Joe runs 10 m north, 20 m south, 9m south, and then 15 m north. What is Joe's
displacement?

(a) The required magnitude of the electric field when the point charge is an electron is 5.57 x 10⁻¹¹ N/C.

(b) The required magnitude of the electric field when the point charge is an proton is 1.02 x 10⁻⁷ N/C.

The magnitude of electric field is given by the following equation.

F = qE

But F = mg

mg = qE

E = mg/q

where;

E = (9.11 x 10⁻³¹ x 9.8)/(1.602 x 10⁻¹⁹)

E = 5.57 x 10⁻¹¹ N/C

E = (1.67 x 10⁻²⁷ x 9.8)/(1.602 x 10⁻¹⁹)

E = 1.02 x 10⁻⁷ N/C

Thus, the required vertical electric field is greater when the charge is proton.

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The stress on the wire is determined as  6.37 x 10⁶ N/m².

The stress on the wire is calculated as follows;

σ = F/A

where;

Let the diameter of the wire = 2 mm

Area = πr²

Area = π(1 x 10⁻³)²

Area = 3.142 x 10⁻⁶ m²

Stress = 20/(3.142 x 10⁻⁶)

Stress = 6.37 x 10⁶ N/m²

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(a) The equilibrant C for force of vector A and B is 3.43 N.

(b) The equilibrant C for fx of vector A and B is 2.1 N.

(c) The equilibrant  C, for fy of vector A and B is 2.12 N.

An equilibrant force is a single force that will bring other bodies into equilibrium.

Vector A: mass = 0.2 kg, θ = 20⁰

Vector B: mass = 0.15 kg, θ = 80⁰

Fx = mg cosθ

Fy = mg sinθ

where;

Force of A due to its weight

F(A) = mg

F(A) = 0.2 x 9.8 = 1.96 N

Fx = (0.2 x 9.8) cos(20) = 1.84 N

Fy = (0.2 x 9.8) sin(20) = 0.67 N

R = √(0.67² + 1.84²)

R = 1.96 N

Force of B due to its weight

F(B) = mg

F(B) = 0.15 x 9.8

F(B) = 1.47 N

Fx = (0.15 x 9.8) cos(80) = 0.26 N

Fy = (0.15 x 9.8) sin(80) = 1.45 N

R = √(0.26² + 1.45²)

R= 1.47 N

Equilibrant force:

Force, C = 1.96 N + 1.47 N

Force, C = 3.43 N

Equilibrant FX:

Fx, C = Fx(A) + Fx(B)

Fx, C = 1.84 N + 0.26 N = 2.1 N

Equilibrant FY:

Fy, C = Fy(A) + Fy(B)

Fy, C =0.67 N + 1.45 N = 2.12 N

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The total mass of the galaxy is 443.4 Solar mass

Orbital velocity ([tex]v[/tex]) = [tex]\sqrt{\frac{MG}{R} }[/tex]

where M= weight of galaxy

G= gravitational constatnt = [tex]6.674*10^-^1^1[/tex] (given)

R = distance from centre = [tex]41200[/tex] Light years (given)= [tex]4.12*9.5*10^1^6[/tex]  km (1 ly= [tex]9.5*10^3[/tex] billion km)

v= orbital velocity = [tex]275[/tex]  [tex]km/s[/tex] (given)

∴ According to the formula

[tex](2.75*10^2)^2[/tex] = [tex]\frac{M*6.674*10^-^1^1}{4.12*9.5*10^1^6}[/tex]

⇒ [tex]7.56*10^4*4.12*9.5*10^1^6=M*6.674*10^-^1^1[/tex] (cross multiplying and expanding)

⇒ [tex]29.59*10^2^1=M*6.674*10^-^1^1[/tex]

⇒ [tex]\frac{29.59*10^2^1*10^1^1}{6.674}=M[/tex]

⇒ [tex]4.434*10^3^2=M[/tex]

1 solar mass = [tex]1.989*10^3^0 kg[/tex]

⇒ Mass in solar mass =443.4 Solar mass

⇒ M = 443.4 Solar mass

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Answer:

full explanation and answer is on the picture

Answer:

V = ?

C = 3 × 10^8

wavelength = 2.10 × 10^-7

wavelength = C/V

V = C/wavelength

V = 3 × 10^8/2.10 × 10^-7

V = 1.43 × 10^ 15

Taking into account the definition of calorimetry,  the mass of the coffee is 253.33 g.

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

Sensible heat is defined as the amount of heat that a body absorbs or releases without any changes in its physical state (phase change).

So, the equation that allows to calculate heat exchanges is:

Q = c× m× ΔT

where:

In this case, you know:

Replacing in the equation that allows to calculate heat exchanges is:

190 cal = 1 [tex]\frac{cal}{gC}[/tex]× m× 0.75 C

Solving:

m= 190 cal÷ (1 [tex]\frac{cal}{gC}[/tex]× 0.75 C)

m=253.33 g

Finally, the mass of the coffee is 253.33 g.

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We need Net resistance

solve from right words

[tex]\\ \rm\Rrightarrow R_1=20+10=30\Omega[/tex]

[tex]\\ \rm\Rrightarrow \dfrac{1}{R_2}=\dfrac{1}{30}+\dfrac{1}{50}=\dfrac{5+3}{150}[/tex]

[tex]\\ \rm\Rrightarrow R_2=\dfrac{150}{8}=18.75\Omega[/tex]

[tex]\\ \rm\Rrightarrow \dfrac{1}{R_3}=\dfrac{1}{30}+\dfrac{1}{40}=\dfrac{4+3}{120}[/tex]

[tex]\\ \rm\Rrightarrow R_3=\dfrac{120}{7}=17.14\Omega[/tex]

[tex]\\ \rm\Rrightarrow R_{net}=17.14+18.75=35.89\Omega[/tex]

Use ohm's law

[tex]\\ \rm\Rrightarrow I=\dfrac{V}{R}[/tex]

[tex]\\ \rm\Rrightarrow I=\dfrac{9}{35.89}[/tex]

[tex]\\ \rm\Rrightarrow I\approx 0.25A[/tex]

The weight of an object on pluto will be 15.56. Mass multiplied by the gravitational acceleration gives weight.

Gravitation is a natural law by which all things with all matter are attracted towards one another. Gravity is responsible for large-scale structures present in the Universe.

By dividing the object's weight on Earth by 9.8 m/s², as illustrated below, one may calculate the object's mass.

m = 250 N / 9.8 m/s²

m = 25.51 kg

Multiply the acquired mass by Pluto's gravitational acceleration (g);

W = (25.51 kg) x (0.61 m/s²)  

W = 15.56 N

Thus, the item will only be 15.56 N in weight.

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Answer: B.

15.6 newtons

Explanation: edmentum

The minimum stopping distance when the car is moving at 32.0 m/s is 348.3 m.

The acceleration of the car before stopping at the given distance is calculated as follows;

v² = u² + 2as

when the car stops, v = 0

0 = u² + 2as

0 = 15² + 2(76.5)a

0 = 225 + 153a

-a = 225/153

a = - 1.47 m/s²

If the same force is applied, then acceleration is constant.

v² = u² + 2as

0 = 32² + 2(-1.47)s

2.94s = 1024

s = 348.3 m

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Answer:

D. trying to eat healthier foods

Explanation:

Answer:

kg

Explanation:

the highr and jebad to kilogram mizans

Answer:

w=m×g

Explanation:

w is given and take g as 9.8m/s^2

Answer:

71m/s

Explanation:

when you convert it you get 158.822 miles per hour

Answer:

Something which occupies space and have mass.

(a) The equation for the work done in stretching the spring from x1 to x2 is ¹/₂K₂Δx².

(b) The work done, in stretching the spring from x1 to x2 is 11.25 J.

(c) The work, necessary to stretch the spring from x = 0 to x3 is 64.28 J.

The work done in stretching the spring is calculated as follows;

W = ¹/₂kx²

W(1 to 2) = ¹/₂K₂Δx²

where;

W(1 to 2)  =  ¹/₂(250)(0.65 - 0.35)²

W(1 to 2)  = 11.25 J

W(0  to 3) = ¹/₂k₁x₁² + ¹/₂k₂x₂² + ¹/₂F₃x₃

where;

W(0  to 3) = ¹/₂(660)(0.35)² + ¹/₂(250)(0.65 - 0.35)² + ¹/₂(105)(0.89 - 0.65)

W(0  to 3) = 64.28 J

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A cryptologist is not a type of scientist. The correct answer is a cryptologist and the correct option is (D).

Experts in the study of methods for secure communication and data security through encryption and decryption are known as cryptologists.

Even though they are considered scientists in the broadest sense, they are not typically classed with the other natural scientists (geologists, meteorologists, and botanists) stated in the alternatives.

Due to its emphasis on the methods and mathematical principles used in encryption and decryption, cryptology is more intimately related to computer science and mathematics.

Therefore, The correct answer is a cryptologist and the correct option is (D).

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Answer:

4.2m/s

Explanation:

0.50x8.4=4.2m/s

hi brainly user! ૮₍ ˃ ⤙ ˂ ₎ა

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[tex]\large \bold {ANSWER}[/tex]

Considering that the pulley is fixed, the force applied should be equal to the weight of the object - of 400N.

[tex]\large \bold {EXPLANATION}[/tex]

Pulleys or pulleys are mechanical tools used to assist in the movement of objects and bodies. There are two types of pulleys: fixed and movable. While the fixed pulley changes the direction of force, the moving pulley helps to decrease the force needed to move the object or body in question.

As the statement only tells us a pulley, we must consider that it is fixed, because generally when it is mobile, this information is highlighted in the question.

In this way, a fixed pulley only changes the direction of the applied force. Thus, the force must have the same magnitude as the weight of the object to be moved. If the bucket weighs 400N, the force applied to the pulley must be 400N.

Therefore, having a fixed pulley, the force applied must be equal to the weight of the object, and will be 400N.

The forces a piano tuner applies to stretch the steel piano wire willl be 1.4 × 10¹¹ N.

Force is defined as the push or pulls applied to the body. Sometimes it is used to change the shape, size, and direction of the body. Force is defined as the product of mass and acceleration. Its unit is Newton.

The application of a force may be used to describe the steel as an elastic element within a certain range of applied force.

Given data;

Young modulus, E=2.10 × 10¹¹ N/m²

Cross-sectional area,A

Final length,[tex]\rm L_f = 1.45 m = 1450 \ mm[/tex]

Initial length,[tex]\rm L_i = 7.5 mm[/tex]

[tex]\rm F = \frac{(L_f-L_i)(E)(A)}{L_1} \\\\ \rm F = \frac{(1450 - 7.5)(2.0 \times 0^{11})(0.746)}{1450} \\\\ F = 1.4 \times 10^{11} \ N[/tex]

Hence, the force a piano tuner applies to stretch the steel piano wire willl be 1.4 × 10¹¹ N.

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The minimum coefficient of friction between the track and the mouse's feet that would allow a turn at this speed without sleeping is 1.97.

The minimum coefficient of friction is calculated as follows;

F = μmg = mac

μmg = mac

where;

μg = ac

μg = v²/r

μ = v²/rg

where;

μ = (1.39²)/(0.1 x 9.8)

μ = 1.97

Thus, the minimum coefficient of friction between the track and the mouse's feet that would allow a turn at this speed without sleeping is 1.97.

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The magnitude of the cannonball's velocity after 1.23 s will be 59.46 m/sec.

The change of distance with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.

The given data in the problem is

u is the initial velocity of canonball= 47.4 m/sec

g is the acceleration of free fall = 9.81 m/sec²

v is the velocity after 1.23 s

According to Newton's first equation of motion,

[tex]\rm v=u +gt \\\\ v= 47.4 \ m/sec + 9.81 \ m/sec^2 \times 1.23 sec \\\\\ v=59.46 \ m/sec[/tex]

Hence, the magnitude of the cannonball's velocity after 1.23 s will be 59.46 m/sec.

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Answer:

See the answer Explain why aircraft are carefully designed so that parts do not resonate. Expert Answer This virtually takes place, however maximum usually in small piston-engined airplanes, in particular dual-engined airplanes. The resonant frequency of the fuselage of a small plane goes to have numerous nodes, withinside the low loads of hertz.

• The speed change between T and S is greater between S and R.

• The speed of m at X is  greater that at Q.

• The size of the total force on m at P is less at U.

• The mechanical energy of m at P is equal to that at V.

• The size of the total force on m at S is greater at P.

• The velocity of m at X is equal to that at V.

The mechanical energy is the sum of kinetic energy and the potential energy of an object at any instant of time.

M.E = KE +PE

Given is a small mass m slides with negligible friction down an incline at an angle of 25.76° with respect to the horizontal. It then drops down to a horizontal surface and bounces elastically back up as shown.

The picture shows the position of the mass at equal time intervals starting from rest at T. The height of the mass at X is the same as at V.

Between T to S and S to R, the mass is under constant acceleration. Time taken to move from T to S is greater than S to R. Thus, the speed change between T and S is greater than between S and R.

At Q, there is only a horizontal velocity component, but at X. the speed will be greater and has both vertical and horizontal component. Thus, the speed of m at X is greater than that at Q.

Force is given as the rate of change of momentum with time. At U, change in momentum is large compared to P. Thus, the size of the total force on m at P is less at U.

There is no friction acting on the system. So the energy remains conserved. Mechanical energy at P = V.

The force on mass m at S is only the gravity force. The remaining forces are cancelled by the normal force. Thus, size of the total force on m at S is greater at P.

The energy is conserved at each point of motion of mass. If X and V are at same height, they have same potential energy and so their kinetic energy. Thus, velocity of m at X is equal to that at V.

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Answer:(note that values will be on top in small)

cobalt :- 60Co

potassium :- 40K

neon :- 24Ne

lead :- 208Pb

The two factors that affect current carrying capacity of an accumulator will be accumulator size and ambient temperature

Utilizing the compressible and decompressible properties of nitrogen gas, an accumulation vessel is used to store hydraulic pressure.

These key deciding elements are:

1. Accumulator Size: The current capacity increases as the circumference of the conductor increases.

2. Ambient Temperature: The greater the ambient temperature, the less heat is needed to raise the insulation's maximum operating temperature.

3. Accumulator identification:

4. Conditions for Installation:

Hence, the two factors that affect current carrying capacity of an accumulator will be accumulator size and ambient temperature

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The variable on the y-axis should be the running speed of hamsters since it is the dependent variable.

The dependent variable is the variable whose value fluctuates depending on other variables in an experiment.

In this case, the running speed is the dependent variable because it depends on the weight of the hamster.

The dependent variable is generally graphed on the Y axes, whereas the independent variable is graphed on the X-axes.

In conclusion, the variable on the y-axis should be the running speed of hamsters since it is the dependent variable.

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Answer:

decreases

Explanation:

because all the particles are wanting to say together and hit each other in a small area

Answer:

0.6192

Explanation:

1.44 x 0.430=0.6192

Answer:

Explanation:

Half-life problems are modeled as exponential equations.  The half-life formula is [tex]P=P_o\left (\dfrac{1}{2} \right)^{\frac{t}{k}}[/tex] where [tex]P_o[/tex] is the initial amount, [tex]k[/tex] is the length of the half-life, [tex]t[/tex] is the amount of time that has elapsed since the initial measurement was taken, and [tex]P[/tex] is the amount that remains at time [tex]t[/tex].

[tex]P=14.2\left (\dfrac{1}{2} \right)^{\frac{t}{8.0252}}[/tex]

Deriving the half-life formula

If one forgets the half-life formula, one can derive an equivalent equation by recalling the basic an exponential equation, [tex]y=a b^{t}[/tex], where [tex]t[/tex] is still the amount of time, and [tex]y[/tex] is the amount remaining at time [tex]t[/tex].  The constants a and b can be solved for as follows:

Knowing that amount initially is 14.2g, we let this be time zero:

[tex]y=a b^{t}[/tex]

[tex](14.2)=ab^{(0)}[/tex]

[tex]14.2=a *1[/tex]

[tex]14.2=a[/tex]

So, [tex]a=14.2[/tex], which represents out initial amount of the substance, and our equation becomes: [tex]y=14.2 b^{t}[/tex]

Knowing that the "half-life" is 8.0252 days (note that the unit here is "days", so times for all future uses of this equation must be in "days"), we know that the amount remaining after that time will be one-half of what we started with:

[tex]\left(\frac{1}{2} *14.2 \right)=14.2 b^{(8.0252)}[/tex]

[tex]\dfrac{7.1}{14.2}=\dfrac{14.2 b^{8.0252}}{14.2}[/tex]

[tex]0.5=b^{8.0252}[/tex]

[tex]\sqrt[8.0252]{\frac{1}{2}}=\sqrt[8.0252]{b^{8.0252}}[/tex]

[tex]\sqrt[8.0252]{\frac{1}{2}}=b[/tex]

Recalling exponent properties, one could find that  [tex]\left ( \frac{1}{2} \right )^{\frac{1}{8.0252}}=b[/tex], which will give the equation identical to the half-life formula.  However, recalling this trivia about exponent properties is not necessary to solve this problem.  One can just evaluate the radical in a calculator:

[tex]b=0.9172535661...[/tex]

Using this decimal approximation has advantages (don't have to remember the half-life formula & don't have to remember as many exponent properties), but one minor disadvantage (need to keep more decimal places to reduce rounding error).

So, our general equation derived from the basic exponential function is:

[tex]y=14.2* (0.9172535661)^t[/tex]  or [tex]y=14.2*(0.5)^{\frac{t}{8.0252}}[/tex] where y represents the amount remaining at time t.

Solving for the amount remaining

With the equation set up, substitute the amount of time it takes to cross the Pacific to solve for the amount remaining:

[tex]y=14.2* (0.9172535661)^{(31.8)}[/tex]          [tex]y=14.2*(0.5)^{\frac{(31.8)}{8.0252}}[/tex]

[tex]y=14.2* 0.0641450581[/tex]                    [tex]y=14.2*(0.5)^{3.962518068}[/tex]

[tex]y=0.9108598257[/tex]                              [tex]y=14.2* 0.0641450581[/tex]

                                                        [tex]y=0.9108598257[/tex]

Since both the initial amount of Iodine, and the amount of time were given to 3 significant figures, the amount remaining after 31.8days is 0.911g.