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inn [45]
1 year ago
7

A mechanic changing the spark plugs in a car notes that the instruction manual calls for a torque with a magnitude of

Physics
1 answer:
Papessa [141]1 year ago
5 0

The magnitude (in N) of the force she must exert on the wrench is 150.1 N.

<h3>Force exerted by the wrench</h3>

The force exerted by the wrench is calculated using torque formula as follows;

torque, τ = F x r x sinθ

where;

  • F is the applied force
  • r is the perpendicular distance if force applied

F =  τ /(r sinθ)

F = (39) / (0.3 sin 60)

F = 150.1 N

Thus, the magnitude (in N) of the force she must exert on the wrench is 150.1 N.

Learn more about torque here: brainly.com/question/14839816

#SPJ1

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Marissaâs car accelerates uniformly at a rate of +2.60 m/s². How long does it take for Marissaâs car to accelerate from a speed
DIA [1.3K]

Answer:

The time taken by the car to accelerate from a speed of 24.6 m/s to a speed of 26.8 m/s is 0.84 seconds.

Explanation:

Given that,

Acceleration of the car, a=+2.6\ m/s^2

Initial speed of the car, u = 24.6 m/s

Final speed of the car, v = 26.8 m/s

We need to find the time taken by the car to accelerate from a speed of 24.6 m/s to a speed of 26.8 m/s. The acceleration of an object is given by :

t=\dfrac{v-u}{a}

t=\dfrac{(26.8-24.6)\ m/s}{2.6\ s}

t = 0.84 seconds

So, the time taken by the car to accelerate from a speed of 24.6 m/s to a speed of 26.8 m/s is 0.84 seconds. Hence, this is the required solution.                                    

4 0
3 years ago
Greg is in a car at the top of a roller-coaster ride. The distance, d, of the car from the ground as the car descends is determi
Talja [164]
The solution to the problem is as follows:

 <span>Average = 80 
So Sum = 80 * 5 = 400 
Mode = 88, so two results are 88 (if three results were 88, then the median would be 88). 
Three results are 81, 88, and 88. 
That leaves 143. We could still have one 81 score, so that leaves the lowest score as 62. 

Greg is in a car at the top of a roller-coaster ride. The distance, d, of the car from the ground as the car descends is determined by the equation d = 144 - 16t2, where t is the number of seconds it takes the car to travel down to each point on the ride. How many seconds will it take Greg to reach the ground? 
d = 144 - 16t2 
0 = 144 - 16t2 
16t^2=144 
t^2=9 
t=3</span>
3 0
3 years ago
A sports car accelerates from 0 to 25 meters per second in 4 seconds. What is its acceleration?
WITCHER [35]

Answer:

6.25 ms²

Explanation:

..................

3 0
2 years ago
A 57-kg woman holds a 6-kg package as she stands within an elevator which briefly accelerates upward at a rate of 0.15g. Determi
Temka [501]

Answer:

R = 710.7N

L = 67.689 N

During gravity fall L = R = 0 N

Explanation:

So the acceleration that the elevator is acting on the woman (and the package) in order to result in a net acceleration of 0.15g is

g + 0.15g = 1.15g

The force R that the elevator exerts on her feet would be product of acceleration and total mass (Newton's 2nd law):

a(m + M) = 1.15g(57 + 6) = 1.15*9.81*63 = 710.7N

The force L that she exerts on the package would be:

am = 1.15g *6 = 1.15*9.81*6 = 67.689N

When the system is falling, all have a net acceleration of g. So the acceleration that the elevator exerts on the woman (and the package) is 0, and so are the forces L and R.

7 0
3 years ago
A sample of monatomic ideal gas occupies 5.00 L at atmospheric pressure and 300 K (point A). It is warmed at constant volume to
leonid [27]

Answer:

(a) 0.203 moles

(b) 900 K

(c) 900 K

(d) 15 L

(e) A → B, W = 0, Q = Eint = 1,518.91596 J

B → C, W = Q ≈ 1668.69974 J Eint = 0 J

C → A, Q = -2,531.5266 J, W = -1,013.25 J, Eint = -1,518.91596 J

(g) ∑Q = 656.089 J, ∑W =  655.449 J, ∑Eint = 0 J

Explanation:

At point A

The volume of the gas, V₁ = 5.00 L

The pressure of the gas, P₁ = 1 atm

The temperature of the gas, T₁ = 300 K

At point B

The volume of the gas, V₂ = V₁ = 5.00 L

The pressure of the gas, P₂ = 3.00 atm

The temperature of the gas, T₂ = Not given

At point C

The volume of the gas, V₃ = Not given

The pressure of the gas, P₃ = 1 atm

The temperature of the gas, T₂ = T₃ = 300 K

(a) The ideal gas equation is given as follows;

P·V = n·R·T

Where;

P = The pressure of the gas

V = The volume of the gas

n = The number of moles present

R = The universal gas constant = 0.08205 L·atm·mol⁻¹·K⁻¹

n = PV/(R·T)

∴ The number of moles, n = 1 × 5/(0.08205 × 300) ≈ 0.203 moles

The number of moles in the sample, n ≈ 0.203 moles

(b) The process from points A to B is a constant volume process, therefore, we have, by Gay-Lussac's law;

P₁/T₁ = P₂/T₂

∴ T₂ = P₂·T₁/P₁

From which we get;

T₂ = 3.0 atm. × 300 K/(1.00 atm.) = 900 K

The temperature at point B, T₂ = 900 K

(c) The process from points B to C is a constant temperature process, therefore, T₃ = T₂ = 900 K

(d) For a constant temperature process, according to Boyle's law, we have;

P₂·V₂ = P₃·V₃

V₃ = P₂·V₂/P₃

∴ V₃ = 3.00 atm. × 5.00 L/(1.00 atm.) = 15 L

The volume at point C, V₃ = 15 L

(e) The process A → B, which is a constant volume process, can be carried out in a vessel with a fixed volume

The process B → C, which is a constant temperature process, can be carried out in an insulated adjustable vessel

The process C → A, which is a constant pressure process, can be carried out in an adjustable vessel with a fixed amount of force applied to the piston

(f) For A → B, W = 0,

Q = Eint = n·cv·(T₂ - T₁)

Cv for monoatomic gas = 3/2·R

∴ Q = 0.203 moles × 3/2×0.08205 L·atm·mol⁻¹·K⁻¹×(900 K - 300 K) = 1,518.91596 J

Q = Eint = 1,518.91596 J

For B → C, we have a constant temperature process

Q = n·R·T₂·㏑(V₃/V₂)

∴ Q = 0.203 moles × 0.08205 L·atm/(mol·K) × 900 K × ln(15 L/5.00 L) ≈ 1668.69974 J

Eint = 0

Q = W ≈ 1668.69974 J

For C → A, we have a constant pressure process

Q = n·Cp·(T₁ - T₃)

∴ Q = 0.203 moles × (5/2) × 0.08205 L·atm/(mol·K) × (300 K - 900 K) = -2,531.5266 J

Q = -2,531.5266 J

W = P·(V₂ - V₁)

∴ W = 1.00 atm × (5.00 L - 15.00 L) = -1,013.25 J

W = -1,013.25 J

Eint = n·Cv·(T₁ - T₃)

Eint = 0.203 moles × (3/2) × 0.08205 L·atm/(mol·K) × (300 K - 900 K) = -1,518.91596 J

Eint = -1,518.91596 J

(g) ∑Q = 1,518.91596 J + 1668.69974 J - 2,531.5266 J = 656.089 J

∑W = 0 + 1668.69974 J -1,013.25 J = 655.449 J

∑Eint = 1,518.91596 J + 0 -1,518.91596 J = 0 J

5 0
3 years ago
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