The Sun appears to move relative to the stars. For example, right now the Sun is in front of the constellation Leo, but in a month it will have moved to Virgo. Using the length of the year (365 days), calculate how fast (degrees/day) this motion is.

Answer:

Explanation:

The SI system of units is the system which is Standard International system. It is used internationally.

The SI units of the fundamental quantities are given below

Mass - Kilogram

Length - metre

Time - second

temperature - Kelvin

Amount of substance - mole

Electric current - Ampere

Luminous Intensity - Candela

So, The SI unit of work is Joule

SI unit of acceleration is m/s^2

SI unit of power is watt

SI unit of momentum is kg m /s

SI unit of force is newton

Thus, the last option is incorrect.

Answer:

(177.94 ± 3.81) cm^2

Explanation:

l + Δl = 21.7 ± 0.2 cm

b + Δb = 8.2 ± 0.1 cm

Area, A = l x b = 21.7 x 8.2 = 177.94 cm^2

Now use error propagation

[tex]\frac{\Delta A}{A}=\frac{\Delta l}{l}+\frac{\Delta b}{b}[/tex]

[tex]\frac{\Delta A}{A}=\frac{0.2}{21.7}+\frac{0.1}{8.2}[/tex]

[tex]\Delta A=177.94 \times \left ( 0.0092 + 0.0122 \right )=3.81[/tex]

So, the area with the error limits is written as

A + ΔA = (177.94 ± 3.81) cm^2

Answer:

total number of electron in 1 litter is 3.34 × [tex]10^{26}[/tex] electron

Explanation:

given data

mass per mole = 18 g/mol

no of electron = 10

to find out

how many electron in 1 liter of water

solution

we know molecules per gram mole is 6.02 ×[tex]10^{23}[/tex] molecules

no of moles is 1

so

total number of electron in water is = no of electron ×molecules per gram mole × no of moles

total number of electron in water is = 10 × 6.02 ×[tex]10^{23}[/tex] × 1

total number of electron in water is = 6.02×[tex]10^{24}[/tex] electron

and

we know

mass = density × volume    ..........1

here we know density of water is 1000 kg/m

and volume = 1 litter = 1 × [tex]10^{-3}[/tex] m³

mass of 1 litter = 1000 × 1 × [tex]10^{-3}[/tex]

mass = 1000 g

so

total number of electron in 1 litter =  mass of 1 litter × [tex]\frac{molecules per gram mole}{mass per mole}[/tex]

total number of electron in 1 litter =  1000 × [tex]\frac{6.02*10{24}}{18}[/tex]

total number of electron in 1 litter is 3.34 × [tex]10^{26}[/tex] electron

Answer:

h = 4271.43 m

Explanation:

given,

Volume of the water = 1 m³

temperature decrease by = 10°C

heat removed from water

Q = m c ΔT                            

Q = ρ V c ΔT                            

   = 1000 × 1 × 4186 × 10

   = 4.186 × 10⁷ J

energy is used to do work to move the water against its weight

Q = force  × displacement

4.186 × 10⁷ J =  m g × h                    

4.186 × 10⁷ J =  1000 × 1 × 9.8 × h                

h = 4271.43 m                                

hence, the change in height of is equal to h = 4271.43 m

Answer:

a) m = 993 g

b) E = 6.50 × 10¹⁴ J

Explanation:

atomic mass of hydrogen = 1.00794

4 hydrogen atom will make a helium atom = 4 × 1.00794 = 4.03176

we know atomic mass of helium = 4.002602

difference in the atomic mass of helium = 4.03176-4.002602 = 0.029158

fraction of mass lost = [tex]\dfrac{0.029158}{4.03176}[/tex]= 0.00723

loss of mass for 1000 g = 1000 × 0.00723 = 7.23

a) mass of helium produced = 1000-7.23 = 993 g (approx.)

b) energy released in the process

E = m c²

E = 0.00723 × (3× 10⁸)²

E = 6.50 × 10¹⁴ J

Answer:

(a) 992.87 g

(b) [tex]6.419\times 10^{14} J[/tex]

Solution:

As per the question:

Mass of Hydrogen converted to Helium, M = 1 kg = 1000 g

(a) To calculate mass of He produced:

We know that:

Atomic mass of hydrogen is 1.00784 u

Also,

4 Hydrogen atoms constitutes 1 Helium atom

Mass of Helium formed after conversion:

[tex]4\times 1.00784 = 4.03136 u[/tex]

Also, we know that:

Atomic mass of Helium is 4.002602 u

The loss of mass during conversion is:

4.03136 - 4.002602 = 0.028758 u

Now,

Fraction of lost mass, M' = [tex]\frac{0.028758}{4.03136} = 0.007133 u[/tex]

Now,

For the loss of mass of 1000g = [tex]0.007133\times 1000[/tex] = 7.133 g

Mass of He produced in the process:

[tex]M_{He} = 1000 - 7.133 = 992.87 g[/tex]

(b) To calculate the amount of energy released:

We use Eintein' relation of mass-enegy equivalence:

[tex]E = M'c^{2}[/tex]

[tex]E = 0.007133\times (3\times 10^{8})^{2} = 6.419\times 10^{14} J[/tex]

Answer:

[tex]T_2=13.5\ N[/tex]

Explanation:

Given that,

Speed of transverse wave, v₁ = 20 m/s

Tension in the string, T₁ = 6 N

Let T₂ is the tension required for a wave speed of 30 m/s on the same string. The speed of a transverse wave in a string is given by :

[tex]v=\sqrt{\dfrac{T}{\mu}}[/tex]........(1)

T is the tension in the string

[tex]\mu[/tex] is mass per unit length

It is clear from equation (1) that :

[tex]v\propto\sqrt{T}[/tex]

[tex]\dfrac{v_1}{v_2}=\sqrt{\dfrac{T_1}{T_2}}[/tex]

[tex]T_2=T_1\times (\dfrac{v_2}{v_1})^2[/tex]

[tex]T_2=6\times (\dfrac{30}{20})^2[/tex]

[tex]T_2=13.5\ N[/tex]

So, the tension of 13.5 N is required for a wave speed of 30 m/s. Hence, this is the required solution.

Final answer:

The tension required for a wave speed of 30.0 m/s on a string with an initial tension of 6.00 N is 900 N.

Explanation:

To find the tension required for a wave speed of 30.0 m/s, we can use the equation: wave speed = square root of (tension/linear mass density).

Given that the initial wave speed is 20.0 m/s and tension is 6.00 N, we can rearrange the equation to solve for tension using the new wave speed.

Substituting the values, we have: 30.0 m/s = sqrt(tension/linear mass density). After squaring both sides of the equation, we get: 900 = tension/linear mass density

Since the linear mass density remains constant, the tension required for a wave speed of 30.0 m/s would be 900 N.

Answer:

R=3818Km

Explanation:

Take a look at the picture. Point A is when you start the stopwatch. Then you stand, the planet rotates an angle α and you are standing at point B.

Since you travel 2π radians in 24H, the angle can be calculated as:

[tex]\alpha =\frac{2*\pi *t}{24H}[/tex]  t being expressed in hours.

[tex]\alpha =\frac{2*\pi *11.9s*1H/3600s}{24H}=0.000865rad[/tex]

From the triangle formed by A,B and the center of the planet, we know that:

[tex]cos(\alpha )=\frac{r}{r+H}[/tex]  Solving for r, we get:

[tex]r=\frac{H*cos(\alpha) }{1-cos(\alpha) } =3818Km[/tex]

Using the formula for acceleration, a = (v^2) / 2d, and the given velocity and acceleration distance, the acceleration of the tennis ball during Lisicki's serve was approximately 11368 m/s². The force exerted by the racket is generally higher than the force due to gravity during this action.

To calculate the acceleration of the tennis ball during Sabine Lisicki's serve, we can use the formula for acceleration: a = (v^2) / 2d, where 'v' corresponds to the final velocity and 'd' represents the distance over which the ball accelerated.

In this case, the final velocity 'v' is 211 km/h (converted to m/s gives us approximately 58.6 m/s), and Lisicki's racquet was in contact with the ball, causing it to accelerate over a distance 'd' of 0.15 m. Plugging these values into the formula gives us: a = (58.6 m/s)^2 / 2(0.15 m), which equals about 11368 m/s².

The average force exerted by the racket can be understood through the equation F = ma, a case of Newton's second law of motion. However, in this scenario, the force due to gravity is negligible as it's much smaller than the force exerted by the racket. The main focus here is the force exerted by the racket which made such a high acceleration possible.

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Answer:

Explanation:

We know that 3.20 grams of salt in 9.10 grams of water gives us a saturated solution at 25°C. To find how many grams of salt will gives us a saturated solution in 100 grams of water at the same temperature, we can use the rule of three.

[tex]\frac{3.20 \g \ salt}{9.10 \ g \ water} = \frac{x \ g \ salt}{100 \ g \ water}[/tex]

Working it a little this gives us :

[tex] x = 100 \ g \ water * \frac{3.20 \g \ salt}{9.10 \ g \ water} [/tex]

[tex] x = 35.16 \ g \ salt [/tex]

So, the solubility of the salt is 35.16 (g/100 g of water).

Using the rule of three, we got:

[tex]\frac{3.20 \g \ salt}{9.10 \ g \ water} = \frac{25 \ g \ salt}{x \ g \ water}[/tex]

Working it a little this gives us :

[tex] x = \frac{25 \ g \ salt}{ \frac{3.20 \g \ salt}{9.10 \ g \ water}} [/tex]

[tex] x = 71.09 g \ water [/tex]

So, it would take 71.09 grams of water to dissolve 25 grams of salt.

Using the rule of three, we got that for 15.0 grams of water the salt dissolved will be:

[tex]\frac{3.20 \g \ salt}{9.10 \ g \ water} = \frac{x \ g \ salt}{15.0 \ g \ water}[/tex]

Working it a little this gives us :

[tex] x = 15.0 \ g \ water * \frac{3.20 \g \ salt}{9.10 \ g \ water} [/tex]

[tex] x = 5.27\ g \ salt [/tex]

This is the salt dissolved

The percentage of salt dissolved is:

[tex]percentage \ salt \ dissolved = 100 \% * \frac{g \ salt \ dissolved}{g \ salt}[/tex]

[tex]percentage \ salt \ dissolved = 100 \% * \frac{ 5.27\ g \ salt }{ 10.0 \ g \ salt}[/tex]

[tex]percentage \ salt \ dissolved = 52.7 \% [/tex]

a. The solubility (in g salt/100 g of water) of the salt is 35.16 g/100 g water. b. Amount of water it would take to dissolve 25 g of this salt is 71.10 g. c. If 10.0 g of the salt is mixed with 15.0 g of water, 52.7% of the salt will dissolve.

To solve the problem, we need to determine the solubility of the salt at 25°C using the given data:

Part a: Solubility

Part b: Amount of Water Needed to Dissolve 25 g of Salt

Part c: Percentage of Salt Dissolved

Therefore, 52.7% of the 10.0 g of salt will dissolve in 15.0 g of water.

Answer:

[tex]v= v_{student} ( 1 + \frac{d_{bus \ to \ the \ student}}{d_{student \ to \ puddle}} )[/tex]

Explanation:

We know that speed is:

[tex]v= \frac{distance}{time}[/tex].

So, to find the speed for the bus, we need to know:

Lets call [tex]v_{student}[/tex] the speed of the student, and [tex]d_{student \ to \ puddle}[/tex] the distance from the student to the puddle.

We can obtain the time taking

[tex]v_{student}= \frac {d_student \ to \ puddle}{t}[/tex]

as t must be the time that the student will take to reach the puddle:

[tex]t= \frac {d_{student \ to \ puddle}}{v_{student}}[/tex]

The bus is at a distance [tex]d_{bus \ to \ the \ student}[/tex] behind the student, so, the total distance that the bus must travel to the puddle is:

[tex]d_{bus \ to \ the \ puddle} = d_{bus \ to \ the \ student} + d_{student \ to \ puddle}[/tex]

Taking all this togethes, the formula must be:

[tex]v= \frac{d_{bus \ to \ the \ puddle}}{t}[/tex].

[tex]v= \frac{d_{bus \ to \ the \ student} + d_{student \ to \ puddle}}{\frac {d_{student \ to \ puddle}}{v_{student}}}[/tex].

[tex]v= v_{student} \frac{d_{bus \ to \ the \ student} + d_{student \ to \ puddle}}{d_{student \ to \ puddle}}[/tex].

[tex]v= v_{student} (\frac{d_{bus \ to \ the \ student}}{d_{student \ to \ puddle}} + \frac{d_{student \ to \ puddle}}{d_{student \ to \ puddle}} )[/tex].

[tex]v= v_{student} ( 1 + \frac{d_{bus \ to \ the \ student}}{d_{student \ to \ puddle}} )[/tex].

And this is the formula we are looking for.

Taking the values from the problem, we find

[tex]v= 1 \frac{m}{s} ( 1 + \frac{2 m}{1 m})[/tex]

[tex]v= 1 \frac{m}{s} ( 1 + 2)[/tex]

[tex]v= 3 \frac{m}{s}[/tex]

Answer: the right answer is C

Explanation:

Answer:

Double the charge to 2Q

Explanation:

The magnitude of a electric field caused by a charge Q at a distance R can be find by the following expression:

[tex]E = K\frac{Q}{R^2}[/tex]

Where K is the Coulomb constant.

As the relationship between the charge and the electric fiel is proportional, by simply doubling the charge, you would double the magnitude of the electric field.

Notice that the distance affects the magnitude of the electric field by the inverse square. So if you half the distance, the magnitude of the field will quadruple.

To double the electric field's magnitude at point P from a point charge Q at distance R, double the charge to 2Q while keeping the distance R unchanged. Option B is the correct option.

To double the magnitude of the electric field at point P due to a point charge Q a distance R away from P, you can follow Coulomb's law. Electric field (E) is directly proportional to the charge (Q) and inversely proportional to the square of the distance (R) between the point charge and the observation point.

If you double the charge to 2Q while keeping the distance R the same, the new electric field magnitude at point P will be double the original E. This is because E ∝ Q. Other options like reducing the distance to R/2 or R/4, or doubling the distance to 2R, would not result in a doubling of the field magnitude but rather lead to different field strengths according to the inverse square law, and altering both the charge and distance simultaneously is not necessary to achieve this specific goal. Option B is the correct option.

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The appropriate question is :

The electric field at point P due to a point charge Q a distance R away from P has magnitude E. In order to double the magnitude of the field at P, you could The electric field at point P due to a point charge Q a distance R away from P has magnitude E. In order to double the magnitude of the field at P, you could double the charge to 2Q and at the same time reduce the distance to R/2 OR

A) reduce the distance to R/4.

B) double the charge to 2Q.

C)reduce the distance to R/2.

D) double the distance to 2R.

Answer:

[tex]v_{max}=52.38\frac{m}{s}[/tex]

[tex]v_{100}=33.81[/tex]

Explanation:

the maximum speed is reached when the drag force and the weight are at equilibrium, therefore:

[tex]\sum{F}=0=F_d-W[/tex]

[tex]F_d=W[/tex]

[tex]kv_{max}^2=m*g[/tex]

[tex]v_{max}=\sqrt{\frac{m*g}{k}} =\sqrt{\frac{70*9.8}{0.25}}=52.38\frac{m}{s}[/tex]

To calculate the velocity after 100 meters, we can no longer assume equilibrium, therefore:

[tex]\sum{F}=ma=W-F_d[/tex]

[tex]ma=W-F_d[/tex]

[tex]ma=mg-kv_{100}^2[/tex]

[tex]a=g-\frac{kv_{100}^2}{m}[/tex] (1)

consider the next equation of motion:

[tex]a = \frac{(v_{x}-v_0)^2}{2x}[/tex]

If assuming initial velocity=0:

[tex]a = \frac{v_{100}^2}{2x}[/tex] (2)

joining (1) and (2):

[tex]\frac{v_{100}^2}{2x}=g-\frac{kv_{100}^2}{m}[/tex]

[tex]\frac{v_{100}^2}{2x}+\frac{kv_{100}^2}{m}=g[/tex]

[tex]v_{100}^2(\frac{1}{2x}+\frac{k}{m})=g[/tex]

[tex]v_{100}^2=\frac{g}{(\frac{1}{2x}+\frac{k}{m})}[/tex]

[tex]v_{100}=\sqrt{\frac{g}{(\frac{1}{2x}+\frac{k}{m})}}[/tex] (3)

[tex]v_{100}=\sqrt{\frac{9.8}{(\frac{1}{2*100}+\frac{0.25}{70})}}[/tex]

[tex]v_{100}=\sqrt{\frac{9.8}{(\frac{1}{200}+\frac{1}{280})}}[/tex]

[tex]v_{100}=\sqrt{\frac{9.8}{(\frac{3}{350})}}[/tex]

[tex]v_{100}=\sqrt{1,143.3}[/tex]

[tex]v_{100}=33.81[/tex]

To plot velocity as a function of distance, just plot equation (3).

To plot velocity as a function of time, you have to consider the next equation of motion:

[tex]v = v_0 +at[/tex]

as stated before, the initial velocity is 0:

[tex]v =at[/tex] (4)

joining (1) and (4) and reducing you will get:

[tex]\frac{kt}{m}v^2+v-gt=0[/tex]

solving for v:

[tex]v=\frac{ \sqrt{1+\frac{4gk}{m}t^2}-1}{\frac{2kt}{m} }[/tex]

Plots:

Answer:

Explanation:

Charge on honeybee  q = 23 x 10⁻¹²

Force due to electric field  E = E x q

= 100 x 23 x 10⁻¹²

= 23 x 10⁻¹⁰ N

Gravitational force on the honeybee

= m g = .09x 10⁻³ x 9.8

= .882x 10⁻³ N

Ratio of electric field and gravitational field

23 x 10⁻¹⁰ / .882x 10⁻³

26.07 x 10⁻⁷

= 26.07

Answer:

1497×10⁵ km

Explanation:

Speed of light in vacuum = 3×10⁵ km/s

Time taken by the light of the Sun to reach the Earth = 8 min and 19 s

Converting to seconds we get

8×60+19 = 499 seconds

Distance = Speed × Time

[tex]\text{Distance}=3\times 10^5\times 499\\\Rightarrow \text{Distance}=1497\times 10^5\ km[/tex]

1 AU = 1497×10⁵ km

The Sun is 1497×10⁵ km from Earth

Answer:  

2.26 mm.

Explanation:

According to Rayleigh criterion , angular rosolution of eye is given by the expression

Angular resolution ( in radian ) = 1.22 λ / D

λ wave length of light, D is diameter of the eye

Given

angular resolution in degree = 1.67 x 10⁻²

= 1.67 x 10⁻² x π / 180 radian ( 180 degree = π radian )

= 29.13 x 10⁻⁵ radian

λ = 540 x 10⁻⁹ m

Put these values in the expression

29.13 x 10⁻⁵ = 1.22 x 540 x 10⁻⁹ / D

D = [tex]\frac{1.22\times540\times10^{-9}}{29.13\times10^{-5}}[/tex]

D = 2.26 mm.

The effective diameter of an eye's optical system that corresponds to a resolving power of one arcminute, calculated using Rayleigh's criterion and the given values of resolving power and wavelength of light, is approximately 2.27 mm.

The effective diameter of an eye's optical system that corresponds to a resolving power of one arcminute can be determined using Rayleigh's criterion for diffraction limit and the given values of the resolving power and wavelength of light.

Rayleigh's criterion states that the minimum resolvable angle for a diffraction-limited system is θmin = 1.22*λ/D where λ is the wavelength of light and D is the diameter of the optical system. Given that the resolving power of the eye is 1.67×10^−2 degrees and the wavelength of light is 540 nm, we can rearrange Rayleigh's formula to solve for D.

Converting 1.67×10^−2 degrees to radians gives us 0.00029 rad. Plugging in the values into Rayleigh's formula and solving for D gives us D = 1.22*λ/θmin. Substituting λ=540*10^−9 m and θmin=0.00029 into the equation, we get D = 2.27 mm to three significant figures.

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Answer:

No

Explanation:

As we know that the electric field nullity does not define that the electric potential will be zero at that point.

Answer:

angular acceleration = - 217.5 rad/s²

Explanation:

given data

angular speed = 3600 rev/min

rotate = 50 revolution

to find out

angular acceleration

solution

we know here no of rotation n =  3600 rev/min i.e 60 rev/s

so initial angular velocity will be  ω(i) = 2π× n

ω(i) = 2π× 60 = 376.9 rad/s

and

final angular velocity will be ω(f) = 0

so  

angular displacement will be  = 2π × 52 = 326.56 rad

and angular acceleration calculated as

angular acceleration = [tex]\frac{\omega(f)^2-\omega(i)^2}{2*angular displacement}[/tex]

put here value

angular acceleration = [tex]\frac{-376.9^2}{326.56}[/tex]

angular acceleration = - 217.5 rad/s²

Answer:

It would have been longer.

Explanation:

Lets assume the Sun angle = θ

Distance between Syene and Alexandria = D

Circumference of Earth = C

As per Eratosthenes' calculations,

[tex]\frac{\theta}{360} =\frac{D}{C}[/tex]

From the above equation it is evident that if the circumference decreases value of θ will increase which implies that the shadow length would be longer as compared to that on the Earth.

Answer:

[tex]Volume=160 cm^3[/tex]

Explanation:

The volume of the cylinder is given by the area of the base multiplied by the height.

In this case:

Height:

[tex]h = 10 cm\\[/tex]

Base area = area of the square ([tex]area=side*side=side^2[/tex])

the side of the square is:

[tex]side=4cm[/tex]

thus, the area of the base:

[tex]area=(4cm)^2 = 16 cm^2[/tex]

Now we multiply this quantities, to find the volume:

[tex]Volume= 16 cm^2*10cm=160 cm^3[/tex]

Answer:

Explanation:

Knowing the magnitude and directions relative to the x axis, we can find the Cartesian representation of the vectors using the formula

[tex]\vec{A}= | \vec{A} | \ ( \ cos(\theta) \ , \ sin (\theta) \ )[/tex]

where [tex]| \vec{A} |[/tex] its the magnitude and θ.

So, for our vectors, we will have:

[tex]\vec{D}= 3.00 m \ ( \ cos(315) \ , \ sin (315) \ )[/tex]

[tex]\vec{D}=  ( 2.121 m , -2.121 m )[/tex]

and

[tex]\vec{E}= 4.50 m \ ( \ cos(53.0) \ , \ sin (53.0) \ )[/tex]

[tex]\vec{E}= ( 2.71 m , 3.59 m )[/tex]

Now, we can take the sum of the vectors

[tex]\vec{R} = \vec{D} + \vec{E}[/tex]

[tex]\vec{R} = ( 2.121 \ m , -2.121 \ m ) + ( 2.71 \ m , 3.59 \ m )[/tex]

[tex]\vec{R} = ( 2.121 \ m  + 2.71 \ m , -2.121 \ m + 3.59 \ m ) [/tex]

[tex]\vec{R} = ( 4.831 \ m , 1.469 \ m ) [/tex]

This is R in Cartesian representation, now, to find the magnitude we can use the Pythagorean theorem

[tex]|\vec{R}| = \sqrt{R_x^2 + R_y^2}[/tex]

[tex]|\vec{R}| = \sqrt{(4.831 m)^2 + (1.469 m)^2}[/tex]

[tex]|\vec{R}| = \sqrt{23.338 m^2 + 2.158 m^2}[/tex]

[tex]|\vec{R}| = \sqrt{25.496 m^2}[/tex]

[tex]|\vec{R}| = 5.049 m[/tex]

To find the direction, we can use

[tex]\theta = arctan(\frac{R_y}{R_x})[/tex]

[tex]\theta = arctan(\frac{1.469 \ m}{4.831 \ m})[/tex]

[tex]\theta = arctan(0.304)[/tex]

[tex]\theta = 16\°54'33''[/tex]

As we are in the first quadrant, this is relative to the x axis.

Answer:259.80 N

Explanation:

Given

mass of ball=3 kg

ball velocity =10 m/s

angle made by ball with plane of wall [tex]\theta [/tex]=60

Momentum change in Y direction remains same and there is change only in x direction

therefore

initial momentum[tex]=mvsin\theta [/tex]

=30sin60

Final momentum=-30sin60

Change in momentum is =30sin60+30sin60

=60sin60

and Impulse = change in momentum

Fdt=dP

where F=force applied

dP=change in momentum

[tex]F\times 0.2=60sin60[/tex]

[tex]F\times 0.2=51.96[/tex]

F=259.80 N

The average force exerted by the wall on the 3.00 kg steel ball after it bounces off is 150 N. We calculated this using the principle of conservation of momentum and Newton's second law.

The subject of this question is Physics, specifically, the topic of motion and force. To answer this question, we should apply the law of conservation of momentum and Newton's second law (Force = change in Momentum / change in Time).

Firstly, the ball is projected against the wall at an angle of 60 degrees, but we are concerned with the component of velocity perpendicular to the wall. This means we will consider the velocity component of the ball towards the wall, which is 10 cos(60), giving us 5.0 m/s. The incoming momentum of the ball can then be calculated as the mass times the velocity (3.0 kg * 5.0 m/s = 15 kg*m/s).

Since the ball bounces off with the same speed at the same angle, its outgoing momentum is -15 kg*m/s. The change in momentum is therefore Outgoing momentum - Incoming momentum = -15 kg*m/s - 15 kg*m/s = -30 kg*m/s. The force exerted by the wall on the ball equals the Change in momentum divided by the time it takes for the change to occur (-30 kg*m/s / 0.200 s = -150 N). Given that force is a vector and we are asked for the magnitude of the force, the answer is 150 N.

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Answer:

Explanation:

We shall convert the movement of grasshopper in vector form. Suppose the grass hopper is initially sitting at the origin or (00) position .

It went 40 cm due west so

D₁ = -40 i

It then moves 26 cm 32 ° south of west so

D₂ = - 26 Cos32i - 26 Sin32 j = - 22 i -13.77 j

Then it moves 19 cm 50° south of east

D₃ = 19 Cos 50 i - 19 Sin 50 j = 12.2 i - 14.55 j

Then it moves 18 cm 60° north of east

D₄ = 18 Cos 60 i + 18 Sin 60 j = 9 i + 15.58 j

Total displacement = D₁ +D₂+D₃+D₄

= - 40i -22 i - 13.77 j + 12.2 i - 14.55 j + 9 i + 15.58 j

= - 40.8 i - 12.74 j

Magnitude of displacement D

D² = ( 40.8 )² + ( 12.74)²

D = 42 .74 cm

If ∅ be the required angle

Tan∅ = 12.74 / 40.80 = .31

∅ = 17 ° positive angle with respect to due west.

Answer:

2899.24 N

Explanation:

W = Weight of pilot

r = Radius

v = Velocity

g = Acceleration due to gravity = 9.81 m/s²

[tex]mg=m\frac{v^2}{r}\\\Rightarrow v=\sqrt{gr}\\\Rightarrow v=\sqrt{9.81\times 200}\\\Rightarrow v=44.3\ m/s[/tex]

Speed of the aeroplane at the top of the loop is 44.3 m/s

Now, v = 280 km/h = 280/3.6 = 77.78 m/s

Apparent weight

[tex]A=W+\frac{W}{g}\frac{v^2}{r}\\\Rightarrow A=710+\frac{710}{9.81}\times \frac{77.78^2}{200}\\\Rightarrow A=2899.24\ N[/tex]

Apparent weight at the bottom of the loop is 2899.24 N

Answer:

Explanation:

Ball is thrown downward:

initial velocity, u = - 20 m/s (downward)

height, h = - 60 m

Acceleration due to gravity, g = - 9.8 m/s^2 (downward)

(a) Let the speed of the ball as it hits the ground is v.

Use third equation of motion

[tex]v^{2}=u^{2}+2as[/tex]

[tex]v^{2}=(-20)^{2}+2\times 9.8 \times 60[/tex]

v = 39.69 m/s

(b) Let t be the time taken

Use First equation of motion

v = u + a t

- 39.69 = - 20 - 9.8 t

t = 2 second

Now the ball is thrown upwards:

initial velocity, u = 20 m/s (upward)

height, h = - 60 m

Acceleration due to gravity, g = - 9.8 m/s^2 (downward)

(c) Let the speed of the ball as it hits the ground is v.

Use third equation of motion

[tex]v^{2}=u^{2}+2as[/tex]

[tex]v^{2}=(-20)^{2}+2\times 9.8 \times 60[/tex]

v = 39.69 m/s

(d) Let t be the time taken

Use First equation of motion

v = u + a t

- 39.69 = + 20 - 9.8 t

t = 6.09 second

Answer:

The acceleration is -9.8 m/s²

Explanation:

Hi there!!

When you throw a ball upward, there is a downward acceleration that makes the ball return to your hand. This acceleration is produced by gravity.

The average acceleration is calculated as the variation of the speed over time. In this case, we know the time and the initial and final speed. Then:

acceleration = final speed - initial speed/ elapsed time

acceleration = -4.3 m/s - 4.3 m/s / 0.88 s

acceleration = -9.8 m/s²  

The average acceleration vector would be 0 m/s² due to the upward (negative) and downward (positive) accelerations cancelling. However, the average magnitude of acceleration regardless of direction is 9.8 m/s², which is simply the acceleration due to gravity.

To answer your question about the average acceleration vector of a ball that was thrown upward, we can use the concept of free fall in physics. When you throw a ball upward, it initially slows down due to earth's gravitational pull until it stops at its highest point, then it starts accelerating downward due to gravity, until it reaches your hand again.

The acceleration when the ball is going upward will be the opposite of the acceleration when it's coming downward because they are in opposite directions but magnitude will be the same. We can assume the acceleration due to gravity as -9.8 m/s² (negative indicating upward direction) and when it's coming down it will be 9.8 m/s² (positive indicating downward direction).

So over the course of its flight (0.88s), the average acceleration would be ((-9.8)+(9.8))/2 = 0 m/s². However, this may not be what you're looking for as this is the averaged vector sum of the accelerations upward and downward.

If you are asking about the average magnitude of acceleration regardless of direction (in magnitude) it would be the average of the absolute values of the accelerations, which would be the gravitational acceleration g = 9.8 m/s².

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Answer:

The gauge pressure is [tex]1.8\times10^{5}\ N/m^2[/tex]

Explanation:

Given that,

Gauge pressure of car tires [tex]P_{1}=2.40\times10^{5}\ N/m^2[/tex]

Temperature [tex]T_{1}=35.0^{\circ}C = 35.0+273=308 K[/tex]

Dropped temperature [tex]T_{2}= -42.0^{\circ}C=273-42=231 K[/tex]

We need to calculate the gauge pressure P₂

Using relation pressure and temperature

[tex]\dfrac{P_{1}}{T_{1}}=\dfrac{P_{2}}{T_{2}}[/tex]

Put the value into the formula

[tex]\dfrac{2.40\times10^{5}}{308}=\dfrac{P_{2}}{231}[/tex]

[tex]P_{2}=\dfrac{2.40\times10^{5}\times231}{308}[/tex]

[tex]P_{2}=180000 = 1.8\times10^{5}\ N/m^2[/tex]

Hence, The gauge pressure is [tex]1.8\times10^{5}\ N/m^2[/tex]

Final answer:

The electron's average velocity is found to be (0 m/s, 316,000 m/s, -146,000 m/s), and after another 9 microseconds, it will be at the position (0.02 m, 4.464 m, -2.104 m).

Explanation:

To calculate the average velocity of an electron, we use the formula:

Vavg = (rf - ri) / Δt, where rf is the final position, ri is the initial position, and Δt is the time interval between the positions.

Given the initial position (0.02 m, 0.04 m, -0.06 m) and the final position (0.02 m, 1.62 m, -0.79 m), with a time difference of 5 microseconds (μs), which is 5 x 10-6 seconds:

The position change in vector form is

Δr = (0.02 m - 0.02 m, 1.62 m - 0.04 m, -0.79 m - (-0.06 m))

= (0 m, 1.58 m, -0.73 m).

Thus, the average velocity is

Vavg = Δr / Δt

= (0 m, 1.58 m, -0.73 m) / (5 x 10-6 s)

= (0 m/s, 316,000 m/s, -146,000 m/s)

The electron's new position after another 9 μs, moving with the same average velocity, is calculated by:

rnew = rf + Vavg × Δtnew

Here, Δtnew is 9 μs, which is 9 x 10-6 seconds, so:

rnew = (0.02 m, 1.62 m, -0.79 m) + (0 m/s, 316,000 m/s, -146,000 m/s) × (9 x 10-6 s)

= (0.02 m, 1.62 m + (2.844 m), -0.79 m - (1.314 m))

= (0.02 m, 4.464 m, -2.104 m).

Answer:

V1=2.5ft3

V2=1ft3

n=1.51

Explanation:

PART A:

the volume of each state is obtained by multiplying the mass by the specific volume in each state

V=volume

v=especific volume

m=mass

V=mv

state 1

V1=m.v1

V1=2lb*1.25ft3/lb=2.5ft3

state 2

V2=m.v2

V2=2lb*0.5ft3/lb=   1ft3

PART B:

since the PV ^ n is constant we can equal the equations of state 1 and state 2

P1V1^n=P2V2^n

P1/P2=(V2/V1)^n

ln(P1/P2)=n . ln (V2/V1)

n=ln(P1/P2)/ ln (V2/V1)

n=ln(15/60)/ ln (1/2.5)

n=1.51

Answer:

ΔV = ±0.175 cm

Explanation:

The equation for volume is

V = π/4 * d^2 * h

All the measurements are multiplied. To propagate uncertainties in multiplication we add the relative uncertainties together.

The relative uncertainty of the diameter is:

εd = Δd/d

εd = 0.0005/5.1 = 0.000098

The relative uncertainty of the height is:

εh = Δh/h

εh = 0.005/37.6 = 0.00013

Then, the relative uncertainty of the volume is:

εV = 2 * εd + εh

εV = 2 * 0.000098 + 0.00013 = 0.000228

Then we get the absolute uncertainty of the volume, for that we need the volume:

V = π/4 * 5.1^2 * 37.6 = 768.1 cm^3

So:

ΔV = ±εV * V

ΔV = ±0.000228 * 768.1 = ±0.175 cm