a fixed amount of ideal gas is held in a rigid container that expands negligibly when heated. at 20 the gas pressure is p. if we added enough heat to increase the temperature from 20 to 40, the pressure will be

Question:

A 10 gauge copper wire carries a current of 15 A. Assuming one free electron per copper atom, calculate the drift velocity of the electrons. (The cross-sectional area of a 10-gauge wire is 5.261 mm².)

Answer:

3.22 x 10⁻⁴ m/s

Explanation:

The drift velocity (v) of the electrons in a wire (copper wire in this case) carrying current (I) is given by;

v = [tex]\frac{I}{nqA}[/tex]

Where;

n = number of free electrons per cubic meter

q =  electron charge

A =  cross-sectional area of the wire

First let's calculate the number of free electrons per cubic meter (n)

Known constants:

density of copper, ρ = 8.95 x 10³kg/m³

molar mass of copper, M = 63.5 x 10⁻³kg/mol

Avogadro's number, Nₐ = 6.02 x 10²³ particles/mol

But;

The number of copper atoms, N, per cubic meter is given by;

N = (Nₐ x ρ / M)          -------------(ii)

Substitute the values of Nₐ, ρ and M into equation (ii) as follows;

N = (6.02 x 10²³ x 8.95 x 10³) / 63.5 x 10⁻³

N = 8.49 x 10²⁸ atom/m³

Since there is one free electron per copper atom, the number of free electrons per cubic meter is simply;

n = 8.49 x 10²⁸ electrons/m³

Now let's calculate the drift electron

Known values from question:

A = 5.261 mm² = 5.261 x 10⁻⁶m²

I = 23A

q = 1.6 x 10⁻¹⁹C

Substitute these values into equation (i) as follows;

v = [tex]\frac{I}{nqA}[/tex]

v = [tex]\frac{23}{8.49*10^{28} * 1.6 *10^{-19} * 5.261*10^{-6}}[/tex]

v = 3.22 x 10⁻⁴ m/s

Therefore, the drift electron is 3.22 x 10⁻⁴ m/s

Answer:

The new angular speed is [tex]w = 6 \ rev/s[/tex]

Explanation:

From the  question we are told that

      The angular velocity of the spin is  [tex]w_o = 3 \ rev/s[/tex]

       The  original moment of inertia is  [tex]I_o[/tex]

        The new moment of inertia is  [tex]I =\frac{I_o}{2}[/tex]    

Generally angular momentum is mathematically represented as

      [tex]L = I * w[/tex]

Now according to the law of conservation of momentum, the initial momentum is equal to the final momentum hence the angular momentum is constant so

         [tex]I * w = constant[/tex]

=>       [tex]I_o * w _o = I * w[/tex]

where w is the new angular speed  

  So  

          [tex]I_o * 3 = \frac{I_o}{2} * w[/tex]

=>        [tex]w = \frac{3 * I_o}{\frac{I_o}{2} }[/tex]

=>         [tex]w = 6 \ rev/s[/tex]

Answer:

95.5 m

Explanation:

The displacement is the position of the ending point relative to the starting point.

In this case, the magnitude of the displacement is the diameter of the circular track.

d = 300 m / π

d ≈ 95.5 m

Answer:

Acceleration of the asteroid relative to the spacecraft = 2ε[tex]E^{2}[/tex]A/m

Explanation:

The maximum electric field in the beam = 2E

cross-sectional area of beam = A

The intensity of an electromagnetic wave with electric field is

I = cε[tex]E_{0} ^{2}[/tex]/2

for [tex]E_{0}[/tex] = 2E

I = 2cε[tex]E^{2}[/tex]    ....equ 1

where

I is the intensity

c is the speed of light

ε is the permeability of free space

[tex]E_{0}[/tex]  is electric field

Radiation pressure of an electromagnetic wave on an absorbing surface is given as

P = I/c

substituting for I from above equ 1. we have

P = 2cε[tex]E^{2}[/tex]/c = 2ε[tex]E^{2}[/tex]    ....equ 2

Also, pressure P = F/A

therefore,

F = PA    ....equ 3

where

F is the force

P is pressure

A is cross-sectional area

substitute equ 2 into equ 3, we have

F = 2ε[tex]E^{2}[/tex]A

force on a body = mass x acceleration.

that is

F = ma

therefore,

a = F/m

acceleration of the asteroid will then be

a = 2ε[tex]E^{2}[/tex]A/m

where m is the mass of the asteroid.

Answer:

The two masses are 3.39 Kg and 1.75 Kg

Explanation:

The gravitational force of attraction between two bodies is given by the formula;

F = Gm₁m₂/d²

where G is the gravitational force constant = 6.67 * 10⁻¹¹ Nm²Kg⁻²

m₁ = mass of first object; m₂ = mass of second object; d = distance of separation between the objects

Further calculations are provided in the attachment below

Answer:

389 kg

Explanation:

The computation of mass is shown below:-

[tex]T = 2\pi \sqrt{\frac{m}{k} }[/tex]

Where K indicates spring constant

m indicates mass

For the new time period

[tex]T^' = 2\pi \sqrt{\frac{m'}{k} }[/tex]

Now, we will take 2 ratios of the time period

[tex]\frac{T}{T'} = \sqrt{\frac{m}{m'} }[/tex]

[tex]\frac{1.50}{2.00} = \sqrt{\frac{0.500}{m'} }[/tex]

[tex]0.5625 = \sqrt{\frac{0.500}{m'} }[/tex]

[tex]m' = \frac{0.500}{0.5625}[/tex]

= 0.889 kg

Since mass to be sum that is

= 0.889 - 0.500

0.389 kg

or

= 389 kg

Therefore for computing the mass we simply applied the above formula.

The mass added to the object to change the period to 2.00 s is 0.389 kg and this can be determined by using the formula of the time period.

Given :

A 0.500-kg mass suspended from a spring oscillates with a period of 1.50 s.

The formula of the time period is given by:

[tex]\rm T = 2\pi\sqrt{\dfrac{m}{K}}[/tex]   ---- (1)

where m is the mass and K is the spring constant.

The new time period is given by:

[tex]\rm T'=2\pi\sqrt{\dfrac{m'}{K}}[/tex]   ---- (2)

where m' is the total mass after the addition and K is the spring constant.

Now, divide equation (1) by equation (2).

[tex]\rm \dfrac{T}{T'}=\sqrt{\dfrac{m}{m'}}[/tex]

Now, substitute the known terms in the above expression.

[tex]\rm \dfrac{1.50}{2}=\sqrt{\dfrac{0.5}{m'}}[/tex]

Simplify the above expression in order to determine the value of m'.

[tex]\rm m'=\dfrac{0.5}{0.5625}[/tex]

m' = 0.889 Kg

Now, the mass added to the object to change the period to 2.00 s is given by:

m" = 0.889 - 0.500

m" = 0.389 Kg

For more information, refer to the link given below:

https://brainly.com/question/2144584

Answer:

c.they have opposite charges.

Explanation:

because the protons have a positive charge and the neutrons have no charge.

Answer:

τ = 0.26 N.m

Explanation:

First we find the moment of inertia of the wheel, by using the following formula:

I= mr²

where,

I = Moment of Inertia = ?

m = mass of wheel = 2.3 kg

r = radius of wheel = 50 cm/2 = 25 cm = 0.25 m

Therefore,

I = (2.3 kg)(0.25 m)²

I = 0.115 kg.m²

Now, we find the angular acceleration of the wheel:

α = (ωf - ωi)/t

where,

α = angular acceleration = ?

ωf = final angular velocity = (120 rpm)(2π rad/1 rev)(1 m/60 s) = 12.56 rad/s

ωi = Initial Angular Velocity = 0 rad/s

t = time = 5.5 s

Therefore,

α = (12.56 rad/s - 0 rad/s)/(5.5 s)

α = 2.28 rad/s²

Now, the torque is given as:

Torque = τ = Iα

τ = (0.115 kg.m²)(2.28 rad/s²)

τ = 0.26 N.m

Answer:

Yes.

Explanation:

Law of relativity in relation to light states that the speed of light in a vacuum does not depend on all the motion of the observers and that all motion must be defined relative to a frame of reference and that space and time are relative, rather than absolute concepts. This was formulated by Albert Einstein in 1905.

Light travels more slowly in gas than in air because it interacts with atoms of glass that made it way through it and the refractive index of glass is more than air. This does contradict the theory of relativity as the speed of lights travel slower in glass because it's motion is slow and it is not relative.

Answer: after 1.75 seconds

Explanation:

The only force acting on the ball is the gravitational force, so the acceleration will be:

a = -9.8 m/s^2

the velocity can be obtained by integrating over time:

v = -9.8m/s^2*t + v0

where v0 is the initial velocity; v0 = -7.95 m/s.

v = -9.8m/s^2*t - 7.95 m/s.

For the position we integrate again:

p = -4.9m/s^2*t^2 - 7.95 m/s*t + p0

where p0 is the initial position: p0 = 29m

p =  -4.9m/s^2*t^2 - 7.95 m/s*t + 29m

Now we want to find the time such that the position is equal to zero:

0 = -4.9m/s^2*t^2 - 7.95 m/s*t + 29m

Then we solve the Bhaskara's equation:

[tex]t = \frac{7.95 +- \sqrt{7.95^2 +4*4.9*29} }{-2*4.9} = \frac{7.95 +- 25.1}{9.8}[/tex]

Then the solutions are:

t = (7.95 + 25.1)/(-9.8) = -3.37s

t = (7.95 - 25.1)/(-9.8) = 1.75s

We need the positive time, then the correct answer is 1.75s

Answer:

The waiting time is  [tex]t_w = 3 .47 \ s[/tex]

Explanation:

From the question we are told that

       The  height of the hot air balloon above the ground is  [tex]d = 50 \ m[/tex]

         The distance of the balloon from the target is  [tex]l = 100 \ m[/tex]

        The  velocity of the balloon is  [tex]v = 15 \ m/s[/tex]

Generally the time it will take to reach the ground  is

          [tex]t = \sqrt{2 * \frac{d}{g} }[/tex]

substituting values

         [tex]t = \sqrt{2 * \frac{50}{9.8} }[/tex]

        [tex]t = 3.2 \ s[/tex]

The distance that is covered at time with the given velocity is mathematically evaluated as  

            [tex]z = v * t[/tex]

substituting values

           [tex]z = 15 * 3.2[/tex]

           [tex]z = 48 \ m[/tex]

This implies that for the balloon moving at a velocity (v) to hit the target  it must be dropped at this distance (z)

Now the distance the balloonist has to wait before dropping in order to hit the target is  

        [tex]A = d - z[/tex]

substituting values

      [tex]A = 100 - 48[/tex]

      [tex]A = 52 \ m[/tex]

This implies that the time the balloonist has to wait is  

      [tex]t_w = \frac{A}{v}[/tex]

substituting values

      [tex]t_w = \frac{52}{15}[/tex]

      [tex]t_w = 3 .47 \ s[/tex]

Answer:

It's more likely that the trailer is heavily loaded

Explanation:

Due to the fact that the frequency is proportional to the square root of the force constant and inversely proportional to the square root of the mass, it is very likely that the truck would be heavily loaded because the force constant would be the same whether the truck is empty or heavily loaded.

Answer:

The gray spheres is negatively charged while the red is positively charged

Explanation:

This is because theelectrophorus becomes less positive once it pulls some electrons away from the red sphere, but, the electrophorus is replaced on the slab and recharged by grounding it before it proceeds to charge the grey sphere, thereby giving it electrons and making it negatively charged

Answer:

The gray sphere has a positive charge and the red sphere has a positive charge.

Answer:

1ft per second

Explanation:

Physics is on my side!!!!!!!!!!

Answer:

Velocity.

Explanation:

Projectile motion is characterized as the motion that an object undergoes when it is thrown into the air and it is only exposed to acceleration due to gravity.

As per the question, 'any change in the initial velocity of the projectile(object having gravity as the only force) would lead to a change in the range as well as the maximum height of the projectile.' To illustrate numerically:

Horizontal range: As per expression:

R= ([tex]u^{2}[/tex]*sin2θ)/g

the range depending on the square of the initial velocity.

Maximum height: As per expression:

H= ([tex]u^{2}[/tex] * [tex]sin^{2}[/tex]θ )/2g

the maximum distance also depends upon square of the initial velocity.

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Answer:

a) Energy stored in the capacitor, [tex]E = 1.0125 *10^{-3} J[/tex]

b) Q = 45 µC

c) C' = 1.5 μF

d)  [tex]E = 6.75 *10^{-4} J[/tex]

Explanation:

Capacitance, C = 1 µF

Charge on the plates, Q = 45 µC

a) Energy stored in the capacitor is given by the formula:

[tex]E = \frac{Q^2}{2C} \\\\E = \frac{(45 * 10^{-6})^2}{2* 1* 10^{-6}}\\\\E = \frac{2025 * 10^{-6}}{2}\\\\E = 1012.5 *10^{-6}\\\\E = 1.0125 *10^{-3} J[/tex]

b) The charge on the plates of the capacitor will  not change

It will still remains, Q = 45 µC

c)  Electric field is non zero over (1-1/3) = 2/3 of d

From the relation V = Ed,

The voltage has changed by a factor of 2/3

Since the capacitance is given as C = Q/V  

The new capacitance with the conductor in place, C' = (3/2) C

C' = (3/2) * 1μF

C' = 1.5 μF

d) Energy stored in the capacitor with the conductor in place

[tex]E = \frac{Q^2}{2C} \\\\E = \frac{(45 * 10^{-6})^2}{2* 1.5* 10^{-6}}\\\\E = \frac{2025 * 10^{-6}}{3}\\\\E = 675 *10^{-6}\\\\E = 6.75 *10^{-4} J[/tex]

Complete Question

The distance between the objective and eyepiece lenses in a microscope is 19 cm . The objective lens has a focal length of 5.5 mm .

What eyepiece focal length will give the microscope an overall angular magnification of 300?

Answer:

The  eyepiece focal length is  [tex]f_e = 0.027 \ m[/tex]

Explanation:

From the question we are told that

    The focal length is  [tex]f_o = 5.5 \ mm = -0.0055 \ m[/tex]

This negative sign shows the the microscope is diverging light

     The  angular magnification is [tex]m = 300[/tex]

     The  distance between the objective and the eyepieces lenses is  [tex]Z = 19 \ cm = 0.19 \ m[/tex]

Generally the magnification is mathematically represented as

        [tex]m = [\frac{Z - f_e }{f_e}] [\frac{0.25}{f_0} ][/tex]

Where [tex]f_e[/tex] is the eyepiece focal length of the microscope

  Now  making [tex]f_e[/tex] the subject  of the formula

         [tex]f_e = \frac{Z}{1 - [\frac{M * f_o }{0.25}] }[/tex]

substituting values

        [tex]f_e = \frac{ 0.19 }{1 - [\frac{300 * -0.0055 }{0.25}] }[/tex]

         [tex]f_e = 0.027 \ m[/tex]

Answer:

-    Force on A = 0.870N

-    charge of the object B = q = 2.1 μC

    charge of the object A = 2q = 4.2 μC

-    a = 0.966 m/s^2

Explanation:

- In order to determine the force on the object A, you take into account the third Newton law, which states that the force experienced by A has the same magnitude of the force experienced by B, but with an opposite direction.

Then, the force on A is 0.870N

- In order to calculate the charge of both objects, you use the following formula:

[tex]F_e=k\frac{q_Aq_B}{r^2}[/tex]         (1)

k: Coulomb's constant = 8.98*10^9 Nm^2/C^2

r: distance between the objects = 30.0cm = 0.30m

A has twice the charge of B. If the charge of B is qB=q, then the charge of A is qA=2qB = 2q.

You replace the expression for qA and qB into the equation (1), solve for q, and replace the values of the parameters.

[tex]F_e=k\frac{(2q)(q)}{r^2}=2k\frac{q^2}{r^2}\\\\q=\sqrt{\frac{r^2Fe}{2k}}\\\\q=\sqrt{\frac{(0.30m)^2(0.870N)}{2(8.98*10^9Nm^2/C^2)}}=2.1*10^{-6}C\\\\q=2.1\mu C[/tex]

Then, you have:

charge of the object B = q = 2.1 μC

charge of the object A = 2q = 4.2 μC

- In order to calculate the acceleration of A, you use the second Newton law with the electric force, as follow:

[tex]F_e=ma\\\\a=\frac{F_e}{m}[/tex]

m: mass of the object A = 900g = 0.900kg

[tex]a=\frac{0.870N}{0.900kg}=0.966\frac{m}{s^2}[/tex]

The acceleration of A is 0.966m/s^2

Answer and Explanation:

There is no probability of obtaining such a circuit of closed track current carrying wire whose field of magnitude displays i.e.  [tex]B \alpha \frac{1}{r^2}[/tex]

The magnetic field is a volume of vectors

And [tex]\phi\ bds = 0[/tex]. This ensures isolated magnetic poles or magnetic charges would not exit

Therefore for a closed path,  we never received magnetic field that followed the [tex]B \alpha \frac{1}{r^2}[/tex] it is only for the simple current-carrying wire for both finite or infinite length.

Answer:

Explanation:

Check the diagram attached below for the diagram.

Let the weight of the rock be W and the mass of the meter stick be M. Note that the mass of the meter stick will be placed at the middle of the meter stick i.e at the 50cm mark

Using the principle of moment to calculate the weight of the rock. It states that the sum of clockwise moments is equal to the sum of anti clockwise moment.

Moment = Force * perpendicular distance

The meterstick acts in the clockwise direction while the rock acys in the anti clockwise direction

Clockwise moment = 1kg * 25 = 25kg/cm

Anticlockwise moment = W * 25cm = 25W kg/cm

Equating both moments of forces

25W = 25

W = 25/23

W = 1 kg

The mass of the rock is also 1 kg

Answer:

The rubber band will be stretched 0.02 m.

The work done in stretching is 0.11 J.

Explanation:

Force 1 = 44 N

extension of rubber band = 0.080 m

Force 2 = 11 N

extension = ?

According to Hooke's Law, force applied is proportional to the extension provided elastic limit is not extended.

F = ke

where k = constant of elasticity

e = extension of the material

F = force applied.

For the first case,

44 = 0.080K

K = 44/0.080 = 550 N/m

For the second situation involving the same rubber band

Force = 11 N

e = 550 N/m

11 = 550e

extension e = 11/550 = 0.02 m

The work done to stretch the rubber band this far is equal to the potential energy stored within the rubber due to the stretch. This is in line with energy conservation.

potential energy stored = [tex]\frac{1}{2}ke^{2}[/tex]

==> [tex]\frac{1}{2}* 550* 0.02^{2}[/tex] = 0.11 J

Answer:

The angular momentum of the crate is [tex]M_{C} V_{1} d[/tex]

Explanation:

mass of the crate = [tex]M_{C}[/tex]

speed of forklift = [tex]V_{1}[/tex]

The distance between the center of the mass and the point A = d

Recall that the angular moment is the moment of the momentum.

[tex]L = P*d[/tex]    ..... equ 1

where L is the angular momentum,

P is the momentum of the system,

d is the perpendicular distance between the crate and the point on the axis about which the momentum acts. It is equal to d from the image

Also, we know that the momentum P is the product of mass and velocity

P = mv      ....equ 2

in this case, the mass = [tex]M_{C}[/tex]

the velocity = [tex]V_{1}[/tex]

therefore, the momentum P = [tex]M_{C}[/tex][tex]V_{1}[/tex]

we substitute equation 2 into equation 1 to give

[tex]L = M_{C} V_{1} d[/tex]

Answer:

a)  v_correcaminos = 22.95 m / s ,  b)  x = 512.4 m ,

c) v = (45.83 i ^ -109.56 j ^) m / s

Explanation:

We can solve this exercise using the kinematics equations

a) Let's find the time or the coyote takes to reach the cliff, let's start by finding the speed on the cliff

         v² = v₀² + 2 a x

they tell us that the coyote starts from rest v₀ = 0 and its acceleration is a=15 m / s²

         v = √ (2 15 70)

         v = 45.83 m / s

with this value calculate the time it takes to arrive

        v = v₀ + a t

        t = v / a

        t = 45.83 / 15

        t = 3.05 s

having the distance to the cliff and the time, we can find the constant speed of the roadrunner

         v_ roadrunner = x / t

         v_correcaminos = 70 / 3,05

         v_correcaminos = 22.95 m / s

b) if the coyote leaves the cliff with the horizontal velocity v₀ₓ = 45.83 m / s, they ask how far it reaches.

Let's start by looking for the time to reach the cliff floor

            y = y₀ + [tex]v_{oy}[/tex] t - ½ g t²

in this case y = 0 and the height of the cliff is y₀ = 100 m

          0 = 100 + 45.83 t - ½ 9.8 t²

          t² - 9,353 t - 20,408 = 0

we solve the quadratic equation

         t = [9,353 ±√ (9,353² + 4 20,408)] / 2

         t = [9,353 ± 13] / 2

         t₁ = 11.18 s

        t₂ = -1.8 s

Since time must be a positive quantity, the answer is t = 11.18 s

we calculate the horizontal distance traveled

        x = v₀ₓ t

        x = 45.83 11.18

        x = 512.4 m

c) speed when it hits the ground

         vₓ = v₀ₓ = 45.83 m / s

we look for vertical speed

         v_{y} = [tex]v_{oy}[/tex] - gt

         v_{y} = 0 - 9.8 11.18

         v_{y} = - 109.56 m / s

         v = (45.83 i ^ -109.56 j ^) m / s

Answer:

Coulomb's law is:

[tex]F = \frac{1}{4*pi*e0} *(q1*q2)/r^2[/tex]

First, force has units of Newtons, the charges have units of Coulombs, and r, the distance, has units of meters, then, working only with the units we have:

N = (1/{e0})*C^2/m^2

then we have:

{e0} = C^2/(m^2*N)

And we know that N = kg*m/s^2

then the dimensions of e0 are:

{e0} = C^2*s^2/(m^3)

(current square per time square over cubed distance)

And knowing that a Faraday is:

F = C^2*S^2/m^2

The units of e0 are:

{e0} = F/m.

Answer:

 θ  = 180

Explanation:

When an electric dipole is placed in an electric field, there is a torque due to the electric force

           τ = p x E

by rotating the dipole there is a change in potential energy

        ΔU = ∫ τ dθ

        ΔU = p E (cos θ₂ - cos θ₁)

when the dipole starts from an angle to the equilibrium position for θ = 0

          ΔU = pE (cos θ  - cos 0)

           cos θ  = 1 + DU / pE)

let's apply this expression to our case, the change in potential energy is ΔU = -0.2J

let's calculate

          cos θ  = 1 -0.2 / 0.1

          cos θ  = -1

           θ  = 180

Answer:

A: An upward force balances the downward force of gravity on the skydiver.

Explanation:

on edge! hope this helps!!~ (⌒▽⌒)☆

Answer:

Since the angular acceleration of the objects will be proportional to the torque (due to gravity) acting on them and they will all experience the same torque their accelerations will be inversely proportional to their moments of inertia:

I disk = 1/2 M R^2

I sphere = 2/5 M R^2

I shell = 2/3 M R^2

Thus the sphere will experience the greatest angular acceleration and reach the bottom first, and then be followed by the disk and the shell.

By conservation of energy they will all have the same kinetic energy when they reach the bottom of the ramp.

(a) The ranking of the objects in order of how they will finish the race is

solid sphere > disk > hollow spherical shell

(b) The ranking of the objects in order of kinetic energy is

solid sphere > disk > hollow spherical shell

The moment of inertia of each object is calculated as follows;

The angular momentum of the objects is calculated as follows;

[tex]L =I \omega \\\\\omega = \frac{L}{I}[/tex]

The object with the least moment of inertia is will have the highest speed.

The ranking of the objects in order of how they will finish the race;

solid sphere > disk > hollow

The kinetic energy of the objects is calculated as follows;

[tex]K.E = \frac{1}{2} I \omega ^2[/tex]

The ranking of the objects in order of kinetic energy;

solid sphere > disk > hollow

Learn more here:https://brainly.com/question/15076457

Answer:

Explanation:

Given data

length of cube= 0.2015 m

density = 19300 kg/m^3.

But the volume of cube is given as [tex]l*l*l= l^3[/tex]

[tex]volume -of- cube= 0.2015*0.2015*0.2015= 0.00818 m^3[/tex]

The density is expressed as = mass/volume

[tex]mass=19300*0.00818= 157.87m^3[/tex]

Answer:

335.79cm/s

Explanation:

When a transverse wave of wavelength λ is produced during the vibration of a wire, the frequency(f), and the speed(v) of the wave pulses are related to the wavelength as follows;

v = fλ        ------------------(ii)

From the question;

f = 6.3Hz

λ = 53.3cm

Substitute these values into equation (i) as follows;

v = 6.3 x 53.3

v = 335.79cm/s

Therefore, the speed of the wave pulses along the wire is 335.79cm/s

Answer:

[tex]v_{2}=3.5 m/s[/tex]

Explanation:

Using the conservation of energy we have:

[tex]\frac{1}{2}mv^{2}=mgh[/tex]

Let's solve it for v:

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

So the speed at the lowest point is [tex]v=7 m/s[/tex]

Now, using the conservation of momentum we have:

[tex]m_{1}v_{1}=m_{2}v_{2}[/tex]

[tex]v_{2}=\frac{1*7}{2}[/tex]

Therefore the speed of the block after the collision is [tex]v_{2}=3.5 m/s[/tex]

I hope it helps you!