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BartSMP [9]
3 years ago
12

Just for fun, a person jumps from rest from the top of a tall cliff overlooking a lake. in falling through a distance h, she acq

uires a certain speed v. assuming free-fall conditions, how much farther must she fall in order to acquire a speed of 2v? express your answer in terms of h.
b. would the answer to part
a. be different if this event were to occur on another planet where the acceleration due to gravity had a value other than 9.80 m/s2? explain.
Physics
1 answer:
RideAnS [48]3 years ago
4 0
For this we need to use several formulas:
v = a*t
This simply means that after 1 second we will have some speed and if we double time, speed will double as well.

But, relation time-traveled distance isn't linear. Traveled distance we calculate:
s = 1/2*a*t^2                we can say that h=s
so if we double time, traveled distance will increase 4 times because of square relation between time and traveled distance.

New traveled distance (h2) will be: h2=4*h

b)  Because distance depends on gravity acceleration in free fall h2 will certainly change but the relation h2=4*h will remain the same.
You might be interested in
horizontal force pushes block up a 20.0 incline with an initial speed 12.0 m//s. a) how high up the plane does slide before comi
Annette [7]

Answer:

Explanation:

We shall apply conservation of mechanical energy .

initial kinetic energy = 1/2 m v²

= .5 x m x 12 x 12

= 72 m

This energy will be spent to store potential energy . if h be the height attained

potential energy = mgh , h is vertical height attined by block

= mg l sin20  where l is length up the inclined plane  

for conservation of mechanical energy

initial kinetic energy  = potential energy

72 m = mg l sin20

l = 72 /  g  sin20

= 21.5 m

deceleration on inclined plane = g sin20

= 3.35 m /s²

v = u - at

t = v - u / a

= (12 - 0) / 3.35

= 3.58 s

it will take the same time to come back . total time taken to reach original point = 2 x 3.58

= 7.16 s

7 0
3 years ago
Please fast answerrr thank you​
Vedmedyk [2.9K]

Answer:

50J

Explanation:

At the top you have(A)

KE_a = O

PE_a = 100J

KE + PE = 100J

At the bottom you have (C)

KE_c= 100J

PE_c=0J

KE+PE = 100J

At point C:

You are at half the height.

We know that at H, PE =100J

PE_c = mgH

At C,

PE_c= mg (H/2) *at half the height

*m and g stay the same

Intuitively, the higher you are, the more potential energy you have.

If you decrease the height by a half, your PE will also decrease

At A:

PE_a / (mg) = H

At B:

PE_b / (mg) = H/2

to also get H on the right hand side, multiply by 2

2 (PE_b/ (mg))= H

2PE_b / (mg) = H

Ok, now that we have set up 2 equations (where H is isolated), find PE at B

AT A = AT B *This way you are saying that H = H (you compare both equations)

PE_a / (mg) = 2x PE_b / (mg)

*mg are the same for both cancel them (you can do that because of the = sign)

PE_a =  2PE_b

We know that PE_a = 100J

100J/2 = PE_b

PE at b = 50J

**FIND KE at b

We know that

KE_b + PE_b is always 100J

100J = 50J + KE_b

KE_b = 50J

4 0
2 years ago
A car and a truck start from rest at the same instant, with the car initially at some distance behind the truck. The truck has a
Svet_ta [14]

A) The car overtakes the truck after 7.56 s

B) Initial distance between car and truck: 37.1 m

C) Speed of the truck: 15.9 m/s, speed of the car: 25.7 m/s

D) See graph in attachment

Explanation:

A)

The truck starts from rest and has a constant acceleration, so its position at time t can be written as

x_t(t)=d+\frac{1}{2}a_tt^2

where

d is the initial distance between the truck and the car (the truck starts some distance ahead of the car)

a_t=2.10 m/s^2 is the acceleration of the truck

The car position instead it is given by the equation

x_c(t)=\frac{1}{2}a_ct^2

where

a_c=3.40 m/s^2 is the acceleration of the car

The car overtakes the truck when the truck has moved 60.0 m, so when

x_t(t') = d + 60

Therefore, solving the equation, we find the time t when  this occurs:

d+\frac{1}{2}a_t t'^2 = d+60\\\frac{1}{2}a_tt'^2=60\\t'=\sqrt{\frac{2\cdot 60}{a_t}}=\sqrt{\frac{120}{2.1}}=7.56 s

B)

In order to find the initial distance between the car and the truck (d), we have to calculate first the distance covered by the car during these 7.56 s. It is given by:

x_c(t')=\frac{1}{2}a_c t'^2=\frac{1}{2}(3.40)(7.56)^2=97.2 m

This means that after 7.56 s, when the car reaches the truck, the car has covered 97.2 m while the truck has covered 60 m. However, their positions are now equal, so we can write:

x_c(t')=x_t(t')

And by solving the equation, we find the value of d, the initial distance between car and truck:

\frac{1}{2}a_c t'^2 = d + \frac{1}{2}a_t t'^2\\d=\frac{1}{2}(a_c-a_t)t'^2 = \frac{1}{2}(3.40-2.10)(7.56)^2=37.1 m

C)

In order to find the speed of each vehicle, we use the following suvat equation:

v=u+at

where

u is the initial velocity

a is the acceleration

t is the time

For the truck, we have:

u = 0

a_t = 2.10 m/s^2

So its speed after t = 7.56 s is

v_t = 0+(2.10)(7.56)=15.9 m/s

For the car, we have

u = 0

a_c=3.40 m/s^2

So its speed after t = 7.56 s is

v_c=0+(3.40)(7.56)=25.7 m/s

D)

Find the graph required in attachment.

On the x-axis, it is represented the time in seconds. On the y-axis, it is represented the position in meters.

Both curves are in the shape of a parabola since the motion of both vehicles is an accelerated motion.

The curve that starts at -37.1 m is the curve representing the car: in fact, the car starts behind the truck by 37.1 m. The curve that starts from x = 0, t= 0 is that of the truck.

The two curves meets when t = 7.56 s: at that time, the two vehicles have reached the same position, and we see that occurs when x = 60 m, which means that this happens when the truck has covered 60 meters.

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

8 0
4 years ago
As shown in the figure below, a bullet is fired at and passes through a piece of target paper suspended by a massless string. Th
NikAS [45]

Answer:

M = 0.730*m

V = 0.663*v

Explanation:

Data Given:

v_{bullet, initial} = v\\v_{bullet, final} = 0.516*v\\v_{paper, initial} = 0\\v_{paper, final} = V\\mass_{bullet} = m\\mass_{paper} = M\\Loss Ek = 0.413 Ek

Conservation of Momentum:

P_{initial} = P_{final}\\m*v_{i} = m*0.516v_{i} + M*V\\0.484m*v_{i} = M*V .... Eq1

Energy Balance:

\frac{1}{2}*m*v^2_{i} = \frac{1}{2}*m*(0.516v_{i})^2 + \frac{1}{2}*M*V^2 + 0.413*\frac{1}{2}*m*v^2_{i}\\\\0.320744*m*v^2_{i} = M*V^2\\\\M = \frac{0.320744*m*v^2_{i} }{V^2}  ....... Eq 2

Substitute Eq 2 into Eq 1

0.484*m*v_{i} = \frac{0.320744*m*v^2_{i} }{V^2} *V  \\0.484 = 0.320744*\frac{v_{i} }{V} \\\\V = 0.663*v_{i}

Using Eq 1

0.484m*v_{i} = M* 0.663v_{i}\\\\M = 0.730*m

7 0
3 years ago
A stone was dropped off a cliff and hit the ground with a speed of 96 ft/s. What is the height of the cliff? (Use 32 ft/s2 for t
nadezda [96]
The initial velocity of the stone is 0 ft/s. Given the initial velocity (Vi), final velocity (Vf), and acceleration due to gravity (g), the distance may be calculated through the equation,
                                     d = ((Vf)² - (Vi)²) / 2g
Substituting the known values,
                                     d = ((96 ft/s)² - 0))/ (2x32.2)
The value of d is 143.10 ft. 
4 0
3 years ago
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