All categories

2 years ago

Which of these has the most kinetic energy

Physics

2 answers:

Vesnalui [34]2 years ago

7 0

B. A 50g fish swimming in a fish tank.

garri49 [273]2 years ago

4 0

Answer:

A 50 kg boulder suspended from a cliff.

Explanation:

In this case, that totally depends on the velocity of the object, In case of car, that's just parked so the velocity is zero and a fish can't swim with higher velocity than a boulder which is suspended from a cliff.

Hope it helps!<3

You might be interested in

Consider a large tank holding 1000 L of pure water into which a brine solution of salt begins to owat a constant rate of 6L/min.

lbvjy [14]

Answer:

T = 693.147 minutes

Explanation:

The tank is being continuously stirred. So let the salt concentration of the tank at some time t be x in units of kg/L.

Therefore, the total salt in the tank at time t = 1000x kg

Brine water flows into the tank at a rate of 6 L/min which has a concentration of 0.1 kg/L

Hence, the amount of salt that is added to the tank per minute = $(6\times0.1)kg/min=0.6kg/min$

Also, there is a continuous outflow from the tank at a rate of 6 L/min.

Hence, amount of salt subtracted from the tank per minute = 6x kg/min

Now, the rate of change of salt concentration in the tank = $\frac{dx}{dt}$

So, the rate of change of salt in the tank can be given by the following equation,

$1000\frac{dx}{dt} =0.6-6x$

or, $\int\limits^{0.05}_0 {\frac{1000}{0.6-6x} } \, dx =\int\limits^T_0 {} \, dt$

or, T = 693.147 min (time taken for the tank to reach a salt concentration

of 0.05 kg/L)

3 0

2 years ago

A boy throws a 15 kg ball at 4.7 m/s to a 65 kg girl who is stationary and standing on a skateboard. After catching the ball, th

jek_recluse [69]

Answer:

$a)v_{f}=0.88m/s$

Explanation:

To solve this problem we use the Momentum's conservation Law, before and after the girl catch the ball:

$\\ p_{1}=p_{2}\\m_{ball}*v_{o.ball}+m_{girl}*v_{o.girl} = m_{ball}*v_{f.ball} + m_{girl}*v_{f.girl}$ (1)

At the beginning the girl is stationary:

$v_{o.girl}=0m/s$ (2)

If the girl catch the ball, both have the same speed:

$v_{f.girl}=v_{f.ball}=v_{f}$ (3)

We replace (2) and (3) in (1):

$m_{ball}*v_{o.ball} = (m_{ball}+m_{girl})*v_{f} \\$

We can now solve the equation for v_{f}:

$v_{f}=\frac{m_{ball}*v_{o.ball}}{(m_{ball}+m_{girl})}=\frac{15*4.7}{15+65}=0.88m/s$

4 0

2 years ago

If a 110 kg go-cart traveling at a velocity of 13.41 m/s has a collision with an impulse of 615 Nxs, what is the

mafiozo [28]

Answer:

5.59 m/s

Explanation:

We are given;

Mass = 110 kg

Initial velocity: u = 13.41 m/s

Force = 615 N

Time(t) = 1 s

Now, the formula for force is;

Force = mass x acceleration

Thus;

615 = 110 × acceleration

\Acceleration(a) = 615/110 = 5.591 m/s²

Now, using Newton's first law of motion, we can find acceleration (a). Thus;

v = u + at

v = 13.41 + (5.591 × 1)

v ≈ 19 m/s

So,the change in velocity is;

Final velocity(v) - Initial velocity(u) = 19 - 13.41 = 5.59 m/s

6 0

2 years ago

The square loop shown in the figure moves into a 0.80T magnetic field at a constant speed of 10m/s. The loop has a resistance of

PolarNik [594]

The question ask to find and calculate the induced current in the loop as a function time and the best answer would be that the induced current in the loop is 0.08 amperes. I hope you are satisfied with my answer and feel free to ask for more if you have clarifications and further questions

7 0

2 years ago

Read 2 more answers

A particle moves along a straight line with a velocity in meters per second given by v = 12 - 3t + 8t, where t is in seconds. Wh

elena-14-01-66 [18.8K]

Answer:

Explanation:

Given the velocity of a particle modeled by the equation

v = 13-3t+8t² where t is in seconds

Given t = 0 and position s = +5m

A) To get the position as a function of time, we will integrate the function with respect to t ad shown;

v = 13-3t+8t²

S = ∫13-3t+8t² dt

S = 13t-13t²/2+8t³/3 + C

at t = 0 and S = +5m

5 = 13(0)-13(0)²/2+8(0)³/3+C

5 = 0-0+0+C

C = 5

Substituting c = 5 into the displacement function

S = 13t-13t²/2+8t³/3 + C

S = 13t-13t²/2+8t³/3 + 5

B) acceleration is the change in velocity with respect to time.

a = dv/dt

Given v = 13-3t+8t²

a= dv/dt = -3+16t

a = 16-3t

C) acceleration at t = 6s is derived by plugging in t = 6 into the resulting equations in (B)

a = 16-3t

a = 16-3(6)

a = 16-18

a = -2m/s²

D) net displacement from t = 0 to t = 6s

At t = 0:

S(0) = 13(0)-13(0)²/2+8(0)³/3 + 5

S(0) = 0+5

S(0) = 5m

At t = 6s

S(6) = 13(6)-13(6)²/2+8(6)³/3 + 5

S(6) = 78-234+576+5

S(6) = 425m

Net displacement from t = 0s to t = 6s is s(6)-s(0)

= 425-5

= 420m

E) Total distance travelled D = S(6)+S(0)

= 425+5

= 430m

F) Average velocity = ∆S/∆t

Average velocity = S(6)-S(0)/6-0

Average velocity = 425-5/6

Average velocity = 420/6

Average velocity = 70m/s

4 0

2 years ago