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jeka57 [31]
3 years ago
11

A truck accelerates uniformly from rest to a speed of 9.2m/s over a distance of 26.2 m. Determine the acceleration of the truck.

Physics
1 answer:
Anastasy [175]3 years ago
5 0

Answer: 1.62m/s^2

Explanation: Using V^2=U^2+2a*s

V=9.2m/s, U=0, d = 26.2m

V^2=2a*s

a = V^2/2s

a = 9.2^2/2*26.2

a = 1.62m/s^2

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Bezzdna [24]

Common symbol of the volume (L)

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2 years ago
A person gives a shopping cart an initial push along a horizontal floor to get it moving, and then lets go. The cart travels for
mash [69]

Answer:

Only a backward force is acting, no forward force.

Explanation:

  • Once released from the initial push, in absence of friction, the shopping cart would continue moving forward at a constant speed forever.
  • As it would move at a constant speed, no net force would be acting on it.
  • So, if it is gradually slowing, there must  be a net force producing an acceleration in a direction opposite to the movement.
  • This force is the kinetic friction force, and is the only force acting on the cart in the horizontal direction.
  • As any friction force, opposes to the relative movement between the cart and the horizontal floor, which means that is directed backward.
  • This is consistent with the direction of the acceleration of the cart.
8 0
3 years ago
What is a work out time setting? (Gym)
tigry1 [53]

Answer:

It determines how long you do a certain workout.

3 0
3 years ago
A wedge with an inclination of angle θ rests next to a wall. A block of mass m is sliding down the plane. There is no friction b
Softa [21]

Answer:

  The net force on the block  F(net)  = mgsinθ).

   Fw =mg(cosθ)(sinθ)

Explanation:

(a)

Here, m is the mass of the block, n is the normal force, \thetaθ is the wedge angle, and Fw  is the force exerted by the wall on the wedge.

Since the block sliding down, the net force on the block is along the plane of the wedge that is equal to horizontal component of weight of the block.

                    F(net)  = mgsinθ

The net force on the block  F(net)  = mgsinθ).

The direction of motion of the block is along the direction of net force acting on the block. Since there is no frictional force between the wedge and block, the only force acting on the block along the direction of motion is mgsinθ.

(b)

From the free body diagram, the normal force n is equal to mgcosθ .

                           n=mgcosθ

The horizontal component of normal force on the block is equal to force

                           Fw=n*sin(θ) that exerted by the wall on the wedge.

Substitute mgcosθ for n in the above equation;

                           Fw =mg(cosθ)(sinθ)

Since, there is no friction between the wedge and the wall, there is component force acting on the wall to restrict the motion of the wedge on the surface and that force is arises from the horizontal component for normal force on the block.

6 0
3 years ago
A plate drops onto a smooth floor and shatters into three pieces of equal mass.Two of the pieces go off with equal speeds v at r
Firlakuza [10]

Answer:

Speed of the this part is given as

v_3 = \sqrt2 v

Also the direction of the velocity of the third part of plate is moving along 135 degree with respect to one part of the moving plate

Explanation:

As we know by the momentum conservation of the system

we will have

P_1 + P_2 + P_3 = P_i

here we know that

P_1 = P_2

the momentum of two parts are equal in magnitude but perpendicular to each other

so we will have

P_1 + P_2 = \sqrt{P^2 + P^2}

P_1 + P_2 = \sqrt2 mv

now from above equation we have

P_3 = -(P_1 + P_2)

mv_3 = -(\sqrt 2 mv)

v_3 = \sqrt2 v

Also the direction of the velocity of the third part of plate is moving along 135 degree with respect to one part of the moving plate

6 0
3 years ago
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