A 38.0 kg satellite has a circular orbit with a period of 1.30 h and a radius of 7.90 × 106 m around a planet of unknown mass. If the magnitude of the gravitational acceleration on the surface of the planet is 30.0 m/s2, what is the radius of the planet

Final answer:

The magnitude of the total acceleration of the top of the racket during the serve, considering both tangential and radial components, is approximately 270.5 m/s^2.

Explanation:

To find the magnitude of the total acceleration of the top of the racket, we need to consider both tangential and radial (centripetal) acceleration components. The tangential acceleration (at) is directly provided by the angular acceleration (α), and is calculated by multiplying the angular acceleration by the radius (r), so at = α × r. Here, α = 150 rad/s2 and r = 1.30 m, giving at = 195 m/s2. The radial acceleration (ar), also known as centripetal acceleration, depends on the angular speed (ω) and the radius (r), and is calculated with the formula ar = ω2 × r. Given ω = 12.0 rad/s and r = 1.30 m, we find ar = 187.2 m/s2. Finally, the total acceleration (a) is the square root of the sum of the squares of at and ar, resulting in a = √(at2 + ar2). This gives us a = √(1952 + 187.22) m/s2, and by calculating, we find the magnitude of the total acceleration to be approximately 270.5 m/s2.

Answer:

Final speed in reverse direction will be 11 m/s

Explanation:

As per impulse momentum theorem we know that

Change in momentum of the object is product of force and time

so we have

[tex]P_f - P_i = F\Delta t[/tex]

now we have

[tex]P_f = mv_f[/tex]

[tex]P_i = mv_i[/tex]

[tex]\Delta t = 250 ms[/tex]

F = 21 N

now we have

[tex]0.38(v_f) - 0.38(-2.8) = 21(250\times 10^{-3})[/tex]

[tex]v_f = 13.82 - 2.8[/tex]

[tex]v_f = 11 m/s[/tex]

Answer:

energy decreases

Explanation:

The energy of a wave is directly proportional to the square of amplitude of a wave.

Amplitude is defined as the maximum displacement of the wave particle from its mean position.

So, as the amplitude goes on decreasing, the energy of the wave is also decreasing.

If the amplitude of a wave decreases gradually as the wave travels down, the energy of the wave increases.

A wave can be defined as a disturbance in a medium that progressively transport energy from a source to another location, especially without the transportation of matter.

In Science, some of the characteristics of a wave include the following:

Generally, the energy of a wave is directly proportional to the square of wave amplitude and the square of its frequency. Thus, the energy of the wave increases when the amplitude of a wave decreases gradually as the wave travels down.

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Answer:

Part a)

[tex]f = -48 cm[/tex]

Since focal length is negative so its a diverging lens

Part b)

[tex]h_i = 6.375 mm[/tex]

Since the magnification is position for diverging lens so it is ERECT

Explanation:

Part a)

As we know by lens formula

[tex]\frac{1}{d_i} + \frac{1}{d_o} = \frac{1}{f}[/tex]

[tex]-\frac{1}{12cm} + \frac{1}{16} = \frac{1}{f}[/tex]

[tex]f = -48 cm[/tex]

Part b)

Since focal length is negative so its a diverging lens

Part c)

As we know that

[tex]\frac{h_i}{h_o} =\frac{d_i}{d_o}[/tex]

[tex]\frac{h_i}{8.5 mm} = \frac{12}{16}[/tex]

[tex]h_i = 6.375 mm[/tex]

Part d)

Since the magnification is position for diverging lens so it is ERECT

Part e)

Answer:

(a) 64.58 N

(b) 43125 N/C

Explanation:

(a) q1 = 796 uC = 796 x 10^-6 C, q2 = 481 nC = 481 x 10^-9 C,

r = 23.1 cm = 0.231 m

Force, F = k q1 q2 / r^2

F = (9 x 10^9 x 796 x 10^-6 x 481 x 10^-9) / (0.231)^2

F = 64.58 N

(b) F = 6.9 x 10^-15 N

E = F / q

E = (6.9 x 10^-15) / (1.6 x 10^-19) = 43125 N/C

Answer:

The distance is 11 m.

Explanation:

Given that,

Friction coefficient = 0.24

Time = 3.0 s

Initial velocity = 0

We need to calculate the acceleration

Using newton's second law

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

Using formula of friction force

[tex]F= \mu m g[/tex]....(II)

Put the value of F in the equation (II) from equation (I)

[tex]ma=\mu mg[/tex]....(III)

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

Put the value in the equation (III)

[tex]a=0.24\times9.8[/tex]

[tex]a=2.352\ m/s^2[/tex]

We need to calculate the distance,

Using equation of motion

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

[tex]s=0+\dfrac{1}{2}2.352\times(3.0)^2[/tex]

[tex]s=10.584\ m\ approx\ 11\ m[/tex]

Hence, The distance is 11 m.

The truck can travel a maximum distance of 10.584 meters in 3.0 seconds without the box sliding.

To determine the maximum distance the truck can travel in 3.0 s without the box sliding, we start by calculating the maximum acceleration the box can experience before it starts to slide.

This is given by the static friction force, which is the product of the coefficient of static friction  and the normal force (N). Assuming the box’s weight is W = mg, the normal force is N = mg.

The maximum static friction force (f) is:

f = μ N = μ mg

Given μs = 0.24, we need to ensure:

f > ma

Therefore, 0.24mg = ma, solving for a gives us:

a = 0.24g = 0.24 × 9.8 m/s² ≈ 2.352 m/s²

Next, we need to calculate the maximum distance using the kinematic equation for constant acceleration:

d = ut + (1/2)at²

Where:

Substituting these values, we get:

d = 0 + (1/2) × 2.352 m/s² × (3.0 s)² = 0.5 × 2.352 × 9 = 10.584 m

Thus, the maximum distance the truck can travel without the box sliding is approximately 10.584 meters.

Answer:

Magnetic field, B = 0.11 i (in k direction)

Explanation:

It is given that,

Velocity of the proton, [tex]v=4.5\times 10^6\ m/s[/tex]

Magnetic force, [tex]F=8\times 10^{-14}\ N[/tex] (due south)

The magnetic force acting on the electron is given by :

[tex]F=qvB[/tex]

(-j) = (i) × (B)

[tex]B=\dfrac{F}{qv}[/tex]

q is the charge on proton

[tex]B=\dfrac{8\times 10^{-14}\ N}{1.6\times 10^{-19}\ C\times 4.5\times 10^6\ m/s}[/tex]

B = 0.11

So, the magnitude of 0.11 T is acting on the proton in and is acting directed upward above the plane. Hence, this is the required solution.

Final answer:

To calculate the magnitude of the magnetic field that causes a force on a moving proton, we use the rearranged Lorentz force law, with inputs of proton charge, velocity, and the force experienced. The direction of the magnetic field is determined by the right-hand rule, given the directions of the force and proton's velocity.

Explanation:

The student's question involves the effect of a magnetic field on a moving charged particle, specifically a proton, which falls under the subject of Physics. The Lorentz force law states that the force (F) on a charged particle is directly proportional to the charge (q), the velocity (v), and the magnetic field (B), and is given by the equation F = q(v x B), where v x B denotes the cross product of the velocity vector and the magnetic field vector. The force is perpendicular to both the velocity and the magnetic field.

To find the magnitude of the magnetic field, we can rearrange this equation to B = F / (q * v * sin(θ)), where θ is the angle between the velocity of the proton and the direction of the magnetic field. In the case where force is maximum, the angle is 90 degrees, and the sine of 90 degrees is 1. Given the maximum magnetic force (8.0 × 10⁻¹⁴ N), the charge of a proton (approximately 1.6 × 10⁻¹⁹ C), and the velocity of the proton (4.5 × 10⁶ m/s), we can calculate the magnitude of the magnetic field. As the force experienced by the proton is directed due south and the velocity is due east, the magnetic field must be pointing downwards to produce a force in that direction according to the right-hand rule.

Answer:

a) 0.41 m/s

b) 0.51 m/s

Explanation:

(a)

M = total mass of wagon, rider and the rock = 93.1 kg

V = initial velocity of wagon = 0.456 m/s

m = mass of the rock = 0.292 kg

v = velocity of rock after throw = 15.4 m/s

V' = velocity of wagon after rock is thrown

Using conservation of momentum

M V = m v + (M - m) V'

(93.1) (0.456) = (0.292) (15.4) + (93.1 - 0.292) V'

V' = 0.41 m/s

b)

M = total mass of wagon, rider and the rock = 93.1 kg

V = initial velocity of wagon = 0.456 m/s

m = mass of the rock = 0.292 kg

v = velocity of rock after throw = - 15.4 m/s

V' = velocity of wagon after rock is thrown

Using conservation of momentum

M V = m v + (M - m) V'

(93.1) (0.456) = (0.292) (- 15.4) + (93.1 - 0.292) V'

V' = 0.51 m/s

0.005366 seconds is the time will the current flowing in this circuit be 55.0% of the peak current.

In a simple AC circuit with a resistor connected to an AC voltage source, the current flowing through the circuit can be calculated using Ohm's law:

I(t)=ΔV(t)/R

Where: I(t) is the current at time

ΔV(t) is the instantaneous voltage at time t

R is the resistance of the resistor.

Given that the source voltage is [tex]\bigtriangleup V=\bigtriangleup V_{max}sin(2 \pi ft)[/tex], and you want to find the time  t at which the current is 55% of the peak current, you need to find the time when [tex]I(t)=0.55\times I_{peak}.[/tex]

Find the peak current [tex]I_{peak}[/tex].

The peak current corresponds to the current when the source voltage is at its peak value  [tex]\bigtriangleup V_{max}[/tex] ​ .

Using Ohm's law:

[tex]I_{peak}=\frac{\bigtriangleup V_{max}}{R}[/tex]

The time t when the current  I(t) is 55% of the peak current:

[tex]I(t)=0.55 \times I_{peak}[/tex]

Substitute the expressions for I(t) and  [tex]I_{peak}[/tex].

[tex]\frac{\bigtriangleup V(t)}{R}=0.55 \times \frac{\bigtriangleup V_{max}}{R}[/tex]

2πft=arcsin(0.55)

t=arcsin(0.55)/2πf

t= 0.5716/2×3.14×16.9

t=0.5716/106.132

t=0.0053s

Hence, at approximately 0.005366 seconds after the start of the AC cycle (or 5.366 ms 5.366ms), the current flowing in the circuit will be 55.0% of the peak current.

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Answer:

The correct answer is d. tension pneumothorax.

Explanation:

The increasing build-up of air that is in the pleural space is what we call the tension pneumothorax and this happens due to the lung laceration that lets the air to flee inside the pleural space but it does not return.

Tension pneumothorax is the condition where air that enters the pleural space during inspiration but is unable to exit during expiration, causing a collapse of the lung and compression of surrounding structures.

Air that enters the pleural space during inspiration but is unable to exit during expiration creates a condition called tension pneumothorax. In this condition, there is a buildup of air in the pleural space, leading to a collapse of the lung and compression of surrounding structures. It is a medical emergency that requires immediate treatment.

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Answer:

FALSE

Explanation:

As we know that equation for charging of a capacitor is given by

[tex]q = CV(1 - e^{-\frac{t}{\tau}})[/tex]

now at t = 0 we can say there is no charge on the capacitor

now as the time will increase the value of charge will increase on the plates of capacitor

now after t = 1 time constant

we will have

[tex]q = CV(1 - e^{-1})[/tex]

[tex]q = 0.63 CV[/tex]

so after one time constant the capacitor will charged upto 63% of its maximum value.

Answer:

The value of [tex]v_{x}^2[/tex] is [tex]\dfrac{KT}{M}[/tex].

Explanation:

A gas is equilibrium at T kelvin.

Mass = M

We know that,

The average square of the velocity in the x,y and z direction are equal.

[tex]\bar{v}_{x}^2=\bar{v}_{y}^{2}=\bar{v}_{z}^{2}[/tex]

[tex]v_{rms}^2=v_{x}^{2}+v_{y}^{2}+v_{z}^{2}[/tex]

[tex]v_{rms}^2=3v_{x}^2[/tex]

[tex]v_{x}^{2}=\dfrac{v_{rms}^2}{3}[/tex]

Equation of ideal gas

[tex]PV=RT[/tex]

[tex]P=\dfrac{nKT}{V}[/tex]

Here, R = nK

We know that,

[tex]\dfrac{nKT}{V}=\dfrac{nM}{3V}v_{rms}^2[/tex]

[tex]v_{rms}^2=\dfrac{3KT}{M}[/tex]....(I)

Put the value of [tex]v_{rms}^2[/tex] in the equation (I)

[tex]v_{x}^2=\dfrac{1}{3}\times\dfrac{3KT}{M}[/tex]

[tex]v_{x}^2=\dfrac{KT}{M}[/tex]

Hence, The value of [tex]v_{x}^2[/tex] is [tex]\dfrac{KT}{M}[/tex].  

Answer:

Yes it is related

Explanation:

Consider 2 points in a rectangular co-ordinate system

[tex](x_{1},y_{1}),(x_{2},y_{2})[/tex]

The distance between them can be found by

d = [tex]\sqrt{(x_{1}-x_{2})^{2}-(y_{2}-y_{1})^{^{2}}}[/tex]

This is clear from the figure attached below

Answer:

The rate of current in the solenoid  is 0.398 A/s

Explanation:

Given that,

Electric field [tex]E = 4.0\ \mu V/m[/tex]

Distance = 2.0 cm

Radius = 3.0 cm

Number of turns per unit length = 800

We need to calculate the rate of current

Using formula of electric field for solenoid

[tex]E = \dfrac{x}{2}\mu_{0}n\dfrac{dI}{dt}[/tex]

Where, x = distance

n = number of turns per unit length

E = electric field

r = radius

Put the value into the formula

[tex]4.0\times10^{-6}=\dfrac{2.0\times10^{-2}}{2}\times4\pi\times10^{-7}\times800\times\dfrac{dI}{dt}[/tex]

[tex]\dfrac{dI}{dt}=\dfrac{4.0\times10^{-6}\times2}{2.0\times10^{-2}\times4\pi\times10^{-7}\times800}[/tex]

[tex]\dfrac{dI}{dt}=0.397\ A/s[/tex]

Hence, The rate of current in the solenoid  is 0.398 A/s.

Answer:

The coefficient of kinetic friction [tex]\mu= 0.16989[/tex]

Explanation:

From Newton's second law

[tex]\sum\overset{\rightarrow}{F}=m\cdot\overset{\rightarrow}{a}[/tex]

If the velocity is constant, that means the summation of all forces must be equal to zero. Draw the free-body diagram to obtain the sums of forces in x and y. It must include the Friction Force, in the opposite direction of the displacement, the weight ([tex]W=mg=390*9.81=3825.9N[/tex]), the Normal Force, which is the is the consequence of Newton's third law and the forces from the two workers.

The sum in y is:

[tex]\sum F_{y}=F_{N}-3825.9=0[/tex]

Solving for the [tex]F_{N}[/tex]:

[tex]F_{N}=$ $3825.\,\allowbreak9N[/tex]

The sum in x is:

[tex]\sum F_{x}=450+200-F_{f}=0[/tex]

Solving for the [tex]F_{f}[/tex]:

[tex]$F_{f}=650.0N[/tex]

The formula of the magnitude of the Friction force is

[tex]F_{f}=\mu F_{N}[/tex]

That means the coefficient of friction is:

[tex]\mu=\frac{F_{f}}{F_{N}}=\frac{650.0}{3825.\,\allowbreak9}=\allowbreak0.16989[/tex]

Answer:

3.43 x 10⁵ N/m

Explanation:

M = mass of the car alone = 2000 kg

m = mass of the four people combined = 315 kg

x = compression of the spring when there is car alone

x' = compression of the spring when there is car and four people = x + 0.009

k = spring constant

When there is car, the weight of the car is balanced by the spring force and the force equation is given as

k x = Mg

k x = (2000)(9.8)

k x = 19600                                                eq-1

When there is car and four people, the weight of the car and four people is balanced by the spring force and the force equation is given as

k x' = (M + m) g

k (x + 0.009) = (2000 + 315) (9.8)

k x + 0.009 k = (2000 + 315) (9.8)

using eq-1

19600 + 0.009 k = 22687

k = 3.43 x 10⁵ N/m

Final answer:

The effective force constant of the springs k = 343000 N/m

Explanation:

To solve for the effective force constant of the springs in a car's suspension system, we can apply Hooke's Law, which is stated as F = -kx. Here, F represents the restoring force supplied by the springs, k is the spring constant or force constant, and x is the displacement of the springs from their equilibrium position.

When the four people with a combined mass of 315 kg sit in the car, their weight causes an additional displacement of 0.90 cm, or 0.009 m.

Since weight acts as the restoring force F, we can calculate it as F = mg, where m is the total mass and g is the acceleration due to gravity (9.80 m/s²). So, F = (315 kg)(9.80 m/s²) = 3087 N.

Using Hooke's Law, we can rearrange the formula to solve for k: k = F/x. Plugging in the values gives us k = 3087 N / 0.009 m, which provides the spring constant for the suspension system.

The effective force constant of the springs k = 343000 N/m

Answer:

The deflection of the spring is 34.56 mm.

Explanation:

Given that,

Diameter = 10 mm

Number of turns = 10

[tex]Radius_{mean} = 60\ mm[/tex]

[tex]Diameter_{mean} = 120\ mm[/tex]

Load = 200 N

We need to calculate the deflection

Using formula of deflection

[tex]\delta=\dfrac{8pD^3n}{Cd^4}[/tex]

Put the value into the formula

[tex]\delta=\dfrac{8\times200\times(120)^3\times10}{80\times10^{3}\times10^4}[/tex]

[tex]\delta =34.56\ mm[/tex]

Hence, The deflection of the spring is 34.56 mm.

Answer:

A) 1.2

B) 1.1

Explanation:

A)

F = force required to start the box moving = 36.0 N

m = mass of the box = 3 kg

[tex]F_{g}[/tex]  = weight of the box = mg = 3 x 9.8 = 29.4 N

[tex]F_{n}[/tex]  = normal force acting on the box by the floor

normal force acting on the box by the floor is given as

[tex]F_{n}[/tex]  = [tex]F_{g}[/tex] = 29.4

[tex]F_{s}[/tex]  = Static frictional force = F = 36.0 N

[tex]\mu _{s}[/tex] = Coefficient of static friction

Static frictional force is given as

[tex]F_{s}[/tex]  = [tex]\mu _{s}[/tex] [tex]F_{n}[/tex]

36.0  = [tex]\mu _{s}[/tex] (29.4)

[tex]\mu _{s}[/tex] = 1.2

B)

a = acceleration of the box = 0.54 m/s²

F = force applied = 36.0 N

[tex]f_{k}[/tex] = kinetic frictional force

[tex]\mu _{k}[/tex] = Coefficient of kinetic friction

force equation for the motion of the box is given as

F - [tex]f_{k}[/tex] = ma

36.0 - [tex]f_{k}[/tex] = (3) (0.54)

[tex]f_{k}[/tex] = 34.38 N

Coefficient of kinetic friction is given as

[tex]\mu _{k}=\frac{f_{k}}{F_{n}}[/tex]

[tex]\mu _{k}=\frac{34.38}{29.4}[/tex]

[tex]\mu _{k}[/tex] = 1.1

Answer:

The electric field is [tex]-2.14\times10^{5}\ N/C[/tex]

Explanation:

Given that,

Distance = 2.3 m

Charge [tex]q= -20.0\ \muC[/tex]

We need to calculate the electric field

Using formula of electric field

[tex]E=\dfrac{\sigma}{2\epsilon_{0}}[/tex]

Where, [tex]\sigma=\dfrac{Q}{A}[/tex]

Q = charge

A = area

Put the value into the formula

[tex]E=\dfrac{Q}{2A\epsilon_{0}}[/tex]

[tex]E=\dfrac{-20.0\times10^{-6}}{2\times2.3\times2.3\times8.85\times10^{-12}}[/tex]

[tex]E=-213599.91\ N/C[/tex]

[tex]E=-2.14\times10^{5}\ N/C[/tex]

Negative sign shows the direction of the electric field.

The direction of electric field is toward the plates.

Hence, The electric field is [tex]-2.14\times10^{5}\ N/C[/tex]

Final answer:

The electric field 1.0 cm above a thin conducting plate with a side length of 2.3 m and a total charge of -20.0 µC is -213,559.32 N/C, indicating a downward direction.

Explanation:

A thin conducting plate 2.3 m on a side is given a total charge of -20.0 µC. To find the electric field 1.0 cm above the plate, we use the concept of surface charge density (σ) and the formula for the electric field near an infinite plane sheet of charge, E = σ / (2ε0), where ε0 is the vacuum permittivity (ε0 = 8.85 x [tex]10^-^1^0[/tex] C2/N·m²). The surface charge density is σ = Q / A, with Q being the charge and A the area of the plate.

The area of the plate, A, is 2.3 m × 2.3 m = 5.29 m2. Therefore, σ = (-20.0 x 10-6 C) / (5.29 m2) = -3.78 x [tex]10^-^6[/tex] C/m². Substituting σ into the electric field equation yields E = -3.78 x [tex]10^-^6[/tex] C/m² / (2 × 8.85 x [tex]10^-^1^2[/tex] C2/N·m2) = -213,559.32 N/C, indicating the direction is downward due to the negative sign.

Answer:

For proton: 2592 m/s In the same direction of electric field.

For electron: 4752000 m/s In the opposite direction of electric field.

Explanation:

E = 500 N/C, t = 54 ns = 54 x 10^-9 s,

Acceleration = Force /mass

Acceleration of proton, ap = q E / mp

ap = (1.6 x 10^-19 x 500) / (1.67 x 10^-27) = 4.8 x 10^10 m/s^2

Acceleration of electron, ae = q E / me

ae = (1.6 x 10^-19 x 500) / (9.1 x 10^-31) = 8.8 x 10^13 m/s^2

For proton:

u = 0, ap = 4.8 x 10^10 m/s^2, t = 54 x 10^-9 s

use first equation of motion

v = u + at

vp = 0 + 4.8 x 10^10 x 54 x 10^-9 = 2592 m/s In the same direction of electric field.

For electron:

u = 0, ae = 8.8 x 10^13 m/s^2, t = 54 x 10^-9 s

use first equation of motion

v = u + at

vp = 0 + 8.8 x 10^13 x 54 x 10^-9 = 4752000 m/s In the opposite direction of electric field.

Answer:

[tex]\vec \tau = 31.9 N[/tex]

Explanation:

As we know that torque due to a force is given by the formula

[tex]\vec \tau = \vec r \times \vec F[/tex]

here we know that force is exerted due to weight of the mass hold in his hand

so we have

F = mg = 59 N

Now Lever arm is the perpendicular distance on the line of action of force from the axis about which the system is rotated

so here we can say

[tex]r = Lcos\theta [/tex]

[tex]r = 0.56 cos15 = 0.54 m[/tex]

now we have

[tex]\vec \tau = (0.54)(59)[/tex]

[tex]\vec \tau = 31.9 Nm[/tex]

Answer:

The value of 2.2 radians in degree is 126.11°.

Explanation:

Given that,

Angle = 2.2 radians

We need to calculate the value in degrees

We know that,

[tex]1\ radians = \dfrac{180}{\pi}[/tex]

[tex]2.2\ radians = \dfrac{180}{3.14}\times2.2[/tex]

[tex]2.2\ radians = 126.11^{\circ}[/tex]

Hence, The value of 2.2 radians in degree is 126.11°.

Answer:

Ek= 35.392 kJ or 35.392 10³ J

Explanation:

Formula for Kinetic energy:

Ek= [tex]\frac{1}{2}[/tex] × m × [tex]v^{2}[/tex]

since m= 65kg and v= 33m/s

Ek= [tex]\frac{1}{2}[/tex]× 65×[tex]33^{2}[/tex]

Ek= 35.392 kJ or 35.392 10³ J

Final answer:

The kinetic energy of a 65 kg cheetah running at its top speed of 33 m/s is calculated using the formula KE = 1/2 mv², resulting in a kinetic energy of 35442.5 Joules.

Explanation:

The question asks for the kinetic energy of a 65 kg cheetah running at its top speed, which is given to be 33 m/s. To calculate kinetic energy, we use the formula KE = 1/2 mv², where m is the mass of the object (in this case, the cheetah) and v is the velocity of the object. Substituting the given values, we get:

KE = 1/2 × 65 kg × (33 m/s)²

KE = 1/2 × 65 × 1089

KE = 1/2 × 70885

KE = 35442.5 Joules

Therefore, the kinetic energy of a 65 kg cheetah running at its top speed of 33 m/s is 35442.5 Joules.

Answer:

[tex]a = 5.83 \times 10^{-4} m/s^2[/tex]

Explanation:

Since the system is in international space station

so here we can say that net force on the system is zero here

so Force by the astronaut on the space station = Force due to space station on boy

so here we know that

mass of boy = 70 kg

acceleration of boy = [tex]1.50 m/s^2[/tex]

now we know that

[tex]F = ma[/tex]

[tex]F = 70(1.50) = 105 N[/tex]

now for the space station will be same as above force

[tex]F = ma[/tex]

[tex]105 = 1.8 \times 10^5 (a)[/tex]

[tex]a = \frac{105}{1.8 \times 10^5}[/tex]

[tex]a = 5.83 \times 10^{-4} m/s^2[/tex]

The magnitude of the acceleration of the space station is mathematically given as

[tex]a = 5.83 * 10^{-4} m/s^2[/tex]

Question Parameter(s):

The International Space Station has a mass of 1.8 × 105 kg. A 70.0-kg.

The station pushes off one wall of the station so she accelerates at 1.50 m/s2.

Generally, the equation for the  Force is mathematically given as

F = ma

Therefore

F = 70(1.50)

F= 105 N

In conclusion, the space station force

F = ma

105 = 1.8 * 10^5 a

a = 5.83 * 10^{-4} m/s^2

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Answer and Explanation:

By measuring in millimeter we can decrease the associated error with the measurement because when we measure in smaller unit the measurement is more precise  and accurate rather than when we measure in larger unit so when we measure 325 millimeter instead of 32.5 cm then there is chance of less error produced.

Answer:

Explanation:

Let E be the strength of electric field and m be the mass of plastic sphere.

charge on plastic sphere, q = - 3 n C = - 3 x 10^-9 C

The force due to electric field on the charged plastic sphere = q E

The weight of the plastic sphere = m g

where, g is the acceleration due to gravity.

Now the force due to electrostatic field is balanced by teh weight of the plastic sphere.

q E = m g

E = m g / q

E = m x 9.8 / (3 x 10^-9) = 3.26 x 10^9 x m

Substitute the value of m and get the value E.

The magnitude of the vertical electric field required to balance the weight of the charged plastic sphere with a mass of 2.1 g and a charge of -3.0 nC is [tex]6.86 * 10^6 N/C[/tex].

To find the magnitude of the vertical electric field that will balance the weight of a charged plastic sphere, we need to equate the electrostatic force on the sphere due to the electric field with the weight of the sphere.

Given information:

- Mass of the plastic sphere, m = 2.1 g = [tex]2.1 * 10^-3 kg[/tex]

- Charge on the plastic sphere, q = -3.0 nC = [tex]-3.0 * 10^-9 C[/tex]

- Acceleration due to gravity, g = [tex]9.8 m/s^2[/tex]

- Coulomb's constant, k = 1/(4πε₀) =[tex]9.0 * 10^9 N.m^2/C^2[/tex]

Step 1: Calculate the weight of the plastic sphere.

Weight of the sphere = mg

Weight = [tex](2.1 * 10^-3 kg)[/tex] × [tex](9.8 m/s^2)[/tex]

Weight = 2.058 × [tex]10^{-2}[/tex] N

Step 2: Equate the electrostatic force on the sphere to its weight.

Electrostatic force = qE

Weight = qE

[tex]2.058 * 10^{-2} N = (-3.0 * 10^{-9} C) * E[/tex]

Step 3: Rearrange the equation to find the electric field magnitude E.

Equation to find the electric field magnitude E.

E = (2.058 × [tex]10^{-2}[/tex] N) / (-3.0 × [tex]10^{-9}[/tex]C)

[tex]E = 6.86 * 10^6 N/C[/tex]

Complete question:

what's the magnitude of a vertical electric field that will balance the weight of a plastic sphere of mass 2.1g that has been charged to -3.0nC? (k=[tex]1/4\pi[/tex], [tex]E_0[/tex]=[tex]9.0*10^9 N*m^2/C^2[/tex])

Answer:

34.01 A

Explanation:

[tex]P_{heater}[/tex] = Power of the electric heater = 1430 W

[tex]P_{toaster}[/tex] = Power of the toaster = 890 W

[tex]P_{grill}[/tex] = Power of the electric grill = 1760 W

V = Voltage of the battery connected in parallel to appliances = 120 volts

Using ohm's law, current drawn by heater is given as

[tex]i_{heater} = \frac{P_{heater}}{V}[/tex]

[tex]i_{heater} = \frac{1430}{120}[/tex]

[tex]i_{heater} [/tex] = 11.92 A

Using ohm's law, current drawn by toaster is given as

[tex]i_{toaster} = \frac{P_{toaster}}{V}[/tex]

[tex]i_{toaster} = \frac{890}{120}[/tex]

[tex]i_{toaster} [/tex] = 7.42 A

Using ohm's law, current drawn by grill is given as

[tex]i_{grill} = \frac{P_{grill}}{V}[/tex]

[tex]i_{grill} = \frac{1760}{120}[/tex]

[tex]i_{grill} [/tex] = 14.67 A

Total circuit drawn is given as

[tex]i_{total} [/tex] = [tex]i_{heater} [/tex] + [tex]i_{toaster} [/tex] + [tex]i_{grill} [/tex]

[tex]i_{total} [/tex] = 11.92 + 7.42 + 14.67

[tex]i_{total} [/tex] = 34.01 A

Answer:

R = 50.016 cm

Explanation:

Coefficient of thermal expansion for steel is given as

[tex]\alpha = 13 \times 10^{-6} per ^0C[/tex]

now as we know that the change in the length due to thermal expansion depends of change in temperature and initial length of the object

so here we will have

[tex]\Delta R = R_o \alpha \Delta T[/tex]

here we know that

[tex]R_o = \frac{50}{2} cm = 25 cm[/tex]

[tex]\Delta T = 50 K[/tex]

now we will have

[tex]\Delta R = (25)(13 \times 10^{-6})(50)[/tex]

[tex]\Delta R = 0.016 cm[/tex]

so final radius of the disc will be R = 50.016 cm

Answer:

7174.09 N

Explanation:

Area = 7 ft x 4 x ft = 28 ft^2 = 2.60129 m^2

Pressure = 0.4 psi = 2757.9 Pa

Force = Pressure x Area

Force = 2757.9 x 2.60129 = 7174.09 N

Answer:

0.0557 mm / s

Explanation:

r = 1.25 mm = 1.25 x 10^-3 m, i = 3.7 A, n = 8.46 x 10^28 per cubic metre,

e = 1.6 x 10^-19 C

Let vd be the drift velocity

use the formula for the current in terms of drift velocity

i = n e A vd

3.7 = 8.46 x 10^28 x 1.6 x 10^-19 x 3.14 x 1.25 x 1.25 x 10^-6 x vd

vd = 5.57 x 10^-5 m/s

vd = 0.0557 mm / s