All categories

3 years ago

Work out the kinetic energy of a 2.5 kg remote-controlled car that is moving at 2 m/s.

Physics

1 answer:

lbvjy [14]3 years ago

8 0

Answer: 5 joules

Explanation:

mass=m=2.5kg

Velocity=v=2m/s

Kinetic energy=ke

ke=(m x v x v)/2

ke=(2.5 x 2 x 2)/2

Ke=10/2

Ke=5

Kinetic energy=5 joules

You might be interested in

Samantha is checking the weather for her upcoming trip to Mexico City. The weather forecast predicts a high-pressure system for

Zanzabum

Calm, sunny days with wind moving away from the center.

5 0

3 years ago

Read 2 more answers

. Prior to Remy's trip to Cleveland, his uncle tells him about this amazing barbecue restaurant there and raves about

Andrews [41]

whats the restraunt called

6 0

3 years ago

A gry is an old English measure for length, defined as 1/10 of a line, where line is another old English measure for length, def

Valentin [98]

Answer:

0.2448 point²

Explanation:

1 gry = 1/10 line

1 line = 1/12 inch

=> 1 gry in inches = 1/10 * 1/12 = 1/120 inch

=> 1 inch = 120 gry

1 point = 1/72 inch

=> 1 inch = 72 points

Therefore,

120 gry = 72 points

=> 1 gry = 3/5 point

Therefore,

1 gry² = (3/5)² point²

1 gry² = 9/25 point²

This means that 0.68 gry² will be:

0.68 gry² = 0.68 * 9/25 point²

=> 0.68 gry² = 0.2448 point²

6 0

3 years ago

In a laundromat, during the spin-dry cycle of a washer, the rotating tub goes from rest to its maximum angular speed of 2.2 rev/

Hunter-Best [27]

Answer:

$n_{T} = 31.68\,rev$

Explanation:

The angular acceleration is:

$\ddot n_{1} = \frac{2.2\,\frac{rev}{s} -0\,\frac{rev}{s} }{8.8\,s}$

$\ddot n_{1} = 0.25\,\frac{rev}{s^{2}}$

And the angular deceleration is:

$\ddot n_{2} = \frac{0\,\frac{rev}{s}-2.2\,\frac{rev}{s} }{20\,s}$

$\ddot n_{2} = -0.11\,\frac{rev}{s^{2}}$

The total number of revolutions is:

$n_{T} = n_{1} + n_{2}$

$n_{T} = \frac{\left(2.2\,\frac{rev}{s} \right)^{2}-\left(0\,\frac{rev}{s} \right)^{2}}{2\cdot \left(0.25\,\frac{rev}{s^{2}} \right)} + \frac{\left(0\,\frac{rev}{s} \right)^{2}-\left(2.2\,\frac{rev}{s} \right)^{2}}{2\cdot \left(-0.11\,\frac{rev}{s^{2}} \right)}$

$n_{T} = 31.68\,rev$

4 0

4 years ago

Estimate the constant rate of withdrawal (in m3 /s) from a 1375 ha reservoir in a month of 30 days during which the reservoir le

kap26 [50]

Answer:

Explanation:

1 ha = 10⁴ m²

1375 ha = 1375 x 10⁴ m² = 13.75 x 10⁶ m²

In flow in a month = .5 x 10⁶ x 30 m³ = 15 x 10⁶ m³

Net inflow after all loss = 18.5 - 9.5 - 2.5 cm = 6.5 cm = .065 m

Net inflow in volume = 13.75 x 10⁶ x .065 m³= .89375 x 10⁶ m³

Let Q be the withdrawal in m³

Q - 15 x 10⁶ - .89375 x 10⁶ = 13.75 x 10⁶ x .75 = 10.3125 x 10⁶

Q = 26.20 x 10⁶ m³

rate of withdrawal per second

= 26.20 x 10⁶ / 30 x 24 x 60 x 60

= 26.20 x 10⁶ / 2.592 x 10⁶

= 10.11 m³ / s

6 0

3 years ago