As a diligent physics student, you carry out physics experiments at every opportunity. At this opportunity, you carry a 1.05 m 1.05 m long rod as you jog at 3.27 m/s 3.27 m/s , holding the rod perpendicular to your direction of motion. What is the strength of the magnetic field that is perpendicular to both the rod and your direction of motion and that induces an EMF of 0.275 mV 0.275 mV across the rod? Express the answer in milliteslas.

Final answer:

The acceleration of object Y is 2.5 m/s/s which is less than the acceleration of object X which is 3.0 m/s/s.

Explanation:

The acceleration of an object is calculated using the formula: a = (v_f -v_i) / t, where v_f is the final velocity, v_i is the initial velocity and t is time. For object Y, the final velocity is 25m/s, the initial velocity is 0m/s and the time is 10s, hence its acceleration will be (25-0)/10 = 2.5 m/s/s. This indicates that Object Y's acceleration is less compared to Object X's acceleration of 3.0 m/s/s.

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a. Approximately 1.5 revolutions per minute are needed for the "artificial gravity" acceleration to be 9.80 m/s²

b. approximately 1 revolutions per minute are needed to simulate the acceleration due to gravity on the Martian surface.

Given that,

Humans living in outer space experience weightlessness, which can be a challenge.

One solution is to design a space station that spins at a constant rate.

This spinning creates "artificial gravity" at the outside rim of the station.

The diameter of the space station is 800 m.

The desired acceleration for artificial gravity is 9.80 m/s².

(a) To find the number of revolutions per minute,

Determine the tangential velocity at the rim of the space station.

Use the formula for centripetal acceleration:

a = (v²) / r

Where,

a is the desired acceleration (9.80 m/s²),

v is the tangential velocity at the rim of the space station (which we need to find),

r is the radius of the space station (half of the diameter, 400 m).

Rearranging the formula, we have:

Plugging in the given values, we get:

Now, to find the number of revolutions per minute,

Convert the tangential velocity to the circumference of the space station:

C = 2πr

So, the circumference is:

C = 2π x 400 m

C = 2513.27 m

Now, calculate the number of revolutions per minute by dividing the tangential velocity (62.60 m/s) by the circumference:

n = v/C

n = 62.60/2513.27  

n = 0.0249 rev/s

To convert this to revolutions per minute, multiply by 60:

n = 0.0249 rev/s x 60 s/min

n = 1.49 rev/min

Therefore, approximately 1.5 revolutions per minute are needed for the "artificial gravity" acceleration to be 9.80 m/s² in the space station with a diameter of 800 m.

(b) To achieve the desired acceleration due to gravity (3.70 m/s²), Put:

a = 3.70 m/s²

r = 400 m

We have:

Plugging in the values, we get:

Now, calculate the number of revolutions per minute required to achieve this tangential velocity.

Using the circumference formula: C = 2πr

The circumference is:

C = 2π x 400 m

C = 2513.27 m

Now, calculate the number of revolutions per minute by dividing the tangential velocity (38.47 m/s) by the circumference:

n = v/C

n = 38.47 m/s / 2513.27 m

n = 0.0153 rev/s

To convert this to revolutions per minute, multiply by 60:

n = 0.0153 x 60  

n = 0.918 rev/min

Therefore, approximately 1 revolutions per minute are needed to simulate the acceleration due to gravity on the Martian surface (3.70 m/s²) in the waiting area of the space station.

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Final answer:

To create artificial gravity on a space station, you can spin it about its center. To calculate the number of revolutions per minute needed for a desired acceleration, use the formula a = rω^2. For the diameter of 800 m, the revolutions per minute needed to achieve a gravitational acceleration of 9.80 m/s^2 is approximately 2.63. If simulating the gravity on the Mars surface with an acceleration of 3.70 m/s^2, the revolutions per minute required would be around 1.77.

Explanation:

(a) To calculate the number of revolutions per minute needed for artificial gravity acceleration to be 9.80 m/s2, we can use the formula:

a = rω2

where a is the acceleration, r is the radius of the space station, and ω is the angular velocity. Since the diameter of the space station is 800 m, the radius would be 400 m. Rearranging the formula, we get:

ω = sqrt(a/r)

Substituting the values, we have:

ω = sqrt(9.80/400) ≈ 0.22 rad/s

Now, we can convert the angular velocity to revolutions per minute:

Revolution per minute = (ω × 60) / (2π)

Substituting the value of ω, we get:

Revolution per minute ≈ (0.22 × 60) / (2π) ≈ 2.63 revolutions per minute

(b) To simulate the acceleration due to gravity on the Martian surface (3.70 m/s2), we can use the same formula and follow similar steps as before. Substituting a = 3.70 m/s2 and r = 400 m, we can calculate ω. Converting it to revolutions per minute, we get:

Revolution per minute ≈ (ω × 60) / (2π) ≈ 1.77 revolutions per minute

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53 yards per houris a speed value and when converted gives 320544 inches per week.

What is speed value?

The speed value is defined by the distance traveled by the object with reference to the time. It is a scalar quantity, with a magnitude but no direction. The speed is calculated using the formula:

Speed = Distance ÷ Time

It can be measured in, meters per second, miles per hour, kilometers per hour, etc.

From the known data:
7 days per week
24 Hours per day
3 ft per yard
12 inches per feet

The conversion of yards per hour into inches per week is given as:

= 53 yards per hour × 3 feet per yard × 12 inches per foot × 24 hour per day × 7 day per week

= 53 × 3 × 12 × 24 × 7
= 320,544 inches per week

Therefore, the unit is converted and is 320,544 inches per week.

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The pressure exerted on your thumb by the thumbtack is approximately 3.818 x 10^5 Pascal (Pa).

To calculate the pressure on your thumb, we can use the formula:

Pressure = Force / Area

Force (F) = 30 N

Radius of the thumbtack head (r) = 5 mm = 0.005 m

First, we need to calculate the area of the thumbtack head. Since the head is circular, the area can be found using the formula for the area of a circle:

Area =

Area =

Area ≈

Now we can calculate the pressure:

Pressure = Force / Area

Pressure = 30 N / 7.854 x 10^(-5) m^2

Pressure ≈ Pa

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Answer:

(c) 2

Explanation:

Heat transfer across the walls due to conduction is given by:

where,

q = heat transfer rate

K = thermal conductivity

A = Area

ΔT = change in temperature

L = thickness

For wall A:

For wall B:

Because the change of temperature and area of walls are the same. Dividing both terms:

using values given in the question:

Therefore, the correct answer is:

(c) 2