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2 years ago

A frog leaps up from the ground and lands on a step 0.1 m above the ground 2 s later. We want to find the

Physics

1 answer:

mash [69]2 years ago

4 0

Answer:

$\Delta x = v_0 t + \frac{1}{2}at^2$

Explanation:

To solve this problem, we can use the following suvat equation:

$\Delta x = v_0 t + \frac{1}{2}at^2$

where

$\Delta x$ is the vertical displacement of the frog

$v_0$ is the initial vertical velocity

t is the time

a is the acceleration

We have chosen this formula because apart from $v_0$ , all the other quantities are known. In fact:

$\Delta x =0.1 m$ is the vertical displacement

t = 2 s is the total time of flight

$a=g=-9.8 m/s^2$ is the acceleration due to gravity (negative because it is downward)

Therefore, solving for $v_0$ , we find the initial velocity of the frog:

$v_0 = \frac{\Delta x-\frac{1}{2}at^2}{t}=\frac{0.1-\frac{1}{2}(-9.8)(2)^2}{2}=9.85 m/s$

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Draw the vector C⃗ =1.5A⃗ −3B⃗ . The length and orientation of the vector will be graded. The location of the vector is not impo

Nutka1998 [239]

I made the drawing in the attached file.

I included two figures.

The upper figure shows the effect of:

- multiplying vector A times 1.5.
It is drawn in red with dotted line.

- multiplying vector B times - 3 .
It is drawn in purple with dotted line.

In the lower figure you have the resultant vector: C = 1.5A - 3B.

The method is that you translate the tail of the vector -3B unitl the point of the vector 1,5A, preserving the angles.

Then you draw the arrow that joins the tail of 1,5A with the point of -3B after translation.

The resultant arrow is the vector C and it is drawn in black dotted line.

Download pdf

7 0

2 years ago

Read 2 more answers

A spacecraft of the Trade Federation flies past the planet Coruscant at a speed of 0.610 c. A scientist on Coruscant measures th

mamaluj [8]

Answer:

the length of the now stationary spacecraft = 89.65m

Explanation:

In contraction equation, Length contraction L is the shortening of the measured length of an object moving relative to the observer’s frame.

Thus, it has a formula;

L = L_o(√(1 - (v²/c²))

Where in this question;

L = 71m and v = 0.610 c

Thus;

71 = L_o (√(1 - ((0.61c)²/c²))

c² will cancel out to give;

71 = L_o (√(1 - 0.61²)

71 = 0.792L_o

L_o = 71/0.792

L_o = 89.65m

6 0

2 years ago

You are sitting in your car at rest at a traffic light with a bicyclist at rest next to you in the adjoining bicycle lane. As so

grigory [225]

Answer:

Explanation:

Time duration during which acceleration exists in bicycle =

23 / 12 = 1.91 s

Time duration during which acceleration exists in car

= 47 / 8 = 5.875 s

Distance covered by bicycle during acceleration ( t = 1.91 s )

= 1/2 x 12 x (1.91)²

= 21.88 mi

Distance covered by car during this time ( t = 1.91 s )

= 1/2 x 8 x (1.91)²

7.64 mi ,

velocity of car after 1.91 s

= 8 x 1.91 = 15.28 mi/h

Let after time 1.91 , time taken by them to meet each other be t

Total distance covered by cycle = total distance covered by car

21.88 + 23 t = 7.64 + 15.28t + 4 t²

21.88 = 7.64 - 7.72t +4 t²

4 t² -7.72 t -14.24 = 0

t = 2.83 s

Total time taken

= 2.83 + 1.91

= 4.74 s

So after 4.74 s they will meet each other.

b ) Maximum distance occurs when velocity of both of them becomes equal .

Velocity after 1.91 s of bicycle

12 x 1.91 = 23 mi/h

Velocity after 1.91 s of car

8 x 1.91 = 15.28 mi/h . Let after time t , the velocity of car becomes 23

15.28 + 8t = 23

t = .965 s

So after time .965 s , car has velocity equal to that of bicycle.

The bicycle will travel a distance of

= 21.88 + .965 x 23 = 44.075 mi

car will travel a distance of

7.64 + 15.28 x .965 + .5 x 8 x .965²

= 7.64 + 14.75 + 3.72

= 26.11 mi

Distance between car and bicycle

= 44.075 - 26.11 = 17.965 mi

= 17.965 x 1760

= 31618.4 ft.

5 0

2 years ago

Tex, an 85.0 kilogram rodeo bull rider is thrown from the bull after a short ride. The 520. kilogram bull chases after Tex at 13

Julli [10]

The question above can be answered through using the concept of Conservation of Momentum which may be expressed as,
m1v1 + m2v2 = mTvT
where m1 and v1 are mass and initial velocity of Tex, 2s are that of the bull, and the Ts are the total. Then substituting,
(85 kg)(3 m/s) + (520 kg)(13 m/s) = (520 + 85)(vT)
The value of vT obtained from above equation is 11.6 m/s

3 0

2 years ago

30) A force produces power P by doing work W in a time T. What power will be produced by a force that does six times as much wor

schepotkina [342]

Answer:

A) 12P

Explanation:

The power produced by a force is given by the equation

$P=\frac{W}{T}$

where

W is the work done by the force

T is the time in which the work is done

At the beginning in this problem, we have:

W = work done by the force

T = time taken

So the power produced is

$P=\frac{W}{T}$

Later, the force does six times more work, so the work done now is

$W'=6W$

And this work is done in half the time, so the new time is

$T'=\frac{T}{2}$

Substituting into the equation of the power, we find the new power produced:

$P'=\frac{W'}{T'}=\frac{6W}{T/2}=12\frac{W}{T}=12P$

So, 12 times more power.

4 0

2 years ago