vector A has a magnitude of 8 unit make an angle of 45° with posetive x axis vector B also has the same magnitude of 8 unit along negative x axis find the magnitude of A+B?​

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!                                    

The free body diagram for the block of mass M consists of four forces:

• the block's weight, Mg, pointing downward

• the normal force of the table pushing upward on the block, also with magnitude Mg

• kinetic friction with magnitude µMg = 0.2 Mg, pointing to the left

• tension of magnitude T pulling the block to the right

For the block of mass m, there are only two forces:

• its weight, mg, pulling downward

• tension T pulling upward

The m-block will pull the M-block toward the edge of the table, so we take the right direction to be positive for the M-block, and downward to be positive for the m-block.

Newton's second law gives us

T - 0.2Mg = Ma

mg - T = ma

where a is the acceleration of either block/the system. Adding these equations together eliminates T and we can solve for a :

mg - 0.2 Mg = (m + M) a

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

a = 1.96 m/s²

Then the tension in the string is

T = m (g - a)

T = 78.4 N

Answer:

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Explanation:

The 2 capacitors in the middle are connected in parallel so simply add their capacitance together:

[tex]5.0\:\mu\text{F} + 8.0\:\mu\text{F} = 13.0\:\mu \text{F}[/tex]

Now we have 3 capacitors connected in series so their equivalent capacitance [tex]C_{eq}[/tex] is

[tex]\dfrac{1}{C_{eq}} = \dfrac{1}{10.0\:\mu \text{F}} + \dfrac{1}{13.0\:\mu \text{F}} + \dfrac{1}{9.0\:\ mu \text{F}} [/tex]

or

[tex]C_{eq} = 3.5\:\mu \text{F}[/tex]

Answer:

the pressure exerted by the object is 392,000 N/m²

Explanation:

Given;

mass of the object, m = 8 kg

area of your finger, A = 2 cm² = 2.0 x 10⁻⁴ m²

acceleration due to gravity, g = 9.8 m/s²

The pressure exerted by the object is calculated as;

[tex]Pressure = \frac{F}{A} = \frac{mg}{A} = \frac{8 \times 9.8}{2\times 10^{-4}} = 392,000 \ N/m^2[/tex]

Therefore, the pressure exerted by the object is 392,000 N/m²

Answer:

stress = 98000 N/m^2

strain = 3.92 x 10^-6

change in height = 0.0196 mm

Explanation:

Height, h = 5 m

Area, A = 3 m²

mass, m = 30000 kg

Stress is defined as the force per  unit area.

[tex]stress = \frac{mg}{A}\\\\stress = \frac{30000\times 9.8}{3}\\\\stress = 98000 N/m^2[/tex]

Young's modulus of concrete is Y = 2.5 x 10^10 N/m^2

Young's modulus is defined as the ratio of stress to the strain.

[tex]Y = \frac{stress}{strain}\\\\2.5\times 10^{10}= \frac{98000}{strain}\\\\strain = 3.92\times 10^{-6}[/tex]

let the change in height is h'.

Strain is defined as the ratio of change in height to the original height.

[tex]3.92\times 10^{-6} = \frac{h'}{5}\\\\h' = 1.96\times 10^{-5}m = 0.0196 mm[/tex]

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

Answer:

SUPER TYPHOON (STY), a tropical cyclone with maximum wind speed exceeding 220 kph or more than 120 knots.

Answer:

The force on the second piston is 5246.5 N .

Explanation:

Area of first piston, a = 3.002 cm^2

Area of second piston, A = 315 cm^2

Force on first piston, f = 50 N

let the force of the second piston is F.

According to the Pascal's law

[tex]\frac{f}{a} = \frac{F}{A}\\\\\frac{50}{3.002}=\frac{F}{315}\\\\F = 5246.5 N[/tex]

Explanation:

The area of a circle of radius r is given by

[tex]A = \pi r^2[/tex]

Taking the derivative of A with respect to time t, we get

[tex]\dfrac{dA}{dt} = 2\pi r \dfrac{dr}{dt}[/tex]

We also know that

[tex]\dfrac{dr}{dt} = 2\:\text{m/s}\:\text{at}\:r = 39\:\text{m}[/tex]

[tex]\dfrac{dA}{dt} = 2\pi (39\:\text{m})(2\:\text{m/s})= 490\:\text{m}^2\text{/s}[/tex]

Complete Question

Four small masses of 0.2 kg each are connected by light rods 0.4m long to form a square. What is the moment of inertia of this object for an axis through the middle of the square and parallel to two sides.

Answer:

[tex]I=0.032kgm^2[/tex]

Explanation:

From the question we are told that:

Mass[tex]m=0.2kg[/tex]

Length [tex]l=0.4m[/tex]

Generally the equation for Inertia is mathematically given by

 [tex]I=md^2[/tex]

 [tex]I=0.8*0.20(\frac{0.40}{2})^2[/tex]

 [tex]I=0.032kgm^2[/tex]

Answer:

a. 282.84 cm/s b. 100 cm/s c. 400 cm/s d. 200 cm/s

Explanation:

The speed of the wave v = √(T/μ) where T = tension and μ = mass per unit length = m/l where m = mass of string and l = length of string.

So, v = √(T/μ)

v = √(T/m/l)

v = √(Tl/m)

a. The string's tension is doubled?

If the tension is doubled, T' = 2T the new speed is

v' = √(T'l/m)

v' = √(2Tl/m)

v' = √2(√Tl/m)

v' = √2v

v' = √2 × 200 cm/s

v' = 282.84 cm/s

b. The string's mass is quadrupled (but its length is unchanged)?

If the mass is quadrupled, m' = 4m the new speed is

v' = √(Tl/m')

v' = √(Tl/4m)

v' = (1/√4)(√Tl/m)

v' = v/2

v' = 200/2 cm/s

v' = 100 cm/s

c. The string's length is quadrupled (but its mass is unchanged)?

If the length is quadrupled, l' = 4l the new speed is

v' = √(Tl'/m)

v' = √(T(4l)/m)

v' = √4)(√Tl/m)

v' = 2v

v' = 200 × 2 cm/s

v' = 400 cm/s

d. The string's mass and length are both quadrupled?

If the length is quadrupled, l' = 4l and mass quadrupled, m' = 4m, the new speed is

v' = √(Tl'/m')

v' = √(T(4l)/4m)

v' = √(Tl/m)

v' = v

v' = 200 cm/s

Answer:

the centripetal acceleration at a point that is 0.0897 m from the center of the disc is 1143.8 m/s²

Explanation:

Given the data in the question;

centripetal acceleration a[tex]_c[/tex]₁ = 241 m/s²

radius r₁ = 0.0189 m

radius r₂ = 0.0897 m

centripetal acceleration a[tex]_c[/tex]₂ = ? m/s²

since the rotational period will be the same for the two disk,

we use the centripetal acceleration formula a[tex]_c[/tex] = (4π²r/T²) to find the rotational period for the first disk.

a[tex]_c[/tex]₁ = (4π²r₁/T²)

make T² subject of formula

T² = 4π²r₁ / a[tex]_c[/tex]₁

we substitute

T² = ( 4 × π² × 0.0189 )  / 241  

T² = 0.00309602528 s²

Now we use the same formula to find a[tex]_c[/tex]₂

a[tex]_c[/tex]₂ = ( 4π²r₂ / T² )

we substitute

a[tex]_c[/tex]₂ = ( 4 × π² × 0.0897 )  / 0.00309602528

a[tex]_c[/tex]₂ = 1143.8 m/s²

Therefore, the centripetal acceleration at a point that is 0.0897 m from the center of the disc is 1143.8 m/s²

Answer:

It is a example of physical change

kylydljty many true dvx*&;'*+$_5+

Solution :

It is given that a car ran off from a cliff and it crashes on canyon floor. Now the system of a car as well as the earth together have a  [tex]\text{ net momentum of zero}[/tex] when the car crashes on the canyon floor, thus reducing the momentum of the car to zero. The earth also stops its upward motion and it also reduces the momentum to zero.

By the work-energy theorem, the total work done on the mass by the spring is equal to the change in the mass's kinetic energy:

W = ∆K

and the work done by a spring with constant k as it gets compressed a distance x is -1/2 kx ²; the work it does is negative because the restoring force of the spring points opposite the direction in which it's getting compressed.

So we have

-1/2 k (0.15 m)² = 0 - 1/2 (2.0 kg) (3.0 m/s)²

Solve for k to get k = 800 N/m.

Answer:

(a) t = 0 s

(b) t = 0 s, 30 s, 55 s

(c) t = 40 s to t = 60 s

(d) t = 10 s to t = 15 s

(e) a = 6 m/s^2

Explanation:

(a) The car is at starting position at t = 0 s and v = 0 m/s.

(b) The velocity of car is zero when the time is t = 0 s, 30 s and 55 s.

(c) from t = 40 s to 60 s the car is moving in the negative direction.

(d) The fastest speed is 60m/s from t = 10 s to t = 15 s.

(e) The slope of the velocity time graph gives acceleration.

a = (60 - 0) / (10 - 0) = 6 m/s^2

Answer:

1 month

Explanation:

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.

This above answer helps a lot.

Answer:

The answer is below

Explanation:

Specific heat capacity is an intensive property of a material. The specific heat of a material is the amount of energy required to raise the temperature of one unit mass m of material by one unit of temperature.

a) Temperature is inversely proportional to specific heat capacity. If the same amount of heat is applied to all three balls, the ball that will reach the highest temperature is the ball with the least specific heat capacity.

Hence lead will have the highest temperature since it has the least specific heat capacity.

b) The quantity of heat is directly proportional to the specific heat capacity. Hence if all balls experience the same temperature change, the ball that have the most energy will be that with the highest specific heat capacity.

Hence aluminum will have the most heat since it has the highest specific heat capacity.

Answer:

5.78amps

Explanation:

Given data

Time t= 57 seconds

Charge Q= 330C

Current I= ??

The expression for the electric current is given as

Q= It

Substituting we have

330= I*57

I= 330/57

I=5.78 amps

Hence the current is 5.78amps

Answer:

The charged particle will follow a circular path.

Explanation:

Formula for the magnetic force is;

F = qvb sin θ

Where;

where;

q = the charge

v = the velocity

B = the magnetic field

θ = the angle between the velocity and magnetic field

We are told that velocity vector is perpendicular to the magnetic field lines. Thus, angle is 90.

So sin θ = sin 90 = 1

Thus,

F = qvB

Now, since the velocity vector is perpendicular to the magnetic field line,it also means from flemmings right hand rule, that the magnetic force is as well perpendicular to both of them.

Therefore, we have:

- a force that is always perpendicular to the velocity and as well constant in magnitude since magnitude of velocity or magnetic field does not change.

What this statement implies is that the force is acting as a centripetal force, and therefore, the charged particle will be kept in a uniform circular motion.

Answer:

Option B.

Explanation:

Remember that we can think on any movement as a sum of a movement in the y-axis, the movement in the x-axis, and the movement in the z-axis. And these are not related, this means that, for example, the movement in x does not affect the movement in y.

So, when we analyze the problem of "how long takes an object to hit the ground"

We do not care for the horizontal motion of the object, we only care for the vertical motion of the object.

So, if an object is dropped, and another has a given initial velocity in the x-axis, in both cases the initial velocity in the y-axis will zero.

And in both cases, the only vertical force acting on the balls will be the gravitational force (so both objects will have the same vertical acceleration and the same vertical initial velocity) with this, we already know that the vertical motion of both objects will be exactly the same.

So, both objects will hit the ground at the same time.

(notice that here we are ignoring things like air resistance and other complex forces)

So here the correct option is b:  Neither. Their vertical motion is the same, so they will reach the ground at the same time.

Explanation:

option b is the correct one

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]