An electron enters a region of uniform electric field with an initial velocity of 50 km/s in the same direction as the electric field, which has magnitude E = 50 N/C, (a) what is the speed of the electron 1.5 ns after entering this region? (b) How far does the electron travel during the 1.5 ns interval?

Answer:

The car travels 800m

Explanation:

We will need two formulas to solve this question:

[tex]V_{f}=V_{0}+at\\X=V_{0}t+\frac{at^{2}}{2}[/tex]

We can obtain the the time by using the final velocity formula and replacing with known values:

[tex]V_{f}=V_{0}+at\\\\0\frac{m}{s}=80\frac{m}{s}-4\frac{m}{s^{2}}t\\\\4\frac{m}{s^{2}}t=80\frac{m}{s}\\t=20 s[/tex]

Since we know the time, the question will be answered by inputing all the variables needed in the distance formula:

[tex]X=V_{0}t+\frac{at^{2}}{2}\\\\X=80\frac{m}{s}(20s)-\frac{4(20s)^{2}}{2}=800m[/tex]

Final answer:

The race car will travel 800 meters before coming to a stop after deploying a drag parachute causing a deceleration of 4 m/s^2 from an initial speed of 80 m/s.

Explanation:

To calculate how far the race car will travel before it stops after deploying a drag parachute, we can use the kinematic equation that relates initial velocity, final velocity, acceleration, and distance traveled. Considering that the deceleration is constant, the kinematic equation we can use is v2 = u2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance traveled.

Since the car comes to a stop, v = 0 m/s. The initial speed is u = 80 m/s and the deceleration provided by the drag parachute is a = -4 m/s2 (negative because it's deceleration). Plugging these values into the equation gives us:

0 = (80 m/s)2 + 2(-4 m/s2)s

Solving for s, we find:

s = (80 m/s)2 / (2 * 4 m/s2)

s = 6400 m2/s2 / 8 m/s2

s = 800 meters

Therefore, the race car will travel 800 meters before coming to a stop after the drag parachute is deployed.

Answer:

Precise and Reproducible

Explanation:

For a system or a process to be useful a measurement standard for any physical quantity, it must be:

Answer:

precise

reproducible

Explanation:

Answer:

[tex]\eta = 0.411[/tex]

Explanation:

As we know that efficiency is defined as the ratio of output useful work and the input energy to the engine

So here we know that the

input energy = 441.3 kJ

energy rejected = 259.8 kJ

so we have

[tex]Q_1 - Q_2 = W[/tex]

[tex]W = 441.3 kJ - 259.8 kJ = 181.5 kJ[/tex]

now efficiency is defined as

[tex]\eta = \frac{W}{Q_1}[/tex]

[tex]\eta = \frac{181.5}{441.3}[/tex]

[tex]\eta = 0.411[/tex]

Answer:

A. 33.77 m/s

B. 6.20 s

Explanation:

Frame of reference:

Gravity g=-9.8 m/s^2; Initial position (roof) y=0; Final Position street y= -21 m

Initial velocity upwards v= 27 m/s

Part A. Using kinematics expression for velocities and distance:

[tex]V_{final}^{2}=V_{initial}^{2}+2g(y_{final}-y_{initial})\\V_{f}^{2}=27^{2}-2*9.8(-21-0)=33.77 m/s[/tex]

Part B. Using Kinematics expression for distance, time and initial velocity

[tex]y_{final}=y_{initial}+V_{initial}t+\frac{1}{2} g*t^{2}\\==> -0.5*9.8t^{2}+27t+21=0\\==> t_1 =-0.69 s\\t_2=6.20 s[/tex]

Since it is a second order equation for time, we solved it with a calculator. We pick the positive solution.

The speed of the rock just before it hits the street is approximately 20.3 m/s, and the time elapsed from when the rock is thrown until it hits the street is approximately 5.5 seconds.

Part A: To determine the speed of the rock just before it hits the street, we can use the concept of conservation of energy. When the rock is at the top of its trajectory, its potential energy is at its maximum and its kinetic energy is at its minimum. When the rock is just before hitting the street, its potential energy is at its minimum (zero) and its kinetic energy is at its maximum. Using the equations for potential energy and kinetic energy, we can calculate the speed of the rock:

Initial potential energy = final potential energy + final kinetic energy

mgh = 1/2m*v^2

where m is the mass of the rock, g is the acceleration due to gravity, h is the height of the building, and v is the final velocity of the rock.

Since the mass of the rock cancels out, we have:

gh = 1/2v^2

Plugging in the values, g = 9.8 m/s^2 and h = 21.0 m, we can solve for v:

v = sqrt(2gh)

v = sqrt(2*9.8*21.0)

v = sqrt(411.6)

v = 20.3 m/s

Therefore, the speed of the rock just before it hits the street is approximately 20.3 m/s.

Part B: To calculate the time elapsed from when the rock is thrown until it hits the street, we can use the equation for time of flight of a vertically thrown object:

t = 2v/g

where t is the time of flight, v is the initial upward velocity of the rock, and g is the acceleration due to gravity.

Plugging in the values, v = 27.0 m/s and g = 9.8 m/s^2, we can solve for t:

t = 2*27.0/9.8

t = 5.5 s

Therefore, the time elapsed from when the rock is thrown until it hits the street is approximately 5.5 seconds.

Answer:

Wavelength = 736.67 nm

Explanation:

Given

Energy of the photon = 2.70 × 10⁻¹⁹ J

Considering:

[tex]Energy=h\times frequency[/tex]

where, h is Plank's constant having value as 6.63 x 10⁻³⁴ J.s

The relation between frequency and wavelength is shown below as:

c = frequency × Wavelength

Where, c is the speed of light having value = 3×10⁸ m/s

So, Frequency is:

Frequency = c / Wavelength

So,  Formula for energy:

[tex]Energy=h\times \frac {c}{\lambda}[/tex]

Energy = 2.70 × 10⁻¹⁹ J

c = 3×10⁸ m/s

h = 6.63 x 10⁻³⁴ J.s

Thus, applying in the formula:

[tex]2.70\times 10^{-19}=6.63\times 10^{-34}\times \frac {3\times 10^8}{\lambda}[/tex]

Wavelength = 736.67 × 10⁻⁹ m

1 nm = 10⁻⁹ m

So,

Wavelength = 736.67 nm

Answer:

18.21 m

41.1 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

Part A

[tex]v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-17.9^2}{2\times -8.8}\\\Rightarrow s=18.21\ m[/tex]

On a dry day she would have to start braking 18.21 m away from the intersection

Part B

[tex]v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-17.9^2}{2\times -3.9}\\\Rightarrow s=41.1\ m[/tex]

On a wet day she would have to start braking 41.1 m away from the intersection

Answer:

D. pre schematic

Explanation:

During the _pre schematic____ developmental stage children are drawing their thoughts instead of merely observations.

Pre schematic stage in children is the age at which they drawing, symbols, spatial figures to express themselves. Their becomes more complex and full of lines as they age. They use single base lines, multiple base lines and fold up views.

Answer:

The magnetic field is [tex]2.11\times10^{-5}\ T[/tex]

Explanation:

Given that,

Speed [tex]v=1.30\times10^{6}\ m/s[/tex]

Radius = 0.350 m

We need to calculate the magnetic field

Using formula of magnetic field

[tex]B =\dfrac{mv}{qr}[/tex]

Where, m = mass of electron

v = speed of electron

q = charge of electron

r = radius

Put the value into the formula

[tex]B=\dfrac{9.1\times10^{-31}\times1.30\times10^{6}}{1.6\times10^{-19}\times0.350}[/tex]

[tex]B=2.11\times10^{-5}\ T[/tex]

Hence, The magnetic field is [tex]2.11\times10^{-5}\ T[/tex]

Answer:

56 m/s

Explanation:

The electirc force applied on the particle by the field will be

F = q * E

F = 3*10^-3 * 80 = 0.24 N

This force will cause an acceleration:

F = m * a

a = F/m

a = 0.24 / 0.02 = 12 m/s^2

The equation for speed under constant acceleration is:

V(t) = V0 + a*t

V(3) = 20 + 12 * t = 56 m/s

The final speed will be  56 m/s.

Answer:

a. a = [tex]6.41 m/s^2[/tex]

b. T = -0.81 N

Explanation:

Given,

Let 'a' be the acceleration of the blocks and 'T' be the tension in the string.

From the f.b.d. of the lighter block,

[tex]w_1sin\theta\ -\ T\ -\ f_1\ =\ \dfrac{w_1a}{g}\\\Rightarrow T\ =\ w_1sin\theta\ -\ \dfrac{w_1a}{g}\ -\ f_1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,eqn (1)[/tex]

From the f.b.d. of the heavier block,

[tex]w_2sin\theta\ +\ T\ -\ f_2\ =\ \dfrac{w_2a}{g}\\\Rightarrow T\ =\ \dfrac{w_2a}{g}\ -\ w_2sin\theta\ -\ f_2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,eqn (2)[/tex]

From eqn (1) and (2), we get,

[tex]w_1sin\theta\ -\ \dfrac{w_1a}{g}\ -\ f_1\ =\ \dfrac{w_2a}{g}\ -\ w_2sin\theta\ -\ f_2\\\Rightarrow w_1gsin\theta \ -\ w_1a\ -\ \mu_1w_1gcos\theta\ =\ w_2a\ -\ w_2gsin\theta\ -\ \mu_2 w_2gcos\theta\\[/tex]

[tex]\Rightarrow a\ =\ \dfrac{g(w_1sin\theta\ -\ \mu_1w_1cos\theta\ +\ w_2sin\theta\ +\ \mu_2w_2cos\theta)}{w_1\ +\ w_2}\\\Rightarrow a\ =\ \dfrac{g[sin\theta(w_1\ +\ w_2)\ +\ cos\theta(\mu_2w_2\ -\ \mu_1w_1)]}{w_1\ +\ w_2}\\[/tex]

[tex]\Rightarrow a\ =\ \dfrac{9.81\times [sin30^o\times (3.0\ +\ 7.0)\ +\ cos30^o\times (0.31\times 7.0\ -\ 0.13\times 3.0)]}{3.0\ +\ 7.0}\\\Rightarrow a\ =\ 6.4\ m/s.[/tex]

part (b)

From the eqn (2), we get,[tex]T\ =\ \dfrac{w_2a}{g}\ -\ w_2sin\theta\ -\ \mu_2w_2cos\theta\\\Rightarrow T\ =\ \dfrac{7.0\times 6.41}{9.81}\ -\ 7.0\times sin30^o\ -\ 0.31\times 7.0\times cos30^o\\\Rightarrow T\ =\ -0.81\ N[/tex]

Answer:

(a) 2.793 m/s^2

(b) 0.335 N

Explanation:

Let a be the acceleration in the system and T be the tension in the string.

W1 = 3 N

W2 = 7 N

m1 = 3 /1 0 = 0.3 kg

m2 = 7 / 10 = 0.7 kg

θ = 30°

μ1 = 0.13, μ2 = 0.31

Let f1 be the friction force acting on block 1 and f2 be the friction force acting on block 2.

By the laws of friction

f1 = μ1 x N1

Where, N1 be the normal reaction acting on block 1.

So, f1 = 0.13 x W1 Cosθ = 0.13 x 3 x cos 30 = 0.337 N

By the laws of friction

f2 = μ2 x N2

Where, N2 be the normal reaction acting on block 2.

So, f2 = 0.31 x W2 Cosθ = 0.31 x 7 x cos 30 = 1.88 N

Apply Newton's second law for both the blocks

W1 Sin30 - T - f1 = m1 a

3 Sin 30 - T - 0.337 = 0.3 x a

1.163 - T = 0.3 a ..... (1)

W2 Sin30 + T - f2 = m2 a

7 Sin30 + T - 1.88 = 0.7 x a

1.62 + T = 0.7 a ..... (2)

By solving equation (1) and (2) we get

a = 2.793 m/s^2

(b) Put the value of a in equation (2), we get

1.62 + T = 0.7 x 2.793

T = 0.335 N

Explanation:

Given that,

Distance, s = 47 m

Time taken, t = 8.6 s

Final speed of the truck, v = 2.3 m/s

Let u is the initial speed of the truck and a is its acceleration such that :

[tex]a=\dfrac{v-u}{t}[/tex].............(1)

Now, the second equation of motion is :

[tex]s=ut+\dfrac{1}{2}at^2[/tex]

Put the value of a in above equation as :

[tex]s=ut+\dfrac{1}{2}\times \dfrac{v-u}{t}\times t^2[/tex]

[tex]s=\dfrac{t(u+v)}{2}[/tex]

[tex]u=\dfrac{2s}{t}-v[/tex]

[tex]u=\dfrac{2\times 47}{8.6}-2.3[/tex]

u = 8.63 m/s

So, the original speed of the truck is 8.63 m/s. Hence, this is the required solution.

Answer:

Explanation:

Given

Velocity of helicopter relative to air 12.5 m/s to west

In vector form

[tex]V_{ha}=-12.5\hat{i}[/tex]

where [tex]V_h[/tex]= velocity of helicopter relative to ground

Also the velocity of air

[tex]V_a=4.55\hat{j}[/tex]

[tex]V_{ha}=V_h-V_a[/tex]

[tex]V_h=V_{ha}+V_a[/tex]

[tex]V_h=-12.5\hat{i}+4.55\hat{j}[/tex]

Speed relative to east[tex]=\sqrt{12.5^2+4.55^2}[/tex]

=13.302

For Direction

[tex]tan\theta =\frac{4.55}{-12.5}[/tex]

[tex]180-\theta =20[/tex]

[tex]\theta =160^{\circ}[/tex] relative to east

For 1 km travel it takes

[tex]t=\frac{1000}{13.302}=75.17 s[/tex]

For a mass-spring system with specified parameters, the damping constant is found to be 4.999 Ns/m. The amplitude reduces exponentially over time. The critical damping constant, defining the transition from underdamped to overdamped behavior, is 5 Ns/m.

This problem pertains to a system of mass springs undergoing damped oscillations. The damping constant (b) can be calculated using the formula b = 2*m*ωd. Here ωd signifies the damped frequency and is given by ωd = ω * (1 - γ/2), where ω is the natural frequency and γ is the damping ratio of 0.2% = 0.002. The natural frequency ω = sqrt(k/m) = sqrt(12.5/0.5) = 5 rad/s. Substituting these values in the equation gives b = 2*0.5*4.999 = 4.999 Ns/m.

The amplitude (A) of a damped oscillation decreases exponentially with time due to energy loss, described by A(t) = A0 * e^(-γωt/2), wherein A0 is the initial amplitude.

The critical damping constant (bc), describing the boundary between underdamped and overdamped systems, is equal to 2*sqrt(m*k) = 2*sqrt(0.5*12.5) = 5 Ns/m.

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Answer:

B. a conversion factor

Explanation:

The expression "1 in. = 2.54 cm" is called a conversion factor.

With this expression inches can be converted to centimeters.

Inversely the expression can also be used to convert centimeters to inches

By rearranging the equation we get

[tex]\frac{1}{2.54}\ inches=1\ cm\\\Rightarrow 1\ cm=0.394\ inches[/tex]

The statement which shows the equal nature of two different expressions is called an equation

Rule of thumb is a general approximation of a result of a test or experiment.

Hi your answer is B on edge

Answer:

altitude of satellite from earth surface is 7140 km

Explanation:

given data:

mass of satellite = 5000kg

altitude of satellite from earth surface can be calculated by using given relation

[tex]p^2 = \frac{(4* \pi^2)(a^3)}{(G*M)}[/tex]

where,

p -  satellite's period in seconds,

a - satellite's altitude in meters,

G - gravitational constant (6.67 x 10^-11 m³/kgs²) and

M -  mass of Earth 5.98 x 10^24 kg

solving for altitude we have

[tex]a^3 =\fracp{(G*M*p^2)}{(4*\pi^2)}[/tex]

[tex]a^3 = 3.637 * 10^{20}[/tex]

a = 7,138,000 meters = 7,140 km

altitude of satellite from earth surface is 7140 km

Answer:

D. measurement and geometry

Explanation:

Building helps children develop their _ measurement and geometry__ skills. The measurement and geometry skill of the children by observing buildings.

Geometry is a method or tool for understanding the relations among shapes and spatial properties. Children can develop spatial reasoning and can visualize shapes in different positions (orientation) when observing a building.

Answer:

Speed of the wave, v = 2.4 m/s

Explanation:

Given that,

Frequency of vibrations, f = 4 Hz

Wavelength of the transverse wave, [tex]\lambda=60\ cm=0.6\ m[/tex]

We need to find the speed of the waves along the wire. Let it is equal to v. So, the relationship is given by :

[tex]v=f\times \lambda[/tex]

[tex]v=4\times 0.6[/tex]

v = 2.4 m/s

So, the speed of waves along the wire is 2.4 m/s. Hence, this is the required solution.

Final answer:

The speed of the transverse wave on the wire, given a frequency of 4.00 Hz and a wavelength of 60 cm, is calculated to be 2.40 meters per second (m/s) using the formula Speed = Frequency × Wavelength.

Explanation:

To determine the speed of a wave, you can use the formula:

Speed = Frequency × Wavelength

In the given scenario, the frequency is 4.00 Hz, and the wavelength is 60 cm, which is equivalent to 0.60 meters (since 1 meter = 100 cm). Thus, the speed of the wave on the wire can be calculated as follows:

Speed = 4.00 Hz × 0.60 m

Speed = 2.40 m/s

Therefore, the speed of the transverse wave along the wire is 2.40 meters per second (m/s).

Answer:

- 32.17 fts/s in the imperial system, -9.8 m/s in the SI

Explanation:

We know that acceleration its the derivative of velocity with respect to time, this is (in 1D):

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

So, if we wanna know the change in velocity, we can take the integral:

[tex]v(t_f) - v(t_i) = \int\limits^{t_f}_{t_i} {\frac{dv}{dt} \, dt = \int\limits^{t_f}_{t_i} a \, dt[/tex]

Luckily for us, the acceleration in this problem is constant

[tex]a \ = \ - g[/tex]

the minus sign its necessary, as downward direction is negative. Now, for a interval of 1 second, we got:

[tex]t_f = t_i + 1 s[/tex]

[tex]v(t_i + 1 s) - v(t_i) =  \int\limits^{t_i + 1 s}_{t_i} a \, dt[/tex]

[tex]v(t_i + 1 s) - v(t_i) =  \int\limits^{t_i + 1 s}_{t_i} (-g ) \, dt[/tex]

[tex]v(t_i + 1 s) - v(t_i) =  [-g t]^{t_i + 1 s}_{t_i} [/tex]

[tex]v(t_i + 1 s) - v(t_i) =  -g (t_i+1s) + g t_i [/tex]

[tex]v(t_i + 1 s) - v(t_i) =  -g t_i + -g 1s + g t_i[/tex]

[tex]v(t_i + 1 s) - v(t_i) =  -g 1s [/tex]

taking g in the SI

[tex]g=9.8 \frac{m}{s^2}[/tex]

this is:

[tex]v(t_i + 1 s) - v(t_i) =  - 9.8 \frac{m}{s} [/tex]

or, in imperial units:

[tex]g=32.17 \frac{fts}{s^2}[/tex]

this is:

[tex]v(t_i + 1 s) - v(t_i) =  - 32.17 \frac{fts}{s} [/tex]

Answer and Explanation:

The inertial reference frame is one with constant velocity or non-accelerated frame of reference.

The value of acceleration and velocity change will vary in the two frames and will not be same.

As in case, we observe the acceleration and velocity of a moving train from the platform and the one observed in the train itself will be different.

In case of energy, it is dependent on the frame of reference but the energy change is independent of the frame of reference.

Answer:

Explanation:

We need to supply 5 Nm to the bolt.

If we have a force of 25 N in the direction (2*i - 3*j), the torque is related to the length of the wrench.

T = F * d

d = T / F

d = 5 / 25 = 0.2 m

Assuming the wrench is perpendicular to the force.

The vector of the force is:

[tex]F = 25 N \frac{2i - 3j}{\sqrt{2^2 + (-3)^2}} = 13.9i - 20.8[/tex]

This is a countr-clockwise direction. If the bolt is right handed (most are) we are tightening it.

Answer:

The frequency of the emitted EM wave by He-Ne laser is [tex]4.74\times 10^{14} Hz[/tex]

Given:

Power emitted by He-Ne laser, P = 1.5 mW = [tex]1.5\times 10^{- 3}[/tex]

Wavelength, [tex]\lambda = 632.8 nm = 632.8\times 10^{- 9} m[/tex]

Solution:

Now, to calculate the frequency, [tex]\vartheta[/tex] of the EM wave emitted:

Energy associated with 1 photon = Power of one photon per sec = [tex]\frac{hc}{\lambda}[/tex]

Therefore, power associated with 'N' no. of photons, P = [tex]N\frac{hc}{\lambda}[/tex]

where

[tex]h = 6.626\times 10^{-34} m^{2}kg/s[/tex] = Planck's constant

Now,

[tex]1.5\times 10^{- 3} = N\frac{6.626\times 10^{-34}\times 3\times 10^{8})}{632.8\times 10^{- 9}}[/tex]

N = [tex]4.77\times 10^{15}[/tex]

Also, we know that:

[tex]c = \vartheta \lambda[/tex]

Thus

[tex]\vartheta = \frac{c}{\lambda} = \frac{3\times 10^{8}}{632.8\times 10^{- 9}}[/tex]

[tex]\vartheta = 4.74\times 10^{14} Hz[/tex]

Answer:

Explanation:

Given

Cheetah  speed=21.8 m/s in 2.55 sec

i.e. its [tex]a=\frac{21.8}{2.55}=8.55 m/s^2[/tex]

Cheetah top speed=28.1 m/s

And 1 m/s is equal to 2.23694 mph

therefore 28.1 m/s is [tex]2.236\times 28.1=62.858 mph[/tex]

Time taken to reach top speed

v=u+at

[tex]28.1=0+8.55\times t[/tex]

[tex]t=\frac{28.1}{8.55}=3.286 s [/tex]

Distance traveled during this time

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

[tex]28.1^2=2\times 8.55\times s[/tex]

[tex]s=\frac{789.61}{2\times 8.55}=46.176 m[/tex]

If cheetah sees a rabbit 122 m away

time taken to reach rabbit

[tex]s=ut+\frac{1}{2}at^2[/tex]

[tex]122=0+\frac{1}{2}\times 8.55\times t^2[/tex]

[tex]t^2=\frac{244}{8.55}[/tex]

[tex]t=\sqrt{28.53}=5.34 s[/tex]

Answer:

Explanation:

given,

mass of the object = 280 g = 0.28 kg

time period = 0.270 s

total energy of the system  = 4.75 J

                 [tex]\dfrac{1}{2}\ m\ V^2 = 4.75 J[/tex]

maximum speed of the object V =     [tex]\sqrt{ \dfrac{2 \times 4.75}{0.28} }[/tex]

                                              V= 5.82 m / s

(b)   force constant of the spring K = m ω²

where ω = angular frequency = 2π / T

          T= time period = 0.25 s

ω = 25.13 rad / s

K = 0.28 × 25.13²

K = 176.824 N / m

(c). Amplitude of motion A =    [tex]\dfrac{V}{\omega}[/tex]

                                         =     [tex]\dfrac{5.82}{25.13}[/tex]

A = 0.232 m

Answer:

L =4166.66 N   and  D = 46296.29 N and T = 96047.34 N

Explanation:

given,

flight speed = 200 ft/s

turn rate = 5 deg/s

weight of aircraft = 50000 lb

L/D = 10

for equilibrium in the horizontal position

w - L cos ∅ - D sin ∅ = 0 .............(1)

D cos ∅ - L sin ∅ = 0................(2)

L/D = tan ∅ = 10

∅ = 84.29°

50000 - L cos  84.29°- D sin  84.29°= 0

D cos  84.29° - L sin 84.29° = 0

on solving the above equation we get

L =4166.66 N   and  D = 46296.29 N

thrust force calculation:

T = W sin ∅ + D

  = 50000×sin 84.29 + 46296.29

T = 96047.34 N

Answer:

16 % and 9 %

Explanation:

Accurate reading = 10 N

reading of first balance = 8.4 N

reading of second balance = 9.1 N

Percentage difference in first reading

= [tex]\frac{True reading - reading of first balance}{True reading}\times 100 %[/tex]

= [tex]\frac{10-8.4}{10}\times 100 %[/tex] = 16 %

Percentage difference in second reading

= [tex]\frac{True reading - reading of second balance}{True reading}\times 100 %[/tex]

= [tex]\frac{10-9.1}{10}\times 100 %[/tex] = 9 %

Answer:

F = 22572N

Explanation:

We will first calculate the acceleration needed to achieve landing speed equals zero:

[tex]V_{f}^{2}=V_{o}^{2}+2*a*\Delta Y[/tex]

[tex]0=18^{2}+2*a*(-165)[/tex]   Solving for a:

[tex]a=0.98 m/s^{2}[/tex]

Now that we have the required acceleration, we need to check all forces acting on the craft:

[tex]F_{thrust} - W = m*a[/tex]   where W = 11400N; m = 11400Kg; a = 0.98m/s2

Solving the equation:

[tex]F_{thrust}=22572N[/tex]

The magniture of thrust required to reduce the lunar craft's velocity to zero can be calculated using the principles of kinematics and forces. The change in kinetic energy is calculated and equated with the work done by the retro-rocket's thrust, in order to find the required thrust.

The physics involved in this problem is related to kinematics and forces. To calculate the amount of thrust needed, we first need to find the change in kinetic energy. The initial kinetic energy is 0.5 * 11400 kg * (18 m/s)^2 and the final kinetic energy is zero since the craft is at rest. Therefore, the change in kinetic energy is -0.5 * 11400 kg * (18 m/s)^2.

By the work-energy theorem, the work done by the retro-rocket's thrust is equal to the change in kinetic energy. Since work done is also equal to force (thrust) multiplied by distance, we can equate these two expressions to find the thrust: Thrust * 165 m = -0.5 * 11400 kg * (18 m/s)^2. Solve for thrust to get the magnitude of thrust needed.

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Answer:

[tex]Q=3.47\times 10^{-15}\ C[/tex]

Explanation:

Given that,

Number of extra electrons, n = 21749

We need to find the net charge on the metal ball. Let Q is the net charge.

We know that the charge on an electron is [tex]q=1.6\times 10^{-19}\ C[/tex]

To find the net charge if there are n number of extra electrons is :

Q = n × q

[tex]Q=21749\times 1.6\times 10^{-19}\ C[/tex]

[tex]Q=3.47\times 10^{-15}\ C[/tex]

So, the net charge on the metal ball is [tex]3.47\times 10^{-15}\ C[/tex]. Hence, this is the required solution.

Answer:

the frequency heard by the observer is equal to 2677 Hz

Explanation:

given,      

velocity of the observer = 17 m/s

speed of the sound = 343 m/s    

velocity of the source = 0 m/s    

frequency emitted from the source  = 2550 Hz              

[tex]f = f_0(\dfrac{v-v_0}{v-v_s})[/tex]              

[tex]f = 2550\times (\dfrac{343+17}{343-0})[/tex]

velocity of observer is negative as it is approaching the source.                   f = 2676.38 Hz ≈ 2677 Hz                    

hence, the frequency heard by the observer is equal to 2677 Hz

Answer:

The car is traveling at [tex]8.1366\frac{m}{s}[/tex]

Explanation:

The known variables are the following:

[tex]V_{0} = 56 \frac{km}{h} = 15.56 \frac{m}{s}\\ D=25m\\ t=2.11s\\ V_{f}=?[/tex]

First, from the equation of motion we find the deceleration:

[tex]D=V_{0}*t+\frac{1}{2} a*t^{2} \\ a=\frac{2(D-V_{0})}{t^{2} } \\ a=3.5182\frac{m}{s^2}[/tex]

Then, with the equation for the speed:

[tex]V_{f}=V_{o}+a*t\\ V_{f}=8.1366\frac{m}{s}[/tex]

The speed of the car at impact is approximately 25.72 m/s.

To find the speed of the car at impact, we first need to calculate the deceleration. We can use the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time taken.

Given that the initial velocity (u) is 56.0 km/h, the distance (s) is 25.0 m, and the time taken (t) is 2.11 s, we can rearrange the equation to solve for a: a = (v - u) / t.

Plugging in the values, we have a = (0 - 56.0 km/h) / 2.11 s = -26.5 m/s². Since the car is decelerating, the acceleration is negative.

Now, to find the final velocity (v) at impact, we can use the equation v² = u² + 2as. Rearranging the equation, we have v² = u² + 2(-26.5 m/s²)(25.0 m) = u² - 2(26.5 m/s²)(25.0 m).

Plugging in the values, we have v² = (56.0 km/h)² - 2(26.5 m/s²)(25.0 m).

Converting the initial velocity (u) to m/s, we get u = 56.0 km/h * (1000 m/1 km) * (1 h/3600 s) = 15.6 m/s.

Substituting the values, we have v² = (15.6 m/s)² - 2(26.5 m/s²)(25.0 m) = 0.16 m²/s² - 26.5 m²/s² * 25.0 m = 0.16 m²/s² - 662.5 m²/s² = -662.34 m²/s².

Taking the square root of both sides, we get v ≈ √(-662.34 m²/s²) ≈ -25.72 m/s.

Since speed is a positive quantity, the speed of the car at impact is approximately 25.72 m/s.

Answer:

c)[tex]F_{net} = 0.640 kN[/tex]

Explanation:

As we know that resultant force is the net force that is acting on the system

As per Newton's II law we know that net force is product of mass and acceleration

so we will have

[tex]F_{net} = ma[/tex]

here we know

m = 80 kg

for circular motion acceleration is given as

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

[tex]a_c = \frac{40^2}{200} = 8 m/s^2[/tex]

now we have

[tex]F_{net} = 80 \times 8[/tex]

[tex]F_{net} = 640 N[/tex]

[tex]F_{net} = 0.640 kN[/tex]