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sleet_krkn [62]
3 years ago
13

A moving car has kinetic energy. If it speeds up until it is going four times faster than before, how much kinetic energy does i

t have in comparison?
Physics
1 answer:
Eva8 [605]3 years ago
7 0
<h2>Ratio of final kinetic energy to initial kinetic energy is 16.</h2>

Explanation:

Kinetic energy , KE = 0.5mv²

Here car speeds up to four times the initial speed, we need to find ratio of final kinetic energy to initial kinetic energy.

Final speed = 4 x Initial speed = 4v

Initial KE = 0.5mv²

Final KE = 0.5 x m x (4v)²

Final KE = 16 x 0.5 x m x v²

Final KE = 16 x Initial KE

Ratio of final kinetic energy to initial kinetic energy is 16.

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A 3.1 kg ball is dropped from the top of a 38 m tall building. What is the speed of the ball when it is halfway from the buildin
Archy [21]

Answer:

19.3m/s

Explanation:

Use third equation of motion

v^2-u^2=2gh

where v is the velocity at halfway, u is the initial velocity, g is gravity (9.81m/s^2) and h is the height at which you'd want to find the velocity

insert values to get answer

v^2-0^2=2(9.81m/s^2)(38/2)\\v^2=9.81m/s^2 *38\\v^2=372.78\\v=\sqrt[]{372.78} \\v=19.3m/s

4 0
2 years ago
Which situation is the best analogy for the doppler effect?
Rudiy27
The best scenario to describe the doppler effect would be listening to the siren of a passing ambulance or fire truck

then it is coming towards you, the pitch is higher, it gets higher as it approaches and peaks as it gets right in front of you. then it drop at once when it passes you and continues to drop till it fades away. this is a classic descrption of the doppler effect
8 0
3 years ago
14. If the spring constant of a simple harmonic oscillator is doubled, by what factor will the mass of the system need to change
klio [65]

Lets se

And

\\ \rm\Rrightarrow T=2\pi\sqrt{\dfrac{m}{k}}

\\ \rm\Rrightarrow \sqrt{k}T=2\pi\sqrt{m}

So

\\ \rm\Rrightarrow k\propto m

If spring constant is doubled mass must be doubled

8 0
2 years ago
The owner of a company that manufactures drinking cups decides it would be impressive to build an inground swimming pool that is
garri49 [273]

Answer:

The depth is 5.15 m.

Explanation:

Lets take the depth of the pool = h m

The atmospheric pressure ,P = 101235 N/m²

The area of the top = A m²

The area of the bottom =  a m²

Given that A= 1.5 a

The force on the top of the pool = P A

The total pressure on the bottom = P + ρ g h

ρ =Density of the  water = 1000 kg/m³

The total pressure at the bottom of the pool =  (P + ρ g h) a

The bottom and the top force is same

(P + ρ g h) a = P A

P a +ρ g h a =  P A

ρ g h a =  P A - P a

h=\dfrac{P ( A-a)}{\rho g a}

h=\dfrac{P ( 1.5 a-a)}{\rho g a}

h=\dfrac{P ( 1.5- 1)}{\rho g}

h=\dfrac{101235 ( 1.5- 1)}{1000\times 9.81}\ m

h=5.15 m

The depth is 5.15 m.

7 0
2 years ago
The constant forces F1 = 8 + 29 + 32 N and F2 = 48 - 59 - 22 N act together on a particle during a displacement from the point A
steposvetlana [31]

Answer:

- 600 J

Explanation:

A (20, 15, 0 ) m

B (0, 0, 7) m

\overrightarrow{F_{1}}=8\widehat{i}+29\widehat{j}+32\widehat{k}

\overrightarrow{F_{2}}=48\widehat{i}-59\widehat{j}-22\widehat{k}

Net force

\overrightarrow{F}=\overrightarrow{F_{1}}+\overrightarrow{F_{2}}

\overrightarrow{F}}=\left ( 8+48 \right )\widehat{i}+\left ( 29-59 \right )\widehat{j}+\left ( 32-22 \right )\widehat{k}

\overrightarrow{F}}=56\widehat{i}-30\widehat{j}+10\widehat{k}

\overrightarrow{S}=\overrightarrow{OB}-\overrightarrow{OA}

\overrightarrow{S}=\left ( 0-20 \right )\widehat{i}+\left ( 0-15 \right )\widehat{j}+\left ( 7-0 \right )\widehat{k}

\overrightarrow{S}=-20\widehat{i}-15\widehat{j}+7\widehat{k}

Work done is defined as

W = \overrightarrow{F}.\overrightarrow{S}

W = \left ( 56\widehat{i}-30\widehat{j}+10\widehat{k} \right ).\left (-20\widehat{i}-15\widehat{j}+7\widehat{k}  \right )

W = -1120 + 450 + 70

W = - 600 J

3 0
2 years ago
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