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Pavel [41]
3 years ago
12

If a car stops suddenly, you feel "thrown forward." We’d like to understand what happens to the passengers as a car stops. Imagi

ne yourself sitting on a very slippery bench inside a car. This bench has no friction, no seat back, and there’s nothing for you to hold onto.(a) identify all of the forces action on you as thecar travels at a perfectly steady speed on level ground.(b) repeat part A with the car slowing down(c) describe what happens to you as the car slowsdown(d) suppose now that the bench is not slippery. as the carslows down, you stay on the bench and dont slide off. what force isresponsible for you deceleration?
Physics
1 answer:
stellarik [79]3 years ago
5 0
<h2>A)</h2>

The forces acting on you are:

  • The gravitational pull of the Earth (an others Celestial objects)
  • The Normal Force that balance the gravitational pull and points upward.

That is. There is not need for any other force, cause the car its going at constant speed, so the acceleration its zero, as the net force its mass multiplied by acceleration, the net force is zero.

<h3>B)</h3>

The forces acting on you are:

  • The gravitational pull of the Earth (an others Celestial objects)
  • The Normal Force that balance the gravitational pull and points upward.

That is. Again. The car its slowing down. But, it cant apply a force in you against the direction of movement to slow you down.

<h3>C)</h3>

As you can't be slowed down, you will go forward at the same speed you had when the car went steady, and will crash against whatever its in front of you.

D)

The friction force, will pull you against the direction of movement, and slow you down will the car slows down.

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What are the change in internal energy if 250 j of hear is added to a system of 80 j of work is done by the system
Sphinxa [80]

Answer:

Explanation:

dU= dq+w

dU is change in internal energy of the system

dq is the amount heat added or released by the system which be positive or negative respectivelý

And w is the amount of work done by the system or on the system which will be positive or negative respectively.

Hence,

dU= 250+80= 330 J

The change will be positive

5 0
3 years ago
Consider the following kinds of electromagnetic waves:
marysya [2.9K]

Answer:

According to wavelengths in descending order: C, B, D, E and A

According to frequencies in descending order: A, E, D, B and C

According to speeds in vacuum: A =B =C =D =E

Explanation:

in the EM spectrum, radio waves have the longest wavelength while gamma rays have the shortest wavelength.

All EM waves travel with the same speed of 3*10^8m/s in a vacuum.

7 0
3 years ago
You throw a ball straight up. You are extra strong feeling today. It takes 11 seconds for the ball to come back down.
ollegr [7]

Answer:

A. 148.23 m

B. 2.75 m/s

Explanation:

The following data were obtained from the question:

Time of flight (T) = 11 s

Maximum height (h) =?

Initial velocity (u) =?

Next, we shall determine the time taken for the ball to get to the maximum height. This can be obtained as follow:

Time of flight (T) = 11 s

Time (t) to reach the maximum height =.?

T = 2t

11 = 2t

Divide both side by 2

t = 11/2

t = 5.5 s

NOTE: Time to reach the maximum height is the same as the time taken for the ball to fall back to the plane of projection.

A. Determination of the maximum height to which the ball was thrown.

Time (t) to reach maximum height = 5.5 s

Acceleration due to gravity (g) = 9.8 m/s²

Maximum height (h) =?

h = ½gt²

h = ½ × 9.8 × 5.5²

h = 4.9 × 30.25

h = 148.23 m

B. Determination of the initial velocity.

Maximum height (h) reached = 148.23 m

Acceleration due to gravity (g) = 9.8 m/s²

Initial velocity (u) =?

u² = h/2g

u² = 148.23 / (2 × 9.8)

u² = 148.23 / 19.6

Take the square root of both side

u = √(148.23 / 19.6)

u = 2.75 m/s

5 0
3 years ago
Two forces are acting on an object. The first force has magnitude F1=33.4 N and is pointing at an angle of θ1=23.8 clockwise fro
marishachu [46]

Answer:

Fe= 28.2 N : Magnitude of the equilibrant (Fe)

β = 18.34° , clockwise from the positive x axis

Explanation:

Concept of the equilibrant

It is called equilibrant  to a force with the same magnitude and direction as the resulting one (in case it is non-zero) but in the opposite direction. Adding vectorially to all the forces (that is to say the resulting one) with the equilibrant you get zero

To solve this problem we decompose the forces given into x-y components to find the resulting force:

Look at the attached graphic

F₁= 33.4 N  , θ₁=23.8° clockwise from the positive y axis (y+)

F₁x= 33.4 *sin23.8° = 13.48 N

F₁y= 33.4 *cos23.8° =30.6 N

F₂=46.1 N ,  θ₂=28.8 counterclockwise from the negative x axis (x-)

F₂x= -46.1 *cos28.8° = -40.4 N

F₂y=  -46.1 *sin28.8° =  -22.2 N

Components of the resultant in x-y R(x,y)

Rx= 13.48 N -40.4 N = - 26.92 N

Ry= 30.6 N  -22.2 N =  + 8.4 N

Components of the equilibrant in x-y Fe(x,y)

Fex= +26.92 N

Fey=  - 8.4 N

Magnitude of the equilibrant (Fe)

F_{e} = \sqrt{(F_{ex})^{2}+{(F_{ey})^{2}  }

F_{e} = \sqrt{(26.92)^{2}+(8.4)^{2}  }

Fe= 28.2 N

Angle the equilibrant makes with the x axis ( β)

\beta = tan^{-1} (\frac{F_{ey} }{F_{ex} } )

\beta = tan^{-1} (\frac-8.4 }{26.92 } )

β = -18.34°                  

β = 18.34° , clockwise from the positive x axis

8 0
3 years ago
A projectile of mass 9.6 kg is launched from the ground with an initial velocity of 12.4 m/s at angle of 54° above the horizonta
Temka [501]

Answer:

The location is at (3.535, 1.162) m

Solution:

As per the question:

Mass of the projectile, m = 9.6 kg

Initial velocity, v = 12.4 m/s

Angle, \theta = 54^{\circ}

Mass of one fragment, m = 6.5 kg

Time taken by the fragment, t = 1.42 s

Height of the fragment, y = 5.9 m

Horizontal distance, x = 13.6 m

Now,

To determine the location of the second fragment:

Horizontal Range, R = \frac{v^{2}sin2\theta}{g}

R = \frac{12.4^{2}sin2(54)}{9.8} = 14.92\ m

Time of flight, t' = \frac{2vsin\theta}{g} = \frac{2\times 12.4sin108}{9.8}= 2.406\ s

Now, for the fragments:

Mass of the other fragment, m' = M - m = 9.6 - 6.5 = 3.1 kg

Distance traveled horizontally:

s_{x} = vcos\theta = 12.4cos54^{\circ}\times 1.42 = 10.35\ m

Distance traveled vertically:

s_{y} = vcos\theta - \frac{1}{2}gt^{2}

s_{y} = 12.4sin54^{\circ}\times 1.42 -  \frac{1}{2}\times 9.8\times 1.42^{2} = 14.25 - 9.88 = 4.37\ m

Now,

s_{x} = \frac{mx + m'x'}{M}

10.35= \frac{6.5\times 13.6 + 3.1x'}{9.6}

x' = 3.535 m

Similarly,

s_{y} = \frac{my + m'y'}{M}

4.37= \frac{6.5\times 5.9 + 3.1y'}{9.6} = 1.162\ m

The location of the other fragment is at (3.535, 1.162)

5 0
3 years ago
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