Two 1kg masses as distance D as part each exerted a gravitational attraction Force on the other one. If 1kg is now add to each mass the gravitational attraction exerted by each would be​

The rubber band travels horizontally 7.0 cm in second time.

Given parameters:

the rubber band is stretched = 2.0 cm.

And, the rubber band travels = 3.5 cm.

Secondly,  the rubber band is stretched = 4.0 cm.

As force is directly proportional with stretching and landing at a horizontal distance,  it travels horizontally in second time

= 4.0 cm x 3.5 cm/ 2.0 cm.

= 7.0 cm.

The vertical movement can't be measured as a new gravitational force comes in play in this case.

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The maximum number of revolutions per minute the fan can run at before she will be flung off is  1,982.24 rev/min.

The maximum angular speed of the fan is calculated by applying the following kinematic equation.

From Newton's second law of motion,

F = ma

F = mv/t

F = m(ωr/t)

where;

ω/t = F/mr

ω/t = (0.0137 N) / (0.00022 kg x 0.3 m)

ω/t = 207.58 rad/s

ω/t = (207.58 rad/s)  x (1 rev / 2π rad) x  (60 s / min)

ω/t = 1,982.24 rev/min

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We first apply the data to the problem.

Data:

[tex] \bold{V = 90km/h}[/tex]

[tex] \bold{T = 20s}[/tex]

[tex] \bold{D = ?}[/tex]

Now, we convert km/h to m/s.

Conversion:

[tex] \bold{90km/h * (1000m/1km) * (1h/3600s)}[/tex]

[tex] \boxed{ \boxed{ \bold{V = 25m/s}}}[/tex]

Then, we apply the formula that is.

Formula:

[tex] \bold{D = V * T}[/tex]

To determine the distance traveled we develop the problem.

Developing:

[tex] \bold{D = (25m/s) * (20s)}[/tex]

[tex] \boxed{\boxed{ \bold{D = 500m}}}[/tex]

Its speed is 25 meters per second and the distance traveled is 500 meters.

Answer:

[tex]0.60\; {\rm m \cdot s^{-1}}[/tex], assuming that the friction between the skates and the ice is negligible.

Explanation:

Under the assumptions, the total momentum of the two skaters will be conserved. In other words, the sum of the momentum of the two skaters will be the same before and after the collision.

When an object of mass [tex]m[/tex] moves at velocity [tex]v[/tex], the momentum [tex]p[/tex] of that object will be [tex]p = m\, v[/tex].

The [tex]m_{b} = 50\; {\rm kg}[/tex] skater was initially moving with a velocity of [tex]v_{b} = 1.5\; {\rm m\cdot s^{-1}}[/tex]. The momentum of this skater will be:

[tex]m_{b}\, v_{b} = 50\; {\rm kg }\times 1.5\; {\rm m\cdot s^{-1}} = 75\; {\rm kg \cdot m \cdot s^{-1}}[/tex].

Since the [tex]m_{a} = 75\; {\rm kg}[/tex] skater was initially not moving (velocity is [tex]v_{b} = 0\: {\rm m\cdot s^{-1}}[/tex],) the momentum of that skater will be:

[tex]m_{a}\, v_{a} = 75\; {\rm kg }\times 0\; {\rm m\cdot s^{-1}} = 0\; {\rm kg \cdot m \cdot s^{-1}}[/tex].

Thus, the total momentum of the two skaters was [tex]75\; {\rm kg \cdot m \cdot s^{-1}}[/tex] before the collision.

Since the two skaters held on to each other, the two will travel at the same velocity after the collision. Let [tex]v[/tex] denote this velocity. The total momentum of the two skaters after the collision will be [tex]m_{a}\, v + m_{b}\, v = (m_{a} + m_{b})\, v[/tex].

Under the assumptions, momentum will be conserved in the collision. Hence, the total momentum after collision [tex](m_{a} + m_{b})\, v[/tex] should be equal to the total momentum before the collision, [tex]75\; {\rm kg \cdot m \cdot s^{-1}}[/tex]. In other words:

[tex](m_{a} + m_{b})\, v = 75\; {\rm kg \cdot m\cdot s^{-1}}[/tex].

Rearrange this equation and solve for the velocity [tex]v[/tex] of the two skaters after the collision:

[tex]\begin{aligned}v &= \frac{75\; {\rm kg \cdot m\cdot s^{-1}}}{m_{a} + m_{b}} \\ &= \frac{75\; {\rm kg \cdot m\cdot s^{-1}}}{75\; {\rm kg} + 50\; {\rm kg}} \\ &= 0.60\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].

In other words, the two skaters will travel at approximately [tex]0.60\; {\rm m\cdot s^{-1}}[/tex] after the collision.

The net work done on the block over the given distance is 39.4d (joules)

The net work done on the block over the given distance is calculated by applying the following equation as shown below;

W(net) = F(net) x d

where;

F(net) = F - μmgcosθ

where;

F(net) = 50 N - (0.25 x 5 x 9.8 x cos30)N

F(net) = 50 N - 10.6 N

F(net) = 39.4 N

The net work done on the block over the given distance is calculated as;

W = 39.4 N x d

where;

W = 39.4d (joules)

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5/4 m/s^{2} would be its new acceleration .

What is centrifugal acceleration ?

The simplest case of circular motion is uniform circular motion, where an object moves in a circular trajectory with constant velocity. Note that the linear velocity of an object in circular motion is constantly changing because, unlike velocity, it is constantly changing direction. From kinematics we know that acceleration is a change in velocity, either in magnitude or direction, or both. Therefore, an object in uniform circular motion will always be accelerated even if the magnitude of its velocity is constant. You can feel this acceleration every time you get in the car while cornering. Keeping the handle steady while turning and moving at a constant speed creates a smooth circular motion. What you will notice is a feeling of skidding (or skidding depending on speed) out of the center of the turn.

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The system acceleration for the pulley having masses m1 = 0.7 kg and m2 = 0.4 kg  Is 35.93 [tex]m/s^2[/tex].

Newton's second law provides a detailed explanation of how a force can alter a body's motion. It is believed that a body's momentum varies over time at a rate equal to the force acting on it in both direction and amplitude. The combined effects of a body's mass and velocity determine its momentum.

Given:

The mass, m₁ = 0.7 kg,

m₂ = 0.4 kg,

Calculate the system acceleration by the following formula,

a = g(m1 − m2)/(m1 + m2)

Here a is the acceleration.

Substitute the values,

a = 9.8 * (0.7 - 0.4 ) / (0.7 + 0.4)

a = 35.93 [tex]m/s^2[/tex]

Therefore, the system acceleration for the pulley is 35.93 [tex]m/s^2[/tex].

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The force that Superman must exert is -1.33 * 10⁶ N

The force that must be exerted by Superman is 37.6 % of the weight of the train.

The force the train exerts on superman is 1.33 * 10⁶ N

The average acceleration of the train is determined from the equation of motion as follows:

v² = u² + 2as

where

v is final velocity = 0 m/s

u is initial velocity = 120 km/h

To convert km/h to m/s, we multiply km/h by 1000/3600

u = 120 * 1000/3600

u = 33.3 m/s

a = ?

s = 150 m

a = v² - u² / 2s

a = 0 - 33.3² / 2 * 150

a = 3.696 m/s²

Force that must be exerted will be:

Force = 3.6 × 10⁵ * 3.696 m/s²

Force = -1.33 * 10⁶ N

b. Weight of train =  3.6 × 10⁵ * 9.81

Weight of train = 3.53 * 10⁶ N

The ratio of the force and the weight of the train = (-1.33 * 10⁶ / 3.53 * 10⁶) * 100%

The ratio of the force and the weight of the train = 37.6%

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The answer is 70.6°

Answer:

0 J

Explanation:

KE = .5(m)(v2-v1)

if velocity is constant v2 = v1, then v2 - v1 = 0

therefore if (v2-v1) = 0, then KE must also be equal to 0

So, this is the diagram of the described situation, the strings are attached to a rigid support at their one end and the other end is connected to the particle.

Let, the string making an angle of 45 with the horizontal has a tension of T 1 and

the one making an angle of 30 has a tension of T2.

Now,

what we have done is that we have taken the vertical and

horizontal components of the tension in the two strings.

Now,considering the system to be in equilibrium,

see the diagram,the vertical component of the strings i.e

(T 1 sin 45 & T 2 sin 30) together will act to balance the weight of

the particle(mg)

So,we can say, T 1 sin 45 + T 2 sin 30 = mg. ...(1)

From horizontal equilibrium,we can say, T 1 cos 45 must balance T 2 cos 30 ,

as no other force is present in horizontal direction.

So,T 1 cos 45 = T 2 cos 30. ...(2)

Given,

mg = 10 ⋅ 10 = 100N

So, solving equation 1 & 2 we get,

T 1 = 89.65N and, T 2 = 73.20N

The amount of work required to move the rock of mass 95 kg up a ramp of 100 m is 93100 J.

Work can be fined as the product of force and distance.

To calculate the amount of work required to get the rock up, we use the formula below.

Formula:

Where:

From the question,

Given:

Substitute these values into equation 1

Hence, the amount of work required is 93100 J.

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Answer:

120

Explanation:

because it is horizontal

An object of mass 3.0 kg starts from rest and moves along the x-axis. A net horizontal force is applied to the object in the +x direction. The Force-time graph is shown below. The net impulse delivered by the applied force is 102 joule.

A force is an influence that has the power to alter an object's motion. An object with mass can change its velocity, or accelerate, as a result of a force. An obvious way to describe force is as a push or a pull. A force is a vector quantity since it has both magnitude and direction.

Force is used to describe a body's tendency to modify or change its state as a result of an external cause. When force is applied, the body can also alter its size, shape, and direction. kicking a ball, pushing and pulling on the door, or kneading dough are a few examples.

We know the Impulse = Area of the Graph

= ( 6 × 4) + 0.5 × 2 × 6

= 30 sec

For First 4 sec

Acceleration = 6 ÷ 3

= 2 m/sec²

S = 0.5 × 2 × 4²

= 16 m

Therefore, work Done = 6 × 16

= 96 J

Then, Average force

= ( 6 - 0 ) ÷ 2

= 3 N

Acceleration = 1 m/sec²

S = 0.5 × 1 × 2²

= 2 m

Work = 3 × 2

= 6 J

Total Work = 96 + 6

= 102 J

Thus, The net impulse delivered by the applied force is 102 joule.

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The gravitational potential energy change during this climb is 83,790 J.

The change in the gravitational potential energy of the bike racer is calculated as follows;

ΔP.E = mgh

where;

The vertical height of the road is calculated as follows;

h = L sinθ

where;

h = 1500 x sin(4.3)

h = 112.5 m

ΔP.E = 76 x 9.8 x 112.5

ΔP.E = 83,790 J

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Answer:

A

Explanation:

A constant velocity means the position graph has a constant slope. It's a straight line sloping up.

The momentum of the sprinter of mass and velocity of 68 kg and 8 m/s respectively is  544 kgm/s

Momentum can be defined as the product of mass and velocity of a body.

To calculate the linear momentum of the sprinter, we use the formula below.

Formula:

Where:

From the question,

Given:

Substitite the values into equation 1

Hence, the momentum of the sprinter is 544 kgm/s.

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The linear momentum of the sprinter with a mass of 68 kg that reaches a speed of 8 m/s during a race is 544kgm/s.

The momentum of a body in motion is the tendency of a body to maintain its inertial motion. It is the product of its mass and velocity, or the vector sum of the products of its masses and velocities.

The linear momentum of a body can be calculated as follows:

p = m × v

According to this question, a sprinter has a mass of 68kg and reaches a speed of 8 m/s during a race. The momentum can be calculated as follows:

p = 68kg × 8m/s

p = 544kgm/s

Therefore, 544kgm/s is the linear momentum of the sprinter.

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Answer:

Explanation:

m₀ = 100 g

m = 6.25 g

T = 33 s

_____________

t - ?

Radioactive decay constant:

λ = ln 2 / T = 0.693 / 33  ≈ 0.021 s⁻¹

Law of radioactive decay:

[tex]m=m_{0} e^{-\lambda t} \\[/tex]

[tex]e^{-\lambda t} = \frac{m}{m_0} \\[/tex]

[tex]ln (e^{-\lambda t}) = ln(\frac{m}{m_0} ) \\[/tex]

[tex]-\lambda t = ln (\frac{m}{m_0})[/tex]

[tex]t = -\frac{ln(\frac{m}{m_0}) }{\lambda}[/tex]

t = - ln (6.25/100) / 0.021 ≈ 132 s

Answer: Hawking radiation is a result of doing quantum mechanics outside a black hole! In short, a stationary observer outside a black hole, according to quantum mechanics, measures a precisely thermal spectrum of radiation originating from the black hole. What form this radiation takes depends on what quantum fields inhabit the universe, but this crucial result would hold in any case; photons or no photons!

Ultimately I don’t think that there’s really any way to visualise what’s going on. For me I just take it as a consequence of doing quantum mechanics near a black hole *. I suppose that I think (and believe that physicists should perhaps most correctly think) of Hawking radiation as a process, not a “noun”. Some people have cooked up explanations involving virtual particle pairs popping out of the vacuum on the horizon, where one exits and one falls in etc. and the outgoing guy is the Hawking radiation but I don’t really buy this, chiefly because I don’t think that virtual particles exist. Hawking’s original calculation doesn’t care about any of this anyway.

Black holes in nature, naively certainly do appear to violate the conservation of information . This is precisely because of the nature of Hawking radiation: it’s exactly thermal. This means that, no matter what goes in, the same spectrum of radiation emerges. Evidently, it’s then naively impossible to reconstruct the information pertaining to the in-falling stuff from what is, in this sense, just whitenoise. Some argue that if you do the calculation carefully, you can in principle recover this information but as far as I’m aware there’s no consensus as to whether or not this is possible.

As far as things falling into the horizon are concerned (such as your photons), there are two camps. One (I agree with them) maintains that due to Einstein's equivalence principle (loosely speaking, that no local patch of spacetime is in any sense “special”), an observer falling through the event horizon wouldn’t notice anything particularly special (though it’d probably be an interesting light-show!), they’d just go right on in. It’s only when they approached the singularity that things would get tense: the tidal forces near the singularity in the centre would ultimately tear any matter to shreds! The second camp support the firewall argument, which claims that the paradox is solved by an impassible wall of extremely high-energy quanta inhabiting the region just behind the horizon. As a relativist I’m less sympathetic to this view since I’d rather like to hope that solving the information paradox doesn’t require giving up Einstein’s equivalence principle (upon which Einstein’s theory of General Relativity is based). Nonetheless, if this is true, anything entering the black hole horizon would be immediately destroyed by this “firewall”.

*The reason that this happens is rather involved, and, most elegantly, requires understanding the path-integral formulation of quantum mechanical states. At the very least, what one can do is to see that the creation/annihilation operators acting on the vacuum state in Minkowski space correspond to those in a thermal state from the point of view of an observer at a fixed distance outside the black hole (and therefore a uniformally accelerating observer). This latter approach is rather ugly but tractable with some work. This is all a consequence of the fact that, in general, the ontology of quantum mechanical states are dependent on a choice of coordinates. In other words, what one observer calls a “vacuum state”, another observer (with a different coordinate system) might call an “excited state”. Indeed, the example at hand is precisely an instance of this.

Explanation:

The short summary of the concept of dark matter and dark energy is given here.

An estimated 85% of the universe's mass is assumed to be made up of dark matter, a hypothetical type of stuff.  Because it does not appear to interact with the electromagnetic field—that is, it does not absorb, reflect, or emit electromagnetic radiation—dark matter is referred to as being "dark," making it challenging to detect.

Numerous astrophysical observations support the existence of dark matter, including gravitational effects that cannot be described by the gravity theories currently in use without the presence of more matter than can be observed. Because of this, the majority of scientists believe that dark matter is prevalent in the universe and has significantly influenced both its structure and evolution.

Dark energy is an undiscovered type of energy that has the largest effects on the cosmos according to physical cosmology and astronomy. Supernova observations provided the first direct proof for its existence by demonstrating that the cosmos is expanding faster than ever rather than at a fixed rate. It is necessary to understand the universe's origins and initial elements in order to comprehend its evolution.

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The longer a wire is, the higher the resistance will be.

Neither length (L) nor cross-sectional area (A) influences Rh. It totally depends on the type of material used to make the wire. It is somewhat dependent on temperature variation, but this variation is negligible and is disregarded for most purposes.

Let Resistivity be Rh and Resistance be R.

L for the wire's length and A for its cross-section.

This formula relates to resistance.

R=Rh*L/A

Rh, L, and A are the three parameters that determine resistance.

The resistance increases as the wire lengthen.

Less resistance is there the larger the cross-sectional area. (Put A in the denominator; the numerator is L.)

Only take into account the Length variable. As was previously said, shorter wires have lower resistance and vice versa.

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1. The fraction of the neutron's kinetic energy transferred to the atomic nucleus is 8% or 0.08.

2. The final kinetic energy of the neutron KE₂ (neutron), is 6.42 x 10⁻¹³ J.

Kinetic energy is the energy possessed by a body by virtue of its motion.

The fraction of the neutron's kinetic energy that is transferred to the atomic nucleus is calculated as follows:

The final velocity of the atom is found using the principle of conservation of linear momentum.

initial momentum of the neutron = final momentum of the atom

m₁u₁  = m₂u₂

where;

m₁ = mass of the neutron

u₁ = initial velocity of the neutron

m₂ = mass of the atomic nucleus

u₂ = final velocity of the atomic nucleus

m₂ = 112.4 m₁

u₂  = m₁u₁ / m₂

u₂  = m₁u₁ / (12.4 m₁)

u₂  = 0.08 u₁

The initial kinetic energy of the neutron is determined as follows:

KE₁ = ¹/₂m₁u₁²

The final kinetic energy of the atomic nucleus is determined as follows:

KE₂ =  ¹/₂m₂u₂²

KE₂ =  ¹/₂(12.4 m₁)(0.08u₁)²

KE₂  = 0.08 (¹/₂m₁u₁²)

KE₂ = 0.08 (KE₁)

The fraction of the neutron's kinetic energy transferred to the atomic nucleus is determined as follows:

Fraction = 0.08 (KE₂) / KE₁

Fraction= 0.08

Fraction = 8 %

The final kinetic energy of the neutron is calculated as follows;

KE₂ (neutron) = (1 - 0.08) x (6.97 x 10⁻¹³ J)

KE₂ (neutron) = 6.42 x 10⁻¹³ J

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Answer:

Explanation:

A = 6i - 4j + 8k

B = 2i + 4j - 14k

A + B = (6+2)i + (-4+4)j + (8-14)k

A + B = 8i +0j -6k

A + B = 8i - 6k

(1/2)(A + B) = 4i - 3k

d = √ (4² + 3²) = 5

e = (4/5)i -(3/4)j

Answer:

Explanation:

Given:

F = Fₓ·i + Fy·j

S = Sₓ·i + Sy·j

Fₓ = 3 N

Fy = - 2 N

Sₓ = 5 m

Sy = 2 m

_________

A - ?   - Work

α - ? - Angle

F = 3·i - 2·j

S = 5·i + 2·j

The work is numerically equal to the scalar product of the displacement force

A = (F·S) = 3·5 + (-2)·2 = 15 - 4 = 11 J

Modules:

| F | = √ (3² + (-2)² ) = √ (9 + 4) = √ 13

| S | = √ (5² + 2² ) = √ (25 + 4) = √ 29

Angle:

cos α = (F·S) / ( |F| · |S| ) = 11 / ( √13 · √29) ≈ 0,5665

α ≈ 55.5°

The acceleration of the car is 57.04 [tex]m/s^{2}[/tex]

It is given that a sports car accelerates from rest to 103 km/h in 6.5 s.

So, Initial velocity (u) = 0 m/s   { since the car starts from rest }

Final velocity (v) = 103 km/hr = 103 * 18/5 m/s = 370.8 m/s

Now, the time taken (t) = 6.5 seconds.

Acceleration is the measure of how fast the object's velocity changes with respect to time.

Now, Acceleration (a) = Change in velocity/Time taken

So,

[tex]a = \frac{v -u}{t}[/tex]

Putting all the values, we get

[tex]a = \frac{370.8 - 0}{6.5} =\frac{370.8}{6.5} = 57.04[/tex]

Hence, the acceleration of the car is 57.04 [tex]m/s^{2}[/tex]

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You are travelling towards the back of the bus. If your walking speed is slower than the speed of the moving bus (which it generally is), your motion relative to a point on the ground will be travelling in the same direction as the moving bus, but at a slower rate.

What is speed?

Speed can be defined as the distance travelled by an object in relation to the time it takes to travel that distance. In other words, it measures how rapidly an object moves but does not provide direction. When we get direction with speed, we term it "velocity."

The SI unit system is most typically used to express speed. Speed is measured in metres per second, or m/s, because distance is measured in metres and time is recorded in seconds.

Speed is measured in metres per second (ms-1) and is symbolised by the letter "s."

The distance travelled is denoted by d and measured in meters (m).

The distance travelled per unit of time is referred to as speed. It is the rate at which an object moves. The magnitude of the velocity vector is represented by the scalar quantity speed. It lacks a sense of direction. A higher speed indicates that an object is travelling faster. Lower speed indicates that it is travelling more slowly. It has zero speed if it is not moving at all.

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Answer: C

Explanation:

Answer:

Prokaryotes such as E. coli, Archaea as well as eukaryotes such as euglena reproduce through binary fission.

Explanation:

Answer:

Prokaryotes such as E. coli, Archaea as well as eukaryotes such as euglena reproduce through binary fission.

Explanation:

If the absolute temperature of an ideal diatomic gas is quadrupled, then, the average speed of molecules is doubled.

[tex][V_avg=\sqrt(8k *4T)/\pi m],[V_avg=2\sqrt(8kT) \pi m],[V_avg=2v_avg][/tex]

Speed (often abbreviated as "s") is a scalar variable that describes how much an object's location changes over time or how much it changes per unit of time. The average speed of an item in a period of time is the distance traveled by the object divided by the duration of the period.

Time divided by distance are the speed-related metrics.. The meter per second (m/s) is the SI unit of speed, while the kilometer per hour (kph) is the most often used unit of speed in daily life.

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When an object transported from earth to the moon, its mass remains same but weight becomes 1/6th that on earth..

In physics, mass is a quantitative measurement of inertia, a basic characteristic of all matter.

It essentially refers to a body of matter's resistance to changing its speed or location in response to the application of a force. The change caused by an applied force is smaller the more mass a body has.

The amount of a body's weight indicates how much gravity is pulling on it. Weight is calculated using the method w = mg. Since weight is a force, it has the same SI unit as a force, which is the Newton (N).

So, mass of an object is intrinsic property of the object - it remains same in earth and in moon. But weight is the amount of gravitational  attraction force. Moon has nearly 1/6th gravitational attraction field. So, the weight of the object on moon will be 1/6th that on earth.

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The coefficeint of static friction, given that the 12 Kg sled is lying on the hill with an incline of 21 degrees is 0.38

We know that the coefficient of static friction is related to frictional force according to the following formula:

Frictional force (N) = coefficient of friction (μ) × normal reaction (N)

F = μN

μ = F / N

For inclined plane, we have:

F = mgSineθ

N = mgCosθ

Thus,

μ = mgSineθ / mgCosθ

Recall

Sineθ / Cosθ = Tanθ

μ = Tanθ

Where

Now, we shall determine the coefficient of static friction as follow:

μ = Tanθ

μ = Tan21

μ = 0.38

Thus, the coefficient of static friction is 0.38

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