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3 years ago

HELP ME

Physics

2 answers:

STatiana [176]3 years ago

5 0

Answer:

Explanation:

Electromagnetic and gravitational forces have equations that clearly describe them.

Electromagnetic Fe = K q₁ q₂ / r²

Gravitational Fg = G m₁ m₂ / r²

Analyzing these expressions a bit we can see some things in which they are EQUAL:

• They depend on the inverse of the squared distance

• They are long range, they become zero In the infinite

• They are proportional to a system constant, in one case the charge and in the other the mass

We can also observe some DIFFERENCES:

• Gravitational always attractive, the electric can be attractive or repulsive depending on the charge

• When we calculate their magnitudes the electric force in 10 40 times greater than the gravitational

* The electric force can be zero if the charge is zero, the gravitational force can never be zero

Katarina [22]3 years ago

3 0

Explanation:

alike

no contact with the object itself

they can attract objects

they're between 2 objects

difference

source of foundations

type of forces

gravity is only for huge objects

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9) A bug of mass 0.020 kg is at rest on the edge of a solid cylindrical disk (M = 0.10 kg, R = 0.10 m) rotating in a horizontal

natima [27]

Answer:

a. ω₂ = 14rad/s

b. ∇K.E = 0.014J

c. The bug does not conserve force while moving on the disk (non-conservative force).

Explanation:

Mass of the bug (m) = 0.02kg

Mass of the cylindrical disk (M) = 0.10kg

Radius of the disk (r) = 0.10m

Initial angular velocity ω₁ = 10rad/s

final angular velocity ω₂ = ?

To calculate the new angular velocity, we relate it to the conservation of angular momentum of the system I.e when the bug was at the edge of the disk and when it is located at the centre of the disk.

I = Mr² / 2

I₁ = Mr₂ / 2 + mr²

I₁ = moment of inertia when the bug was at the edge

I₁ = [(0.10 * 0.10²) / 2 ] + (0.02 * 0.1²)

I₁ = 0.0005 + 0.0002

I₁ = 7.0*10⁻⁴kgm²

I₂ = moment of inertia when yhe bug was at the center of the disk.

I₂ = Mr² / 2

I₂ = (0.01 * 0.01²) 2

I₂ = 0.0005kgm²

for conservation of angular momentum,

I₁ω₁ = I₂ω₂

solve for ω₂

ω₂ = (I₁ * ω₁) / I₂

ω₂ = (7.0*10⁻⁴ * 10) / 5.0*10⁻⁴

ω₂ = 14 rad/s

b. the change in kinetic energy of the system is

∇K = K₂ - K₁

∇K = ½I₂*ω₂² - ½I₁*ω₁²

∇K = ½(I₂*ω₂² - I₁ω₁²)

∇K = ½[(5.0*10⁻⁴ * 14²) - (7.0*10⁻⁴ * 10²)]

∇k = ½(0.098 - 0.07)

∇K = ½ * 0.028

∇K = 0.014J

The cause of the decrease and increase in kinetic energy is because the bug uses a non-conservative force. To conserve the mechanical energy of a system, all the forces acting in it must be conservative.

The work W produced by this force brings the difference in kinetic energy of the system

W = K₂ - K₁

6 0

3 years ago

The normal boiling point of water is 100.0 °c and its molar enthalpy of vaporization is 40.67 kj/mol. what is the change in entr

SashulF [63]

ΔS= nΔHvap/T,

Where, ΔS = Change in entropy, n = moles of water = 39.3/18 = 2.188 moles, ΔHvap = 40.67kJ/mol = 40670 J/mol, T = Temperature (K) = 100+272.15 = 373.15 K

Therefore,
ΔS = (2.188*40670)/373.15 = 237.96 J/K

6 0

3 years ago

A rock is thrown off a cliff at an angle of 53° with respect to the horizontal. The cliff is 100 m high. The initial speed of th

Nadusha1986 [10]

(a) 129.3 m

The motion of the rock is a projectile motion, consisting of two indipendent motions along the x- direction and the y-direction. In particular, the motion along the x- (horizontal) direction is a uniform motion with constant speed, while the motion along the y- (vertical) direction is an accelerated motion with constant acceleration $g=-9.8 m/s^2$ downward.

The maximum height of the rock is reached when the vertical component of the velocity becomes zero. The vertical velocity at time t is given by

$v(t) = v_0 sin \theta +gt$

where

$v_0 = 30 m/s$ is the initial velocity of the rock

$\theta=53^{\circ}$ is the angle

t is the time

Requiring $v(t)=0$ , we find the time at which the heigth is maximum:

$0=v_0 sin \theta + gt\\t=\frac{-v_0 sin \theta}{g}=-\frac{(30)(sin 53^{\circ})}{(-9.8)}=2.44 s$

The heigth of the rock at time t is given by

$y(t) = h+(v_0 sin\theta) t + \frac{1}{2}gt^2$

Where h=100 m is the initial heigth. Substituting t = 2.44 s, we find the maximum height of the rock:

$y=100+(30)(sin 53^{\circ})(2.44)+\frac{1}{2}(-9.8)(2.44)^2=129.3 m$

(b) 44.1 m

For this part of the problem, we just need to consider the horizontal motion of the rock. The horizontal displacement of the rock at time t is given by

$x(t) = (v_0 cos \theta) t$

where

$v_0 cos \theta$ is the horizontal component of the velocity, which remains constant during the entire motion

t is the time

If we substitute

t = 2.44 s

Which is the time at which the rock reaches the maximum height, we find how far the rock has moved at that time:

$x=(30)(cos 53^{\circ})(2.44)=44.1 m$

(c) 7.58 s

For this part, we need to consider the vertical motion again.

We said that the vertical position of the rock at time t is

$y(t) = h+(v_0 sin\theta) t + \frac{1}{2}gt^2$

By substituting

y(t)=0

We find the time t at which the rock reaches the heigth y=0, so the time at which the rock reaches the ground:

$0=100+(30)(sin 53^{\circ})t+\frac{1}{2}(-9.8)t^2\\0=100+23.96t-4.9t^2$

which gives two solutions:

t = -2.69 s (negative, we discard it)

t = 7.58 s --> this is our solution

(d) 136.8 m

The range of the rock can be simply calculated by calculating the horizontal distance travelled by the rock when it reaches the ground, so when

t = 7.58 s

Since the horizontal position of the rock is given by

$x(t) = (v_0 cos \theta) t$

Substituting

$v_0 = 30 m/s\\\theta=53^{\circ}$

and t = 7.58 s we find:

$x=(30)(cos 53^{\circ})(7.58)=136.8 m$

(e) (36.1 m, 128.3 m), (72.2 m, 117.4 m), (108.3 m, 67.4 m)

Using the equations of motions along the two directions:

$x(t) = (v_0 cos \theta) t$

$y(t) = h+(v_0 sin\theta) t + \frac{1}{2}gt^2$

And substituting the different times, we find:

$x(2.0 s)=(30)(cos 53^{\circ})(2.0)=36.1 m$

$y(2.0 s)= 100+(23.96)(2.0)-4.9(2.0)^2=128.3 m$

$x(4.0 s)=(30)(cos 53^{\circ})(4.0)=72.2 m$

$y(4.0 s)= 100+(23.96)(4.0)-4.9(4.0)^2=117.4 m$

$x(6.0 s)=(30)(cos 53^{\circ})(6.0)=108.3 m$

$y(6.0 s)= 100+(23.96)(6.0)-4.9(6.0)^2=67.4 m$

3 0

3 years ago

A 25 kg child is bouncing on a trampoline, which can be treated as an ideal spring with spring constant 2900 N/m. If the trampol

White raven [17]

Answer:

Explanation:

Energy store in the compressed trampoline = potential energy of at the maximum height ignoring friction

energy stored in the trampoline = $\frac{1}{2} kx^{2}$ where k is spring constant and x is the distance compressed = 0.5 × 2900 × 0.38² = 209.38 J

209.38 J = 25 × 9.8 × h

h maximum height attain = 209.38 J / ( 25 × 9.8) = 0.855 m

5 0

3 years ago

Draw a ray diagram to show the formation of an image when the object is placed beyond the ‘c’of a concave mirror.

ICE Princess25 [194]

the object is real inverted and diminishes

6 0

4 years ago