The section of a longitudinal wave in which the particles are densely packed together

A 50.0-kg person steps on a scale in an elevator. The scale reads 5000 N. What is the magnitude of the acceleration of the eleva

zaharov [31]

The magnitude of the acceleration of the elevator is 90m/s^2

Due to Newton's Law ∑ Forces in direction of motion is equal to mass

multiplied by the acceleration

We have here two forces 5000 N in direction of motion and the weight of the person in opposite direction of motion

The weight of the person is his mass multiplied by the acceleration of gravity

So ,

W = mg , where m is the mass and g is the acceleration of gravity

m = 50 kg and g = 9.8 m/s²

Substitute these values in the rule above

W = 50 × 9.8 = 490 N

The scale reads 5000 N

F = 5000 N , W = 490 N , m = 50kg

F - W = ma

5000 - 490 = 50 a

4510 = 50 a

Divide both sides by 50

a = 90.2 m/s²

Hence the acceleration is 90m/s^2

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5 0

2 years ago

A centimeter in milimetersr difiers from each other by a fact of 100 ,1000, 1 or 10

umka21 [38]

Answer:

The correct answer is - 1.

Explanation:

The one cubic centimeter is equal to the one mililitter of a liquid. It is same de to the fact that volume of water that fits in a one cubic centimeter is equal to the liquid present in one mililiter of a liquid.

1 cm² = 1 ml

1 cm³ = 1 ml

Thus, the use of the both unit is interchangebly.

So , correct answer is - 1.

7 0

3 years ago

The portion of electromagnetic spectrum is occupied by signal is called

Alex Ar [27]

Answer:

the portion of electromagnetic spectrum is occupied by signal is called bandwidth

Explanation:

Bandwidth is the part of electromagnetic spectrum which is occupied by the signal. It is also called as the range of frequency over which the receiver or any other electronic circuit works. Bandwidth is the amount of the data been transmitted from one point to other within the network in the specific amount of the time. It is expressed as the bit rate and are measured as bits per seconds (bps). Bandwidth is the sum of the connections.

3 0

3 years ago

The relationship between power, force, and velocity is ______?

Kipish [7]

Answer: P = FV becuase all of them are equal in strength.

Explanation: In the straight forward cases where a constant force moves an object at constant velocity, the power is just P = Fv. In a more general case where the velocity is not in the same direction as the force, then the scalar product offorce and velocity must be used. P = Power, F = Force, V = Velocity.

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5 0

1 year ago

Read 2 more answers

An object with charge q = −6.00×10−9 C is placed in a region of uniform electric field and is released from rest at point A. Aft

sergeinik [125]

Answer:

a) 80 V

b) The magnitude of the electric field is 100 N/C and the direction of the electric field is from point B to point A where the electric field is toward the negative charge

Explanation:

Given :

We are given an object with charge q = -6.00 x I0^-9 C starts moving from the rest at point A, which means its kinetic energy at point A is zero ( $K_{A}$ = 0) to the point B at distance l = 0.500m where its kinetic energy is ( $K_{B}$ = 5.00 x 10^-7J) . Also, the electric potential of q at point A is VA = + 30.0 v.

Required :

(a) We are asked to find the electric potential VB

(b) We want to determine the magnitude and the direction of the electric field E.

Solution

(a) We are given the values for VA, $K_{B}$ and q so we want to find a relationship between these three parameters and VB to get the value of VB.

As we have two states, at points A and B , where the charge moved from A to B due to the applied electric field. The mechanical energy of the object is conservative during this travel, and we can apply eq(1) in this situation:

$K_{A} +U_{A} =K_{B} +U_{B}$ .........................................(1)

Where $K_{A}$ = 0 and the potential energy U of the charge is given by U = q V

where V is the electric potential. So, equation (1) will be in the form :

0+qVA= $K_{B}$ +qVB (Divide by q)

VA= $K_{B}$ /q + VB (solve for VB)

VB=VA- $K_{B}$ /q .......................................(2)

We get the relation between VB, VA and $K_{B}$ , now we can plug our values for VA, $K_{B}$ and q into equation (2) to get VB

VB=VA- $K_{B}$ /q

=30V-(5.00 x 10^-7J)/(-6.00 x I0^-9)

=80 V

(b) After we calculated VB we can use equation a to get the electric field E that applied to the charge q, where the potential difference between the two points equals the integration of the electric field multiplied by the distance l between the two points

VA-VB = $\int\limits^1_0 {E} \, dl$ ...................................(a)

$=E\int\limits^1_0 {} \, dl$

VA-VB=E $l$ (solve for E)

E= VA-VB/ $l$ ..................................(3)

Now let us plug our values for VA, Vs and l into equation (3) to get the electric field E

E= VA-VB/ $l$

=-100 N/C

The magnitude of the electric field is 100 N/C and the direction of the electric field is from point B to point A where the electric field is toward the negative charge

5 0

3 years ago