I need help with this physics question.

Answer: hello  below is the missing part of your question

A mass m = 10 kg rests on a frictionless table and accelerated by a spring with spring constant k = 5029 N/m. The floor is frictionless except for a rough patch. For this rough path, the coefficient of friction is μk = 0.49. The mass leaves the spring at a speed v = 3.4 m/s.

answer

x = 0.0962 m

Explanation:

First step :

Determine the length of the rough patch/spot

F = Uₓ (mg)

and  w = F.d = Uₓ (mg)  * d

hence;

d( length of rough patch) = w / Uₓ (mg) = 46.55 / (0.49 * 10 * 9.8) = 0.9694 m

next :

work done on unstretched spring length

Given that block travels halfway i.e. d = 0.9694 / 2 = 0.4847 m

w' = Uₓ (mg)  * d

    = 0.49 * 10 * 9.81 * 0.4847 = 23.27 J

also given that the Elastic energy of spring = work done ( w')

1/2 * kx^2 = 23.27 J

x = [tex]\sqrt{\frac{2*23.27}{5029} }[/tex]  = 0.0962 m

Explanation:

The period T of a simple pendulum is given by

[tex]T = 2 \pi \sqrt{\dfrac{l}{g}}[/tex]

Doubling the length of the pendulum gives us a new period T'

[tex]T' = 2 \pi \sqrt{\dfrac{l'}{g}} = 2 \pi \sqrt{\dfrac{2l}{g}}[/tex]

[tex]\:\:\:\:\:\:\:= \sqrt{2} \left(2 \pi \sqrt{\dfrac{l}{g}} \right)[/tex]

[tex]\:\:\:\:\:\:\:= \sqrt{2}\:T = \sqrt{2}(3.5\:\text{s})= 4.95\:\text{s}[/tex]

Answer:

A) 8GMm/d^2

Explanation:

We are given that

[tex]m_1=M[/tex]

[tex]m_2=3M[/tex]

[tex]m_3=m[/tex]

Distance between m1 and m2=d

Distance of object of mass m from m1 and m2=d/2

Gravitational force formula

[tex]F=\frac{Gm_1m_2}{r^2}[/tex]

Using the formula

Force acting between m and M is given by

[tex]F_1=\frac{GmM}{d^2/4}[/tex]

Force acting between m and 3M is given by

[tex]F_2=\frac{Gm(3M)}{d^2/4}[/tex]

Now, net force acting on  object of mass is given by

[tex]F=F_2-F_1[/tex]

[tex]F=\frac{Gm(3M)}{d^2/4}-\frac{GmM}{d^2/4}[/tex]

[tex]F=\frac{12GmM}{d^2}-\frac{4GmM}{d^2}[/tex]

[tex]F=\frac{12GmM-4GmM}{d^2}[/tex]

[tex]F=\frac{8GmM}{d^2}[/tex]

Hence, the magnitude of the force on the object with mass m=[tex]\frac{8GmM}{d^2}[/tex]

Option A is correct.

(b) Use Newton's second law. The net forces on block M are

• ∑ F (horizontal) = T - f = Ma … … … [1]

• ∑ F (vertical) = n - Mg = 0 … … … [2]

where T is the magnitude of the tension, f is the mag. of kinetic friction between block M and the table, a is the acceleration of block M (but since both blocks are moving together, the smaller block m also shares this acceleration), and n is the mag. of the normal force between the block and the table.

Right away, we see n = Mg, and so f = µn = 0.2Mg.

The net force on block m is

• ∑ F = mg - T = ma … … … [3]

You can eliminate T and solve for a by adding [1] to [3] :

(T - 0.2Mg) + (mg - T ) = Ma + ma

(m - 0.2M) g = (M + m) a

a = (10 kg - 0.2 (20 kg)) (9.8 m/s²) / (10 kg + 20 kg)

a = 1.96 m/s²

We can get the tension from [3] :

T = m (g - a)

T = (10 kg) (9.8 m/s² - 1.96 m/s²)

T = 78.4 N

(c/d) No time duration seems to be specified, so I'll just assume some time t before block M reaches the edge of the table (whatever that time might be), after which either block would move the same distance of

1/2 (1.96 m/s²) t

(e) Assuming block M starts from rest, its velocity at time t is

(1.96 m/s²) t

(f) After t = 1 s, block M reaches a speed of 1.96 m/s. When the string is cut, the tension force vanishes and the block slows down due to friction. By Newton's second law, we have

∑ F = -f = Ma

The effect of friction is constant, so that f = 0.2Mg as before, and

-0.2Mg = Ma

a = -0.2g

a = -1.96 m/s²

Then block M slides a distance x such that

0² - (1.96 m/s²) = 2 (-1.96 m/s²) x

x = (1.96 m/s²) /  (2 (1.96 m/s²))

x = 0.5 m

(I don't quite understand what is being asked by the part that says "calculate the time taken to contact block M and pulley" …)

Meanwhile, block m would be in free fall, so after 1 s it would fall a distance

x = 1/2 (-9.8 m/s²) (1 s)

x = 4.9 m

Assuming no heat lost to the surrounding,

-5⁰C ice → 0⁰C ice

Specific heat capacity of ice = 2.0 x 10³ J/kg/⁰C

Q = mc∆θ

Q = 5(2.0 x 10³) x (0-(-5))

Q = 50000J

0⁰C ice → 0⁰C water

Specific latent heat of fusion of ice = 3.34 x 10⁵J/kg

Q = mLf

Q = 5(3.34 x 10⁵)

Q = 1670000J

0⁰C water → 100⁰C water

Specific heat capacity of water = 4.2 x 10³ J/kg/⁰C

Q = mc∆θ

Q = 5(4.2 x 10³) x (100-0)

Q = 2100000J

100⁰C water → 100⁰C steam

Specific latent heat of vaporization of water = 2.26 x 10⁶ J/kg

Q = mLv

Q = 5(2.26 x 10⁶)

Q = 11300000J

Total amount of heat required

= 50000 + 1670000 + 2100000 + 11300000

= 15120000J

Answer: The correct statement is:

--> The voltage across each resistor is the same.

Explanation:

RESISTORS are defined as the components of an electric circuit which are capable of creating resistance to the file of electric current in the circuit. They work by converting electrical energy into heat, which is dissipated into the air. These resistors can be divided into two according to their arrangements in the electric cell. It include:

--> Resistors in parallel and

--> Resistors in series

RESISTORS are said to be in parallel when two or more resistance or conductors are connected to common terminals so that the potential difference ( voltage) across each conductor IS THE SAME but with different current flow through each of them. Also, Individual resistances diminish to equal a smaller total resistance rather than add to make the total.

Answer:

1 eV = 1.60 * 10^-19 J      work done in accelerating electron thru 1 V

KE (total energy) = 1350 ^ 1 eV     (note proton goes from +  to -)

KE = 1.60 * 10^-19 * 1350 = 2.16 * 10^-16 Joules

1/2 m v^2 = KE = 2.16 * 10^-16 J

v^2 = 4.32 * 10E-16 / 1.67 * 10-27 = 2.59 * 10^11

v = 5.09 * 10^5 m/s

The proton's kinetic energy, in joules is  2.16 *[tex]10^{-16}[/tex] J. The proton's velocity is 5.09 * [tex]10^{5}[/tex]m/s.

When an item is moving, its velocity is the rate at which its direction is changing as seen from a certain point of view and as measured by a specific unit of time.

Uniform motion an object is said to have uniform motion when object cover equal distance in equal interval of time within exact fixed direction. For a body in uniform motion, the magnitude of its velocity remains constant over time.

1 eV = 1.60 * [tex]10^{-19} J[/tex]     work done in accelerating electron throw 1 V

K.E (total energy) = 1350 ^ 1 eV     (note proton goes from +  to -)

K.E = 1.60 * [tex]10^{-19}[/tex]J * 1350 = 2.16 * [tex]10^{-16}[/tex] Joules

1/2 m v² = KE = 2.16 *[tex]10^{-16}[/tex] J

Velocity of proton is,

v² = 4.32 * 10[tex]e^{-16}[/tex] / 1.67 * [tex]10{-27}[/tex] = 2.59 * [tex]10^{11}[/tex]

v = 5.09 * [tex]10^{5}[/tex]m/s

The proton's kinetic energy, in joules is  2.16 *[tex]10^{-16}[/tex] J. The proton's velocity is 5.09 * [tex]10^{5}[/tex]m/s.

To learn more about velocity refer the link:

brainly.com/question/18084516  

#SPJ2

Explanation:

static equilibrium means its on the floor or something

so slightly greater than 98 newtons in the upward direction

Answer:

[tex]I_2=30.9A[/tex]

Explanation:

From the question we are told that:

Wire segment [tex]l_s=2.9m[/tex]

Initial Current [tex]I_1=1400A[/tex]

Force [tex]F=2.00N[/tex]

Distance of Wire [tex]d=4.50cm=>0.0450m[/tex]

Generally the equation for Force is mathematically given by

 [tex]F=\frac{\mu_0 * I_1*I_2*l_s}{2 \pi *r}[/tex]

 [tex]F=\frac{4 \pi*10^{-7} *1400 I*I_2*2.9}{2 \pi *0.0450}[/tex]

 [tex]I_2=\frac{22.5*10^-2}{2*10^{-7}*1400*2.6}[/tex]

 [tex]I_2=30.9A[/tex]

Answer:

The electron’s wavefunction has at least one node (i.e., at least one place in space where it goes to zero).

Explanation:

We know that the p-orbitals have nodes. A node is a region where the probability of finding an electron goes down to zero.

P orbitals are oriented along the x,y,z Cartesian axes and are known to have angular nodes along the axes.

Hence, if an electron in a hydrogen atom is in a p state, the electron’s wavefunction has at least one node

Answer:

Due to the applied filed the electrons move in a particular direction.

Explanation:

Initially when the switch is off, the free electrons move here and there in any random directions in the conductor with the random speeds called thermal velocity.  So, tat the net flow is almost zero.

When the battery is connected is switch is ON, the random motion of the electrons aligned in a particular direction due to the force applied by the electric filed, so the net flow is not zero it increases and thus the current flow.

Answer:

978.6084 Newton

Explanation:

Given the following data;

To find the weight in Newtown;

Conversion:

1 lb = 4.448220 N

220 lb = 220 * 4.448220 = 978.6084 Newton

220 lb = 978.6084 Newton

Therefore, the weight of the astronaut in Newton is 978.6084.

Weight can be defined as the force acting on a body or an object as a result of gravity.

Mathematically, the weight of an object is given by the formula;

Weight = mg

Where;

Note:

What

is the the improvetanse of water in our body and how do

Or

open Meet and enter this code

Answer:

Or open Meet and enter this code to the source of definition and how to use it to make

Answer:

Acosθ

Explanation:

The x-component of a vector is defined as :

Magnitude * cosine of the angle

Maginitude * cosθ

The magnitude is represented as A

Hence, horizontal, x - component of the vector is :

Acosθ

Furthermore,

The y-component is taken as the sin of the of the angle multiplied by the magnitude

Vertical, y component : Asinθ

I assume the blocks are pushed together at constant speed, and it's not so important but I'll also assume it's the smaller block being pushed up against the larger one. (The opposite arrangement works out much the same way.)

Consider the forces acting on either block. Let the direction in which the blocks are being pushed by the positive direction.

The 2.0-kg block feels

• the downward pull of its own weight, (2.0 kg) g

• the upward normal force of the surface, magnitude n₁

• kinetic friction, mag. f₁ = 0.30n₁, pointing in the negative horizontal direction

• the contact force of the larger block, mag. c₁, also pointing in the negative horizontal direction

• the applied force, mag. F, pointing in the positive horizontal direction

Meanwhile the 3.0-kg block feels

• its own weight, (3.0 kg) g, pointing downward

• normal force, mag. n₂, pointing upward

• kinetic friction, mag. f₂ = 0.30n₂, pointing in the negative horizontal direction

• contact force from the smaller block, mag. c₂, pointing in the positive horizontal direction (this is the force that is causing the larger block to move)

Notice the contact forces form an action-reaction pair, so that c₁ = c₂, so we only need to find one of these, and we can get it right away from the net forces acting on the 3.0-kg block in the vertical and horizontal directions:

• net vertical force:

n₂ - (3.0 kg) g = 0   ==>   n₂ = (3.0 kg) g   ==>   f₂ = 0.30 (3.0 kg) g

• net horizontal force:

c₂ - f₂ = 0   ==>   c₂ = 0.30 (3.0 kg) g ≈ 8.8 N

Answer: I beleive A

Explanation:

Answer:

A

Explanation:

We can see the light being reflected off the mirror.

Answer:

a=10m/s^2

Explanation:

acceleration= final velocity+ initial velocity/time taken

v-u/t=a

100-0/5=a

100/5=a

a=20m/s^2

case2

100-0/10=a

100/10=a

a=10m/s^2

Don't forget to write the units.

Hope this helps

please mark me as brainliest.

Answer:

running away and launching the anchor that will give a greater speed towards the dock v₄.

Explanation:

To try to bring the boat closer to the dock, several cases can be carried out.

* move inside the ship so that the center of mass changes and since moving away you have a speed v, the ship will approach the dock at a speed v₂,

* Throw the anchor in the opposite direction to the dock so that using the conservation of the moment the boat moves towards it, it moves at a speed v₃

* A combination of the two processes running away and launching the anchor that will give a greater speed towards the dock v₄.

In all cases, the friction must be zero.

All other movements move the ship away from the dock

Answer:

answer is 18.58 because

Answer:

My answer is d.25.1 because

Answer:

7.5 kg

Explanation:

We are given that

[tex]m_1=10 kg[/tex]

Length of plank, l=3 m

Distance of fulcrum from one end of the plank=1 m

[tex]m_2=20 kg[/tex]

We have to find the mass must be on the other end if the plank remains balanced.

Let m be the mass must be on the other end if the plank remains balanced.

In balance condition

[tex]20\times 1=10\times (1.5-1)+m\times (1.5+0.5)[/tex]

[tex]20=10(0.5)+2m[/tex]

[tex]20=5+2m[/tex]

[tex]2m=20-5=15[/tex]

[tex]\implies m=\frac{15}{2}[/tex]

[tex]m=7.5 kg[/tex]

Hence, mass 7.5 kg   must be on the other end if the plank remains balanced.

Answer:

The mass at the other end is 7.5 kg.

Explanation:

Let the mass is m.

Take the moments about the fulcrum.

20 x 1 = 10 x 0.5 + m x 2

20 = 5 + 2 m

2 m = 15

m = 7.5 kg

Answer:

R/2

Explanation:

The potential at a distance r is given by :

[tex]V=\dfrac{kq}{r}[/tex]

Where

k is electrostatic constant

q is the charge

The potential (relative to infinity) due to a point charge is V at a distance R from this charge. So,

[tex]\dfrac{V_1}{V_2}=\dfrac{r_2}{r_1}[/tex]

Put all the values,

[tex]\dfrac{V}{2V}=\dfrac{r_2}{R}\\\\\dfrac{1}{2}=\dfrac{r_2}{R}\\\\r_2=\dfrac{R}{2}[/tex]

Answer:

No, it will not affect the results.

Explanation:

For elastic collisions in an isolated system, when a collision occurs, it means that the systems objects total momentum will be conserved under the condition that there will be no net external forces that act upon the objects.

What that means is that if the pucks start spinning after the collision, we are not told that there was any net external force acting on the puck and thus momentum will be conserved because momentum before collision will be equal to the momentum after the collision.

Answer:

A point on the edge of the wheel will travel 199.563 radians at the given time.

Explanation:

Given;

initial angular velocity of the wheel; [tex]\omega _i = 245 \ rev/\min = 245\ \frac{rev}{\min} \times \frac{2\pi}{1\ rev} \times \frac{1 \ \min}{60 \ s} = 25.66 \ rad/s[/tex]

final angular velocity of the wheel;

[tex]\omega _f = 380 \ rev/\min = 380 \ \frac{rev}{\min} \times \frac{2\pi}{1\ rev} \times \frac{1 \ \min}{60 \ s} = 39.80 \ rad/s[/tex]

radius of the wheel, d/2 = (30 cm ) / 2 = 15 cm = 0.15 m

time of motion, t = 6.1 s

The angular distance traveled by the edge of the wheel is calculated as;

[tex]\theta = (\frac{\omega_f + \omega_i}{2} )t\\\\\theta = (\frac{39.8 + 25.66}{2} )\times 6.1\\\\\theta = 199.653 \ radian[/tex]

Therefore, a point on the edge of the wheel will travel 199.563 radians at the given time.

Answer:

 B = 1.1413 10⁻² T

Explanation:

We use energy concepts to calculate the proton velocity

starting point. When entering the electric field

        Em₀ = U = q V

final point. Right out of the electric field

        em_f = K = ½ m v²

energy is conserved

       Em₀ = Em_f

       q V = ½ m v²

       v = [tex]\sqrt{2qV/m}[/tex]

we calculate

       v = [tex]\sqrt{\frac{ 2 \ 1.6 \ 10^{-19} \ 300}{1.67 \ 0^{-27}} }[/tex]

       v = [tex]\sqrt{632.3353 \ 10^8}[/tex]

       v = 25.15 10⁴ m / s

now enters the region with magnetic field, so it is subjected to a magnetic force

        F = m a

the force is

       F = q v x B

as the velocity is perpendicular to the magnetic field

       F = q v B

acceleration is centripetal

       a = v² / r

we substitute

       qvB =1/2  m v² / r

       B =  v[tex]\frac{m v}{2 q r}[/tex]

we calculate

       B = [tex]\frac{1.67 \ 10^{-27} 25.15 \ 10^4 }{1.6 \ 10^{-19} 0.23}[/tex]

       B = 1.1413 10⁻² T

Answer: a rippling fountain

Explanation: diffuse reflection happens on rough surfaces, so using the process of elimination, that leaves us with b, a rippling fountain (I also just took this test I'm pretty sure I'm right)

Answer:

a)     a = 27.44 m / s²,  b) a = 5.39 m / s², c)  a = 156.8 m / s², cabinet maximum acceleration does not change

Explanation:

a) In this exercise the wheels of the truck rotate to provide acceleration, but the contact point between the ground and the 2 wheels remains fixed, therefore the coefficient of friction for this point is static.

Let's apply Newton's second law

we set a regency hiss where the x axis is in the direction of movement of the truck

Y axis y

        N- W = 0

        N = W = m g

X axis

       2fr = m a

the expression for the friction force is

      fr = μ N

      fr = μ m g

we substitute

      2 μ m g = m /2   a

     a = 4 μ g

      a = 4 0.7 9.8

      a = 27.44 m / s²

b) let's look for the maximum acceleration that can be applied to the cabinet

       fr = m a

       μ N = ma

       μ m g = m a

       a = μ g

       a = 0.55  9.8

       a = 5.39 m / s²

as the acceleration of the platform is greater than this acceleration the cabinet must slip

c) the friction force is in the four wheels as well

With when the truck had two-wheel Thracian the weight of distributed evenly between the wheels, in this case with 4-wheel Thracian the weight must be distributed among all

applying Newton's second law

         4 fr = (m/4) a

         16 mg = (m) a

         a = 16 g

         a = 16 9.8

         a = 156.8 m / s²

cabinet maximum acceleration does not change