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k0ka [10]
3 years ago
7

A cart starts at x = +6.0 m and travels towards the origin with a constant speed of 2.0 m/s. What is it the exact cart position

(in m) 3.0 seconds later?
Physics
1 answer:
Ira Lisetskai [31]3 years ago
5 0

Answer:

At the origin (x' = 0 m)

Explanation:

Note: From the question, when the cart travels towards the origin, the magnitude of its exact position reduces with time.

The formula of speed is given as

S = d/t................. Equation 1

Where S = speed of the cart, d = distance covered by the cart over a certain time. t = time taken to cover the distance.

make d the subject of the equation,

d = St ................. Equation 2

Given: S = 2.0 m/s, t = 3.0 s

Substitute into equation 2

d = 2(3)

d = 6 m.

From the above, the cart covered a distance of 6 m in 3 s.

The exact position of the cart = Initial position-distance covered

x' = x-d............ Equation 3

Where x' = exact position of the cart 3 s later, x = initial position of the cart, d = distance covered by the cart in 3.0 s.

Given: x = +6.0 m, d = 6 m.

Substitute into equation 3

x' = +6-6

x' = 0 m.

Hence the cart will be at 0 m (origin) 3 s later

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You were in a tog-o-war contest which your team won. Write a letter to your friend that tells why your team won the contest. Rem
ollegr [7]

Answer:

Imagine a tug-of-war between you and one

friend. If you are stronger, you apply more

force to the rope. You pull your friend across

the line, and you are the winner! If your friend

is stronger, he might pull you across the line.

Sometimes the forces are equal. Neither you

nor your friend moves across the line. The two

forces are balanced.

3

We say that the net force on an object is the combination of all the forces acting on it. To find the net force of

forces that are acting in the same direction, add them together. For example, if you pull on a box with a force of 25

newtons (N) while your friend pushes the box (in the same direction you are pulling) with a force of 30 N, the net

force applied to the box in that direction is 55 newtons.

4

To find the net force of forces that are acting in opposite directions, subtract the smaller force from the larger

one. If you are pulling on a tug-of-war rope with a force of 40 N, and your friend is pulling with a force of 35 N in

the opposite direction, the net force on the rope is 5 newtons in your direction. You win!

Explanation:

3 0
3 years ago
A pickup truck is traveling down the highway at a steady speed of 30.1 m/s. The truck has a drag coefficient of 0.45 and a cross
Sav [38]

Answer:

The energy that the truck lose to air resistance per hour is 87.47MJ

Explanation:

To solve this exercise it is necessary to compile the concepts of kinetic energy because of the drag force given in aerodynamic bodies. According to the theory we know that the drag force is defined by

F_D=\frac{1}{2}\rhoC_dAV^2

Our values are:

V=30.1m/s

C_d=0.45

A=3.3m^2

\rho=1.2kg/m^3

Replacing,

F_D=\frac{1}{2}(1.2)(0.45)(3.3)(30.1)^2

F_D=807.25N

We need calculate now the energy lost through a time T, then,

W = F_D d

But we know that d is equal to

d=vt

Where

v is the velocity and t the time. However the time is given in seconds but for this problem we need the time in hours, so,

W=(807.25N)(30.1m/s)(3600s/1hr)

W=87.47*10^6J (per hour)

Therefore the energy that the truck lose to air resistance per hour is 87.47MJ

4 0
3 years ago
What are the differences between the practical and the ideal pendulum​
ankoles [38]

lf a heavy point mass is suspended by a weightless, inextensible and perfectly flexible string from a rigid support, then this arrangement is called simple pendulum.

In practice, however, these requirements cannot be fulfilled. So we use a practical pendulum.

A practical pendulum consists of a small metallic solid sphere suspended by a fine silk thread from a rigid support. This is the practical simple pendulum which is nearest to the ideal simple pendulum.

Note :

The metallic sphere is called the bob.

When the bob is displaced slightly to one side from its mean position and released, it oscillates about its mean position in a vertical plane.

4 0
3 years ago
When the batter hit the foul ball, the baseball moved upward to a delighted fan in the top deck.
oee [108]

Answer:

B) Kinetic energy increases, potential energy decreases

Explanation:

In a given system, when a body is at rest, v =0m/s, the kinetic energy is at zero while the potential energy is at maximum. However, when a body is in motion with a velocity = v, the potential energy is at zero while the kinetic energy is at maximum.

Before this happen, the a body at rest (P.E = max) is set on motion, the kinetic energy gradually increases till it converts all the potential energy in the system to kinetic energy and then reverses back when the body goes to rest again.

In this case, before the batter hits the ball, the kinetic energy was at zero while the potential energy was at maximum. However, when he hits the ball and sets it into motion with a velocity V, the potential energy converts to kinetic energy and moves the ball with that energy till it has expanded it and comes to rest.

Potential Energy → Kinetic Energy → Potential Energy.

That's how the system keeps changing.

6 0
3 years ago
Alguém tem a resposta ?
Maru [420]
The answer is D or C
3 0
3 years ago
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