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GarryVolchara [31]
2 years ago
3

A crate of fragile dishes is in the back of a pickup truck. The truck accelerates north from a stop sign, and the crate moves wi

thout slipping.
Does the friction force on the crate point north or south?

Or is the friction force zero?

Explain in complete sentences.
Physics
1 answer:
sergejj [24]2 years ago
8 0

Answer and Explanation:

As the truck accelerates towards north, the crate in the back of the pickup truck is accelerated towards south without slipping but the friction force between the bottom of the crate and the surface of truck is along the direction of the truck and thus the direction of truck is north so the fiction force also would be along north.

HELP

As the truck accelerates towards north, the crate will experience a force towards south, thus accelerating it (crate) due south. The force of friction between the crate and truck floor will thus act towards north direction to oppose the motion of the crate.

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A student throws a 0.22 kg rock horizontally at 20.0 m/s from 10.0 m above the ground. Find the initial kinetic energy of the ro
LekaFEV [45]

Answer:

44J

Explanation:

Given parameters:

Mass of rock  = 0.22kg

Initial velocity  = 20m/s

Distance moved  = 10m

Unknown:

Initial kinetic energy of the rock  = ?

Solution:

To solve this problem, we need to understand that kinetic energy is the energy due to the motion of a body.

It is mathematically expressed as;

     Kinetic energy  = \frac{1}{2} m v²

m is the mass

v is the velocity

   Kinetic energy  =  \frac{1}{2} x 0.22 x 20²   = 44J

6 0
1 year ago
Compare and contrast the strength of the forces between two objects with a mass of 1 kg each, a charge of 10
DochEvi [55]

Answer:

Let's see the similarities between the two forces

* are proportional to the product of a magnitude, mass or charge

* They are inversely proportional to the square of the distance

* They are long-range forces since zero is not made up to an infinite distance. The gravitational force is always attractive, the electrical force can be attractive or repulsive.

The differences in them

* The electric force in much greater than the gravitational force

* The gravitational force is always attractive, the electrical force can be attractive or repulsive.

Explanation:

Let's start by calculating each force.

Gravitational force

             F =G \frac{m_1m_2}{r^2}  

let's calculate

             F = 6.67 10⁻¹¹  1  1 / 1²

             F = 6.67 10⁻¹¹ N

Electric force

             F = k \frac{q_1q_2}{r^2}  

indicates that the charge is q = 10 C

            F = 9 10⁹ 10 10 / 1²

            F = 9 10¹¹ N

Let's see the similarities between the two forces

* are proportional to the product of a magnitude, mass or charge

* They are inversely proportional to the square of the distance

* They are long-range forces since zero is not made up to an infinite distance. The gravitational force is always attractive, the electrical force can be attractive or repulsive.

The differences in them

* The electric force in much greater than the gravitational force

* The gravitational force is always attractive, the electrical force can be attractive or repulsive.

3 0
2 years ago
Two 5.0 mm × 5.0 mm electrodes are held 0.10 mm apart and are attached to 7.5 V battery. Without disconnecting the battery, a 0.
Musya8 [376]

Answer:

A) V = 7.5 V

B) E = 75,000 V/m

C) Q = 16.6 pC

D) V = 7.5 V

E) E = 24,000 V/m

F) Q = 52 pC

Explanation:

Given:

- The Area of plate A = ( 5 x 5 ) mm^2

- The distance between plates d = 0.10 mm

- The thickness of Mylar added t = 0.10 mm

- Voltage supplied by battery V = 7.5 V

Solution:

A) What is the capacitor's potential difference before the Mylar is inserted?

- The potential difference across the two plates is equal to the voltage provided by the battery V = 7.5 V which remains constant throughout.

B) What is the capacitor's electric field before the Mylar is inserted?

- The Electric Field E between the capacitor plates is given by:

                                E = V / k*d

k = 1 (air)                  E = 7.5 / 0.10*10^-3

                                E = 75,000 V/m

C) What is the capacitor's charge Q before the Mylar is inserted?

                                C = k*A*ε / d

k = 1 (air)                   C = ( 0.005^2 * 8.85*10^-12 ) / 0.0001

                                C = 2.213 pF

                                Q = C*V

                                Q = 7.5*(2.213)

                                Q = 16.6 pC

D) What is the capacitor's potential difference after the Mylar is inserted?

- The potential difference across the two plates is equal to the voltage provided by the battery V = 7.5 V which remains constant throughout.

E) What is the capacitor's electric field after the Mylar is inserted?    

- The Electric Field E between the capacitor plates is given by:

                                E = V / k*d

k = 3.13                     E = 7.5 / (3.13)0.10*10^-3

                                E = 24,000 V/m              

F) What is the capacitor's charge after the Mylar is inserted?      

                                C = k*A*ε / d

k = 3.13                    C = 3.13*( 0.005^2 * 8.85*10^-12 ) / 0.0001

                                C = 6.927 pF

                                Q = C*V

                                Q = 7.5*(6.927)

                                Q = 52 pC                                      

6 0
2 years ago
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BabaBlast [244]
I could help
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3 0
2 years ago
A 1500 kg car traveling at 20 m/s suddenly runs out of gas while approaching the valley shown in the figure. The alert driver im
geniusboy [140]

Answer:

v_f = 17.4 m / s

Explanation:

For this exercise we can use conservation of energy

starting point. On the hill when running out of gas

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final point. Arriving at the gas station

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energy is conserved

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we calculate

        v_f ² = 20² + 2 9.8  (10 -15)

        v_f = √302

         v_f = 17.4 m / s

8 0
2 years ago
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