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sweet [91]
3 years ago
15

A grocery cart with a mass of 15 kg is pushed at constant speed along an aisle by a force fp = 12 n which acts at an angle of 17

° below the horizontal. find the work done by each of the forces on the cart if the aisle is 14 m long.
Physics
1 answer:
Korolek [52]3 years ago
8 0

<span>Given:

Mass of cart: 15kg</span>

Aisle length = 14m

Angle = 17° below the horizontal

Force fp = 12 N

 

So, the solution would be like this for this specific problem:

 

<span><span>1)    </span>W(by applied force) = F(applied) x s x cosθ <span>
=>W(a) = 12 x 14 x cos17* = 160.66 J </span></span>
<span><span>
2)    </span>By F(net) = F(applied) - F(friction) <span>
=>As v = constant => a = 0 => F(net) = 0 
=>F(applied) = F(friction) 
<span>=>W(friction) = -F(friction) x s
=>W(friction) = -F(applied) x s 
=>W(f) = -12 x 14 x cos17* = -160.56 J </span></span></span>
<span><span>
3)    </span>0, as the displacement is perpendicular to Force </span>
<span><span>
4)    </span>0, as the displacement is perpendicular to Force</span>  
<span>
To add, the force that is applied<span> to an object by a person or another object is called the applied force.</span></span>
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To model this process, assume two charged spherical conductors are connected by a long conducting wire and a 1.20-mC charge is p
il63 [147K]

Answer:

Part a: The electric potential of each sphere is 1.35x10⁸V

Part b: The electric field at the surface of sphere 1 and 2 is 2.25x10⁹ N/C and 6.75x10⁹ N/C respectively

Explanation:

As the complete question is not given, the similar question is attached herewith. The values are used as indicated in the given question

Let r_1 = 6 cm=0.06 m

r2 = 2 cm = 0.02 m

Q = 1.2 mC

Let q1 and q2 are the charges on each sphere.

q1 + q2 = 1.2 mC -------(1)

In the equilibrium, V1 = V2

k*q1/r1 = k*q2/r2

q1/0.06 = q2/0.02

q1/q2 = 0.06/0.02

q1/q2 = 3 ---------(2)

On solving equation 1 and 2

we get

q1 = 0.9 mC

q2 = 0.3 mC

So

V1 = k*q1/r1 = (9*10^9*0.9*10^-3)/0.06 = 1.35*10^8 Volts

V2 = k*q2/r2 = 9*10^9*0.3*10^-3/0.02 = 1.35*10^8 Volts

So the electric potential of each sphere is 1.35x10⁸V

Part b

Now the electric potential is given as

E1 = k*q1/r1^2 = 9*10^9*0.9*10^-3/0.06^2 = 2.25*10^9 N/C

E2 = k*q2/r2 = 9*10^9*0.3*10^-3/0.02^2 = 6.75*10^9 N/C

So the electric field at the surface of sphere 1 and 2 is 2.25x10⁹ N/C and 6.75x10⁹ N/C respectively

7 0
3 years ago
Calculate the kinetic energy in joules of an automobile weighing 2135 lb and traveling at 55 mph. (1 mile = 1.6093 km, 1 lb = 45
victus00 [196]
<span>Let's convert the speed to m/s: speed = (55 mph) (1609.3 m / mile) (1 hour / 3600 seconds) speed = 24.59 m/s Let's convert the mass to kilograms: mass = (2135 lb) (0.45359 kg / lb) mass = 968.4 kg We can find the kinetic energy KE: KE = (1/2) m v^2 KE = (1/2) (968.4 kg) (24.59 m/s)^2 KE = 292780 joules The kinetic energy of the automobile is 292780 joules.</span>
4 0
3 years ago
What effort force will be required to lift the 20 N object using the pulley above?
Umnica [9.8K]

Answer:

I think it is 5N.

Explanation:

4 0
3 years ago
Air contained in a rigid, insulated tank fitted with a paddle wheel, initially at 300 K, 2 bar, and a volume of 3 m3, is stirred
madam [21]

Answer:

a)P₂ =4 bar

b)W= - 1482.48 KJ

It means that work done on the system.

c)S₂ - S₁ = 3.42 KJ/K

Explanation:

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T₁ = 300 K   ,V₁ = 3 m³  ,P₁=2 bar

T₂ = 600 K ,V₂=V₁ 3 m³

Given that tank is rigid and insulated.It means that volume of the gas will remain constant.

Lets take the final pressure = P₂

For ideal gas  P V = m R T

\dfrac{P_2}{P_1}=\dfrac{T_2}{T_1}

P_2=\dfrac{T_2}{T_1}\times P_1

P_2=\dfrac{600}{300}\times 2

P₂ =4 bar

Internal energy

ΔU = m Cv ΔT

Cv=0.71 KJ/kg.k for air

m=\dfrac{PV}{RT}

m=\dfrac{200\times 3}{0.287\times 300}\ kg

m= 6.96 kg

ΔU= 6.96 x 0.71 x (600 - 300)

ΔU=1482.48 KJ

From first law

Q= ΔU + W

Q= 0  Insulated

W = - ΔU

W= - 1482.48 KJ

It means that work done on the system.

Change in the entropy

S_2-S_1=mC_v\ \ln\dfrac{T_2}{T_1}

S_2-S_1=6.96\times 0.71\ \ln\dfrac{600}{300}

S₂ - S₁ = 3.42 KJ/K

 

5 0
3 years ago
You are going 30 m/s in a car of mass 1000 kg when you reach a red light and stop your car.
Radda [10]

Answer:

Explanation:

The kinetic energy will convert to heat energy (provided the car has friction brakes and not regenerative brakes as might be found on an electric or hybrid) Also<u> assuming level road</u>.

E = ½mv² = ½(1000)30² = 450,000 J

5 0
2 years ago
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