60. The label on a box of cereal gives the mass of cereal in two units: 978 grams and 34.5 oz. Use this information to find a conversion factor between the English and metric units. How many significant figures can you justify in your conversion factor?

Answer:

3260.33 J

Explanation:

[tex]n [/tex]  = number of rods = 5

[tex]L [/tex]  = length of each rod = 0.715 m

[tex]m [/tex]  = mass of rod = 2.51 kg

[tex]I [/tex]  = total moment of inertia

Total moment of inertia is given as

[tex]I = \frac{nmL^{2}}{3}[/tex]

[tex]I = \frac{(5)(2.51)(0.715)^{2}}{3}[/tex]

[tex]I [/tex] = 2.14 kgm²

[tex]w [/tex]  = angular speed = 527 rpm = 55.2 rad/s

Rotational kinetic energy is given as

E = (0.5) [tex]I [/tex]  ([tex]w [/tex] )²

E = (0.5) (2.14) (55.2)²

E = 3260.33 J

Answer:

a) Magnitude of the average velocity from t = 0 to t = 2 s = 2.24 m/s

b) Magnitude of the average velocity from t = 0 to t = 5 s = 3.16 m/s

Explanation:

a) Velocity is rate of change of position.

A particle's position coordinates (x, y) are (1.0 m, 3.0 m) at t = 0

A particle's position coordinates (x, y) are (5.0 m, 5.0 m) at t = 2.0 s

Displacement = (5-1)i + (5-3)j = 4i + 2j

Change in time = 2s

Velocity

      [tex]v=\frac{4i+2j}{2}=2i+j[/tex]

Magnitude of velocity

      [tex]v=\sqrt{2^2+1^2}=2.24m/s[/tex]

b) Velocity is rate of change of position.

A particle's position coordinates (x, y) are (1.0 m, 3.0 m) at t = 0

A particle's position coordinates (x, y) are (14.0 m, 12.0 m) at t = 5.0 s

Displacement = (14-1)i + (12-3)j = 13i + 9j

Change in time = 5s

Velocity

      [tex]v=\frac{13i+9j}{5}=2.6i+1.8j[/tex]

Magnitude of velocity

      [tex]v=\sqrt{2.6^2+1.8^2}=3.16m/s[/tex]

Answer:

The average power is 619 W.

(B) is correct option

Explanation:

Given that,

Weight = 62.0 kg

Height = 4.28 m

Time t = 4.20 s

We need to calculate the work done

Work done by woman

[tex]W=mgh[/tex]

Where,

m = mass

g = acceleration due to gravity

t = time

Put the value into the formula

[tex]W=62\times9.8\times 4.28[/tex]

[tex]W=2600.528 J[/tex]

We need to calculate the power

Using formula of power

[tex]P=\dfrac{W}{t}[/tex]

Where,

W = work done

t = time

Put the value in to the formula

[tex]P=\dfrac{2600.528}{4.20}[/tex]

[tex]P=619\ W[/tex]

Hence, The average power is 619 W.

The average power supplied by the woman is calculated by first determining the work done against gravity to move up the stairs, and then dividing this by the time taken. After performing these calculations, the most accurate answer is approximately 629 W.

The question is asking to calculate the average power supplied by a woman while running up the stairs. Power is defined as the rate at which work is done. In this scenario, the work is the woman's movement against the gravitational force, which is calculated using the formula W = mgh, where m is mass, g is the acceleration due to gravity and h is the height of the stairs. Substituting the given values, we get W = 62.0 Kg * 9.8 m/s² * 4.28 m = 2650.368 Joules. Power is calculated by dividing the work done by the time taken, so P = W/t = 2650.368 J / 4.20 s = 630.8 Watts, which is closest to option (d) 629 W.

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Answer:

a)

600 N

b)

1.2 x 10⁵ Pa

Explanation:

(a)

d₁ = diameter of small piston = 8 cm = 0.08 m

d₂ = diameter of large piston = 40 cm = 0.40 m

F₂ = force applied to large piston = 15000 N

F₁ = force applied to small piston = ?

Using pascal's law

[tex]\frac{F_{1}}{(0.25)\pi d_{1}^{2}} = \frac{F_{2}}{(0.25)\pi d_{2}^{2}}[/tex]

Inserting the values

[tex]\frac{F_{1}}{(0.08)^{2}} = \frac{15000}{(0.40)^{2}}[/tex]

F₁ = 600 N

b)

Pressure applied is given as

[tex]P = \frac{F_{1}}{(0.25)\pi d_{1}^{2}}[/tex]

[tex]P = \frac{(600)}{(0.25)(3.14) (0.08)^{2}}[/tex]

P = 1.2 x 10⁵ Pa

To determine the force needed on the small piston of a hydraulic lift, the ratios of the pistons' areas are used in combination with the load on the large piston, applying Pascal's Principle. The pressure in the hydraulic fluid is then found by dividing the force by the area of the piston where the force is applied.

Given the diameters of the pistons, we can find the areas by using the formula for the area of a circle, A = πr², where r is the radius of the piston.

For the provided diameters of 8.0 cm and 40 cm, the areas of the pistons are:

With a load of 15,000 N on the large piston:

Once we perform these calculations, we can determine the required force to apply to the small piston and the pressure applied to the fluid within the lift.

Answer:

1)  The acceleration of the car = 1.17 m/s²

2) The time required to reach 55 mi/h = 7.59 seconds.

Explanation:

1)  Initial speed of motorist, u = 35 mph = 15.56 m/s

   Final speed of motorist, v = 55 mph = 24.44 m/s

   Distance traveled, s = 500ft = 152.4 m

   We have v² = u² + 2as

                  24.44² = 15.56² + 2 x a x 152.4

                  a = 1.17 m/s²

  The acceleration of the car = 1.17 m/s²  

2) We have v = u + at

                   24.44 = 15.56 + 1.17 x t

                           t = 7.59 s

   The time required to reach 55 mi/h = 7.59 seconds.  

Answer:

A) 5.2 x 10³ N

B) 8.8 x 10³ N

Explanation:

Part A)

[tex]F_{g}[/tex] = weight of the craft in downward direction = tension force in the cable when stationary = 7000 N

[tex]T[/tex] = Tension force in upward direction

[tex]F_{d}[/tex] = Drag force in upward direction = 1800 N

Force equation for the motion of craft is given as

[tex]F_{g}[/tex] - [tex]F_{d}[/tex] - [tex]T[/tex] = 0

7000 - 1800 - [tex]T[/tex] = 0

[tex]T[/tex] = 5200 N

[tex]T[/tex] = 5.2 x 10³ N

Part B)

[tex]F_{g}[/tex] = weight of the craft in downward direction = tension force in the cable when stationary = 7000 N

[tex]T[/tex] = Tension force in upward direction

[tex]F_{d}[/tex] = Drag force in downward direction = 1800 N

Force equation for the motion of craft is given as

[tex]T[/tex]  - [tex]F_{g}[/tex] - [tex]F_{d}[/tex] = 0

[tex]T[/tex] - 7000 - 1800  = 0

[tex]T[/tex] = 8800 N

[tex]T[/tex] = 8.8 x 10³ N

Answer:

(a) 6.42 x 10^-18 N

(b) 9.73 x 10^8 m/s^2

Explanation:

v = 760 m/s, B = 0.034 T, m = 6.6 x 10^-27 kg, q = 3.2 x 10^-19 C, theta = 51 degree

(a) F = q v B Sin theta

F = 3.2 x 10^-19 x 760 x 0.034 x Sin 51

F = 6.42 x 10^-18 N

(b) Acceleration, a = Force / mass

a = (6.42 x 10^-18) / (6.6 x 10^-27)

a = 0.973 x 10^9

a = 9.73 x 10^8 m/s^2

Answer:

Explanation:

Comet is made by dust particles, icy particles, gases etc.

A comet has a fixed time to complete a revolution around the sun.

As a comet comes nearer to the sun, due to the heat of the sun the vapour and the icy particles becomes gases and due to the radial pressure of energy od sun, we observe a tail of comet which has vapours mainly. SI the comet is visible easily.

Answer:

a) There are 100 centimeters in 1 meter.

b) [tex]\texttt{A cm is equal to }\frac{1}{2.54}\texttt{ inch}[/tex]

Explanation:

a) We have the conversion

         1 m = 100 cm

   So there are 100 centimeters in 1 meter.

b) 1 inch = 2.54 cm

    [tex]1cm=\frac{1}{2.54}inch[/tex]

   [tex]\texttt{A cm is equal to }\frac{1}{2.54}\texttt{ inch}[/tex]

Answer:

a) 800 keV

b) 25 km

Explanation:

[tex]Strength\ of\ Electric\ field=2\times 10^6\ V/m\\a)\ Potential\ Difference=Strength\ of\ Electric\ field\times Distance\\\Rightarrow Potential\ Difference=Kinetic\ Energy\ =2\times 10^6\times 0.4\\\therefore Energy=0.8\times 10^6\ eV=800\ keV\\[/tex]

[tex]b)\ Potential\ difference=50\ GeV=50\times 10^9\ eV\\Distance=\frac{Potential\ difference}{electric\ field}\\\Rightarrow Distance=\frac{50\times 10^9}{2\times 10^6}\\\Rightarrow Distance=25\times 10^3\ m\\\therefore Distance=25\ km[/tex]

Answer:

a) 800 keV

b)  24.996 km.

Explanation:

(a) we have

[tex]\large \Delta K.E=q\Delta V[/tex]  .............(1)

where,

[tex]\large \Delta K.E[/tex] = Change in kinetic energy

[tex]q[/tex] = charge of an electron

[tex]\Delta V[/tex] = Potential difference

also

[tex]\large E=\frac{V}{d}[/tex]       .......(2)

E = electric field

d = distance traveled

Now from (1) and (2) we have,

[tex]\large \Delta K. E=qV=qEd[/tex]

substituting the values in the above equation, we get

[tex]\large \Delta K. E=(1.6\times 10^{-19}C)(2\times 10^6V/m)(0.400m)(\frac{1eV}{1.6\times 10^{-19}J})(\frac{1keV}{1000eV})[/tex]

[tex]\large \Delta K. E=800keV[/tex]

Thus, the energy gained by the electron is 800 keV if it is accelerated over a distance of 0.400 m.

(b) Using the equation (1), we have

[tex]\large d=\frac{\Delta K.E}{qE}[/tex]

[tex]\large d=\frac{(50\times 10^9eV)}{(1.6\times 10^{-19C})(2\times 10^6V/m)}(\frac{1.6\times 10^{-19}J}{1eV})[/tex]

or

[tex]\large d=2.4996\times 10^4m[/tex]

or

[tex]\large d=24.996\times 10^3m=24.996km[/tex]

Thus, to gain 50.0 GeV of energy the electron must be accelerated over a distance of 24.996 km.

Answer:

(a) 2.23 m/s

(b) 0.9 m

Explanation:

h = 1 m, t = 0.1 second

horizontal component of initial velocity, ux = 2 m/s

vertical component of initial velocity, uy = 0  

(a) Let v be the velocity after 0.1 seconds. Its vertical component is vy and horizontal component is vx.

The horizontal component of velocity remains constant as in this direction, acceleration is zero.

vx = ux = 2 m/s

Use first equation of motion in Y axis direction.

vy = uy + g t

vy = 0 + 9.8 x 0.1 = 0.98 m/s

Resultant velocity after 0.1 second

v^2 = vx^2 + vy^2

v^2 = 2^2 + 0.98^2

v = 2.23 m/s

(b) Let it takes time t to land.

Use second equation of motion along Y axis

h = uy t + 1/2 g t^2

1 = 0 + 1/2 x 9.8 x t^2

t = 0.45 second

Let it lan at a distance x.

so, x = ux x t

x = 2 x 0.45 = 0.9 m

Answer:

A) weight=mass X gravitational force

10Kg X 10 M/S 2

100 KgM/S2

100Newton

B) Force =Mass X gravitational force

10Kg X 10 M/S2

100 Newton

The weight of a 10kg stone sitting at rest on the ground is 98 Newtons, and the normal force acting on it is also 98 Newtons.

The weight of an object is calculated by multiplying its mass by the acceleration due to gravity. In this case, the mass of the stone is 10kg and  The standard acceleration due to gravity on Earth is approximately 9.8 m/s2.  Therefore, the weight of the stone would be 10 kg x 9.8 m/s^2 = 98 Newtons.

b) The normal force acting on the stone is the force exerted by the surface of the ground on the stone in the upward direction. When the stone is sitting on the ground, the normal force is equal in magnitude and opposite in direction to the weight of the stone. Therefore, the normal force is also 98 N.

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It takes 225 Earth days for Venus to orbit the Sun. Interestingly, Venus spins on its axis so slowly that its day (243 Earth days) is longer than its year. The Sun takes 117 Earth days to return to the same place in Venus' sky.

The time it takes for Venus to make one orbit around the sun, also known as its orbital period, is actually 225 Earth days. This is quite different compared to Earth which takes 365.25 days to orbit the Sun. Additionally, Venus spins on its axis very slowly, with its rotational period being 243 Earth days. As a result, a day on Venus - considering its rotation - is longer than its year! Also, this leads to an unusual phenomenon where the sun takes 117 Earth days to return to the same place in the Venusian sky. This suggests that Venus' rotation and orbit display unique characteristics compared to other planets in our solar system, likely due to factors from its formation.

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Venus completes one orbit around the Sun in approximately 225 Earth days. This is shorter than Earth's orbital period due to Venus's closer distance to the Sun.

Orbit of Venus Around the Sun-

The distance between the Earth and the Sun is approximately 1.50 x 1011 meters, while Venus is closer, at 1.08 x 1011 meters. This difference in distance influences the orbital period of each planet. Venus takes about 225 Earth days to complete one full orbit around the Sun.

Venus has a nearly circular orbit and, being closer to the Sun than Earth, receives almost twice as much light and heat. The elliptical orbits mean that different planets have different orbital periods, with those closer to the Sun having shorter years.

Answer:

The new force F' will be same of the original force F.

Explanation:

Given that,

Charges = 0.3

We need to calculate the force between the charges

Suppose that the distance between the charges is r.

The force between the charges

[tex]F =\dfrac{kq_{1}q_{2}}{r^2}[/tex]

Put the value into the formula

[tex]F=\dfrac{k\times0.3\times0.3}{r^2}[/tex]

[tex]F=\dfrac{k\times(0.09)}{r^2}[/tex]

If the charges are doubled, and the distance between them increased by 100%.

So, The charges are 0.6 and the distance is 2r.

Then,

The force between the charges

[tex]F'=\dfrac{k\times0.6\times0.6}{(2r)^2}[/tex]

[tex]F'=\dfrac{k\times0.09}{r^2}[/tex]

[tex]F'=F[/tex]

Hence, The new force F' will be same of the original force F.

Answer:

Work done = 35467.278 J

Explanation:

Given:

Height of the cone = 4m

radius (r) of the cone = 1.2m

Density of the cone = 600kg/m³

Acceleration due to gravity, g = 9.8 m/s²

Now,

The total mass of the cone (m) = Density of the cone × volume of the cone

Volume of the cone = [tex]\frac{1}{3}\pi r^2 h[/tex]

thus,

volume of the cone = [tex]\frac{1}{3}\pi 1.2^2\times 4[/tex] = 6.03 m³

therefore, the mass of the cone = 600 Kg/m³ × 6.03 m³ = 3619.11 kg

The center of mass for the cone lies at the [tex]\frac{1}{4}[/tex]times the total height

thus,

center of mass lies at,  h' = [tex]\frac{1}{4}\times4=1m[/tex]

Now, the work gone (W) against gravity is given as:

W = mgh'

W = 3619.11kg × 9.8 m/s² × 1 = 35467.278 J

The work against gravity required to build a cone of height 4 m and base of radius 1.2 m out of a material of density 600 kg/m3 is 35,467.278 J.

Given to us

Height of the cone = 4m

The radius of the cone = 1.2 m

Density of the material = 600 kg/m³

We know that the work gone against the gravity is given as,

[tex]W = mgh'[/tex]

where W is the work, m is the mass, g is the acceleration due to gravity, and h' is the center of mass.

Also, the mass of an object is the product of its volume and density, therefore,

[tex]m = v \times \rho[/tex]

We know the value of the volume of a cone,

[tex]m = \dfrac{1}{3}\pi r^2 h \times \rho[/tex]

The center of mass of a cone lies at the center of its base at 1/4 of the total height from the base,

[tex]h' = \dfrac{h}{4}[/tex]

Substitute all the values we will get,

[tex]W = mgh'\\\\W = (v \times \rho) g \times \dfrac{h}{4}\\\\W = (\dfrac{1}{3}\pi r^2h \times \rho) g \times \dfrac{h}{4}\\\\W = (\dfrac{1}{3}\pi (1.2)^2 \times4 \times 600) \times 9.81 \times \dfrac{4}{4}\\\\W = 35467.278\rm\ J[/tex]

Hence, the work against gravity required to build the right circular cone of height 4 m and base of radius 1.2 m out of a lightweight material of density 600 kg/m3 is 35,467.278 J.

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Answer:

The total elongation for the tension member is of 0.25mm

Explanation:

Assuming that material is under a linear deformation then the relation between the stress and the specific elongation is given as:

[tex]\sigma=E*\epsilon[/tex] (1)

Where E is the modulus of elasticity, σ the stress and ε the specific deformation. Also, the total longitudinal elongation can be expressed as:

[tex]\delta L=L*\epsilon[/tex] (2)

Here L is the member extension and δL the change total longitudinal elongation.  

Now if the stress is found then the deformation can be calculated by solving the stress-deformation equation (1). The stress applied sigama is computed dividing the axial load P by the cross-sectional area A:

[tex]\sigma=P/A[/tex]  

[tex]\sigma=22.44N / 1290 mm^2[/tex]  

[tex]\sigma=0.0174 N/mm^2[/tex]  

Solving for epsilon and replacing the calculated value for the stress and the value for the modulus of elasticity:

[tex]=\sigma=E*\epsilon[/tex]

[tex]\epsilon=\sigma/E[/tex]

[tex]\epsilon=0.0174 \frac{N}{mm^2}/\ 204 \frac{N}{mm^2} [/tex]

[tex]\epsilon=8.53*10^-{5}[/tex]

Finally introducing the specific deformation and the longitudinal extension in the equation of total elongation (2):

[tex]\delta L=3048 mm * 8.53*10^{-5} [/tex]  

[tex]\delta L= 0.25 mm [/tex]

Answer: false

Explanation: the air will increase if it is compressed by an adiabatic compressor

Answer:

09m

Explanation:

yes the current flow in clockwise direction

Explanation:

Given that,

Diameter of the metal ring, d = 4.2 cm

Radius, r = 2.1 cm

Initial magnetic field, B = 1.12 T

The magnetic field is decreasing at the rate of, [tex]\dfrac{dB}{dt}=0.24\ T/s[/tex]

Due to change in magnetic field, an emf is induced in it. And hence, electric field is induced. It is given by :

[tex]\int\limits {E.dl}=\dfrac{d}{dt}(\int\limits{B.dA)}[/tex]

[tex]E.(2\pi r)=\dfrac{dB}{dt}(\pi r^2)[/tex]

[tex]E=\dfrac{dB}{dt}\times \dfrac{r}{2}[/tex]

[tex]E=0.24\times \dfrac{2.1\times 10^{-2}}{2}[/tex]

[tex]E=0.00252\ N/C[/tex]

So, the magnitude of the electric field induced in the ring has a magnitude of 0.00252 N/C.

The direction of electric field will be counter clock wise direction as viewed by someone on the south pole of the magnet.

Explanation:

A satellite orbiting the Earth is a characteristic example of the uniform circular motion, where the direction of the velocity vector [tex]\vec{V}[/tex] is perpendicular to the radius [tex]r[/tex] of the trajectory.  Hence, the velocity is changing although the speed is constant.

On the other hand,  acceleration [tex]\vec{a}[/tex] is directed toward the center of the circumference (that is why it is called centripetal acceleration).  

Now, according to Newton's 2nd law, the force [tex]\vec{F}[/tex] is directly proportional and in the same direction as the acceleration:  

[tex]\vec{F}=m.\vec{a}[/tex]  

Therefore, the net force on the satellite resulting from its circular motion points towards the center of the circle (where the Earth is in the Earth-satellite system).

Answer:

42.3 MV

Explanation:

d = diameter of the metal sphere = 2.15 m

r = radius of the metal sphere

diameter of the metal sphere is given as

d = 2r

2.15 = 2 r

r = 1.075 m

Q = charge on sphere = 5.05 mC = 5.05 x 10⁻³ C

Potential near the surface is given as

[tex]V = \frac{kQ}{r}[/tex]

[tex]V = \frac{(9\times 10^{9})(5.05\times 10^{-3})}{1.075}[/tex]

V = 4.23 x 10⁷ volts

V = 42.3 MV

Answer:

The final temperature is 79.16°C.

Explanation:

Given that,

Heat [tex]Q=1.50\times10^{5}\ J[/tex]

Temperature = 20.0°C

Entropy = 465 J/k

We need to calculate the average temperature

Using relation between entropy and heat

[tex]\Delta S=\dfrac{\Delta Q}{T}[/tex]

[tex]T=\dfrac{\Delta Q}{\Delta S}[/tex]

Where, T = average temperature

[tex]\Delta Q[/tex]= transfer heat

[tex]\Delta S[/tex]= entropy

Put the value into the formula

[tex]T=\dfrac{1.50\times10^{5}}{465}[/tex]

[tex]T=322.58\ K[/tex]

We need to calculate the final temperature

Using formula of average temperature

[tex]T = \dfrac{T_{i}+T_{f}}{2}[/tex]

[tex]T_{f}=2T-T_{i}[/tex]....(I)

Put the value in the equation (I)

[tex]T_{f}=2\times322.58-293[/tex]

[tex]T_{f}=352.16\ K[/tex]

We convert the temperature K to degrees

[tex]T_{f}=352.16-273[/tex]

[tex]T_{f}=79.16^{\circ}\ C[/tex]

Hence, The final temperature is 79.16°C.

Answer:

The acceleration of the block is 6.35 m/s².

Explanation:

It is given that,

A force is applied to a block to move it up a 30° incline. The applied force is, F = 65 N

Mass of the block, m = 5 kg

We need to find the acceleration of the block. From the attached figure, it is clear that.

[tex]F_x=ma_x[/tex]

[tex]F\ cos\theta-mg\ sin\theta=ma_x[/tex]

[tex]a_x=\dfrac{F\ cos\theta-mg\ sin\theta}{m}[/tex]

[tex]a_x=\dfrac{65\ N\ cos(30)-5\ kg\times 9.8\ m/s^2\ sin(30)}{5\ kg}[/tex]

[tex]a_x=6.35\ m/s^2[/tex]

So, the acceleration of the block is 6.35 m/s². Hence, this is the required solution.

Acceleration is defined as the rate of change of the velocity of the body. Its unit is m/sec².The magnitude of the acceleration of the block will be 6.35 m/sec².

It is a type of opposition force acting on the surface of the body that tries to oppose the motion of the body. its unit is Newton (N).

Mathematically it is defined as the product of the coefficient of friction and normal reaction.

On resolving the given force and accelertaion in the different components and balancing the equation gets.Components in the x-direction.

The given data in the problem,

F is the applied force =65 N

Θ is the angle of inclined plane=30°

m is the mass of the block= 5 kg

We need to find the acceleration of the block in the x-direction

[tex]\rm F_x=ma_x\\\\\rm F_x=Fcos\theta-mgsin\theta\\\\\rm Fcos\theta-mgsin\theta=ma_x\\\\\rm a_x=\frac{Fcos\theta-mgsin\theta}{m}\\\\ \rm a_x=\frac{65\timescos30^0-mgsin30^0}{5} \\\\\rm a_x=6.35 m/sec^2[/tex]

Hence the magnitude of the acceleration of the block will be 6.35 m/sec².

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Explanation:

It is given that,

Mass of SUV, m = 900 kg

It is moving with a constant speed of 1.44 m/s due south. We need to find the total force on the vehicle. The second law of motion gives the magnitude of force acting on the object. It is given by :

F = m a

Since, [tex]a=\dfrac{dv}{dt}[/tex] i.e. the rate of change of velocity

As SUV is travelling at a constant speed. This gives acceleration of it as zero.

So, F = 0

So, the total force acting on SUV is 0 N. Hence, this is the required solution.

Answer:

The total force acting on the vehicle is 0 N.

Explanation:

Given that,

Mass of SUV = 900 kg

Velocity = 1.44 m/s

SUV is travelling at a constant speed of 1.44 m/s due north.

We need to calculate the force on the vehicle.

Using newton's second law

[tex]F = ma[/tex]....(I)

We know that,

The acceleration is the first derivative of the velocity of the particle.

[tex]a = \dfrac{dv}{dt}[/tex]

Now,put the value of a in equation (I)

[tex]F=m\dfrac{dv}{dt}[/tex]

SUV is traveling at a constant speed it means acceleration is zero.

So, The force will be zero.

Hence, The total force acting on the vehicle is 0 N.

Answer:

d = 0.50 m

Explanation:

Let say the speed at the top and bottom of the window is

[tex]v_1 \: and \: v_2[/tex] respectively

now we have

[tex]d = \frac{v_1 + v_2}{2}t[/tex]

[tex]1.90 = \frac{v_1 + v_2}{2} (0.380)[/tex]

[tex]v_1 + v_2 = 10 [/tex]

also we know that

[tex]v_2 - v_1 = 9.8(0.380)[/tex]

[tex]v_2 - v_1 = 3.72[/tex]

now we have from above equations

[tex]v_2 = 6.86 m/s[/tex]

[tex]v_1 = 3.14 m/s[/tex]

now the distance from which it fall down is given as

[tex]v_f^2 - v_i^2 = 2ad[/tex]

[tex]3.14^2 - 0^2 = 2(9.8)d[/tex]

[tex]d = 0.50 m[/tex]

Answer:

c) Elastic Modulus

Explanation

As we know that when deformation is under elastic limit then stress applied to the given material is proportional to the strain developed in it

So here we can say that since they both are directly proportional to each other so the proportionality constant here is known as Modulus of elasticity.

So we can say it is given as

[tex]stress = E (strain)[/tex]

so now if we draw a graph between between stress and strain then it must be a straight line and the the slope of this straight line is given as

[tex]Slope = \frac{Stress}{Strain}[/tex]

So here correct answer will be

c) Elastic Modulus

Final answer:

The slope of the stress-strain curve in the elastic deformation region is known as the Elastic Modulus or Young's Modulus, which measures the stiffness of the material. The correct answer to the given question is 'c. Elastic Modulus'.

Explanation:

The slope of the stress-strain curve in the elastic deformation region represents the relationship between stress and strain under elastic conditions. This slope is in fact the Elastic Modulus, which is also known as Young's Modulus when it's in tension or compression. It serves as a measure of the stiffness or rigidity of the material, indicating how much stress is required to achieve a certain amount of strain.

In the context of a stress-strain curve, the plastic modulus is associated with plastic deformation, not the elastic region. Poisson's ratio is another material property that describes the ratio of transverse strain to axial strain, and is not the slope of the curve. Hence, the correct answer to the question is 'c. Elastic Modulus'.

Answer:

The magnitude of the average angular acceleration of the disk is 2761.1 rad/s².

Explanation:

Given that,

Angular velocity of optical disk= 1045 rad/s

Angular velocity of particular disk = 646.1 rad/s

Time = 0.234 sec

We need to calculate the average angular acceleration

Using formula of  angular acceleration

[tex]\alpha=\dfrac{\omega_{f}-\omega_{i}}{t}[/tex]

Where, [tex]\omega_{f}[/tex] = final angular velocity

[tex]\omega_{i}[/tex] = Initial angular velocity

t = time

Put the value in the equation

[tex]\alpha = \dfrac{0-646.1}{0.234}[/tex]

[tex]\alpha= -2761.1\ rad/s^2[/tex]

Negative sign shows the angular deceleration.

Hence, The magnitude of the average angular acceleration of the disk is 2761.1 rad/s².

Answer:

The speed of the ambulance is 4.66 m/s.

Explanation:

Given that,

The siren emitting a whine at 1570 Hz

The cyclist pedaling a bike at 2.45 m/s

The cyclist hears a frequency of 1560 Hz

We know that,

Speed of sound wave

[tex]v = 343\ m/s[/tex]

We calculate the speed of the ambulance

Using Doppler effect,

[tex]f'=f\times\dfrac{v+v_{o}}{v+v_{s}}[/tex]

Where,

[tex]f' [/tex]= frequency of ambulance siren

[tex]f [/tex]= cyclist hears the frequency

[tex]v_{s}[/tex]=speed of source

[tex]v_{v}[/tex]= speed of observer

Put the value in to the formula

[tex]v_{s}=f\times\dfrac{v+v_{o}}{f'}-v[/tex]

[tex]v_{s}=1570\times\dfrac{343-2.45}{1560}-343[/tex]

[tex]v_{s}=4.66\ m/s[/tex]

Hence, The speed of the ambulance is 4.66 m/s.

Answer:

Part a)

[tex]k = 1632 J[/tex]

Part b)

[tex]KE = 47 J[/tex]

Part c)

[tex]m = 7.9 kg[/tex]

Part d)

[tex]v = 2.57 m/s[/tex]

Part e)

[tex]KE = 26.1 J[/tex]

Part f)

[tex]PE = 20.9 J[/tex]

Explanation:

Total Mechanical energy is given as

[tex]E = 47.0 J[/tex]

Its maximum displacement from mean position is given as

[tex]A = 0.240 m[/tex]

Part a)

Now from the formula of energy we know that

[tex]E = \frac{1}{2}kA^2[/tex]

[tex]47 = \frac{1}{2}k(0.240)^2[/tex]

[tex]k = 1632 J[/tex]

Part b)

At the mean position of SHM whole mechanical energy will convert into kinetic energy

so it is given as

[tex]KE = 47 J[/tex]

Part c)

As per the formula of kinetic energy we know that

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

[tex]47 = \frac{1}{2}m(3.45^2)[/tex]

[tex]m = 7.9 kg[/tex]

Part d)

As we know by the equation of the speed of SHM is given as

[tex]v = \sqrt{\frac{k}{m}(A^2 - x^2)}[/tex]

[tex]v = \sqrt{\frac{1632}{7.9}(0.24^2 - 0.16^2)}[/tex]

[tex]v = 2.57 m/s[/tex]

Part e)

As we know by the formula of kinetic energy

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

[tex]KE = \frac{1}{2}(7.9)(2.57^2)[/tex]

[tex]KE = 26.1 J[/tex]

Part f)

As per energy conservation we know

KE + PE = Total energy

[tex]26.1 + PE = 47[/tex]

[tex]PE = 20.9 J[/tex]

(a) The spring constant will be k=1632 [tex]\frac{N}{m^2}[/tex]

(b) The KE at equilibrium point =47j

(c) The mass =7.9kg

(d) The velocity block [tex]2.57\frac{m}{sec}[/tex]

(e)The KE of block 0.160m will be 26.1

(f) The PE will be 20.9j

(a) for finding spring constant

KE=47 j

A=0.240m

By using the formula

[tex]E=\dfrac{1}{2} kx^2[/tex]

[tex]47=\dfrac{1}{2} k(0.240)^2[/tex]

[tex]k=1632\frac{N}{m^2}[/tex]

(b) At the mean position the whole mechanical energy will be equal to KE so

KE=47j

(c) The mass of the system

[tex]KE =\dfrac{1}{2} mv^2[/tex]

[tex]47=\dfrac{1}{2} m(3.45^2)[/tex]

[tex]m=7.9kg[/tex]

(d)Now the speed of the block

[tex]v=\sqrt{\dfrac{k}{m} (A^2-x^2)}[/tex]

[tex]v=\sqrt{\dfrac{1632}{7.9} (0.24^2-0.16^2)}[/tex]

[tex]v=2.57\frac{m}{s}[/tex]

(e) The KE of the block

[tex]KE=\dfrac{1}{2} mv^2=\dfrac{1}{2} 7.9(2.57)^2=26.1J[/tex]

(f) The PE of the system

[tex]Total Energy = KE+PE[/tex]

[tex]PE= 47-26.1 =20.9J[/tex]

Thus

(a) The spring constant will be k=1632 [tex]\frac{N}{m^2}[/tex]

(b) The KE at equilibrium point =47j

(c) The mass =7.9kg

(d) The velocity block [tex]2.57\frac{m}{sec}[/tex]

(e)The KE of block 0.160m will be 26.1

(f) The PE will be 20.9j

To know more about the spring-mass system follow

https://brainly.com/question/15540894

Answer:

21828 m

Explanation:

[tex]v_{ground}[/tex] = speed of sound through ground = 6 km/s = 6000 m/s

[tex]v_{air}[/tex] = speed of sound through air = 343 m/s

t = time taken for the vibrations to arrive

t' = time taken for sound to arrive = t + 60

d = distance of the point of explosion

time taken for the vibrations to arrive is given as

[tex]t = \frac{d}{v_{ground}}[/tex]                            eq-1

time taken for the sound to arrive is given as

[tex]t' = \frac{d}{v_{air}}[/tex]

[tex]t + 60 = \frac{d}{v_{air}}[/tex]

using eq-1

[tex]\frac{d}{v_{ground}} + 60 = \frac{d}{v_{air}}[/tex]

[tex]\frac{d}{6000} + 60 = \frac{d}{343}[/tex]

d = 21828 m

Explanation:

It is given that,

Mass of first ball, m₁ = 2 kg

Mass of other ball, m₂ = 3.5 kg

Velocities of both balls, u = 0.9 m/s

(1) The lighter ball rebounds opposite its initial direction, with speed 0.90 m/s. We need to find the final velocity of second ball. Applying the conservation of momentum as :

[tex]2\ kg\times 0.9-3.5\ kg\times 0.9\ m/s=-2\ kg\times 0.9\ m/s+3.5v[/tex]

v is the final velocity of heavier ball.

v = 0.128 m/s

or

v = 0.13 m/s

Initial kinetic energy, [tex]E_i=\dfrac{1}{2}\times (2\ kg+3.5\ kg)\times (0.9\ m/s)^2=2.23\ J[/tex]

Final kinetic energy, [tex]E_f=\dfrac{1}{2}\times 2\ kg\times (0.9\ m/s)^2+\dfrac{1}{2}\times 3.5\ kg\times (0.13\ m/s)^2=0.84\ J[/tex]

Lost in kinetic energy, [tex]\Delta KE=0.84-2.23=-1.39\ J[/tex]

Hence, this is the required solution.