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Goshia [24]
3 years ago
10

Technician A says that a tire will increase or decrease approximately 1 psi for each 10°F change of temperature. Technician B sa

ys that if the vehicle has been driven to warm the tires, you must adjust the air pressure down to the manufacturer's recommended pressure. Who is correct?
A. Tech A
B. Tech B
C. Both Techs A and B
D. Neither Tech A nor B
Physics
1 answer:
bearhunter [10]3 years ago
8 0

Answer:

Technician A

Explanation:

It is seen that a tire pressure will increase or decrease 1 psi for each 10^{\circ} F  change in temperature.

For Technician B vehicle pressure should not be adjusted after tire has been warmed as the warm air may increase the pressure but it will be auto adjusted as  the temperature falls to normal .          

You might be interested in
When you push a child on a swing, your action is most effective when your pushes are timed to coincide with the natural frequenc
OleMash [197]

Answer:

T = 4.48 s

we can see that this time period is independent of the mass of the child so answer would be same if the child mass is different

Explanation:

Natural frequency of a simple pendulum of L length is given as

f = \frac{1}{2\pi}\sqrt{\frac{g}{L}}

so the time period of the oscillation is given as

T = 2\pi \sqrt{\frac{L}{g}}

so we will have

L = 5 m

T = 2\pi\sqrt{\frac{5}{9.81}}

T = 4.48 s

also from above formula we can see that this time period is independent of the mass of the child so answer would be same if the child mass is different

3 0
4 years ago
1) Calculate the torque required to accelerate the Earth in 5 days from rest to its present angular speed about its axis. 2) Cal
disa [49]

Answer:

a) τ = 4.47746 * 10^25 N-m

b) E = 2.06301 * 10^13 J

c) P = 3.25511*10^21 W

Explanation:

Given:

- The radius of earth r = 6.3781×10^6 m

- The angular speed of earth w = 7.27*10^-5 rad/s

- The time taken to reach above speed t = 5 yrs = 1.57784760 * 10^8 s

- The mass of earth m = 5.972 × 10^24 kg

- The inertia of sphere I = 2/5 * m* r^2

Solution:

- The angular acceleration of the earth from rest to w is given by α:

                                α = w / t

                                α = (7.27*10^-5) / (1.57784760 * 10^8)

                                α = 4.60754*10^-13 rad/s^2

- The required torque τ is given by:

                                τ = I*α

                                τ = 2/5 * m* r^2 * α

                           τ = 2/5 *(5.972 × 10^24) * (6.3781×10^6)^2 * (4.60754*10^-13)

                            τ = 4.47746 * 10^25 N-m

- The power required P to turn the earth to the speed w is:

                           P = τ*w

                           P = (4.47746 * 10^25)*(7.27*10^-5)

                           P = 3.25511*10^21 W

- The energy E required is :

                           E = P / t

                           E = (3.25511*10^21) / (1.57784760 * 10^8)

                           E = 2.06301 * 10^13 J

4 0
3 years ago
A 100 g aluminum calorimeter contains 250 g of water. The two substances are in thermal equilibrium at 10°C. Two metallic blocks
ipn [44]

Answer:

A. 1,950 J/kgºC

Explanation:

Assuming that all materials involved, finally arrive to a final state of thermal equilibrium, and neglecting any heat exchange through the walls of the calorimeter, the heat gained by the system "water+calorimeter" must be equal to the one lost by the copper and the unknown metal.

The equation that states how much heat is needed to change the temperature of a body in contact with another one, is as follows:

Q = c * m* Δt

where m is the mass of the body, Δt is the change in temperature due to the external heat, and c is a proportionality constant, different for each material, called specific heat.

In our case, we can write the following equality:

(cAl * mal * Δtal) + (cH₂₀*mw* Δtw) = (ccu*mcu*Δtcu) + (cₓ*mₓ*Δtₓ)

Replacing by the givens , and taking ccu = 0.385 J/gºC and cAl = 0.9 J/gºC, we have:

Qg= 0.9 J/gºC*100g*10ºC + 4.186 J/gºC*250g*10ºC  = 11,365 J(1)

Ql = 0.385 J/gºC*50g*55ºC + cₓ*66g*80ºC = 1,058.75 J + cx*66g*80ºC (2)

Based on all the previous assumptions, we have:

Qg = Ql

So, we can solve for cx, as follows:

cx = (11,365 J - 1,058.75 J) / 66g*80ºC = 1.95 J/gºC (3)

Expressing (3) in J/kgºC:

1.95 J/gºC * (1,000g/1 kg) = 1,950 J/kgºC

3 0
3 years ago
A negative test charge experiences a force to the right as a result of an electric field. Which is the best conclusion to draw
Alona [7]

Answer:a

Explanation:

4 0
3 years ago
A nuclear particle with no charge
pishuonlain [190]
A nuclear particle with no charge is a neutron
7 0
3 years ago
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