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2 years ago

Snowboarder Jump—Energy and Momentum

Physics

1 answer:

soldier1979 [14.2K]2 years ago

4 0

Answer:

THE ANSWER TERMS ARE DEFINED BLOW:-

Explanation:

MOMENTUM- IT IS THE ABILITY TO INCREASE OR DEVELOP CONSTANT FORCE.

KINETIC ENERGY:- IT IS THE ENERGY THAT A PRTICLE POSSES WHEN IT IS ACTUALLY IN MOTION.

POTENTIAL ENERGY:- IT IS THE ENERGY THAT A PARTICLE POSSES WHEN IT ACTUALLY IS IN RESTING STATE.

IN THIS ACIVITY THE SNOWBOARDER IS IN THE MOTION STATE THEREFORE HE POSSES KINETIC ENERGY AND TO MAINTAIN THAT KINEITC ENERG FOR A PERIOD OF TIME,MOMENTUM PLAYS IT'S ROLE.

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A block moves at 5 m/s in the positive x direction and hits an identical block, initially at rest. A small amount of gunpowder h

Anestetic [448]

Answer:

Speed of 1.83 m/s and 6.83 m/s

Explanation:

From the principle of conservation of momentum

$mv_o=m(v_1 + v_2)$ where m is the mass, $v_o$ is the initial speed before impact, $v_1$ and $v_2$ are velocity of the impacting object after collision and velocity after impact of the originally constant object

$5m=m(v_1 +v_2)$

Therefore $v_1+v_2=5$

After collision, kinetic energy doubles hence

$2m*(0.5mv_o)=0.5m(v_1^{2}+v_2^{2})$

$2v_o^{2}=v_1^{2} + v_2^{2}$

Substituting 5 m/s for $v_o$ then

$2*(5^{2})= v_1^{2} + v_2^{2}$

$50= v_1^{2} + v_2^{2}$

Also, it’s known that $v_1+v_2=5$ hence $v_1=5-v_2$

$50=(5-v_2)^{2}+ v_2^{2}$

$50=25+v_2^{2}-10v_2+v_2^{2}$

$2v_2^{2}-10v_2-25=0$

Solving the equation using quadratic formula where a=2, b=-10 and c=-25 then $v_2=6.83 m/s$

Substituting, $v_1=-1.83 m/s$

Therefore, the blocks move at a speed of 1.83 m/s and 6.83 m/s

6 0

1 year ago

Table 2.4 shows how the dispacement of a runner changed during a sprint race. Draw a dispacement-time graph to show this data, a

GalinKa [24]

4. Table 2.4 shows how the displacement of a runner changed
during a sprint race. Draw a displacement–time graph to show
this data, and use it to deduce the runner’s speed in the middle
of the race.
Table 2.4 Data for a sprinter during a race
Displacement
(m)
0 4 10 20 50 80 105
Time (s) 1 2 3 6 9 12

8 0

1 year ago

An ambulance driving 35.0 m/s emits a sound wave with a wavelength of 80.0 centimeters. As it drives away from a hospital, which

katen-ka-za [31]

Apparent frequency heard by the staff: 389 Hz

Explanation:

The phenomenon described in this situation is called Doppler effect.

Doppler effect occurs when there is a source emitting a wave in relative motion with respect an observer. In such situation, the frequency of the wave as perceived by the observer ("apparent frequency") is shifted from the real frequency of the sound ("proper frequency"). In particular:

- The observer perceives a higher frequency if the source is moving towards them

- The observer perceives a lower frequency if the source is moving away from them

The formula to calculate the apparent frequency in the Doppler effect is

$f'=\frac{v\pm v_o}{v\pm v_s}f$

where

f is the proper frequency

f' is the apparent frequency

v is the speed of the wave

$v_o$ is the velocity of the observer (positive if they are moving towards the source, negative if moving away)

$v_s$ is the velocity of the source (positive if it is moving away, negative if moving towards the observer)

First of all, in this problem we have to calculate the proper frequency of the sound wave emitted from the ambulance; we have:

v = 343 m/s (speed of sound wave)

$\lambda=80 cm = 0.80 m$ (wavelength)

So the proper frequency is

$f=\frac{v}{\lambda}=\frac{343}{0.80}=429 Hz$

Now we can calculate the apparent frequency heard by the staff at the hospital when the ambulance moves away; we have:

$v_s = +35.0 m/s$ (velocity of the ambulance)

$v_o = 0$ (velocity of the staff)

Substituting,

$f'=\frac{343+0}{343+35}(429)=389 Hz$

Learn more about frequency and wavelength:

brainly.com/question/5354733

brainly.com/question/9077368

#LearnwithBrainly

4 0

1 year ago

Force F acts between two charges, q1 and q2, separated by a distance d. If q1 is increased to twice its original value and the d

Step2247 [10]

Okay, haven't done physics in years, let's see if I remember this.

So Coulomb's Law states that $F = k \frac{Q_1Q_2}{d^2}$ so if we double the charge on $Q_1$ and double the distance to $(2d)$ we plug these into the equation to find

<span> $F_{new} = k \frac{2Q_1Q_2}{(2d)^2}=k \frac{2Q_1Q_2}{4d^2} = \frac{2}{4} \cdot k \frac{Q_1Q_2}{d^2} = \frac{1}{2} \cdot F_{old}$ </span>

So we see the new force is exactly 1/2 of the old force so your answer should be $\frac{1}{2}F$ if I can remember my physics correctly.

9 0

2 years ago

Read 2 more answers

A rock is dropped from the top of a tall building. The rock's displacement in the last second before it hits the ground is 46 %

olasank [31]

Answer:

height is 69.68 m

Explanation:

given data

before it hits the ground = 46 % of entire distance

to find out

the height

solution

we know here acceleration and displacement that is

d = (0.5)gt² ..............1

here d is distance and g is acceleration and t is time

so when object falling it will be

h = 4.9 t² ....................2

and in 1st part of question

we have (100% - 46% ) = 54 %

so falling objects will be there

0.54 h = 4.9 (t-1)² ...................3

now we have 2 equation with unknown

we equate both equation

1st equation already solve for h

substitute h in the second equation and find t

0.54 × 4.9 t² = 4.9 (t-1)²

t = 0.576 s and 3.771 s

we use here 3.771 s because 0.576 s is useless displacement in the last second before it hits the ground is 46 % of the entire distance it falls

so take t = 3.771 s

then h from equation 2

h = 4.9 t²

h = 4.9 (3.771)²

h = 69.68 m

so height is 69.68 m

6 0

2 years ago