A 300 kg block of dimensions 1.5 m × 1.0 m × 0.5 m lays on the table with its largest face.

Calculate:



Area of the largest face

Answer:

i₀ = V / R_i

Explanation:

For this exercise we use Ohm's law

         V = i R

          i = V / R

the equivalent resistance for

         [tex]\frac{1}{R_{eq}}[/tex] =  ∑ [tex]\frac{1}{R_i}[/tex]

if all the bulbs have the same resistance, there are N bulbs

         [tex]\frac{1}{ R_{eq}} = \frac{N}{R_i}[/tex]

         R_{eq} = R_i / N

we substitute

         i = N V / Ri

where i is the total current that passes through the parallel, the current in a branch is

         i₀ = i / N

         i₀ = V / R_i

Answer:

The speed of the car, v = 21.69 m/s

Explanation:

The diameter is  = 48.01 m

Therefore, the radius of the loop R = 24.005 m

Weight at the top is n = mv^2/R - mg

Since the apparent weight is equal to the real weight.

So, mv^2/R - mg = mg

v = √(2Rg)

v = √[2(24.005 m)(9.8 m/s^2)]

The speed of the car, v = 21.69 m/s

Answer:

The speed is 15.34 m/s.

Explanation:

Diameter, d = 48.01 m

Radius, R = 24.005 m

Let the speed is v and the mass is m.

Here, the weight of the car is balanced by the centripetal force.

According to the question

[tex]m g = \frac{mv^2}{R}\\\\v =\sqrt{24.005\times9.8}\\\\v = 15.34 m/s[/tex]

Answer:

A gold medal has the (minimum) dimensions of:

diameter = 60mm

thickness = 3mm

So we will work with those dimensions.

The medal is then a cyinder of diameter

D = 60mm = 6cm

and height:

H = 3mm = 0.3cm

Remember that the volume of a cylinder is:

V = pi*(D/2)^2*H

where pi = 3.14

Then the volume of a medal is:

V = 3.14*(6cm/3)^2*0.3cm = 3.768 cm^3

The density of the gold in g/cm^3 is:

d = 19.3 g/cm^3

And remember that:

density = mass/volume

So, if the volume is 3.768 cm^3

Then the mass will be:

mass = density*volume =  19.3 g/cm^3*3.768 cm^3 = 72.7 g

So, a single gold medal would weight 72.7 grams

And each gram of gold costs $40

Then the total cost of the gold medal would be:

value = $40*72.7 = $2,908

Now, if yo think that in the Olympics there are 35 sports (a lot with a large number of players) and near 50 disciplines, they need a lot of gold medals.

And each gold medal costs $2,908

So the total cost (only for the gold medals, ignoring the others) would be to high.

This is why the gold medals are made mostly of silver.

Answer:

-1.0778×10⁻¹⁰ N/C

Explanation:

Applying,

E = kq/r²................ equation 1

Where E = elctric field, q = charge, r = distance, k = coulomb's law

From the question,

Given: q = -3.0×10 C, r = 5.0 m

Constant: k = 8.98×10⁹ Nm²/C²

Substitute these values in equation 1

E = (-3.0×10)(8.98×10⁹)/5²

E = -1.0778×10⁻¹⁰ N/C

Hence the electric field on the x-axis is -1.0778×10⁻¹⁰ N/C

Answer:

Look at work

Explanation:

a) I am not sure if you want tangential or centripetal but I will give both

Centripetal acceleration = r*α

Since ω is constant, α is 0 so centripetal acceleration is 0m/s^2

Tangential acceleration = ω^2*r

convert 200rev/min into rev/s

200/60= 10/3 rev/s

a= 100/9*1.2= 120/9= 40/3 m/s^2

b) Rotational Kinetic Energy = 1/2Iω^2

I= mr^2

Plug in givens

I= 43.2kgm^2

K= 1/2*43.2*100/9=2160/9=240J

Answer:

5.04 m

Explanation:

You are told that the homeowner wants to increase their fences by 34 percent meaning Original+ 34 percent. If the original is 100 percent, then the new fence size will be 134 % of the original. You are given the original which is 3.76 meters, to find new fence size 1.34 * 3.76m to get 5.0384 meters, rounded to 5.04 m.

Answer:

5.0384m

Explanation:

% increase = 100 x (Final - Initial / | initial | )

( |~~| Bars indicate absolute value since you can't have a negative height)

Answer:

Its the Federal government

Answer:

a. 8p

Explanation:

We are given that

Radius of hollow sphere , R1=R

Density of hollow sphere=[tex]\rho[/tex]

After compress

Radius of hollow sphere, R2=R/2

We have to find density of the compressed sphere.

We know that

[tex]Density=\frac{mass}{volume}[/tex]

[tex]Mass=Density\times volume=Constant[/tex]

Therefore,[tex]\rho_1 V_1=\rho_2V_2[/tex]

Volume of sphere=[tex]\frac{4}{3}\pi r^3[/tex]

Using the formula

[tex]\rho\times \frac{4}{3}\pi R^3=\rho_2\times \frac{4}{3}\pi (R/2)^3[/tex]

[tex]\rho R^3=\rho_2\times \frac{R^3}{8}[/tex]

[tex]\rho_2=8\rho[/tex]

Hence, the density of  the compressed sphere=[tex]8\rho[/tex]

Option a is correct.

The magnetic field of the wire will be directed towards west. Using right thumb rule one can get the direction of field lines.

Question: A commuter backs her car out of her garage with an acceleration of 1.4 m/s² (a) How long does it take her to reach a speed of 2.00 m/s

Answer:

1.43 s

Explanation:

Applying,

a = (v-u)/t........... Equation 1

Where a = acceleration, v = final velocity, u = initial velocity, t = time

make t the subject of the equation

t = (v-u)/a........... Equation 2

From the question,

Given: v = 2 m/s, u = 0m/s (from rest), a = 1.4 m/s²

Substitute into equation 2

t = (2-0)/1.4

t = 1.43 s

Answer:

1.67 m

Explanation:

The potential energy change of the small ball ΔU equals the work done by the buoyant force, W

ΔU = -W

Now ΔU = mgΔh where m = mass of small ball = ρV where ρ = density of small ball and V = volume of small ball. Δh = h - h' where h = final depth of small ball and h' = initial height of small ball = 5 m. Δh = h - 5

ΔU = mgΔh

ΔU = ρVgΔh

Now, W = ρ'VgΔh'   where ρ = density of water and V = volume of water displaced = volume of small ball. Δh' = h - h' where h = final depth of small ball and h' = initial depth of small ball at water surface = 0 m. Δh' = h - h' = h - 0 = h

So, ΔU = -W

ρVgΔh = -ρ'VgΔh'

ρVg(h - 5) = -ρ'Vgh

ρ(h - 5) = -ρ'h

Since the density of the small ball equals 1/2 the density of water,

ρ = ρ'/2

ρ(h - 5) = -ρ'h

(ρ'/2)(h - 5) = -ρ'h

ρ'(h - 5)/2 = -ρ'h

(h - 5)/2 = -h

h - 5 = -2h

h + 2h = 5

3h = 5

h = 5/3

h = 1.67 m

So, the maximum depth the ball reaches is 1.67 m.

Answer:

2m

Explanation:

wavelength=speed/frequency

=340/170

=2m

Explanation:

Let x = distance of [tex]F_1[/tex] from the fulcrum and let's assume that the counterclockwise direction is positive. In order to attain equilibrium, the net torque [tex]\tau_{net}[/tex] about the fulcrum is zero:

[tex]\tau_{net} = -F_1x + F_2d_2 = 0[/tex]

[tex] -m_1gx + m_2gd_2 = 0[/tex]

[tex]m_1x = m_2d_2[/tex]

Solving for x,

[tex]x = \dfrac{m_2}{m_1}d_2[/tex]

[tex]\:\:\:\:=\left(\dfrac{105.7\:\text{g}}{65.7\:\text{g}} \right)(13.8\:\text{cm}) = 22.2\:\text{cm}[/tex]

Explanation:

700n I think friend .. if worng

The complete question is as follows: At 700 K, [tex]CCl_{4}[/tex] decomposes to carbon and chlorine. The Kp for the decomposition is 0.76.

Find the starting pressure of [tex]CCl_{4}[/tex] at this temperature that will produce a total pressure of 1.1 atm at equilibrium.

Answer: The starting pressure of [tex]CCl_{4}[/tex] is 0.79 atm.

Explanation:

The equation for decomposition of [tex]CCl_{4}[/tex] is as follows.

[tex]CCl_{4}(g) \rightleftharpoons C(s) + 2Cl_{2}(g)[/tex]

Let us assume that initial concentration of [tex]CCl_{4}[/tex] is 'a'. Hence, the initial and equilibrium concentrations will be as follows.

                   [tex]CCl_{4}(g) \rightleftharpoons C(s) + 2Cl_{2}(g)[/tex]

Initial:            a                0          0

Equilibrium:  (a - x)          0          2x

Total pressure = (a - x) + 2x = a + x

As it is given that the total pressure is 1.1 atm.

So, a + x = 1.1

a = 1.1 - x

Now, expression for equilibrium constant for this equation is as follows.

[tex]K_{p} = \frac{P^{2}_{Cl_{2}}}{P_{CCl_{4}}}\\0.76 = \frac{(2x)^{2}}{(a - x)}\\0.76 = \frac{4x^{2}}{1.1 - x - x}\\0.76 = \frac{4x^{2}}{1.1 - 2x}\\x = 0.31 atm[/tex]

Hence, the value of 'a' is calculated as follows.

a + x = 1.1 atm

a = 1.1 atm - x

  = 1.1 atm - 0.31 atm

  = 0.79 atm

Thus, we can conclude that starting pressure of [tex]CCl_{4}[/tex] is 0.79 atm.

Answer:

7.9 [tex]\frac{m}{s^{2} }[/tex]

Explanation:

Take the fact that mass is inversely proportional to accelertation:

m ∝ a

Therefore m = a, but because we are finding the change in acceleration, we would set our problem up to look more like this:

[tex]\frac{m_{1} }{m_{2} } = \frac{a_{2} }{a_{1} } \\[/tex]

Using algebra, we can rearrange our equation to find the final acceleration, [tex]a_{2}[/tex]:

[tex]a_{2} = \frac{a_{1}*m_{1} }{m_{2} } \\[/tex]

Before plugging everything in, since you are being asked to find acceleration, you will want to convert 0.85g to m/s^2. To do this, multiply by g, which is equal to 9.8 m/s^2:

0.85g * 9.8 [tex]\frac{m }{s^{2} }[/tex] = 8.33 [tex]\frac{m }{s^{2} }[/tex]

Plug everything in:

7.9 [tex]\frac{m }{s^{2} }[/tex] = [tex]\frac{ 8.33\frac{m}{s^{2} }*1510kg }{1590kg}[/tex]

(1590kg the initial weight plus the weight of the added passenger)

Answer: 2.4 km/hr

Explanation:

Distance = 600m

Time= 15 minutes = 15 x 60 second/minute = 900 seconds

Speed =  [tex]\frac{Distance}{Time}[/tex]  =  [tex]\frac{600}{900}[/tex]  = [tex]\frac{2}{3}[/tex] m/sec

⇒ [tex]\frac{2}{3}[/tex] x [tex]\frac{18}{5}[/tex] = 2.4 km/hr (1 m/sec = 3.6 km/hr)

Answer:

Hope you find it useful. please correct me if I am wrong

The tension in the cable if the elevator is moving upward with its speed decreasing at a rate of 1.7 m/s² is equal to 1983.67 N.

Tension can be described as a force acting along the length of a medium such as a rope, mainly a force carried by a flexible medium.

Tension can be defined as an action-reaction pair of forces acting at each end of the elements. The tension force is in every section of the rope in both directions, apart from the endpoints. Each endpoint of the rope experience tension and force from the weight attached.

Given the force due to the weight of the elevator = mg = 2400N

m = 2400/9.8 Kg

The elevator deaccelerating while moving upward, a = -1.7 m/s²

According to Newton's 3rd law: T - mg = ma

T - 2400 = (2400/9.8) × (-1.7)

T = 2400 - 416.32

T = 1983.67 N

Learn more about tension, here:

brainly.com/question/28965515

#SPJ5

Answer:

Yes

Explanation:

speed of truck = 20 km/h

Initially the car at rest.

maximum velocity of car = 40 km/h

When the truck and the car collide, the momentum of the truck transferred to car.

So, the car can attain the speed of 40 km/h.

The question is incomplete. The complete question is :

High-speed stroboscopic photographs show that the head of a 200 g golf club is traveling at 60 m/s just before it strikes a  50 g golf ball at rest on a tee. After the collision, the club head travels (in the same direction) at 40 m/s. Find the speed of the golf ball just after impact.

Solution :

We know that momentum = mass x velocity

The momentum of the golf club before impact = 0.200 x 60

                                                                             = 12 kg m/s

The momentum of the ball before impact is zero. So the total momentum before he impact is 12 kg m/s. Therefore, due to the conservation of momentum of the two bodies after the impact is 12 kg m/s.

Now the momentum of the club after the impact is = 0.2 x 40

                                                                                    = 8 kg m/s

Therefore the momentum of the ball is = 12 - 8

                                                                = 4 kg m/s

We know momentum of the ball, p = mass x velocity

                                                     4 = 0.050 x velocity

∴ Velocity =  [tex]$\frac{4}{0.050}$[/tex]

                 = 80 m/s

Hence the speed of the golf ball after the impact is 80 m/s.

Answer:

The drop time ball 1 is less than the drop time of ball 2. A further explanation is provided below.

Explanation:

The net force acting on the ball will be:

⇒ [tex]F_{net}=mg-F_r[/tex]

Here,

F = Force

m = mass

g = acceleration

Now,

According to the Newton's 2nd law of motion, we get

⇒ [tex]F_{net} = ma[/tex]

To find the value of "a", we have to substitute "[tex]F_{net}=ma[/tex]" in the above equation,

⇒ [tex]ma=mg-F_r[/tex]

⇒    [tex]a=g-\frac{F_r}{m}[/tex]

We can see that, the acceleration is greater for the greater mass of less for the lesser mass. Thus the above is the appropriate solution.

Answer:

Both the ball takes equal time to reach to the ground.  

Explanation:

Two balls of same diameter

Let the height is h.

Mass of ball 1 is more than the mass of ball 2.

The second equation of motion is

[tex]h = u t +0.5 gt^2[/tex]

Here, the buoyant force due to air is same. So, the time of fall is independent of the mass.

So, both the ball takes equal time to reach to the ground.  

I see six (6) loops.

I attached a drawing to show where I get six loops from.

Answer:

Young's modulus for the rope material is 20.8 MPa.    

Explanation:

The Young's modulus is given by:

[tex] E = \frac{FL_{0}}{A\Delta L} [/tex]

Where:

F: is the force applied on the wire

L₀: is the initial length of the wire = 3.1 m

A: is the cross-section area of the wire

ΔL: is the change in the length = 0.17 m

The cross-section area of the wire is given by the area of a circle:

[tex] A = \pi r^{2} = \pi (\frac{0.006 m}{2})^{2} = 2.83 \cdot 10^{-5} m^{2} [/tex]

Now we need to find the force applied on the wire. Since the wire is lifting an object, the force is equal to the tension of the wire as follows:

[tex] F = T_{w} = W_{o} [/tex]

Where:

[tex] T_{w} [/tex]: is the tension of the wire

[tex]W_{o} [/tex]: is the weigh of the object = mg

m: is the mass of the object = 1700 kg

g: is the acceleration due to gravity = 9.81 m/s²

[tex] F = mg = 1700 kg*9.81 m/s^{2} = 16677 N [/tex]

Hence, the Young's modulus is:

[tex] E = \frac{16677 N*0.006 m}{2.83 \cdot 10^{-5} m^{2}*0.17 m} = 20.8 MPa [/tex]          

Therefore, Young's modulus for the rope material is 20.8 MPa.                

I hope it helps you!                                    

Answer:

0.000064 cubic meters.

Explanation:

Given the following data;

Length of side = 4 centimeters

Conversion:

100 centimeters = 1 meters

4 cm = 4/100 = 0.04 meters

To find the volume of cuboid;

Mathematically, the volume of a cuboid is given by the formula;

Volume of cuboid = length * width * height

However, when all the sides are equal the formula is;

Volume of cuboid = L³

Volume of cuboid = 0.04³

Volume of cuboid = 0.000064 cubic meters.

Answer:

1.67 N

Explanation:

Applying,

F = u(dm/dt)+m(du/dt)................ Equation 1

Where F = force, m = mass of the vehicle, u = speed.

Since u is constant,

Therefore, du/dt = 0

F = u(dm/dt)............... Equation 2

From the question,

Given: u = 10 m/s, dm/dt = 10 kg/min = (10/60) kg/s

Substitute these values into equation 2

F = 10(10/60)

F = 100/60

F = 1.67 N

Answer:

[tex]\sigma=0.014\ C/m^2[/tex]

Explanation:

Given that,

The radius of sphere, r = 5 cm = 0.05 m

Net charge carries, q = 7.5 µC = 7.5 × 10⁻⁶ C

We need to find the surface charge density on the sphere. Net charge per unit area is called the surface charge density. So,

[tex]\sigma=\dfrac{7.5\times 10^{-6}}{\dfrac{4}{3}\pi \times (0.05)^3}\\\\=0.014\ C/m^2[/tex]

So, the surface charge density on the sphere is [tex]0.014\ C/m^2[/tex].

Explanation:

Given

Acceleration of the pebble is

At t=0, velocity is

considering horizontal motion

[tex]\Rightarrow x=ut+0.5at^2 \\\Rightarrow 11=4.3t+0.5(4.6)t^2\\\Rightarrow 2.3t^2+4.3t-11=0\\\Rightarrow (t-1.4435)(t+3.3131)=0\\\Rightarrow t=1.44\ s\quad [\text{Neglecting negative time}]\\[/tex]

Velocity acquired during this time

[tex]\Rightarrow v_x=4.3+4.6\times 1.44\\\Rightarrow v_x=4.3+6.624\\\Rightarrow v_x=10.92\ s[/tex]

Consider vertical motion

[tex]\Rightarrow v_y=0+7(1.44)\\\Rightarrow v_y=10.08\ m/s[/tex]

Net velocity is

[tex]\Rightarrow v=\sqrt{10.92^2+10.08^2}\\\Rightarrow v=\sqrt{220.85}\\\Rightarrow v=14.86\ m/s[/tex]

Angle made is

[tex]\Rightarrow \tan \theta =\dfrac{10.08}{10.92}\\\\\Rightarrow \tan \theta =0.92307\\\\\Rightarrow \theta =42.7^{\circ}[/tex]

Explanation:

Given that,

Work done to stretch the spring, W = 130 J

Distance, x = 0.1 m

(a) We know that work done in stretching the spring is as follows :

[tex]W=\dfrac{1}{2}kx^2\\\\k=\dfrac{2W}{x^2}\\\\k=\dfrac{2\times 130}{(0.1)^2}\\\\k=26000\ N/m[/tex]

(b) If additional distance is 0.1 m i.e. x = 0.1 + 0.1 = 0.2 m

So,

[tex]W=\dfrac{1}{2}kx^2\\\\W=\dfrac{1}{2}\times 26000\times 0.2^2\\\\W=520\ J[/tex]

So, the new work is more than 130 J.

Answer:

r = 20.22 m

Explanation:

Given that,

Charge,[tex]q=2.5\times 10^{-6}\ C[/tex]

Electric field, [tex]E=55\ N/C[/tex]

We need to find the distance. We know that, the electric field a distance r is as follows :

[tex]E=\dfrac{kq}{r^2}\\\\r=\sqrt{\dfrac{kq}{E}}\\\\r=\sqrt{\dfrac{9\times 10^9\times 2.5\times 10^{-6}}{55}}\\\\r=20.22\ m[/tex]

So, the required distance is 20.22 m.