Two long straight wires each carry a current I out of the figure toward the viewer (Figure 1) .
I O o 4
o 1
o 2 o 5
o 3
I O o6
Part A
Mark the points in which the magnetic field is zero.
Check all that apply.
Check all that apply.
O 6
O 2
O 1
O 5
O 4
O 3

323.5 N is the tension in the cable.

Given

Mass of crate(M) = 175.5 kg

Mass of boom(m) = 94.7 kg

The tension, T in the cable can be calculated by taking moments of force about the central point of marked X.

The Angle of the boom with the horizontal can be calculated by

tanθ = 5/10

θ = tan⁻¹(5/10) = 26.56°

Angle of the boom with horizontal is 26.56°

The angle of cable with horizontal can be calculated by

tan B = 4/10

B = tan⁻¹(4/10) = 21.80°

Angle of cable with horizontal is 21.80°

Taking moments of force about the point X

(Mcosθ + mcosθ) 0.5 = T(sin(θ +B)1

(175.5 × cos 26.56 + 94.7 × cos 26.56 )× 0.5 = T (sin(26.56 + 21.80) X 1

By calculating, we get

Tension(T) = 241.68/0.747

Tension(T) = 323.5 N

Hence, 323.5 N is the tension in the cable.

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Solutions for the three problems are is mathematically given as

(a) First cosmic speed (arbitral velocily)

[tex]v=\sqrt{\frac{G M}{r}}\\v=\sqrt{\frac{6.67 \times 10^{-11} \times 4.74 \times 10^{24}}{5870 \times 10^{3}}}[/tex]

v=7338.9349

(b) Second cosmic speed (escape velo.)

$$

\begin{gathered}

[tex]V=\sqrt{\frac{2 G M}{r}}\\\\V=\sqrt{2} \sqrt{\frac{G M}{r}}\\\\=V\sqrt{2} \times 7338.9349 \\[/tex]

[tex]V=14677.86986 \mathrm{~m} / \mathrm{s}[/tex]

(c) In conclusion, in a circular orbit, the gravitational force is gets balanced by centripetal force

[tex]&m_{q \omega \omega^{2}}=\frac{G M M}{r^{2}} \\&r^{3}=\frac{G M}{\omega^{2}}=\frac{G M}{4 \pi^{2}} T^{2} \\&r^{3}=\frac{6.67 \times 10^{-11} \times 4.74 \times 10^{24} \times(16.6 \times 3600)^{7}}{4 \pi^{2}} \\[/tex]

[tex]&r=30581248.06 \times 10^{4} \mathrm{~m} / \mathrm{s}[/tex]

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Answer:

It will still hover until the spaceship "hits" or exerts a force on it.

Explanation:

Remember, if there is no net force, there is no acceleration or movement.

In this case, our ball is hovering in the spaceship, and in space, we can assume there is no [tex]F_g[/tex], and we can assume there is no [tex]F_N[/tex], nor no forces acting against it.

So, the ball would not move.

However, once the spaceship starts accelerating, the ball would still hover until the spaceship exerts a force on it.

This is because of the same thing as explained above, no forces acting on it, therefore, no acceleration.

Think about it this way.

Imagine you jumped up, then someone threw a ball at you. Now let's imagine you can't move until you hit the floor, meaning that in an ideal situation only  [tex]F_g[/tex] is acting on you. Now again, let's imagine time slows really down for you, but not the ball. Before the ball comes and hits you, you are "hovering" like a ball. But after the ball hits you, you move a little because the ball exerted a force on you.

If you did not understand what I meant above, just forget about it, and think about the fact that if there is a Net force (all the force values added up), then there is acceleration and movement.

Acceleration is the rate of change of velocity. Usually, acceleration means the speed is transforming, but not always. When an object moves in a circular path at a steady speed, it is still accelerating, because the focus of its velocity is changing.

a = 4,552 m / s²,   b)  a = 2,588 m / s²

Newton's second law is

F = ma

a = F / m

in this case the force remains constant

indicate us

* for a mass m₁

a₁ = F/m₁

a₁ = 12, m/ s²

* for a mass m₂

a₂= 3.3 m / s²

acceleration  m = m₂-m₁

substitute

[tex]$\begin{aligned}&a=\frac{F}{m_{2}-m_{1}} \\&1 / a=\frac{m_{2}}{F}-\frac{m_{1}}{F}\end{aligned}$[/tex]

let's calculate

1/a=1/3.3 - 1/12  = 0.21969

 a = 4,552 m / s²

a= 4,552m/s²

Speed ​​estimates the rate of movement of an object, that is, the distance traveled per unit of time. Acceleration calculates the rate of change of velocity, that is, the change in velocity between two different moments

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Answer:a)45cos40 b)45sin40 c) 2.95 d)102

Explanation:

a) [tex]v_{x} = 45 cos40[/tex]

b) [tex]v_{y}=45sin40[/tex]

c) Since air resistance is negligible, the only force acting on the golf ball is gravity. Thus, its vertical acceleration is -g. We know the final velocity must be 0 m/s, because this will be when the golf ball reaches the maximum height and starts to change direction (it falls back to the ground). We also know initial velocity in the vertical direction (see part b). Thus, we can use this equation: [tex]v_{f} = v_{i} + at[/tex].

[tex]0 m/s = 45sin40 + (-9.8m/s^2)t\\t = 2.95s[/tex]

d) The horizontal distance traveled is dependent on (1) how long the ball is in the air and (2) what the horizontal velocity is. (1) was found in part c, and (2) was found in part a.

[tex]x=vt\\x=45cos40(2.95) =102m[/tex]

The acceleration due to gravity is 3 m/s^2 and the work done is -4.3 J.

Now we must use the formula;

h = ut + 1/2gt^2

Since it was dropped from a height u = 0 m/s

h = height

u = initial velocity

g = acceleration due to gravity

t = time

h = 1/2gt^2

g = 2h/t^2

g = 2 * 2.85 /(1.38)^2

g = 5.7/1.9

g = 3 m/s^2

The work that must be done is against gravity hence;

W = -(mgh)

W = - (0.5 kg *  3 m/s^2 * 2.85 m)

W = - 4.3 J

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a)New position vector in vector form= r = 0.233404 î + 2.94056j m

b) Lies in second quadrant at 161.391°

c)Velocity =4.8675 m/s

d)It is moving in a direction making 161.391° with positive x-direction.

e)Acceleration will be centripetal acceleration (8.031 m/s²).

Given:

Mass of the object m = 3.95 kg

ω=1.65 rad/s

Radius of the circle = 2.95 m

a)

new position vector in vector form

=R cos1.65 î  + R sin 1.65 j

= 2.95 cos1.65 î  +2.95 sin1.65 j

= 2.95 x 0.07912 î + 2.95 x 0.9968 j

r = 0.233404 î + 2.94056j

b)

Angular Displacement = θ₀ = 9.10 rad

9.10 radian = 180/π× 9/10 degree

= 521.391°

=521.391°- 360°

=161.391°

This will lie in second quadrant.

Angle made with positive x-axis

=161.391°

c)

Velocity

v = ω R

= 1.65 x 2.95

=4.8675 m/s

d)

It is moving in a direction making 161.391° with positive x-direction.

e)

Acceleration will be centripetal acceleration.

= v²/R

=(4.8675)² / 2.95

=23.6925562 / 2.95

=8.031 m/s²

f) Position, Velocity and Acceleration graph:

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a) New position vector is r = 0.233404 î + 2.94056j m

b) Lies in second quadrant at 161.391°

c) Velocity =4.8675 m/s

d) It is moving in a direction making 161.391° with positive x-direction.

e) Acceleration will be centripetal acceleration (8.031 m/s²).

Given:

Mass of the object m = 3.95 kg

ω=1.65 rad/s

The radius of the circle = 2.95 m

a) new position vector in vector form

 r =R cos1.65 î  + R sin 1.65 j

 r = 2.95 cos1.65 î  +2.95 sin1.65 j

 r = 2.95 x 0.07912 î + 2.95 x 0.9968 j

 r = 0.233404 î + 2.94056j

b) Angular Displacement = θ₀ = 9.10 rad

9.10 radian = 180/π× 9/10 degree

 = 521.391°

 = 521.391°- 360°

 =161.391°

This will lie in the second quadrant.

Angle made with the positive x-axis =161.391°

c) Velocity

   v = ω R

   v = 1.65 x 2.95

   v = 4.8675 m/s

d) It is moving in a direction making 161.391° with positive x-direction.

e) Acceleration will be centripetal acceleration.= v²/R

   =(4.8675)² / 2.95

   =23.6925562 / 2.95

   =8.031 m/s²

f) Position, Velocity, and Acceleration graph:

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Answer:

The type of lens involved is a convex lens and the object is positioned at the center of the curvature of the convex lens.

According to the question, a real, inverted, and same-sized image of the object is formed.

A concave lens is a diverging lens and always forms virtual images. But convex lenses are converging lenses and are the only type of lens that produce real and inverted images of the corresponding objects.

When an object is placed at the center of the curvature of a convex lens, its corresponding image is formed on the opposite side of the convex lens. The image formed is real and inverted.

The distance of the image from the lens is equal to the distance of the object from the lens.

For a convex lens, the distance of the center of curvature from the lens is double the focal length of the lens.

That's why the convex lens and center of curvature are the correct answer to this question.

Explanation:

The speed of the space craft relative to the earth is given as: 0.024c. This is solved using the the equation for time dilation.

Time dilation is the "slowing down" of a clock as determined by an observer in relative motion with regard to that clock under the theory of special relativity.

The formula is given as :

Δt = [Δr]/ √ 1 - (v²/c²)

Thus,

v = c√1 - (Δr/Δt)²

= c √(1 - (3600/3601)²

v = 0.024c

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Answer: A candle 4.75 cm tall is 40.0 cm to the left of a plane mirror. Then, the image formed by the mirror will be to the right of the mirror and the height of this image is same as that of the object.

Explanation: To find the answer, we need to know about the reflection at a plane mirror.

There are some points about the reflection at a plane surface.

Thus, we can conclude that, a candle 4.75 cm tall is 40.0 cm to the left of a plane mirror. Then, the image formed by the mirror will be to the right of the mirror and the height of this image is same as that of the object.

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In response, a plane mirror is 40.0 cm to the left of a candle that is 4.75 cm tall. The mirror's generated image will then be to the right of the mirror and have the same height as the object.

In order to determine the solution, we must understand how a plane mirror reflects light.

As a result, we may say that a flat mirror is 40.0 cm to the left of a candle that is 4.75 cm tall. The mirror's generated image will then be to the right of the mirror and have the same height as the object.

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(1) The greatest height h at which the pilot could shut off the engine if the velocity of the lander relative to the surface at that moment is zero is 0.4 m.

(2) The greatest height h at which the pilot could shut off the engine if the velocity of the lander relative to the surface at that moment is 1.7 m/s downward is 0.25 m.

(3) The greatest height h at which the pilot could shut off the engine if the velocity of the lander relative to the surface at that moment is 1.7 m/s upward is 0.25 m.

v² = u² - 2gh

where;

when the lander's velocity = 0

0 = u² - 2gh

u² = 2gh

h = u²/2g

h = (2.8²)/(2 x 9.8)

h = 0.4 m

h = (u² - v²)/2g

h = (2.8² - 1.7²)/(2 x 9.8)

h = 0.25 m

h = (u² - v²)/2g

h = (2.8² - 1.7²)/(2 x 9.8)

h = 0.25 m

Thus, the greatest height h at which the pilot could shut off the engine if the velocity of the lander relative to the surface at that moment is zero is 0.4 m.

The greatest height h at which the pilot could shut off the engine if the velocity of the lander relative to the surface at that moment is 1.7 m/s downward is 0.25 m.

The greatest height h at which the pilot could shut off the engine if the velocity of the lander relative to the surface at that moment is 1.7 m/s upward is 0.25 m.

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The velocity with which the particle Q is fired is 15m/s upwards.

The vector quantity velocity (v), represented by the equation v = s/t, quantifies displacement (or change in position, s) over the change in time (t). Speed (or rate, r) is a scalar number, denoted by the equation r = d/t, that quantifies the distance traveled (d) over the change in time (t).

Assume the upward direction is positive and the downward direction is negative.

Velocity P, = 40m/s

Distance traveled by P,

Using the first equation of motion for particle P,

v = u + at

⇒ 0 = 40 + (-10)t

⇒ t = 4s

This is the time it takes for P to rise

Now, the maximum height(s) reached by the particle P is,

Using the second equation of motion,

s = ut + 1/2at²

⇒ s = 40×4 + 1/2 × (-10) × 4²

⇒ s = 80m

a) A particle P falls when Q is shot after 4 seconds from the initial time:

V² = U² + 2aH₁

⇒ H₁ = 15² - 0/ 2(-10)

⇒ H₁ = 11.25m

Particles P and Q meet at a distance from the ground (H₂).

Height, H₂ = s - H

⇒ H₂ = 80 - 11.25

         = 68.75m

Particles P and Q meet at a distance of 68.75m from the ground.

v = u + at₁

⇒ 15 = 0 + (-10) t₁

⇒ t₁ = 1.5s

It is equal to P's fall time and Q's rise time.

b) For particle Q

H₂ = u₂t₁ + 1/2at₁

⇒ 11.25 = u₂ × 1.5 + 1/2 (-10) × 1.5

⇒ u₂ = 15 m/s

Therefore,

The velocity with which the particle Q is fired is 15m/s upwards.

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Answer:

unfournatletly

Explanation:

i have no clue sorry to waste ur time ill rather not say a answer that will be incorrect.

Answer:

Pluto

Explanation:

Answer:

D

Explanation:

momentum can be negative as it's a vector quantity. positive momentum indicates that the object is traveling in the positive direction as defined by the coordinate system. negative momentum indicates that the object is moving in the opposite direction (backwards). the momentum equation itself is p = mass x velocity.

and only if you consider velocity a a directed vector, can you have a negative momentum. this equation shows that the momentum is in the same direction as velocity.

so, the sign is only for the direction.

the force is the same (going with the absolute value). a negative momentum is not smaller than a positive momentum.

It is possible that momentum can be negative if an object is moving backward. The correct option is D.

A vector quantity called momentum depends on an object's mass and velocity. It is described as the result of the mass and the velocity of an object.

When an item is moving against a selected positive direction, its velocity will be negative since velocity is a vector variable that comprises both magnitude and direction. As a result, the negative velocity will yield a negative value for momentum when it is calculated.

Thus the correct option is D.

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(a) The time the baseball spends in the air is 0.92 s.

(b) The horizontal distance from the roof edge to the point where the baseball lands on the ground is 3.1 m.

h = vt - ¹/₂gt²

-2.1 = (4.05 x sin 34)t  - ¹/₂(9.8)(t²)

-2.1 = 2.26t - 4.9t²

4.9t² - 2.26t - 2.1 = 0

t = 0.92 s

R = Vx(t)

R = (4.05 x cos 34)(0.92)

R = 3.1 m

Thus, the time the baseball spends in the air is 0.92 s.

The horizontal distance from the roof edge to the point where the baseball lands on the ground is 3.1 m.

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Three moons can fit inside the volume of the sun.

The moon is a non luminous body found in the space. It could cause a solar eclipse when it comes between the sun and the earth.

Since the Earth’s diameter is about 8,000 miles and the Moon’s diameter is about 2,000 miles, to obtain the number of moons that could fit inside the sun we have;

8,000 miles/ 2,000 miles = 3

Hence, three moons can fit inside the volume of the sun.

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(a) The magnetic field the line produced at ground level is 3.75 x 10⁻⁶ T.

(b) The lines magnetic field as a percent of the earth's magnetic field is 7.5%.

(c) No, since this field is much smaller than the earth's magnetic field, it would be expected to have less effect than the earth's field.

B = (μ x I) / (2πr)

B = (4π x 10⁻⁷ x  150) / (2π x 8)

B = 3.75 x 10⁻⁶ T

x = 3.75 x 10⁻⁶ /0.5G

x = (3.75 x 10⁻⁶ ) / (0.5 x 10⁻⁴)

x = 0.075

x = 0.075 x 100% = 7.5%

Thus, we can conclude that, since this field is much smaller than the earth's magnetic field, it would be expected to have less effect than the earth's field.

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Answer:

d

Explanation:

Gravitational force is a type of force that pulls objects to the Earth's surface.

(a) The electric flux on left face is 0.24 Vm and on the right face is 1 Vm.

(b) The total flux is 1.24 Vm.

(c) The total amount of charge that is inside the box is  1.1 x 10⁻¹¹ C.

The area of the left face is calculated as follows;

A1 = 0.03 m x 0.02 m = 0.0006 m²

Ф1 = EA1

Ф1 = (400 V/m)( 0.0006 m²) = 0.24 Vm

Let the dimension of the right face = 5 cm by 2 cm

A2 = 0.05 m x 0.02 m = 0.001 m²

Ф2 = EA2

Ф2 = (1000 V/m)( 0.001 m²) = 1 Vm

Ф = Ф1 + Ф2

Ф = 0.24 Vm + 1 Vm = 1.24 Vm

Ф = Q/ε

Q = εФ

Q = (8.85 x 10⁻¹²)(1.24)

Q = 1.1 x 10⁻¹¹ C

Thus, the electric flux on left face is 0.24 Vm and on the right face is 1 Vm.

The total flux is 1.24 Vm and the total amount of charge that is inside the box is  1.1 x 10⁻¹¹ C.

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The magnitude of the force on the left-hand pole is mathematically given as

f'=0.167N

Q=45 degrees

Generally, the equation for Force is  mathematically given as

F=mg/sinФ

Therefore

F(17.1*10^{-3})*9.8/sin 45

F=0.237N

Considering horizontal axis or plane

f'-fcos=0

Therefore

f'=0.237*cos45

f'=0.167N

In conclusion, the slope

tanФ=30/30

tanФ=1

Ф=45

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3.00 m is the magnitude of the resultant vector

137.1° is the directional angle of the resultant vector.

1) To find the magnitude of the resultant

Discover each vector's individual components first. 

We must consider the signs of the components because the angles in the figure are measured in various ways. 

In this case, both the y component of vector C and the x component of vector A are negative. 

The vectors' components are as follows, using a little trigonometry:  

Magnitude of A = 5 m

[tex]A_{x}[/tex] = - (5.00m) cos20° = -5 ×0.408 = -4.698 m

[tex]A_{y}[/tex] = + (5.00m)sin20° = +5 × 0.342 = +1.710 m

Magnitude of B = 5m

[tex]B_{x}[/tex] = +(5.00m)cos60° = 5 × 0.5 = +2.5m

[tex]B_{y}[/tex] = +(5.00m)sin60° = 5 × √3/2 = +4.33 m

Cx = 0

Cy = -4.00m

The sum of all three vectors, which we refer to as R, produces components.

Rₓ = Aₓ + Bₓ + Cₓ

   = -4.698 + 2.5 + 0

   = -2.198 m

[tex]R_{y}[/tex] = [tex]A_{y} + B_{y} + C_{y}[/tex]

    = +1.71 +4.33 - 4.00

    = 2.040 m

R = [tex]\sqrt{R_{x^{2} }+R_{y^{2} }[/tex]

  = [tex]\sqrt{(-2.198)^{2 } + (2.040)^{2} }[/tex]

  = 3.00 m

2) To find the directional angle of resultant

tan θ = 2.040/-2.198 = -0.928

θ = -42.9°

Such a vector would be in the so-called "fourth quadrant," as it is well known. However, we discovered that R has a negative x component and a positive y component, indicating that such a vector must reside in the "second quadrant.

The calculator accidentally returned an angle that is 180 degrees off, thus we must add 180 degrees to the naïve angle in order to get the right angle.

Therefore, R's actual direction is determined by-

θ = -42.9° + 180° = 137.1°

Hence, the magnitude of resultant vector is 3.00 m and directional angle is 137.1°

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The force required to pull the two hemispheres apart 53696.25N and  the minimum number of horses required is 37 .

( Note: 1 millibar=100 N/m2. One atmosphere is 1013 millibar = 1.013×105 N/m2 )

(1)The contact area between the hemispheres is (π x 0.430²) = 0.5805m²

Pressure difference = (940 - 15) = 925millibars.

(925 x 100) = 92,500N/m^2.

(92500 x 0.5805) = 53696.25N. is the force required to part the hemispheres.

(2)[tex]\frac{53696.25N}{1450N}[/tex] = 37 horses for each side .

37 + 37 = 74 horses will be required.

Force is something which can change the motion of an object, stop it or move it, change its shape or size with it. There are two types of forces, contact forces and non-contact forces. Here, it is a contact force at first, then when the horses come it becomes non-contact force.

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Hello!

a)

To begin, let me first clarify that no section of the wire along the axis of the point 'P' will contribute to the magnetic field (Aka, the straight part of the wire) because the radius vector and current vectors would be parallel.

Now, let's use Biot-Savart's Law:
[tex]dB = \frac{\mu_0}{4\pi }\frac{id\vec{l}\times \hat{r}}{r^2}[/tex]

B = Magnetic field strength (? T)
μ₀ = Permeability of free space (Tm/A)

i = Current in wire ('I' A)
dl = differential length element

r = distance from wire to point P ('R' m, remains constant!)

We can rewrite Biot-Savart as:
[tex]B = \frac{\mu_0}{4\pi } \int \frac{id\vec{l}\times \hat{r}}{r^2}[/tex]

First, let's mind the cross-product.

The angle between the radius vector (Along 'R') and the current vector is ALWAYS 90° since the two vectors are perpendicular along the arc. At a certain point, think about the current as being "tangential" to the differential length and thus perpendicular to the radius.

Therefore, we can disregard the cross-product since sin(90) = 1.

Let's plug in what we already know, replacing 'dl' with 'ds' since this is an arc:

[tex]B = \frac{\mu_0}{4\pi } \int \frac{ids}{R^2}[/tex]

We have a semi-circle, so we are integrating from 0 to π radians.

[tex]B = \frac{\mu_0}{4\pi } \int\limits^{\pi}_{0} {\frac{ids}{R^2}}[/tex]

Simplifying to make the integral easier, we can take constants out of the integral.

[tex]B = \frac{\mu_0i}{4\pi R^2 } \int\limits^{\pi R}_{0} {} \, ds[/tex]

Evaluating:
[tex]B = \frac{\mu_0i}{4\pi R^2} (\pi R- 0) = \boxed{\frac{\mu_0 i}{4R}}[/tex]

b)

Using the current Right-Hand-Rule at the top of the arc, point your thumb to the right. Curl your fingers as if you are gripping the wire over the top and all the way over to the other side of the wire (Where point 'P' would be). Your fingers should point INTO THE PAGE.

The application of total Internal Reflection occurs in a mirage. Option A

1) We know that a charge can be positive or negative. If the charge is negative, we call it an electron. If the charge is positive, we call it a proton. Now we know that the magnitude of charge on each electron is 1.6×10-19 C. Hence, the magnitude of charge on five electrons = 5 * 1.6×10-19 C = 8.0×10-19 C. Option D

2) The details of question 2 are not shown hence the question can not be answered.

3) Looking at the electronic configuration of the element;  1s22s22p63s23p5, it is clear to see that it has the ns2 np5 outermost configuration that is common to halogens thus it;

4) The atom is composed of the protons, neutrons and electrons. Given the nuclide identified as 230/92U we have  92 protons, 138 neutrons and 92 electrons. Option C

5) The application of total Internal Reflection occurs in a mirage. Option A

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Sound will travel fastest in air at 15°C.

The speed of sound in air, given in the range of 100 degrees Celsius and 0 degree Celsius include;

Air at 100°C 387 m/s

Air at 0°C 331 m/s

From the date above, the speed of sound in air increases with increases in temperature. Thus, Sound will travel fastest in air at 15°C.

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Answer: The small spherical planet called "Glob" has a mass of 7.88×1018 kg and a radius of 6.32×104 m. An astronaut on the surface of Glob throws a rock straight up. The rock reaches a maximum height of 1.44×103 m, above the surface of the planet, before it falls back down.

1) the initial speed of the rock as it left the astronaut's hand is 19.46 m/s.

2) A 36.0 kg satellite is in a circular orbit with a radius of 1.45×105 m around the planet Glob. Then the speed of the satellite is 3.624km/s.

Explanation: To find the answer, we need to know about the different equations of planetary motion.

                    [tex]m_g=7.88*10^{18}kg\\r_g=6.32*10^4 m.\\h_{max}=1.44*10^3m.\\[/tex]

                    [tex]v=\sqrt{2gh}[/tex]

                  [tex]g_g=\frac{GM_g}{r_g^2} =\frac{6.67*10^{-11}*7.88*10^{18}}{(6.32*10^4)^2} =0.132m/s^2.[/tex]

                  [tex]v=\sqrt{2*0.132*1.44*10^3} =19.46 m/s.[/tex]

                [tex]v=\sqrt{ \frac{GM}{r}}=\sqrt{\frac{7.88*10^{18}*6.67*10^{-11}}{1.45*10^5 }} =3.624km/s.[/tex]

Thus, we can conclude that,

1) the initial speed of the rock as it left the astronaut's hand is 19.46 m/s.

2) A 36.0 kg satellite is in a circular orbit with a radius of 1.45×105 m around the planet Glob. Then the speed of the satellite is 3.624km/s.

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1) The rock was moving at 19.46 m/s when it first left the astronaut's palm.

2) A 36.0 kg spacecraft is orbiting the planet Glob in a sphere with a radius of 1.45 105 meters. The satellite is moving at 3.624 km/s at that point.

Understanding the planetary motion equations is necessary in order to determine the solution.

                             [tex]v=\sqrt{2gh}[/tex]  

                    [tex]a=\frac{GM}{R^2} =0.132m/s^2[/tex]

                  [tex]v=\sqrt{2*9.8*0.132} =19.46m/s[/tex]

                        [tex]v=\sqrt{\frac{GM}{R} } =3.624km/s\\where, M=7.88*10^{18}kg[/tex]

Consequently, we can say that

1) The rock was moving at 19.46 m/s when it first left the astronaut's palm.

2) A 36.0 kg spacecraft is orbiting the planet Glob in a sphere with a radius of 1.45 105 meters. The satellite is moving at 3.624 km/s at that point.

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Answer:

b

Explanation:

the answer is b because if you do the math it equals to b

The magnitude of the force on the left-hand pole of the thin flexible gold chain of uniform linear density with a mass of 17.1 and, hangs between two 30.0 cm long vertical sticks, which are a distance of 30.0 cm apart will be, 0.167N.

To find the correct answer, we have to know more about the Basic forces that acts upon a body.

                                    [tex]tan\alpha =\frac{30}{30}\\ \alpha =45^0[/tex]

                                 [tex]y-direction\\T_2sin\alpha =mg\\\\T_2=\frac{mg}{sin\alpha } =\frac{17.1*10^{-3}*9.8}{sin45} \\\\T_2=0.236N[/tex]

                                 [tex]x-direction\\T_1-T_2cos\alpha =0\\T_1=T_2cos\alpha \\T_1=0.236*cos45=0.167N[/tex]

Thus, we can conclude that, the magnitude of force on the left-hand pole will be 0.167N.

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The force acting on the left-hand pole of a thin, flexible gold chain of uniform linear density weighing 17.1 grams that is suspended between two vertical sticks that are 30.0 cm apart will be 0.167 N in magnitude.

We need to learn more about the fundamental forces that affect a body in order to arrive at the right answer.

                              [tex]\alpha =tan^{-1}(\frac{30}{30})=45^0[/tex]

                                [tex]T_2sin\alpha =mg\\T_2=\frac{mg}{sin\alpha } =0.236N[/tex]

                                     [tex]T_1=T_2cos\alpha \\T_1=0.167N[/tex]

Thus, we may infer that the force acting on the left-hand pole will have a value of 0.167N.

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Answer:

The Fundamental Quantity is a physical quantity that exists independently and cannot be expressed by any other physical quantity.

Explanation:

Examples of these could be : Length, electric current and mass

Hope this helps! :)

From the calculations, the power expended is 43650 W.

Now we can find the acceleration from;

v = u + at

u = 0 m/s

v =  95 km/h or 26.4 m/s

t =  6.8 s

a = ?

Now

v = at

a = v/t

a = 26.4 m/s/ 6.8 s

a = 3.88 m/s^2

Force = ma = 850-kg * 3.88 m/s^2 = 3298 N

The distance covered is obtained from;

v^2 = u^2 + 2as

v^2 = 2as

s = v^2/2a

s = (26.4)^2/2 * 3.88

s = 696.96/7.76

s = 90 m

Now;

Work = Fs

Work =  3298 N * 90 m = 296820 J

Power =  296820 J/ 6.8 s

= 43650 W

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