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kondaur [170]
3 years ago
14

A 620 N physics student stands on a bathroom scale in an elevator that is supported by a cable. The combined mass of student plu

s elevator is 870 kg. As the elevator starts moving, the scale reads 450 N.
Required:
a. Find the acceleration of the elevator (magnitude and direction).
b. What is the acceleration if the scale reads 670 N?
c. If the scale reads zero, should the student worry? Explain.
d. What is the tension in the cable in parts (a) and (c)?
Physics
1 answer:
Dima020 [189]3 years ago
7 0

Answer:

(a) 9.28 m/s2

(b) 9.03 m/s2

(c) 9.8 m/s2

(d) 450 N, 670 N

Explanation:

mass of elevator + student, m = 870 kg

Reading of scale, R = 450 N

(a) When the elevator goes down, the weight decreases.

Let the acceleration is a.

By the Newton's second law

m g - R = m a

870 x 9.8 - 450 = 870 a

a = 9.28 m/s2

(b) R = 670 N

Let the acceleration is a.

870 x 9.8 - 670 = 870 a

a = 9.03 m/s2

(c) If the scale reads zero, it mean the elevator is falling freely. The acceleration is downwards and its value is 9.8 m/s2.

(d) Tension in cable is 450 N and 670 N.

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A long, straight wire carries a current of 5.20 A. An electron is traveling in the vicinity of the wire. Part A) At the instant
Gre4nikov [31]

A) 2.2\cdot 10^{-19} N

The strength of the magnetic field produced by a current-carrying wire is given by

B=\frac{\mu_0 I}{2\pi r}

where

\mu_0 is the vacuum permeability

I is the current in the wire

r is the distance from the wire

In this situation,

I = 5.20 A

r = 4.60 cm = 0.046 m is the distance of the electron from the wire

Therefore the magnetic field strength at the electron's location is

B=\frac{(4\pi \cdot 10^7)(5.20)}{2\pi (0.046}=2.26\cdot 10^{-5} T

The force exerted on a charged moving particle travelling perpendicular to a magnetic field is given by

F=qvB

where

q is the magnitude of the charge of the particle

v is its velocity

B is the magnetic flux density

For the electron of this problem,

q=1.6\cdot 10^{-19} C is the charge

v=6.10\cdot 10^4 m/s is the speed

B=2.26\cdot 10^{-5} T is the magnetic field

Substituting,

F=(1.6\cdot 10^{-19})(6.10\cdot 10^4)(2.26\cdot 10^{-5})=2.2\cdot 10^{-19} N

B) Same direction as the current in the wire

First of all we have to find the direction of the magnetic field lines, which are concentric around the wire. Assuming the wire carries a current pointing upward, then if we use the right-hand rule:

- The thumb gives the direction of the current -> upward

- The other fingers wrapped give the direction of the field lines -> anticlockwise around the wire (as seen from top)

Now the direction of the force can be found by using the right-hand rule. We have:

- direction of the index finger = direction of motion of the electron (toward the wire, let's assume from east to west)

- middle finger = direction of the magnetic field (to the north)

- Thumb = direction of the force --> downward

However, the electron carries a negative charge, so we must reverse the direction of the force: therefore, the force experienced by the electron will be upward, so in the same direction as the current in the wire.

8 0
4 years ago
What is the temperature change of 1.0 mol of a monotomic gas if its thermal energy is increased by 1.0 J ? Express your answer i
natali 33 [55]

Answer: 0.08K

Explanation:

When temperature changes, the corresponding change in thermal energy of a gas is given by:

ΔE (thermal) = 3/2nRΔT

Defining the parameters:

ΔE (thermal) = Increase in thermal energy of the mono atomic gas = 1.0J

n = number of moles of the gas = 1.0mol

R = Ideal gas constant = 8.314J/mol/K

ΔT = change in temperature. This is what we need to find.

Rearranging the equation to make ΔT subject of the formula,

ΔT= 2 x ΔE (thermal) / (3 x n x R)

Therefore, ΔT = 2 x 1.0J / (3 x 1.0mol x 8.314J/mol/K)

ΔT = 2.0J / 24.942J/K

ΔT = 0.0802K

ΔT = 0.08K

The temperature change of 1.0mol of a monoatomic gas if its thermal energy is increased by 1.0J is 0.08K.

7 0
3 years ago
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Anon25 [30]

Answer:

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19.2 cm = 0.192 m

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57 rpm = 57 * 2π / 60 = 6 rad/s

86 rpm = 86 * 2π / 60 = 9 rad/s

The angular acceleration of the sprocket is the change in angular velocity per unit of time

\alpha = \frac{\Delta \omega}{\Delta t} = \frac{9 - 6}{10.7} = 0.284 rad/s^2

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3 0
4 years ago
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