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Elodia [21]
3 years ago
11

A horizontal force in used to pull a 5 kilogram cart at a constant speed of 5 meters per second across the floor as shown in the

diagram. if the force of friction between the cart and the floor is 10 newtons, the magnitude of the horizontal force along the handle of the cart is?
a. 5N
b. 10 N
c. 25 N
d. 50 N
Physics
1 answer:
ddd [48]3 years ago
5 0

A cart is pulled by horizontal force such that it moves with constant velocity

So here since velocity is constant we can say that its acceleration will be ZERO

now here for zero acceleration we can say

F_{net} = 0

So here we will have

F_{net} = F_{ap} - F_f = 0

here we know that

F_f = 10 N

so we will have

F_{ap} - 10 = 0

F_{ap} = 10 N

so here applied force on handle will be

<em>b. 10 N</em>

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If Juan upgrades from a sports car to a large truck, what will happen to the
Brrunno [24]

Answer:

The force must increase

Explanation:

According to newton's second law "force is the product of mass and acceleration".

      Force  = mass  x acceleration

Now, the mass of the sports car is lesser compared to that of the truck. Therefore, to take both automobiles to the same speed, enough force must be applied by the engine of the truck.

There must be an increase in the force in order to make both automobiles attain the same speed.

5 0
3 years ago
What is the correct answer?
pentagon [3]

Answer:

B) x^2+6x+8

Explanation:

x-4 | x^3+2x^2-16x-32

    -  x^3-4x^2             <-- (x-4)(x^2)

_________________

              6x^2-16x-32

           -  6x^2-24x     <-- (x-4)(6x)

_________________

                          8x-32

                       -  8x-32 <- (x-4)(8)

___________________________

                                 0 | x^2+6x+8

This means the answer is B) x^2+6x+8

             

3 0
3 years ago
A child is riding in a child-restraint chair, securely fastened to the seat of a car. Assume the car has speed 47 km/h when it h
Elina [12.6K]

Answer: F = 1235 N

Explanation: Newton's Second Law of Motion describes the effect of mass and net force upon acceleration: F_{net}=m.a

Acceleration is the change of velocity in a period of time: a=\frac{\Delta v}{\Delta t}

Velocity of the car is in km/h. Transforming it in m/s:

v=\frac{47.10^{3}}{36.10^{2}}

v = 13 m/s

At the moment the car decelerates, acceleration is

a=\frac{13}{0.2}

a = 65 m/s²

Then, force will be

F_{net}=19(65)

F_{net} = 1235 N

The horizontal net force the straps of the restraint chair exerted on the child to hold her is 1235 newtons.

5 0
3 years ago
A thermometer initially reading 212F is placed in a room where the temperature is 70F. After 2 minutes the thermometer reads 125
frez [133]

Answer:

91.3°F

Explanation:

Let T be the temperature of the thermometer at any time

T∞ be the temperature of the room = 70°F

T₀ be the initial temperature of the thermometer = 212°F

And m, c, h are all constants from the cooling law relation

From Newton's law of cooling

Rate of Heat loss by the cake = Rate of Heat gain by the environment

- mc (d/dt)(T - T∞) = h (T - T∞)

(d/dt) (T - T∞) = dT/dt (Because T∞ is a constant)

dT/dt = (-h/mc) (T - T∞)

Let (h/mc) be k

dT/(T - T∞) = -kdt

Integrating the left hand side from T₀ to T and the right hand side from 0 to t

In [(T - T∞)/(T₀ - T∞)] = -kt

(T - T∞)/(T₀ - T∞) = e⁻ᵏᵗ

(T - T∞) = (T₀ - T∞)e⁻ᵏᵗ

Inserting the known variables

(T - 70) = (212 - 70)e⁻ᵏᵗ

(T - 70) = 142 e⁻ᵏᵗ

At t = 2 minute, T = 125°F

125 - 70 = 142 e⁻ᵏᵗ

55/142 = e⁻ᵏᵗ

- kt = In (55/142) = In (0.3873)

- k(2) = - 0.9485

k = 0.4742 /min

At time t = 4 mins

kt = 0.4742 × 4 = 1.897

(T - 70) = 142 e⁻ᵏᵗ

e^(-1.897) = 0.15

T - 70 = 142 × 0.15 = 21.3

T = 91.3°F

7 0
3 years ago
Which option correctly matches the chemical formula of a compound with its name?
77julia77 [94]

Answer:it’s A

Explanation:

4 0
3 years ago
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