation for Hooke's law is F=kx.

To find the spring constant (k), you rearrange the equation for Hooke's Law to give you: k = F/x. Substituting the force (F = 34N) and displacement from equilibrium (x = 0.73m) into the equation gives: k = 34N / 0.73m. Calculating this gives a spring constant of approximately 46.6 N/m. So, none of the given options (A) 47N/m  (B) 38N/m  (C) 53N/m  (D) 25N/m are exactly correct, but Option A is the closest.

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Answer:

4.72 N

Explanation:

The charge density across each plate is given by:

[tex]\sigma = \frac{Q}{A}[/tex]

where

[tex]Q=5\cdot 10^{-6}C[/tex] is the charge on each plate

[tex]A=0.3 m^2[/tex] is the area of each plate

Solving,

[tex]sigma = \frac{5\cdot 10^{-6}C}{0.3 m^2}=1.67\cdot 10^{-5} C/m^2[/tex]

The force exerted by one plate on the other is given by:

[tex]F=\frac{Q\sigma}{2\epsilon_0}[/tex]

where

[tex]Q=5\cdot 10^{-6}C[/tex] is the charge on each plate

[tex]\sigma=1.67\cdot 10^{-5} C/m^2[/tex] is the surface charge density

[tex]\epsilon_0[/tex] is the vacuum permittivity

Substituting,

[tex]F=\frac{(5\cdot 10^{-6} C)(1.67\cdot 10^{-5} C/m^2)}{2(8.85\cdot 10^{-12}F/m)}=4.72 N[/tex]

We can find the force exerted by one plate of a parallel-plate capacitor on the other by utilizing the known quantities and the formula F = Q² / (A * ε), which arises from force definition and electric field for a parallel-plate capacitor.

The magnitude of the force between two charges is given by Coulomb's law: F = k * (Q1 * Q2) / d², where k is the Coulomb's constant, Q1 and Q2 are the charges, and d is the distance. The electric field for a parallel-plate capacitor is E = Q / (A * ε), where ε is the permittivity of free space.

The force exerted on each plate of the capacitor is F = QE which, once substituted, becomes F = Q² / (A * ε) when charge, plate area, and permittivity are known. Given the values of the problem, by substituting into the equation, you can calculate the force between the plates.

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Answer:

7.8 V

Explanation:

When resistors are connected in series, they are connected in the same branch of the circuit. This means that the current flowing each resistor is the same, while the sum of the voltage drops across each resistor is equal to the potential difference of the power supply:

[tex]V= V_1 + V_2 + V_3[/tex]

In this circuit we have:

V = 15 V

[tex]V_1 = 4.1 V[/tex]

[tex]V_2 = 3.1 V[/tex]

So, the voltage drop across resistor 3 is

[tex]V_3 = V-V_1 - V_2 = 15 V - 4.1 V - 3.1 V=7.8 V[/tex]

Final answer:

The exact potential drop across the third resistor, when three resistors are connected in series to a 15-V power supply and the voltage drops across the first two are 4.1V and 3.1V, is 7.8 volts.

Explanation:

The question asks for the exact potential drop across resistor 3 when three resistors are connected in series across a 15-V power supply, and the potential drops across the first two resistors are known. In a series circuit, the total voltage supplied is equal to the sum of individual voltage drops across all components. Given that the voltage drops across resistors 1 and 2 are 4.1 volts and 3.1 volts respectively, the voltage drop across resistor 3 can be calculated by subtracting the sum of the voltage drops across resistors 1 and 2 from the total voltage supplied by the power source.

Therefore, the calculation is:

Total voltage supplied by the power source = 15 volts

Sum of voltage drops across resistor 1 and 2 = 4.1 volts + 3.1 volts = 7.2 volts

Voltage drop across resistor 3 = Total voltage - Sum of voltage drops across resistor 1 and 2 = 15 volts - 7.2 volts = 7.8 volts.

Answer:

4.15 x 10^6 N

Explanation:

Area, A = 1.43 cm^2 = 1.43 x 10^-4 m^2

mass, m = 60.5 kg

Weight, F = m g = 60.5 x 9.8 = 592.9 N

Pressure = Force / Area

P = Weight / Area

P = 592.9 / (1.43 x 10^-4)

P = 4.15 x 10^6 N

Answer:

0.0018 V

Explanation:

According to the law of conservation of energy, the kinetic energy gained by the particle is equal to the electric potential energy lost:

[tex]\frac{1}{2}mv^2 = q\Delta V[/tex]

where

[tex]m=3.8\cdot 10^{-6} kg[/tex] is the mass of the particle

[tex]v=36 m/s[/tex] is the final speed of the particle

q = -2.7 C is the charge

[tex]\Delta V[/tex] is the potential difference between the two points

Solving for [tex]\Delta V[/tex], we find

[tex]\Delta V= \frac{mv^2}{q}=\frac{(3.8 \cdot 10^{-6} kg)(36 m/s)^2}{-2.7 C}=-0.0018 V[/tex]

The particle has been accelerated by this potential difference: since it is a negative charge, it means that the particle has moved from a point at lower potential towards a point of higher potential.

So, since the initial point is A and the final point is B, the result is

[tex]V_B - V_A = 0.0018 V[/tex]

Answer:

1.67 m

Explanation:

Ux = 3 m/s, ax = -2.7 m/s^2, vx =0

Let the distance travelled before stopping along x axis is x.

Use third equation of motion along x axis

Vx^2 = ux^2 + 2 a x

0 = 9 - 2 × 2.7 × x

X = 1.67 m

Answer:

60.4 J

Explanation:

The work done by the gas is given by:

[tex]W=p(V_f-V_i)[/tex]

where

p is the gas pressure

[tex]V_f[/tex] is the final volume of the gas

[tex]V_i[/tex] is the initial volume

We must convert all the quantities into SI units:

[tex]p=860 torr \cdot \frac{1.013\cdot 10^5 pa}{760 torr}=1.146\cdot 10^5 Pa[/tex]

[tex]V_i = 127 mL = 0.127 L = 0.127 dm^3 = 0.127 \cdot 10^{-3}m^2[/tex]

[tex]V_f = 654 mL = 0.654 L = 0.654 dm^3 = 0.654 \cdot 10^{-3}m^2[/tex]

So the work done is

[tex]W=(1.146\cdot 10^5 Pa)(0.654\cdot 10^{-3} m^3-0.127\cdot 10^{-3} m^3)=60.4 J[/tex]

Answer:

[tex]W=60.4 J[/tex]

Work done is 60.4 Joules (J)

Explanation:

Work done 'W' is given by:

[tex]W=P\triangle V[/tex]

Where:

ΔV is the change in Volume.

P is the pressure.

Change in Volume=Final Volume-Initial Volume

Initial Volume= 127 mL

Final Volume= 654 mL

ΔV=654-127 (mL)

ΔV=527 mL

ΔV=0.527 L

Pressure Conversion:

1 atm=760 torr

[tex]P=\frac{860}{760} \\P=1.1315\ atm[/tex]

Now,

[tex]W=P\triangle V[/tex]

[tex]W=1.1315\ atm*0.527\ L[/tex]

[tex]W=0.5963 L.atm[/tex]

In joule (J): (Conversion 1 atm. L=101.325 J)

[tex]W=0.5963\ L.atm*\frac{101.325 J}{1\ L.atm}[/tex]

[tex]W=60.4 J[/tex]

Work done is 60.4 Joules (J)

Answer:

A) 1.1 m/s/s

Explanation:

There exist two forces on the object such that

[tex]F_1[/tex] = 65 N directed 59° clockwise from the positive x-axis

[tex]F_2[/tex] = 35 N at 32° clockwise from the positive y-axis

now we have

[tex]F_1 = 65 cos59\hat i - 65 sin59 \hat j[/tex]

[tex]F_2 = 35 sin32\hat i + 35 cos32 \hat j[/tex]

now the net force on the object is given as

[tex]F_{net} = F_1 + F_2[/tex]

[tex]F_{net} = (65 cos59 + 35 sin32)\hat i + (35cos32 - 65 sin59)\hat j[/tex]

[tex]F_{net} = 52\hat i - 26 \hat j[/tex]

so it's magnitude is given as

[tex]F_{net} = \sqrt{52^2 + 26^2} = 58.15 N[/tex]

now from Newton's II law we have

F = ma

[tex]a = \frac{58.15}{55} = 1.1 m/s^2[/tex]

To find the magnitude of the object's acceleration, resolve the forces into components and use Newton's second law and the Pythagorean theorem.

To find the magnitude of the object's acceleration, we need to first resolve the two forces into their x and y components. For the 65 N force, the x-component is 65 N × cos(59°) and the y-component is 65 N × sin(59°). For the 35 N force, the x-component is 35 N × sin(32°) and the y-component is 35 N × cos(32°).

Next, we add the x-components and the y-components separately to get the net force in the x and y directions. Then we use Newton's second law, F = ma, to find the acceleration in each direction. Finally, we use the Pythagorean theorem to find the magnitude of the acceleration.

By following these steps, we can calculate that the magnitude of the object's acceleration is 1.5 m/s², so the correct answer is C) 1.5 m/s².

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Answer:

Explanation:

We know that the formula for resistance in terms of length and area is given by

R = p l/ a

Where p be the resistivity and a be the area of crossection

a = pi × d^2 / 4

Where d be the diameter

a = 3.14 × (1.628 × 10^-3)^2 / 4

a = 2 × 10^-6 m^2

R = 1.68 × 10^-8 × 24 / (2 × 10^-6)

R = 20.16 × 10^-2 ohm

By use of Ohm's law

V = IR

V = 13 × 20.16 × 10^-2

V = 2.62 Volt

Answer:

Object's acceleration, a = 0.15 m/s²

Explanation:

It is given that,

Net force acting on the object, F = 7.5 N

Mass of an object, m = 47 kg

We need to find the acceleration of the object. It can be calculated using second law of motion as :

F = ma

a = acceleration of the object

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

[tex]a=\dfrac{7.5\ N}{47\ kg}[/tex]  

[tex]a=0.15\ m/s^2[/tex]

So, the acceleration of the object is 0.15 m/s². Hence, this is the required solution.

Answer:

True.

Explanation:

This statament is true.

Answer: I think 8.7 m

Explanation:

Final answer:

A baseball pitched horizontally at 101.0 mi/h will travel 60.5 feet to home plate in about 0.408 seconds; during that time, it will fall approximately 2.68 feet due to gravity.

Explanation:

To calculate how far a baseball would fall vertically when thrown horizontally at 101.0 mi/h, we can divide the problem into two separate motions - horizontal and vertical. Since the vertical and horizontal motions are independent, we can analyze them separately. First, we need to convert the speed to feet per second (1 mi/h = 1.467 ft/s), multiply 101.0 mi/h by 1.467 to get the horizontal velocity in feet per second, and then use that to find the time it takes for the ball to travel the horizontal distance to home plate.

Horizontal distance to home plate = 60.5 ft. Horizontal speed Vh = 101.0 mi/h * 1.467 ft/s/mi/h = 148.137 ft/s. Time to reach home plate t = distance/speed = 60.5 ft / 148.137 ft/s = 0.408 s (approximately).

Next, to find out how far the ball falls vertically, we'll use the formula for the distance an object falls due to gravity: d = 0.5 * g * t2, where g is the acceleration due to gravity. In standard English units, g is approximately 32.2 ft/s2. Plugging in the numbers, we get d = 0.5 * 32.2 ft/s2 * (0.408 s)2 = 2.68 ft (approximately).

Thus, a baseball thrown horizontally at 101.0 mi/h will fall about 2.68 feet by the time it reaches home plate.

The correct answer is that the ball would fall vertically by approximately 2.63 ft by the time it reached home plate, 60.5 ft away.

To solve this problem, we can use the kinematic equation for the vertical motion of the ball under the influence of gravity, assuming no air resistance. The equation is:

[tex]\[ y = v_{0y} t + \frac{1}{2} g t^2 \][/tex]

where:

- y  is the vertical displacement,

- [tex]\( v_{0y} \)[/tex] is the initial vertical velocity (which is 0 m/s in this case since the ball is thrown horizontally),

-  g  is the acceleration due to gravity (9.81 m/s, but we will use 32.2 ft/s for consistency with the imperial units given in the problem),

-  t is the time of flight.

Since the initial vertical velocity [tex]\( v_{0y} \)[/tex] is 0, the equation simplifies to:

[tex]\[ y = \frac{1}{2} g t^2 \][/tex]

We need to find the time  t  it takes for the ball to travel 60.5 ft horizontally. The horizontal velocity [tex]\( v_x \)[/tex] is given as 101 mi/h. We convert this to feet per second:

[tex]\[ v_x = 101 \text{ mi/h} \times \frac{5280 \text{ ft}}{1 \text{ mi}} \times \frac{1 \text{ h}}{3600 \text{ s}} \][/tex]

[tex]\[ v_x = 101 \times 5280 \times \frac{1}{3600} \text{ ft/s} \][/tex]

[tex]\[ v_x = 147.67 \text{ ft/s} \][/tex]

Now we can calculate the time  t it takes to travel 60.5 ft:

[tex]\[ t = \frac{\text{distance}}{\text{velocity}} = \frac{60.5 \text{ ft}}{147.67 \text{ ft/s}} \][/tex]

[tex]\[ t \ = 0.409 \text{ s} \][/tex]

Using this time, we can find the vertical displacement y :

[tex]\[ y = \frac{1}{2} \times 32.2 \text{ ft/s}^2 \times (0.409 \text{ s})^2 \][/tex]

[tex]\[ y = \frac{1}{2} \times 32.2 \times 0.167 \text{ ft} \][/tex]

[tex]\[ y = 16.7 \times 0.167 \text{ ft} \][/tex]

[tex]\[ y \ = 2.63 \text{ ft} \][/tex]

Explanation:

The expression for an Ideal Gas is:  

[tex]P.V=n.R.T[/tex]    (1)

Where:  

[tex]P[/tex] is the pressure of the gas  

[tex]V[/tex] is the volume of the gas  

[tex]n[/tex] the number of moles of gas  

[tex]R[/tex] is the gas constant  

[tex]T[/tex] is the absolute temperature of the gas  

Finding [tex]V[/tex]:

[tex]V=\frac{n.R.T}{P}[/tex]    (2)

If we are told the the amount of gas [tex]n[/tex] and pressure [tex]P[/tex] remain constant, but we increase the temperature [tex]T[/tex] by a factor of three; we will have to rewrite (2) with the new temperature [tex]T_{N}[/tex]:

[tex]T_{N}=3T[/tex]

[tex]V=\frac{n.R.3T}{P}[/tex]    (3)

[tex]V=3\frac{n.R.T}{P}[/tex]    (4)

Now, if we compare (2) with (4), it is clearly noticeable the volume of the gas has increased by a factor of 3.

The volume will triple ( increase by a factor of three )

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The basic formula of pressure that needs to be recalled is:

Pressure = Force / Cross-sectional Area

or symbolized:

[tex]\large {\boxed {P = F \div A} }[/tex]

P = Pressure (Pa)

F = Force (N)

A = Cross-sectional Area (m²)

Let us now tackle the problem !

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In this problem , we will use Ideal Gas Law as follows:

Given:

Initial Temperature of The Gas = T₁

Final Temperature of The Gas = T₂ = 3T₁

Asked:

Final Volume of The Gas = V₂ = ?

Solution:

[tex]\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}[/tex]

[tex]\frac{PV_1}{T_1} = \frac{PV_2}{3T_1}[/tex]

[tex]\frac{V_1}{1} = \frac{V_2}{3}[/tex]

[tex]3(V_1) = 1(V_2)[/tex]

[tex]V_2 = 3V_1[/tex]

[tex]\texttt{ }[/tex]

The volume will triple ( increase by a factor of three )

[tex]\texttt{ }[/tex]

[tex]\texttt{ }[/tex]

Grade: High School

Subject: Physics

Chapter: Pressure

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Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant , Liquid , Pressure, Volume , Ideal , Gas , Law

Answer:

Force of friction, f = 751.97 N

Explanation:

it is given that,

Mass of the car, m = 1100 kg

It is parked on a 4° incline. We need to find the force of friction keeping the car from sliding down the incline.

From the attached figure, it is clear that the normal and its weight is acting on the car. f is the force of friction such that it balances the x component of its weight i.e.

[tex]f=mg\ sin\theta[/tex]

[tex]f=1100\ kg\times 9.8\ m/s^2\ sin(4)[/tex]

f = 751.97 N

So, the force of friction on the car is 751.97 N. Hence, this is the required solution.

In essence, the force of friction that prevents the car from sliding down the incline is equal to the component of the weight of the car down the incline. After performing the necessary calculations, this is found to be roughly 766.9 Newtons.

In response to your question, the force of friction keeping the car from sliding down the incline can be calculated using basic physics principles. Firstly, we need to calculate the component of the gravitational force parallel to the incline, which is given by F_gravity = m * g * sin(theta), where m is the mass of the car, g is the acceleration due to gravity, and theta is the angle of incline.

Substituting the given values, we get F_gravity = 1100 kg * 9.8 m/s² * sin(4°), which approximates to 766.9 N.

Now, since the car is stationary and not sliding down, this suggests that the force of friction equals the component of the weight down the incline. Therefore, the force of friction keeping the car from sliding down is approximately 766.9 Newtons.

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Answer:

a) 8.61 m/s, b) 5.73 m

Explanation:

a) During the collision, momentum is conserved.

mv = (m + M) V

(12.5 g) (86.4 m/s) = (12.5 g + 113 g) V

V = 8.61 m/s

b) After the collision, energy is conserved.

Kinetic energy = Work done by friction

1/2 (m + M) V² = F d

1/2 (m + M) V² = N μk d

1/2 (m + M) V² = (m + M) g μk d

1/2 V² = g μk d

d = V² / (2g μk)

d = (8.61 m/s)² / (2 × 9.8 m/s² × 0.659)

d = 5.73 m

Notice we used the kinetic coefficient of friction.  That's the friction when an object is moving.  The static coefficient of friction is the friction on a stationary object.  Since the bullet/block combination is sliding across the surface, we use the kinetic coefficient.