Calculate the magnitude of the gravitational force exerted by Mercury on a 70 kg human standing on the surface of Mercury. (The mass of Mercury is 3.31023 kg and its radius is 2.4106 m.)

Answers

Answer 1

Answer:

2.66×10⁻⁹ N.

Explanation:

From the question,

Applying newton's law of universal gravitation,

Fg = GMm/r²............................... Equation 1

Where Fg = gravitational force, G = universal constant, M = mass of the mercury, m = mass of the human, r = radius of Mercury

Given: M = 3.31023 kg, M = 70 kg, r = 2.4106

Constant: G = 6.67×10⁻¹¹ Nm²/kg²

Substitute these values into equation 1

Fg = 6.67×10⁻¹¹(70×3.31023)/(2.4106²)

Fg = 2.66×10⁻⁹ N.


Related Questions

help plz! what vibrates in following types of wave motion 1)light wave 2)sound waves 3)x-rays 4)water waves​

Answers

Answer:

I believe it's 2) sound waves

Explanation:

With sound waves, the energy travels along in the same direction as the particles vibrate. This type of wave is known as a longitudinal wave, so named because the energy travels along the direction of vibration of the particles.

A particle move in the xy plane so that its position vector r=bcosQi +bsinQj+ ctk, where b, Q and c are constants. show that the partial move with constant speed.​

Answers

Answer:

The speed of this particle is constantly [tex]c[/tex].

Explanation:

Position vector of this particle at time [tex]t[/tex]:

[tex]\displaystyle \mathbf{r}(t) = b\, \cos(Q)\, \mathbf{i} + b\, \sin(Q) \, \mathbf{j} + c\, t\, \mathbf{k}[/tex].

Write [tex]\mathbf{r}(t)[/tex] as a column vector to distinguish between the components:

[tex]\mathbf{r}(t) = \begin{bmatrix}b\, \cos(Q) \\ b\, \sin(Q) \\ c\, t\end{bmatrix}[/tex].

Both [tex]b[/tex] and [tex]Q[/tex] are constants. Therefore, [tex]b\, \cos(Q)[/tex] and [tex]b \sin (Q)[/tex] would also be constants with respect to [tex]t[/tex]. Hence, [tex]\displaystyle \frac{d}{dt}[b\, \cos(Q)] = 0[/tex] and [tex]\displaystyle \frac{d}{dt}[b\, \sin(Q)] = 0[/tex].

Differentiate [tex]\mathbf{r}(t)[/tex] (component-wise) with respect to time [tex]t[/tex] to find the velocity vector of this particle at time [tex]t\![/tex]:

[tex]\begin{aligned}\mathbf{v}(t) &= \frac{\rm d}{{\rm d} t} [\mathbf{r}(t)] \\ &=\frac{\rm d}{{\rm d} t} \left(\begin{bmatrix}b\, \cos(Q) \\ b\, \sin(Q) \\ c\, t\end{bmatrix}\right) \\ &= \begin{bmatrix}\displaystyle \frac{d}{dt}[b\, \cos(Q)] \\[0.5em] \displaystyle \frac{d}{dt}[b\, \sin(Q)]\\[0.5em]\displaystyle \frac{d}{dt}[c \cdot t]\end{bmatrix} = \begin{bmatrix}0 \\ 0 \\ c\end{bmatrix}\end{aligned}[/tex].

The speed [tex]v[/tex] (a scalar) of a particle is the magnitude of its velocity :

[tex]\begin{aligned}v(t) &= \| \mathbf{v}(t) \| \\ &= \left\|\begin{bmatrix}0 \\ 0 \\ c\end{bmatrix}\right\| \\ &= \sqrt{0^2 + 0^2 + c^2} = c\end{aligned}[/tex].

Therefore, the speed of this particle is constantly [tex]c[/tex] (a constant.)

What is the acceleration of a car that goes from 0 MS to 60 MS and six seconds

Answers

a = Δv/Δt
a = v2-v1/t2-t1
a = 60MS - 0MS / 6 seconds
a = 60 MS/ 6 seconds
a = 10 m/s^2

Which statement best compares potential and kinetic energy?
O Objects always have more potentiał energy than kinetic energy.
O Kinetic energy increases and potential energy decreases when the velocity of an object increases
O Only potential energy decreases when an object's height increases.
O Objects always have more kinetic energy than potential energy.

Answers

Answer:

Kinetic energy increases and potential energy decrease when velocity of an object increase.

39. What is the change in momentum for a 5,000 kg ship in

outer space that experiences no net force over a 1 hr

period?

Answers

Answer:

Change in momentum is zero.

Explanation:

The following data were obtained from the question:

Mass (m) = 5000 kg

Time (t) = 1 h

Net force (F) = 0

Change in momentum =?

Force = Rate of change of momentum

0 = change in momentum

Change in momentum = 0

We can see from the above illustration that the net force is zero. Thus, the change in momentum is also zero.

State three factors affecting pressure in liquids ​

Answers

Answer:

Density of liquid

Depth of liquid

Acceleration due to gravity

How does Physics help you as a student?

Answers

Answer:

The goal of physics is to understand how things work from first principles. ... Courses in physics reveal the mathematical beauty of the universe at scales ranging from subatomic to cosmological. Studying physics strengthens quantitative reasoning and problem solving skills that are valuable in areas beyond physics

Answer:

you get to understand why things happen this way

Explanation:

for example, are you not curious about why when standing in the bus and when the bus stops, you will might feel like you are going to fall ,

why does this happen because....

newton's laws explains it,

inertia causes you to be reluctant to change your initial state of motion due to your mass so you fall because you are still moving at the 'speed of the bus ' , something in like that

hope this helps,

please mark also

A potter’s wheel moves from rest to an angular speed of 0.10 rev/s in 36.5 s.
Assuming constant angular acceleration,
what is its angular acceleration in rad/s2?
Answer in units of rad/s2
.

Answers

Answer:

please find attached pdf

Explanation:

Given that Carbon-14 has a half-life of 5700 years, determine how long it would take for
this reduction to occur.

Answers

Answer:It will take about 3000 years

Explanation:

Which is an example of kinetic energy?

A. The energy stored in
ethanol

B. A ball sitting at the top of a ramp

C. A compressed spring

D. A hockey puck sliding across ice

Answers

D . A hockey puck sliding across ice

D. A hockey puck sliding across ice

Messages from the Perseverance Rover on Mars reach Earth in 11 min. The speed of light is 3.00 x 108 m/s. Using this information, how far is Mars from Earth?

Answers

Answer:

[tex]d=1.98\times 10^{11}\ m[/tex]

Explanation:

Messages from the Perseverance Rover on Mars reach Earth in 11 min i.e. time is 660 s

The speed of light is [tex]3\times 10^8\ m/s[/tex]

We need to find the distance between Mars and Earth. Let the distance be d.

We know that,

Distance = speed × time

So,

[tex]d=660\times 3\times 10^8\\\\d=1.98\times 10^{11}\ m[/tex]

So, Mars is [tex]1.98\times 10^{11}\ m[/tex] from the Earth.

Captain Jack Sparrow has been marooned on an island in the Atlantic by his crew, and decides to builda raft to escape. The wind seems quite steady, and first blows him due east for 11km, and then 6km ina direction 6degrees north of east. Confident that he will eventually find himself in safety, he fallsasleep. When he wakes up, he notices the wind is now blowing him gently 11degrees south of east -but after traveling for 21km, he finds himself back on the island.

Variable Name Min Max Step Sample Value
thetab 5 10 1 6
a 10 20 11 1
b 5 15 1 6
c 20 30 1 21
thetac 10 15 11 1

Required:
How far (in km) did the wind blow him while he was sleeping?

Answers

Answer:

    d₃ = 37,729 km,     θ=  5.1º North of West

Explanation:

This is a velocity addition problem, the easiest way to solve it is to decompose the velocities in a Cartesian system, the x-axis coincides with the West-East direction and the y-axis with the South-North direction

* first displacement is

           d₁ₓ = 11 km

* second offset is

          cos 6 = d₂ₓ / d₂

          sin 6 = d_{2y} / d₂

          d₂ₓ = d₂ cos 6

          d_{2y} = d₂ sin 6

          d₂ₓ = 6 cos 6 = 5.967 km

          d_{2y} = 6 sin 6 = 0.6272 km

* third displacement is unknown

* fourth and last displacement

          cos (-11) = d₄ₓ / d₄

          sin (-11) = d_{4y} / d₄

          d₄ₓ = d₄ cos (-11)

          d_{4y} = d₄ sin (-11)

          d₄ₓ = 21 cos (-11) = 20.61 km

          d_{4y} = 21 sin (-11) = -4.007 km

They tell us that at the end of the tour you are back on the island, so the displacement must be zero

X axis

           x = d₁ₓ + d₂ₓ + d₃ₓ + d₄ₓ

           0 = 11 +5.967 + d₃ₓ + 20.61

           d₃ₓ = -11 - 5.967 - 20.61

           d₃ₓ = -37.577 km

Y axis  

          y = d_{1y} + d_{2y} + d_{3y} + d_{4y}

          0 = 0 + 0.6272 + d_{3y} -4.007

          d_{3y} = 4.007 - 0.6272

          d_{3y} = 3.3798 km

This distance can be given in the form of module and angle

Let's use the Pythagorean theorem for the module

           d₃ = [tex]\sqrt{d_{3x}^2 + d_{3y}^2}[/tex]

           d₃ = [tex]\sqrt{37.577^2 + 3.3798^2}[/tex]

           d₃ = 37,729 km

Let's use trigonometry for the angle

            tan θ = d_{3y} / d₃ₓ

            θ = tan⁻¹ [tex]\frac{d_{3y}}{d_{3x}}[/tex]

            θ = tan-1 (-3.3798 / 37.577)

            θ = 5.1º

Since the y coordinate is positive and the x coordinate is negative, this angle is in the second quadrant, so the direction given in the form of cardinal coordinates is

            θ=  5.1º North of West

The current flow in the light bulb is 0.5A
a.Calculate the amount of electric charge that flow through the bulb in 2 hour
b.If one election carries a
charge 1.6 x 10-14 c Find the number of election through the bulb in 2 hour?

Answers

Answer:

Explanation:

Given that,

The current in the light bulb, I = 0.5 A

(a) We know that,

Electric current = charge/time

or

Q = It

Put t = 2 hours = 7200 s

So,

Q = 0.5 × 7200

Q = 3600 C

(b) Charge on one electron, [tex]Q=1.6\times 10^{-19}\ C[/tex]

Let there are n electrons flow through the bulb in 2 hours.

I = Q/t

Since, Q = ne

So,

I = ne/t

[tex]n=\dfrac{I\times t}{e}\\\\n=\dfrac{0.5\times 7200}{1.6\times 10^{-19}}\\\\n=2.25\times 10^{22}[/tex]

Hence, this is the required solution.

QUCIK!! SOMEONE PLEASE HELP! I’LL MARK BRAINLIEST!!

Answers

Answer:

A. v = √2gh

B. No! The final velocity does not depend on the mass of the car.

C. Yes! the final velocity depends on the steepness of the hill

D. 3.28 m/s

Explanation:

A. Determination of the final velocity.

½mv² = mgh

Cancel out m

½v² = gh

Cross multiply

v² = 2gh

Take the square root of both side

v = √2gh

B. Considering the formula obtained for the final velocity i.e

v = √2gh

We can see that there is no mass (m) in the formula.

Thus, the final velocity does not depend on the mass of the car.

C. Considering the formula obtained for the final velocity i.e

v = √2gh

We can see that there is height (h) in the formula.

Thus, the final velocity depends on the steepness of the hill

D. Determination of the final velocity.

Height (h) = 0.55 m

Acceleration due to gravity (g) = 9.8 m/s²

Velocity (v) =?

v = √2gh

v = √(2 × 9.8 × 0.55)

v = √10.78

v = 3.28 m/s

- .
?



y

(っ◔◡◔)っ ♥ chose the answer with the question marks ♥

Answers

Answer:

okay I'm a bit confused but I like the little emoji dudw

Answer:

?

Explanation:

.

A dog finds a toy at rest on the floor. The dog pushes the toy horizontally on a frictionless floor with a net force of 2.0 Newtons for 3.0 meters. How much kinetic energy does the toy gain? Round your answer to the nearest whole number.​

Answers

Answer:

the kinetic energy gained by the toy is 6J.

Explanation:

Given;

net applied to the toy by dog, F = 2 N

distance moved by the toy, d = 3 m

Apply the principle of work-energy theorem to determine the kinetic energy gained by the toy.

ΔK.E = W

         = F x d

         = 2 x 3

         = 6 J

Therefore, the kinetic energy gained by the toy is 6J.

If you have a 0.125 kg lead piece at
20.0°C, how much heat must you
add to melt it? (Remember, you
must warm it to its melting point
first.)
Material
Lead
Melt Pt (°C)
327
L (1/kg)
2.32.104
Boil Pt (°C) Lv (1/kg)
1750 8.59.105
c (1/(kg*c)
128
(Unit = J)

Answers

Answer:

7,812 J

Explanation:

Using the relation:

Q = mcΔθ

Q = quantity of heat

C = specific heat capacity of lead

Δθ = temperature change (T2 - T1)

M = mass of substance

Q = mass * specific heat * Δθ

Q = 0.125kg * 128 * (327 – 20)

Q = 0.125 * 128 * 307

Q = 4912 J

For melting:

Q = mass * Hf

0.125 * (2.32 * 10^4)

= 2,900 J

Total = 4,912 J + 2,900 J = 7,812 J

Two identical conducting spheres are placed with their centers 0.30 m apart. One is given a charge of 12 X10^-9 C and the other is given a charge of -18 X 10^-9 C. a. Find the electric force exerted on one sphere by the other. b. The sphere are connected by a conducting wire. After equilibrium has occurred, find the electric force between the two spheres.

Answers

Answer:

Explanation:

Force between two charged conducting sphere

= k x Q₁ x Q₂ / r² ,  k is a constant  Q₁ and Q₂ are charges and   r is distance between them .

= 9 x 10⁹ x 12 x 10⁻⁹ x 18 x 10⁻⁹ / .30²

= 21600 x 10⁻⁹

= 2.16 x 10⁻⁵ N .

b )

After the spheres are joined together , there is redistribution of charge and remaining charge will be equally shared by them .

Charge on each sphere = (12 - 18 ) x 10⁻⁹ / 2

= - 3 x 10⁻⁹ C .

Force = 9 x 10⁹ x 3 x 10⁻⁹ x 3 x 10⁻⁹ / .30²

= 900 x 10⁻⁹ N .

Ocean waves crash on the beach at a velocity of 3.5 m/s. If the distance between the crests of each wave is 4 m, find the frequency of the waves.


a. 0.0088 Hz

b. 14.0 Hz

c. 1.14 Hz

d. 0.88 Hz

Answers

Answer:

d

Explanation:

velocity=frequency × wavelength

frequency=speed/wavelength

frequency=3.5÷4

=0.875~0.88

The frequency of the waves is (d) 0.88 Hz. So, correct answer is option (d).

What is frequency of wave?

The frequency of a sinusoidal wave is the number of full oscillations performed by any wave constituent in a unit of time. According to the definition of frequency, if a body is moving periodically, it has completed one cycle after going through a number of situations or postures and then returning to its initial position. Therefore, frequency is a quantity that describes the rate of oscillation and vibration.

Given parameter,

Velocity of the waves = 3.5 m/s.

distance between the crests of each wave, that is, wavelength of the waves = 4 m.

We know that, for a wave transmission,

velocity of wave =frequency of wave  × wavelength of wave

frequency of wave=speed of wave/wavelength of wave

frequency of wave =3.5 m/s ÷4m

=0.875 Hz

≈ 0.88 Hz

Hence, the frequency of the waves is 0.88 Hz.

Learn more about frequency here:

https://brainly.com/question/5102661

#SPJ2

Please help 25 points!

Three waves with frequencies of 1 Hertz (Hz), 3 Hz, and 9Hz travel at the same speed. Which of the following statements is correct?

A. The 1 Hz wave contains the most energy.

B. The crests of all three waves are of equal height.

C. The wavelength of the 9Hz wave is three times that of the 3 Hz wave.

D. The 1 Hz wave has the longest wavelength.

Answers

Answer:

B

Explanation:

The crest of all three waves are of equal height



help please i will mark brainlist!!!

Answers

Answer:

.50 M

Explanation:

5*.50=2.5 + 2*.25=.5 = 3n

6*.50= 3N

Final answer is .50M

A book is moved once around the perimeter of a tabletop with dimensions 1.2 m x 1.8 m.
(a) If the book ends up at its initial position, what is its displacement
(b) What is the distance traveled?​

Answers

Answer:

a) 0m

b) 6m

Explanation:

First, we need to remember:

Displacement: Difference between final and initial position.

Distance traveled: Total distance traveled.

a) If the final position is the same as the initial position, then:

final position = initial position

And we know that:

displacement = final position - initial position = 0

Then the displacement of the book is zero.

b)

We can assume that the book traveled along the perimeter of the table.

The table is a rectangle of width 1.2m and length 1.8m

Remember that for a rectangle of width W and length L, the perimeter is:

P = 2*L + 2*W

Then the perimeter of the table is:

P = 2*1.2m + 2*1.8m = 6m

This means that the distance traveled by the book is 6 meters.

The masses of astronauts are monitored during long stays in orbit, such as when visiting a space station. The astronaut is strapped into a chair that is attached to the space station by springs and the period of oscillation of the chair in a friction-less track is measured.
(a) The period of oscillation of the 10.0 kg chair when empty is 0.750 s. What is the effective force constant of the springs?
(b) What is the mass of an astronaut who has an oscillation period of 2.00 s when in the chair?
(c) The movement of the space station should be negligible. Find the maximum displacement of the 100,000 kg sace station if the astronaut's motion has an amplitude of 0.100 m.

Answers

Answer:

a)  k = 701.8 N / m, b)  m_{ast} = 61.1 kg, c)  v ’= -1.3 10⁻⁴ m / s

Explanation:

a) For this exercise let's use the relationship of the angular velocity

         w = [tex]\sqrt{ \frac{k}{m} }[/tex]

          k = w² m

the angular velocity is related to the period

          w = 2π / T

we substitute

          k = 4 π²    [tex]\frac{m}{T^2}[/tex]

let's calculate

          k = 4 π²   10 /0.75²

          k = 701.8 N / m

b) now repeat the measurement with an astronaut on the chair

         w = [tex]\sqrt{ \frac{k}{m} }[/tex]

where the mass Month the mass of the chair plus the mass of the astronaut

        M = m + [tex]m_{ast}[/tex]

       

          M = k / w²

          w = 2π / T

let's calculate

           w = 2π / 2

            w = π rad / s

           

            M = 701.8 /π²

            M = 71,111 kg

now we use that

          M = m + m_{ast}

          m_{ast} = M - m

          m_{ast} = 71.111 - 10.0

          m_{ast} = 61.1 kg

c) if the astronaut's movement is simple harmonic

          x = A cos wt

therefore the speed is

         v = [tex]\frac{dx}{dt}[/tex]

         v = -Aw sin wt

maximum speed is

          v = - Aw

          v = 0.100 π

          v = 0.31416 m / s

we can suppose that the movement of the space station and the astronaut  is equivalent to division of the same

         

initial instant. Before the move

         p₀ = 0

final instant. When the astronaut is moving

        p_f = M_station v’+ m_{ast} v

the moment is preserved

         p₀ = pf

         0 = M__{station} v ’+ m_{ast} v

         v ’= - [tex]\frac{m_{ast} }{M_{station} } \ v[/tex]

we substitute

         v ’= [tex]\frac{61.1 }{ 100000 } \ 0.31416[/tex]

         v ’= -1.3 10⁻⁴ m / s

the negative sign indicates that the station is moving in the opposite direction from the astronaut

The resistivity of blood is related to its hematocrit, the volume fraction of red blood cells in the blood. A commonly used equation relating the hematocrit h to the blood resistivity rho (in Ω⋅m) is rho=1.32/(1−h)−0.79. In one experiment, blood filled a graduated cylinder with an inner diameter of 0.90 cm. The resistance of the blood between the 1.0 cm and 2.0 cm marks of the cylinder was measured to be 198 Ω.

Required:
What was the hematocrit for this blood?

Answers

Answer:

[tex]0.35598[/tex]

Explanation:

r = Radius = [tex]\dfrac{0.9}{2}=0.45\ \text{cm}[/tex]

R = Resistance = [tex]198\ \Omega[/tex]

A = Area = [tex]\pi r^2[/tex]

l = Length of blood in cylinder = 1 cm

h = Hematocrit of the blood

Resistivity is given by

[tex]\rho=\dfrac{1.32}{1-h}-0.79[/tex]

Resistance is given by

[tex]R=(\dfrac{1.32}{1-h}-0.79)\dfrac{l}{\pi r^2}\\\Rightarrow h=1-\dfrac{1.32}{\dfrac{R\pi r^2}{l}+0.79}\\\Rightarrow h=1-\dfrac{1.32}{\dfrac{198\times \pi\times (0.45\times 10^{-2})^2}{0.01}+0.79}\\\Rightarrow h=0.35598[/tex]

The hematocrit of the blood is [tex]0.35598[/tex].

The spring is unstretched at the position x = 0. under the action of a force p, the cart moves from the initial position x1 = -8 in. to the final position x2 = 5 in. determine (a) the work done on the cart by the spring and (b) the work done on the cart by its weight.

Answers

This question is incomplete, the missing diagram is uploaded along this Answer below.

Answer:

a) the work done on the cart by the spring is 4.875 lb-ft

b) the work done on the cart by its weight is - 3.935 lb-ft

Explanation:

Given the data in the question;

(a) determine the work done on the cart by the spring

we calculate the work done on the cart by the spring as follows;

[tex]W_{spring}[/tex] = 1/2×k( [tex]x^{2} _{1}[/tex] - [tex]x^{2} _{2}[/tex] )

where k is spring constant ( 3 lb/in )

we substitute  

[tex]W_{spring}[/tex] = 1/2 × 3( (-8)² - (5)² )      

[tex]W_{spring}[/tex] = 1/2 × 3( 64 - 25 )

[tex]W_{spring}[/tex] = 1/2 × 3( 39 )

[tex]W_{spring}[/tex] = 58.5 lb-in

we convert to pound force-foot

[tex]W_{spring}[/tex] = 58.5 × 0.0833333 lb-ft

[tex]W_{spring}[/tex] = 4.875 lb-ft

Therefore, the work done on the cart by the spring is 4.875 lb-ft

b) the work done on the cart by its weight

work done by its weight;

[tex]W_{gravity}[/tex] = -mgsin∅( x₂ - x₁ )        

we substitute in of values from the image below;

[tex]W_{gravity}[/tex] = -14 × sin(15°)( 5 - (-8) )  

[tex]W_{gravity}[/tex] = -14 × 0.2588 × 13

[tex]W_{gravity}[/tex] = -47.1  lb-in

we convert to pound force-foot

[tex]W_{gravity}[/tex] = -47.1 × 0.0833333 lb-ft

[tex]W_{gravity}[/tex] = - 3.935 lb-ft

Therefore, the work done on the cart by its weight is - 3.935 lb-ft

a) the work done on the cart by the spring is 4.875 lb-ft.

b) the work done on the cart by its weight is - 3.935 lb-ft.

Calculation of the work done:

a. The work done on the cart by the spring is

= 1/2 × 3( (-8)² - (5)² )      

= 1/2 × 3( 64 - 25 )

= 1/2 × 3( 39 )

= 58.5 lb-in

Now we have to convert to pound force-foot

So,

= 58.5 × 0.0833333 lb-ft

= 4.875 lb-ft

b) Now

work done by its weight;

= -mgsin∅( x₂ - x₁ )        

So,

= -14 × sin(15°)( 5 - (-8) )  

= -14 × 0.2588 × 13

= -47.1  lb-in

Now we convert to pound force-foot

= -47.1 × 0.0833333 lb-ft

= - 3.935 lb-ft

Learn more about spring here: https://brainly.com/question/16060099

A 450.0 kg roller coaster is traveling in a circle with radius 15.0m. Its speed at point A is 28.0m/s and its speed at point B is 14.0 m/s. At point A the cart is already moving with circular motion. a) Draw free bodydiagramsfor the cartatpointsAand B(two separate free body diagrams). b) Calculate the acceleration of the cartat pointsAandB(magnitude and direction). c) Calculate the magnitude of the normal force exerted by the trackson the cartat point A. d) Calculate the magnitude of the normal force exerted by the tracks on the cart at point B.

Answers

Answer:

b)  a = 52.26 m / s², a ’= 13.06 m / s², c) N = 2.79 10⁴ N, d) N = 1.89 10³ N

Explanation:

a) In the attached we can see the free body diagrams for the two positions, position A in the lower part of the circle and position B in the upper part of the circle

b) Let's start at point A

Let's use that the acceleration is centripetal

           a = v² / r

let's calculate

            a = 28² / 15.0

            a = 52.26 m / s²

as they relate it is centripetal it is directed towards the center of the circle, therefore for this point it is directed vertically upwards

Point B

           a ’= 142/15

           a ’= 13.06 m / s²

in this case the acceleration is vertical downwards

c) The values ​​of the normal force

point A

let's use Newton's second law

           ∑ F = m a

           N- W = m a

           N = mg + ma

           N = m (g + a)

           N = 450.0 (9.8 + 52.25)

           N = 2.79 10⁴ N

d) Point B

            -N -W = m (-a)

             N = ma -m g

             N = m (a-g)

             N = 450.0 (14.0 - 9.8)

             N = 1.89 10³ N

how many pennies can 4 folds of a paper hold?

Answers

A lot I don’t really under stand the question

A soccer ball was kicked over the edge of a wall and traveled 35 m horizontally at a speed of 5.6m/s. Calculate the vertical height of the wall.

Answers

Answer:

Are you sure it was soccer ball? Or meine hearts

Explanation:

what is the average velocity of a van that moves from 0 to 60 m east and 20 seconds

Answers

Explanation:

I have a lot to say it was very nice to meet my parents are u doing well I dont want too its been so much I love you so I was like u know I am not a man but you are the auditions I have been in a long long long life is a triangle and a chair for me and my parents think about the way I

Help plsssssssssss I write it 100 time no one answer

Answers

Answer:

1.93×10²⁸ s

Explanation:

From the question given above, the following data were obtained:

Number of electron (e) = 2×10²⁴

Current (I) = 10 A

Time (t) =?

Next, we shall determine the quantity of electricity flowing through pasing through the point. This can be obtained as follow:

1 e = 96500 C

Therefore,

2×10²⁴ e = 2×10²⁴ e × 96500 / 1 e

2×10²⁴ e = 1.93×10²⁹ C

Thus, 1.93×10²⁹ C of electricity is passing through the point.

Finally, we shall determine the time. This can be obtained as follow:

Current (I) = 10 A

Quantity of electricity = 1.93×10²⁹ C

Time (t) =?

Q = it

1.93×10²⁹ = 10 × t

Divide both side by 10

t = 1.93×10²⁹ / 10

t = 1.93×10²⁸ s

Thus, it took 1.93×10²⁸ s for 2×10²⁴ electrons to pass through the point

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