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shusha [124]
2 years ago
7

A car travels north along a straight highway at an average speed of 85 km/h. After driving 2.0 km, the car passes a gas station

and continues along the highway. What is the car’s position relative to the start of its trip 0.25 h after it passes the gas station?
Physics
1 answer:
Sati [7]2 years ago
5 0

Answer:

   x_total = 23250 m

Explanation:

This is a uniform motion exercise, the equation that describes the motion is

         v = x / t

the position is

        x = v t

let's reduce the magnitude to the SI system

        v = 85 km / h (1000 m / 1km) (1h / 3600 s) = 23.61 m / s

         t = 0.25 h (3600 s / 1h) = 900 s

let's calculate

         x₁ = v t₁

         x₁ = 23.61   900

         x₁ = 21250 m

The distance from the origin is

         x_total = x_station + x₁

         x_total = 2000 + 21250

         x_total = 23250 m

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A shot putter releases the shot some distance above the level ground with a velocity of 12.0 m/s, 51.0 ∘above the horizontal. Th
alina1380 [7]

A) Zero

The motion of the shot is a projectile's motion: this means that there is only one force acting on the projectile, which is gravity. However, gravity only acts in the vertical direction: so, there are no forces acting in the horizontal direction. Therefore, the x-component of the acceleration is zero.

B) -9.8 m/s^2

The vertical acceleration is given by the only force acting in the vertical direction, which is gravity:

F=mg

where m is the projectile's mass and g is the gravitational acceleration. Therefore, the y-component of the shot's acceleration is equal to the acceleration due to gravity:

a_y = g = -9.8 m/s^2

where the negative sign means it points downward.

C) 7.6 m/s

The x-component of the shot's velocity is given by:

v_x = v_0 cos \theta

where

v_0 = 12.0 m/s is the initial velocity

\theta=51.0^{\circ} is the angle of the shot

Substituting into the equation, we find

v_x = (12.0 m/s)(cos 51^{\circ})=7.6 m/s

D) 9.3 m/s

The y-component of the shot's velocity is given by:

v_y = v_0 sin \theta

where

v_0 = 12.0 m/s is the initial velocity

\theta=51.0^{\circ} is the angle of the shot

Substituting into the equation, we find

v_y = (12.0 m/s)(sin 51^{\circ})=9.3 m/s

E) 7.6 m/s

We said at point A) that the acceleration along the x-direction is zero: therefore, the velocity along the x-direction does not change, so the x-component of the velocity at the end of the trajectory is equal to the x-velocity at the beginning:

v_x = 7.6 m/s

F) -11.1 m/s

The y-component of the velocity at time t is given by:

v_y(t) = v_y + at

where

v_y = 9.3 m/s is the initial y-velocity

a = g = -9.8 m/s^2 is the vertical acceleration

t is the time

Since the total time of the motion is t=2.08 s, we can substitute this value into the equation, and we find:

v_y(2.08 s)=9.3 m/s + (-9.8 m/s^2)(2.08 s)=-11.1 m/s

where the negative sign means the vertical velocity is now downward.

3 0
2 years ago
A badminton player shuffles his shoes in resins explain the scientific reason behind this​
Kobotan [32]

Answer:

To avoid slipping while playing badminton or to increase friction between the shoes and the badminton court or to enhance the grip.

Explanation:

As the question is general so I will try answer it generally.

As badminton court is usually slippery and flat with no or very little friction and while playing badminton, players have to move very fast from place to place.

So, in order to effectively move faster to take the difficult shots with accuracy badminton players increase the friction between their shoes and the badminton court by shuffling resin to their shoes. It helps to enhance the grip.

4 0
2 years ago
. Emergency rations are to be dropped from a plane to some stranded hikers. The search and rescue plane is flying at an altitude
almond37 [142]

Answer:

35 m/s down

Explanation:

The horizontal speed of the package is 70 m/s.  So the time needed to reach the hikers is:

1000 m / (70 m/s) = 14.28 s

Taking down to be positive, the initial velocity needed is:

Δy = v₀ t + ½ at²

1500 m = v₀ (14.28 s) + ½ (9.8 m/s²) (14.28 s)²

v₀ = 35 m/s

The package must be launched down with an initial velocity of 35 m/s.

3 0
2 years ago
Read 2 more answers
it possible that the net kinetic energy for two objects be zero while the net momentum is zero? Explain.
svp [43]
Of course. That's what you have when both objects are at rest. I'm guessing that you left a word out of the question, and it actually says that the net kinetic energy is NOT zero. In that case, the answer is still 'yes', but you have to think about it for a second.
4 0
2 years ago
What are the real life applications of Hooke's Law?
Anarel [89]

Answer:

1) a rubber band

2) the spring of retractable pen

3) a spring loaded toy gun

Explanation:

Hooke's law states that; provided the elastic limit of a material is not exceeded, the force exerted on an elastic material is directly proportional to its extension. This relationship was first captured by Robert Hooke in 1660 when he asserted that 'as the extension, so is the force!'.

Hooke's law generally deals with elastic or stretchable materials. These materials can be deformed, but returned to their original shapes when the deforming force is removed. This deforming force causes an extension in the material which is directly proportional to the deforming force. That is F= Kx where K is the called the force constant, F is the deforming force and x is the magnitude of extension brought about by the force.

Various real life applications of Hooke's law have been listed in the answer. Any material that makes use of a loaded spiral spring or indeed any kind of elastic material obeys Hooke's law.

5 0
2 years ago
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