.You have always been impressed by the speed of the elevators in your apartment building. You wonder about the maximum acceleration for these elevators during normal operation, so you decide to measure it by using your bathroom scale. While the elevator is at rest on the ground floor,you get in, put down your scale, and stand on it. The scale reads 50 kg. You continue standing on the scale when the elevator goes up, carefully watching the reading. During the trip to the 10th floor, the greatest scale reading was

Answer:

Change in momentum is zero.

Explanation:

The following data were obtained from the question:

Mass (m) = 5000 kg

Time (t) = 1 h

Net force (F) = 0

Change in momentum =?

Force = Rate of change of momentum

0 = change in momentum

Change in momentum = 0

We can see from the above illustration that the net force is zero. Thus, the change in momentum is also zero.

Answer:

Are you sure it was soccer ball? Or meine hearts

Explanation:

Answer:It will take about 3000 years

Explanation:

Answer:

-the ratio of the speed of light

in air to the speed of light in the substance.

-speed of light in air 300,000 km/sec, which decreases when it passes through a transparent substance.

-e.g.. speed of light in substance = 200,000 km/sec, R.I. = 300,000/200,000 = 1.5

Explanation:

Answer: Just a few examples are the tension in the rope on a tether ball, the force of Earth's gravity on the Moon, friction between roller skates and a rink floor, a banked roadway's force on a car, and forces on the tube of a spinning centrifuge. Any net force causing uniform circular motion is called a centripetal force.

Answer:

roller skates and a rink floor, a banked roadway's force on a car, and forces on the tube of a spinning centrifuge

Explanation:

Answer:

10 mm

Explanation:

We'll begin by calculating the spring constant of the spring. This can be obtained as follow:

Extention (e) = 5 mm

Force (F) = 125 N

Spring constant (K) =?

F = Ke

125 = K × 5

Divide both side by 5

K = 125 / 5

K = 25 N/mm

Finally, we shall determine how much the spring will stretch when a 250 N force is applied. This can be obtained as follow:

Force (F) = 250 N

Spring constant (K) = 25 N/mm

Extention (e) =?

F = Ke

250 = 25 × e

Divide both side by 25

e = 250 / 25

e = 10 mm

Thus, the spring will stretch 10 mm when a 250 N force is applied.

Answer:

7,812 J

Explanation:

Using the relation:

Q = mcΔθ

Q = quantity of heat

C = specific heat capacity of lead

Δθ = temperature change (T2 - T1)

M = mass of substance

Q = mass * specific heat * Δθ

Q = 0.125kg * 128 * (327 – 20)

Q = 0.125 * 128 * 307

Q = 4912 J

For melting:

Q = mass * Hf

0.125 * (2.32 * 10^4)

= 2,900 J

Total = 4,912 J + 2,900 J = 7,812 J

Answer:

A. v = √2gh

B. No! The final velocity does not depend on the mass of the car.

C. Yes! the final velocity depends on the steepness of the hill

D. 3.28 m/s

Explanation:

A. Determination of the final velocity.

½mv² = mgh

Cancel out m

½v² = gh

Cross multiply

v² = 2gh

Take the square root of both side

v = √2gh

B. Considering the formula obtained for the final velocity i.e

v = √2gh

We can see that there is no mass (m) in the formula.

Thus, the final velocity does not depend on the mass of the car.

C. Considering the formula obtained for the final velocity i.e

v = √2gh

We can see that there is height (h) in the formula.

Thus, the final velocity depends on the steepness of the hill

D. Determination of the final velocity.

Height (h) = 0.55 m

Acceleration due to gravity (g) = 9.8 m/s²

Velocity (v) =?

v = √2gh

v = √(2 × 9.8 × 0.55)

v = √10.78

v = 3.28 m/s

Answer:

5.7375 seconds

Explanation:

The computation of the time required to reach that maximum velocity is shown below:

Given that

Mass = m =  5100 kg

Net upward force F = 8 × times 10^5 N

Initial speed = V_i = 0

Maximum speed = V = 900 m.s

Based on the above information

Impluse J = m(V - V_i)

= 5100 (900 - 0)

= 459 × 10^4 kg m.s

As we know that

J = FT

So

T = J ÷ F

= (459 × 10^4) ÷ (8 × 10^5)

= 5.7375 seconds

Answer:

Explanation:

Force between two charged conducting sphere

= k x Q₁ x Q₂ / r² ,  k is a constant  Q₁ and Q₂ are charges and   r is distance between them .

= 9 x 10⁹ x 12 x 10⁻⁹ x 18 x 10⁻⁹ / .30²

= 21600 x 10⁻⁹

= 2.16 x 10⁻⁵ N .

b )

After the spheres are joined together , there is redistribution of charge and remaining charge will be equally shared by them .

Charge on each sphere = (12 - 18 ) x 10⁻⁹ / 2

= - 3 x 10⁻⁹ C .

Force = 9 x 10⁹ x 3 x 10⁻⁹ x 3 x 10⁻⁹ / .30²

= 900 x 10⁻⁹ N .

The impulse exerted on the beach ball is 2.75 kgm/s.

The impulse exerted on the exercise ball is - 2.75 kgm/s.

This is the force applied to an object that acts over a period of time.

The impulse exerted on the beach ball is the change in the momentum of the ball and it is calculated as follows;

J = ΔP

J = m(v - u)

J = 0.25(7 - (-4))

J = 0.25(7 + 4)

J = 2.75 kgm/s

The impulse exerted on the exercise ball is equal in magnitude but opposite in direction to the beach ball.

Thus, the impulse exerted on the exercise ball is - 2.75 kgm/s.

Learn more about impulse here: https://brainly.com/question/25700778

Answer:

The speed of this particle is constantly [tex]c[/tex].

Explanation:

Position vector of this particle at time [tex]t[/tex]:

[tex]\displaystyle \mathbf{r}(t) = b\, \cos(Q)\, \mathbf{i} + b\, \sin(Q) \, \mathbf{j} + c\, t\, \mathbf{k}[/tex].

Write [tex]\mathbf{r}(t)[/tex] as a column vector to distinguish between the components:

[tex]\mathbf{r}(t) = \begin{bmatrix}b\, \cos(Q) \\ b\, \sin(Q) \\ c\, t\end{bmatrix}[/tex].

Both [tex]b[/tex] and [tex]Q[/tex] are constants. Therefore, [tex]b\, \cos(Q)[/tex] and [tex]b \sin (Q)[/tex] would also be constants with respect to [tex]t[/tex]. Hence, [tex]\displaystyle \frac{d}{dt}[b\, \cos(Q)] = 0[/tex] and [tex]\displaystyle \frac{d}{dt}[b\, \sin(Q)] = 0[/tex].

Differentiate [tex]\mathbf{r}(t)[/tex] (component-wise) with respect to time [tex]t[/tex] to find the velocity vector of this particle at time [tex]t\![/tex]:

[tex]\begin{aligned}\mathbf{v}(t) &= \frac{\rm d}{{\rm d} t} [\mathbf{r}(t)] \\ &=\frac{\rm d}{{\rm d} t} \left(\begin{bmatrix}b\, \cos(Q) \\ b\, \sin(Q) \\ c\, t\end{bmatrix}\right) \\ &= \begin{bmatrix}\displaystyle \frac{d}{dt}[b\, \cos(Q)] \\[0.5em] \displaystyle \frac{d}{dt}[b\, \sin(Q)]\\[0.5em]\displaystyle \frac{d}{dt}[c \cdot t]\end{bmatrix} = \begin{bmatrix}0 \\ 0 \\ c\end{bmatrix}\end{aligned}[/tex].

The speed [tex]v[/tex] (a scalar) of a particle is the magnitude of its velocity :

[tex]\begin{aligned}v(t) &= \| \mathbf{v}(t) \| \\ &= \left\|\begin{bmatrix}0 \\ 0 \\ c\end{bmatrix}\right\| \\ &= \sqrt{0^2 + 0^2 + c^2} = c\end{aligned}[/tex].

Therefore, the speed of this particle is constantly [tex]c[/tex] (a constant.)

Answer:

the impulse must be the same in these two cases    F t = m ([tex]\sqrt{2g h_f } - \sqrt{2g h_o}[/tex])

Explanation:

For this exercise we use the relationship between momentum and momentum

         I = Δp

         F t = m v_f - m v₀

To know the speed we use the conservation of energy

starting point. Highest point

       Em₀ = U = m g h

fincla point. Just before the crash

      Em_f = K = ½ m v²

energy is conserved

        Em₀ = Em_f

        m g h = ½ m v²

         v = [tex]\sqrt{2gh}[/tex]

we substitute in the impulse relation

     F t = m ([tex]\sqrt{2g h_f } - \sqrt{2g h_o}[/tex])

therefore we can see that as in case the initial and final heights are equal, the impulse must be the same in these two cases

This question is incomplete, the missing image is uploaded along this answer below;

Answer:

a) the required angular acceleration magnitude α is  π rad/s² or 3.14 rad/s²

b) the time at which the contact occur is 8 seconds

Explanation:

Given the data in the question;

first we convert the given angular velocity to rad/s

angular velocity = 240 rpm = ((240/60) × 2π ) = 8π rad/s

so

ωA = 8π rad/s

next we determine angular acceleration at point A; so

ωA = at

8π rad/s = at -------let this be equation

thus, angular acceleration of disk A is ωA and rotates in counter clockwise direction.

Next we determine the velocity of point C;

Vc = rA × ωA

where Vc is velocity at point C, rA is radius of A ( 150/1000)m,  { from the diagram }

so we substitute

Vc = 0.15m × 8π

Vc = 1.2π m/s

for angular velocity at point B;

Vc = rB × ωB

where rB is the radius of B ( 200/1000)m

we substitute

1.2π = 0.2 × ωB

ωB = 1.2π / 0.2

ωB = 6π rad/s

Thus, the wheel B rotates with an angular velocity of 6π rad/s in clock wise direction.

Now,

a) Determine the required angular acceleration magnitude α

we find the the angular acceleration of disk B after 2 seconds, using the expression;

ωB = at

where angular acceleration is a and t is time ( t - 2)

we substitute

ωB = at

6π = a( t - 2) -------- let this be equation 2

now, lets substract equation 1 form equation 2

(6π = a( t - 2)) - (8π = at)

(6π = at - 2a) - ( 8π = at)

-2π = 0 + -2a

2π = 2a

a = 2π/2

a = π rad/s² or 3.14 rad/s²

Therefore, the required angular acceleration magnitude α is  π rad/s² or 3.14 rad/s²

b) determine the time at which the contact occurs;

from equation 1

8π = at

we substitute in the value of a

8π = π × t

t = 8π / π

t = 8 seconds

Therefore, the time at which the contact occur is 8 seconds

Answer: 0.435 m

Explanation:

Given

mass m=20 kg

initial speed u=2 m/s

coefficient of kinetic friction [tex]\mu_k=0.3[/tex]

deceleration which opposes the motion is given by

[tex]\Rightarrow a=g\sin \theta+\mu_kg\cos \theta\\\Rightarrow a=g(\sin \theta +\mu_k\cos \theta)[/tex]

[tex]\Rightarrow a=9.8(\sin 10^{\circ}+0.3\times \cos 10^{\circ})\\\Rightarrow a=4.59\ m/s^2[/tex]

using [tex]v^2-u^2=2as[/tex]

[tex]\Rightarrow s=\dfrac{2^2}{2\times 4.59}=0.435\ m[/tex]

Answer:

a) the total electric potential is 2282000 V

b) the total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V

Explanation:

Given the data in the question and as illustrated in the image below;

a) Determine the total electric potential (in V) at the origin.

We know that; electric potential due to multiple charges is equal to sum of electric potentials due to individual charges

so

Electric potential at p in the diagram 1 below is;

Vp = V1 + V2

Vp = kq1/r1 + kq2/r2

we know that; Coulomb constant, k = 9 × 10⁹ C

q1 = 4.60 uC = 4.60 × 10⁻⁶ C

r1 = 1.25 cm = 0.0125 m

q2 = -2.06 uC = -2.06 × 10⁻⁶ C

location x2 = −1.80 cm; so r2 = 1.80 cm = 0.018 m

so we substitute

Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0125 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.018 )

Vp = (3312000) + ( -1030000 )

Vp = 3312000 -1030000

Vp = 2282000 V

Therefore, the total electric potential is 2282000 V

b)

the total electric potential (in V) at the point with coordinates (0, 1.50 cm).

As illustrated in the second image;

r1² = 0.015² + 0.0125²

r1 = √[ 0.015² + 0.0125² ]

r1 = √0.00038125

r1 = 0.0195

Also

r2² = 0.015² + 0.018²

r2 = √[ 0.015² + 0.018² ]

r2 = √0.000549

r2 = 0.0234

Now, Electric Potential at P in the second image below will be;

Vp = V1 + V2

Vp = kq1/r1 + kq2/r2

we substitute

Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0195 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.0234 )

Vp = 2123076.923 + ( -762962.962 )

Vp = 2123076.923 -792307.692

Vp =  1330769.23 V

Therefore, the total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V

a) The total electric potential is 2282000 V

b) The total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V

The electric potential is defined as the amount of work energy needed to move a unit of electric charge from a reference point to a specific point in an electric field.

Given the data in the question and as illustrated in the image below;

a) Determine the total electric potential (in V) at the origin.

We know that; electric potential due to multiple charges is equal to sum of electric potentials due to individual charges

Electric potential at p in diagram 1 below is;

[tex]V_P=V_1+V_2[/tex]

[tex]Vp = \dfrac{kq_1}{r_1} + \dfrac{kq_2}{r_2}[/tex]

we know that; the Coulomb constant, k = 9 × 10⁹ C

q1 = 4.60 uC = 4.60 × 10⁻⁶ C

r1 = 1.25 cm = 0.0125 m

q2 = -2.06 uC = -2.06 × 10⁻⁶ C

location x2 = −1.80 cm; so r2 = 1.80 cm = 0.018 m

so we substitute

Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0125 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.018 )

Vp = (3312000) + ( -1030000 )

Vp = 3312000 -1030000

Vp = 2282000 V

Therefore, the total electric potential is 2282000 V

b)The total electric potential (in V) at the point with coordinates (0, 1.50 cm).

As illustrated in the second image;

[tex]r_1^2=0.015^2+0.0125^2[/tex]

[tex]r_1 = \sqrt{[ 0.015^2 + 0.0125^2 ][/tex]

[tex]r_1 = \sqrt{0.00038125}[/tex]

[tex]r_1 = 0.0195[/tex]

Also

[tex]r_2^2 = 0.015^2 + 0.018^2[/tex]

[tex]r_2 = \sqrt{0.015^2 + 0.018^2}[/tex]

[tex]r_2 = \sqrt{0.000549[/tex]

[tex]r_2 = 0.0234[/tex]

Now, Electric Potential at P in the second image below will be;

Vp = V1 + V2

[tex]Vp = \dfrac{kq_1}{r_1} + \dfrac{kq_2}{r_2}[/tex]

we substitute

Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0195 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.0234 )

Vp = 2123076.923 + ( -762962.962 )

Vp = 2123076.923 -792307.692

Vp =  1330769.23 V

Therefore, the total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V

To know more about electric potential follow

https://brainly.com/question/25923373

Answer:

Kinetic energy increases and potential energy decrease when velocity of an object increase.

Answer:

b)  a = 52.26 m / s², a ’= 13.06 m / s², c) N = 2.79 10⁴ N, d) N = 1.89 10³ N

Explanation:

a) In the attached we can see the free body diagrams for the two positions, position A in the lower part of the circle and position B in the upper part of the circle

b) Let's start at point A

Let's use that the acceleration is centripetal

           a = v² / r

let's calculate

            a = 28² / 15.0

            a = 52.26 m / s²

as they relate it is centripetal it is directed towards the center of the circle, therefore for this point it is directed vertically upwards

Point B

           a ’= 142/15

           a ’= 13.06 m / s²

in this case the acceleration is vertical downwards

c) The values ​​of the normal force

point A

let's use Newton's second law

           ∑ F = m a

           N- W = m a

           N = mg + ma

           N = m (g + a)

           N = 450.0 (9.8 + 52.25)

           N = 2.79 10⁴ N

d) Point B

            -N -W = m (-a)

             N = ma -m g

             N = m (a-g)

             N = 450.0 (14.0 - 9.8)

             N = 1.89 10³ N

Answer:

1. Temperature= 869.35 K

2. Pressure of combustion = 12994.043 kpa

3. Thrust = 127x10⁶N

Explanation:

this problem has been fully explained in the attachment. please use it to get a clearer explanation of the answer.

1.

The temperature = (273+2400k) - (3800)²/2(4003)

= 2673 - 14440000/8006

= 2673 - 1803.65

= 869.35 K

Approximately 869.4K

2. We first get mach number

= 3800/√1.3(923.8)(869.35)

= 3800/1021.78

= 3.719

Pressure = 100kpa[1+2.07464415]^1.3/0.3

= 12995.043kpa

C. Thrust

Pi/4(3800)²(0.3)²(100x10³)/(923.8)(869.4)

= 12678.621

= 126.781 kN

Thrust is approximately 127kN = 127x10⁶N

Answer:

A - mass. B - Weight

Explanation:

This is because weight varies with the strength of gravity. Mass is just the amount of matter in an object

The simplest definition of a black hole is an object that is so dense that not even light can escape its surface. If we squished the Earth's mass into a sphere with a radius of 9 mm, the escape velocity would be the speed of light. Just a wee-bit smaller, and the escape velocity is greater than the speed of light.

HopeItHelps!

This question is incomplete, the complete question is;

The loaded car of a roller coaster has mass M = 320 kg. It goes over the highest hill with a speed v of 21.4 m/s. The radius of curvature R of the hill is 15.8 m.

(a) What is the force (N) that the track must exert on the car? (positive is up)

(b) What must be the force (N) that the car exerts on a 61 kg passenger?

Answer:

a) the force (N) that the track must exert on the car is -6139.14 N

b) the force (N) that the car exerts on a 61 kg passenger is -1170.27 N

Explanation:

Given the data in the question;

Let N represent the force that the track must exerted on the car

Net force on the car Fnet = Mg + N

so

M × a = Mg + N

N = Ma - Mg

N = Ma - M(v²/R)

we substitute

N = (320kg × 9.8m/s²) - ( 320 × ((21.4m/s)² / 15.8 m) )

N = 3136 - ( 320 × 28.9848 )

N = 3136 - 9275.136

N = -6139.14 N

Therefore, the force (N) that the track must exert on the car is -6139.14 N

b) What must be the force (N) that the car exerts on a 61 kg passenger?

Let N represent the force that the car exerts on 61kg passengers

so

Net force of passengers Fnet = mg + N

Ma = Mg + N

N = Ma - Mg

N = Ma - M(v²/R)

N = (61kg × 9.8m/s²) - ( 61 × ((21.4m/s)² / 15.8 m) )

N = 597.8 - ( 61 × 28.9848)

N = 597.8 - 1768.0728

N = -1170.27 N

Therefore, the force (N) that the car exerts on a 61 kg passenger is -1170.27 N

The centripetal force of the track on the car moving in the circular path is [tex]1.465 \times 10^6 \ N[/tex].

The force (N) that the car exerts on a 61 kg passenger is 597.8 N.

The centripetal force of the track on the car moving in the circular path is calculated as follows;

[tex]F_c = \frac{mv^2}{r}\\\\ F_c = \frac{320 \times 21.4^2}{0.1} \\\\F_c = 1.465 \times 10^6 \ N[/tex]

The force (N) that the car exerts on a 61 kg passenger is equal to the force the passenger exerts on the car based on Newton's third law of motion.

F = mg

F = 61 x 9.8

F = 597.8 N

Learn more about centripetal force here: https://brainly.com/question/20905151

Answer:

75N

Explanation:

a = v/t = 3/2

F = ma = 50(3/2) = 75

Answer:

B

Explanation:

The crest of all three waves are of equal height

Answer:

They are the resultant vector.

Answer:

A) t = 1.249 10⁻² s, B)  f = 80 Hz, C) f = 20 Hz,

D)  slowly advancing an angle of approximately    Δθ = 0.05 rad each flash

E) In each flash it seems to go backward an angle of Δθ = -0.05 rad

Explanation:

A) To make it appear that the blades are immobile, it implies that every time the light turns on, a blade should be in the same position, therefore, as we have 4 blades, they must rotate an angle of 2π/4,  

         θ = π / 2  

         θ = 1.57 rad  

taking the angle let's use the endowment kinematics relations  

          θ = w₀ t + ½ α t²  

in general the fans rotate at constant speed α= 0  

         θ = w₀ t  

         t = θ / w₀  

let's reduce the magnitudes to the SI system  

        w₀ = 20 rev / s (2π rad / 1rev) = 125.66 rad / s  

let's calculate  

        t = 1.57 / 125.66  

        t = 1.249 10⁻² s  

B) the fastest speed for the blades to rotate is when one blade of a complete turn , we use the relationship between the fecuance and the period  

        f = 1 / T  

        f = 1 / 1.25 10⁻²  

       f = 80 Hz

C) we have two possibilities:  

* a yellow dot is placed on each sheet  

In this case the angular velocity of the blade is the same at all points, therefore the results obtained should not change

* a yellow dot is placed on a single sheet.  

Here for the point to remain fixed the angle of rotation must be

       θ= 2π rad  

the time is  

       t = 2π / 125.66  

       t = 5 10⁻² s  

the maximum frequency is  

      f = 1/5 10⁻²  

      f = 20 Hz

D) The copy strobe rotates at f = 19 Hz, the time between each flash is  

      t = 1/19  

      t = 5.26 10⁻² s  

this time is higher, so the angle turned is large  

       θ = w t  

       θ = 125.66 5.26 10⁻²  

       θ = 6.61 rad  

the relationship between this angle and the angle of a circle is  

θ = 1,052

We can see that it is this time the blade rotates 1 complete turns, for this the position of the blade changes us, for the other 0.052 rad the blade rotates a little more than the circumference therefore it seems that it is slowly advancing an angle of approximately  

         Δθ = 0.05 rad each flash  

E) in this case changes the flash speed  

       t = 1/21  

       t = 4.76 10⁻² s  

the angle rotated is  

      θ = 125.66 4.76 10⁻²  

      θ = 5.984 rad  

      θ / 2π = 0.95  

in that case, the blade did not complete the turn, therefore in each flash it seems to go backward an angle

Δθ = -0.05 rad

Answer:34.6 m/s

Explanation: It is asking how long meaning the answer is in time

Answer:

c. There is a buoyant force that is proportional to the volume of your body that is below the level of the water.

Explanation:

Buoyancy can be defined as a force which is created by the water displaced by an object.

Simply stated, buoyancy is directly proportional to the amount of water that is being displaced by an object.

Hence, the greater the amount of water an object displaces; the greater is the force of buoyancy pushing the object up.

The buoyancy of an object is given by the formula;

[tex] Fb = pgV [/tex]

[tex] But, \; V = Ah [/tex]

[tex] Hence, \; Fb = pgAh [/tex]

Where;

Fb = buoyant force of a liquid acting on an object.

g = acceleration due to gravity.

p = density of the liquid.

v = volume of the liquid displaced.

h = height of liquid (water) displaced by an object.

A = surface area of the floating object.

The unit of measurement for buoyancy is Newton (N).

In this scenario, you are standing on the bottom of a lake with your torso above water. Thus, there is a buoyant force that is proportional to the volume of your body that is below the level of the water.

Answer:

[tex]E=30.78\ J[/tex]

Explanation:

The force constant of the spring, k = 76 N/m

The extension in the spring, x = 0.9 m

We need to find the energy is stored in the extended spring. The energy stored in the spring is given by :

[tex]E=\dfrac{1}{2}kx^2\\\\E=\dfrac{1}{2}\times 76\times (0.9)^2\\\\E=30.78\ J[/tex]

So, 30.78 J of energy is stored in the spring.